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University Physics
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SENIOR CONTRIBUTING AUTHORS
SAMUEL J. LING, TRUMAN STATE UNIVERSITY
JEFF SANNY, LOYOLA MARYMOUNT UNIVERSITY
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Table of ContentsPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Unit 1. MechanicsChapter 1: Units and Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.1 The Scope and Scale of Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2 Units and Standards . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.3 Unit Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.4 Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.5 Estimates and Fermi Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.6 Significant Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.7 Solving Problems in Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32Chapter 2: Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.1 Scalars and Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.2 Coordinate Systems and Components of a Vector . . . . . . . . . . . . . . . . . . . . 572.3 Algebra of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 682.4 Products of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77Chapter 3: Motion Along a Straight Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1053.1 Position, Displacement, and Average Velocity . . . . . . . . . . . . . . . . . . . . . . 1063.2 Instantaneous Velocity and Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1103.3 Average and Instantaneous Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . 1163.4 Motion with Constant Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1243.5 Free Fall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1363.6 Finding Velocity and Displacement from Acceleration . . . . . . . . . . . . . . . . . . 142Chapter 4: Motion in Two and Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 1574.1 Displacement and Velocity Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1584.2 Acceleration Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1664.3 Projectile Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1714.4 Uniform Circular Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1824.5 Relative Motion in One and Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . 189Chapter 5: Newton's Laws of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2075.1 Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2085.2 Newton's First Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2135.3 Newton's Second Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2165.4 Mass and Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2275.5 Newton’s Third Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2295.6 Common Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2365.7 Drawing Free-Body Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247Chapter 6: Applications of Newton's Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2656.1 Solving Problems with Newton’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . 2666.2 Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2816.3 Centripetal Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2936.4 Drag Force and Terminal Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302Chapter 7: Work and Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3277.1 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3287.2 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3387.3 Work-Energy Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3417.4 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345Chapter 8: Potential Energy and Conservation of Energy . . . . . . . . . . . . . . . . . . . 3598.1 Potential Energy of a System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3608.2 Conservative and Non-Conservative Forces . . . . . . . . . . . . . . . . . . . . . . . 3688.3 Conservation of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3728.4 Potential Energy Diagrams and Stability . . . . . . . . . . . . . . . . . . . . . . . . . 3778.5 Sources of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382Chapter 9: Linear Momentum and Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . 3979.1 Linear Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3989.2 Impulse and Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4009.3 Conservation of Linear Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4149.4 Types of Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425




9.5 Collisions in Multiple Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4339.6 Center of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4409.7 Rocket Propulsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455Chapter 10: Fixed-Axis Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47510.1 Rotational Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47610.2 Rotation with Constant Angular Acceleration . . . . . . . . . . . . . . . . . . . . . . 48610.3 Relating Angular and Translational Quantities . . . . . . . . . . . . . . . . . . . . . . 49210.4 Moment of Inertia and Rotational Kinetic Energy . . . . . . . . . . . . . . . . . . . . 49710.5 Calculating Moments of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50510.6 Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51310.7 Newton’s Second Law for Rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . 51910.8 Work and Power for Rotational Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 523Chapter 11: Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54511.1 Rolling Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54611.2 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55411.3 Conservation of Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . 56411.4 Precession of a Gyroscope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 571Chapter 12: Static Equilibrium and Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . 58712.1 Conditions for Static Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58812.2 Examples of Static Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59612.3 Stress, Strain, and Elastic Modulus . . . . . . . . . . . . . . . . . . . . . . . . . . . 60912.4 Elasticity and Plasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 619Chapter 13: Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63513.1 Newton's Law of Universal Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . 63613.2 Gravitation Near Earth's Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64113.3 Gravitational Potential Energy and Total Energy . . . . . . . . . . . . . . . . . . . . . 64813.4 Satellite Orbits and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65313.5 Kepler's Laws of Planetary Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66013.6 Tidal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66713.7 Einstein's Theory of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 672Chapter 14: Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69114.1 Fluids, Density, and Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69214.2 Measuring Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70314.3 Pascal's Principle and Hydraulics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70814.4 Archimedes’ Principle and Buoyancy . . . . . . . . . . . . . . . . . . . . . . . . . . 71314.5 Fluid Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71714.6 Bernoulli’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72214.7 Viscosity and Turbulence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 728Unit 2. Waves and AcousticsChapter 15: Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74915.1 Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75015.2 Energy in Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 76015.3 Comparing Simple Harmonic Motion and Circular Motion . . . . . . . . . . . . . . . . 76615.4 Pendulums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77015.5 Damped Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77715.6 Forced Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 780Chapter 16: Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79516.1 Traveling Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79616.2 Mathematics of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80216.3 Wave Speed on a Stretched String . . . . . . . . . . . . . . . . . . . . . . . . . . . 81116.4 Energy and Power of a Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81416.5 Interference of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81916.6 Standing Waves and Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 828Chapter 17: Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85317.1 Sound Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85417.2 Speed of Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85617.3 Sound Intensity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86517.4 Normal Modes of a Standing Sound Wave . . . . . . . . . . . . . . . . . . . . . . . 87317.5 Sources of Musical Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 881


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17.6 Beats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88617.7 The Doppler Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88817.8 Shock Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 895Appendix A: Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 913Appendix B: Conversion Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 917Appendix C: Fundamental Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 921Appendix D: Astronomical Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 923Appendix E: Mathematical Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 925Appendix F: Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 929Appendix G: The Greek Alphabet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 931Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 991




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PREFACE
Welcome to University Physics, an OpenStax resource. This textbook was written to increase student access to high-quality learning materials, maintaining highest standards of academic rigor at little to no cost.
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About University Physics
University Physics is designed for the two- or three-semester calculus-based physics course. The text has been developed to meet the scope and sequence of most university physics courses and provides a foundation for a career in mathematics, science, or engineering. The book provides an important opportunity for students to learn the core concepts of physics and understand how those concepts apply to their lives and to the world around them.
Due to the comprehensive nature of the material, we are offering the book in three volumes for flexibility and efficiency.
Coverage and Scope
Our University Physics textbook adheres to the scope and sequence of most two- and three-semester physics courses nationwide. We have worked to make physics interesting and accessible to students while maintaining the mathematical rigor inherent in the subject. With this objective in mind, the content of this textbook has been developed and arranged to provide a logical progression from fundamental to more advanced concepts, building upon what students have already learned and emphasizing connections between topics and between theory and applications. The goal of each section is to enable students not just to recognize concepts, but to work with them in ways that will be useful in later courses and future careers. The organization and pedagogical features were developed and vetted with feedback from science educators dedicated to the project.


Preface 1




VOLUME I
Unit 1: Mechanics


Chapter 1: Units and Measurement
Chapter 2: Vectors
Chapter 3: Motion Along a Straight Line
Chapter 4: Motion in Two and Three Dimensions
Chapter 5: Newton’s Laws of Motion
Chapter 6: Applications of Newton’s Laws
Chapter 7: Work and Kinetic Energy
Chapter 8: Potential Energy and Conservation of Energy
Chapter 9: Linear Momentum and Collisions
Chapter 10: Fixed-Axis Rotation
Chapter 11: Angular Momentum
Chapter 12: Static Equilibrium and Elasticity
Chapter 13: Gravitation
Chapter 14: Fluid Mechanics


Unit 2: Waves and Acoustics
Chapter 15: Oscillations
Chapter 16: Waves
Chapter 17: Sound


VOLUME II
Unit 1: Thermodynamics


Chapter 1: Temperature and Heat
Chapter 2: The Kinetic Theory of Gases
Chapter 3: The First Law of Thermodynamics
Chapter 4: The Second Law of Thermodynamics


Unit 2: Electricity and Magnetism
Chapter 5: Electric Charges and Fields
Chapter 6: Gauss’s Law
Chapter 7: Electric Potential
Chapter 8: Capacitance
Chapter 9: Current and Resistance
Chapter 10: Direct-Current Circuits
Chapter 11: Magnetic Forces and Fields
Chapter 12: Sources of Magnetic Fields
Chapter 13: Electromagnetic Induction
Chapter 14: Inductance
Chapter 15: Alternating-Current Circuits
Chapter 16: Electromagnetic Waves


VOLUME III
Unit 1: Optics


Chapter 1: The Nature of Light


2 Preface


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Chapter 2: Geometric Optics and Image Formation
Chapter 3: Interference
Chapter 4: Diffraction


Unit 2: Modern Physics
Chapter 5: Relativity
Chapter 6: Photons and Matter Waves
Chapter 7: Quantum Mechanics
Chapter 8: Atomic Structure
Chapter 9: Condensed Matter Physics
Chapter 10: Nuclear Physics
Chapter 11: Particle Physics and Cosmology


Pedagogical Foundation
Throughout University Physics you will find derivations of concepts that present classical ideas and techniques, as wellas modern applications and methods. Most chapters start with observations or experiments that place the material in acontext of physical experience. Presentations and explanations rely on years of classroom experience on the part of long-time physics professors, striving for a balance of clarity and rigor that has proven successful with their students. Throughoutthe text, links enable students to review earlier material and then return to the present discussion, reinforcing connectionsbetween topics. Key historical figures and experiments are discussed in the main text (rather than in boxes or sidebars),maintaining a focus on the development of physical intuition. Key ideas, definitions, and equations are highlighted inthe text and listed in summary form at the end of each chapter. Examples and chapter-opening images often includecontemporary applications from daily life or modern science and engineering that students can relate to, from smart phonesto the internet to GPS devices.
Assessments That Reinforce Key Concepts
In-chapter Examples generally follow a three-part format of Strategy, Solution, and Significance to emphasize how toapproach a problem, how to work with the equations, and how to check and generalize the result. Examples are oftenfollowed by Check Your Understanding questions and answers to help reinforce for students the important ideas of theexamples. Problem-Solving Strategies in each chapter break down methods of approaching various types of problems intosteps students can follow for guidance. The book also includes exercises at the end of each chapter so students can practicewhat they’ve learned.


Conceptual questions do not require calculation but test student learning of the key concepts.
Problems categorized by section test student problem-solving skills and the ability to apply ideas to practicalsituations.
Additional Problems apply knowledge across the chapter, forcing students to identify what concepts and equationsare appropriate for solving given problems. Randomly located throughout the problems are Unreasonable Resultsexercises that ask students to evaluate the answer to a problem and explain why it is not reasonable and whatassumptions made might not be correct.
Challenge Problems extend text ideas to interesting but difficult situations.


Answers for selected exercises are available in an Answer Key at the end of the book.
Additional ResourcesStudent and Instructor Resources
We’ve compiled additional resources for both students and instructors, including Getting Started Guides, PowerPoint slides,and answer and solution guides for instructors and students. Instructor resources require a verified instructor account, whichcan be requested on your openstax.org log-in. Take advantage of these resources to supplement your OpenStax book.
Partner Resources
OpenStax partners are our allies in the mission to make high-quality learning materials affordable and accessible to studentsand instructors everywhere. Their tools integrate seamlessly with our OpenStax titles at a low cost. To access the partnerresources for your text, visit your book page on openstax.org.


Preface 3




About the AuthorsSenior Contributing Authors
Samuel J. Ling, Truman State UniversityDr. Samuel Ling has taught introductory and advanced physics for over 25 years at Truman State University, where he iscurrently Professor of Physics and the Department Chair. Dr. Ling has two PhDs from Boston University, one in Chemistryand the other in Physics, and he was a Research Fellow at the Indian Institute of Science, Bangalore, before joining Truman.Dr. Ling is also an author of A First Course in Vibrations and Waves, published by Oxford University Press. Dr. Ling hasconsiderable experience with research in Physics Education and has published research on collaborative learning methods inphysics teaching. He was awarded a Truman Fellow and a Jepson fellow in recognition of his innovative teaching methods.Dr. Ling’s research publications have spanned Cosmology, Solid State Physics, and Nonlinear Optics.
Jeff Sanny, Loyola Marymount UniversityDr. Jeff Sanny earned a BS in Physics from Harvey Mudd College in 1974 and a PhD in Solid State Physics from theUniversity of California–Los Angeles in 1980. He joined the faculty at Loyola Marymount University in the fall of 1980.During his tenure, he has served as department Chair as well as Associate Dean. Dr. Sanny enjoys teaching introductoryphysics in particular. He is also passionate about providing students with research experience and has directed an activeundergraduate student research group in space physics for many years.
Bill Moebs, PhDDr. William Moebs earned a BS and PhD (1959 and 1965) from the University of Michigan. He then joined their staffas a Research Associate for one year, where he continued his doctoral research in particle physics. In 1966, he acceptedan appointment to the Physics Department of Indiana Purdue Fort Wayne (IPFW), where he served as Department Chairfrom 1971 to 1979. In 1979, he moved to Loyola Marymount University (LMU), where he served as Chair of the PhysicsDepartment from 1979 to 1986. He retired from LMU in 2000. He has published research in particle physics, chemicalkinetics, cell division, atomic physics, and physics teaching.
Contributing Authors
David Anderson, Albion CollegeDaniel Bowman, Ferrum CollegeDedra Demaree, Georgetown UniversityGerald Friedman, Santa Fe Community CollegeLev Gasparov, University of North FloridaEdw. S. Ginsberg, University of MassachusettsAlice Kolakowska, University of MemphisLee LaRue, Paris Junior CollegeMark Lattery, University of WisconsinRichard Ludlow, Daniel Webster CollegePatrick Motl, Indiana University–KokomoTao Pang, University of Nevada–Las VegasKenneth Podolak, Plattsburgh State UniversityTakashi Sato, Kwantlen Polytechnic UniversityDavid Smith, University of the Virgin IslandsJoseph Trout, Richard Stockton CollegeKevin Wheelock, Bellevue College
Reviewers
Salameh Ahmad, Rochester Institute of Technology–DubaiJohn Aiken, University of Colorado–BoulderAnand Batra, Howard UniversityRaymond Benge, Terrant County CollegeGavin Buxton, Robert Morris UniversityErik Christensen, South Florida State CollegeClifton Clark, Fort Hays State UniversityNelson Coates, California Maritime AcademyHerve Collin, Kapi’olani Community CollegeCarl Covatto, Arizona State UniversityAlexander Cozzani, Imperial Valley CollegeDanielle Dalafave, The College of New JerseyNicholas Darnton, Georgia Institute of Technology


4 Preface


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Robert Edmonds, Tarrant County CollegeWilliam Falls, Erie Community CollegeStanley Forrester, Broward CollegeUmesh Garg, University of Notre DameMaurizio Giannotti, Barry UniversityBryan Gibbs, Dallas County Community CollegeMark Giroux, East Tennessee State UniversityMatthew Griffiths, University of New HavenAlfonso Hinojosa, University of Texas–ArlingtonSteuard Jensen, Alma CollegeDavid Kagan, University of MassachusettsJill Leggett, Florida State College–JacksonvilleSergei Katsev, University of Minnesota–DuluthAlfredo Louro, University of CalgaryJames Maclaren, Tulane UniversityPonn Maheswaranathan, Winthrop UniversitySeth Major, Hamilton CollegeOleg Maksimov, Excelsior CollegeAristides Marcano, Delaware State UniversityMarles McCurdy, Tarrant County CollegeJames McDonald, University of HartfordRalph McGrew, SUNY–Broome Community CollegePaul Miller, West Virginia UniversityTamar More, University of PortlandFarzaneh Najmabadi, University of PhoenixRichard Olenick, The University of DallasChristopher Porter, Ohio State UniversityLiza Pujji, Manakau Institute of TechnologyBaishali Ray, Young Harris UniversityAndrew Robinson, Carleton UniversityAruvana Roy, Young Harris UniversityAbhijit Sarkar, The Catholic University of AmericaGajendra Tulsian, Daytona State CollegeAdria Updike, Roger Williams UniversityClark Vangilder, Central Arizona UniversitySteven Wolf, Texas State UniversityAlexander Wurm, Western New England UniversityLei Zhang, Winston Salem State UniversityUlrich Zurcher, Cleveland State University


Preface 5




Preface


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1 | UNITS ANDMEASUREMENT


Figure 1.1 This image might be showing any number of things. It might be a whirlpool in a tank of water or perhaps a collageof paint and shiny beads done for art class. Without knowing the size of the object in units we all recognize, such as meters orinches, it is difficult to know what we’re looking at. In fact, this image shows the Whirlpool Galaxy (and its companion galaxy),
which is about 60,000 light-years in diameter (about 6 × 1017km across). (credit: S. Beckwith (STScI) Hubble Heritage Team,
(STScI/AURA), ESA, NASA)


Chapter Outline
1.1 The Scope and Scale of Physics
1.2 Units and Standards
1.3 Unit Conversion
1.4 Dimensional Analysis
1.5 Estimates and Fermi Calculations
1.6 Significant Figures
1.7 Solving Problems in Physics


Introduction
As noted in the figure caption, the chapter-opening image is of the Whirlpool Galaxy, which we examine in the first sectionof this chapter. Galaxies are as immense as atoms are small, yet the same laws of physics describe both, along with all therest of nature—an indication of the underlying unity in the universe. The laws of physics are surprisingly few, implying anunderlying simplicity to nature’s apparent complexity. In this text, you learn about the laws of physics. Galaxies and atomsmay seem far removed from your daily life, but as you begin to explore this broad-ranging subject, you may soon come to


Chapter 1 | Units and Measurement 7




realize that physics plays a much larger role in your life than you first thought, no matter your life goals or career choice.
1.1 | The Scope and Scale of Physics


Learning Objectives
By the end of this section, you will be able to:
• Describe the scope of physics.
• Calculate the order of magnitude of a quantity.
• Compare measurable length, mass, and timescales quantitatively.
• Describe the relationships among models, theories, and laws.


Physics is devoted to the understanding of all natural phenomena. In physics, we try to understand physical phenomena at allscales—from the world of subatomic particles to the entire universe. Despite the breadth of the subject, the various subfieldsof physics share a common core. The same basic training in physics will prepare you to work in any area of physics and therelated areas of science and engineering. In this section, we investigate the scope of physics; the scales of length, mass, andtime over which the laws of physics have been shown to be applicable; and the process by which science in general, andphysics in particular, operates.
The Scope of Physics
Take another look at the chapter-opening image. The Whirlpool Galaxy contains billions of individual stars as well as hugeclouds of gas and dust. Its companion galaxy is also visible to the right. This pair of galaxies lies a staggering billion
trillion miles (1.4 × 1021mi) from our own galaxy (which is called the Milky Way). The stars and planets that make up
the Whirlpool Galaxy might seem to be the furthest thing from most people’s everyday lives, but the Whirlpool is a greatstarting point to think about the forces that hold the universe together. The forces that cause the Whirlpool Galaxy to act asit does are thought to be the same forces we contend with here on Earth, whether we are planning to send a rocket into spaceor simply planning to raise the walls for a new home. The gravity that causes the stars of the Whirlpool Galaxy to rotate andrevolve is thought to be the same as what causes water to flow over hydroelectric dams here on Earth. When you look upat the stars, realize the forces out there are the same as the ones here on Earth. Through a study of physics, you may gain agreater understanding of the interconnectedness of everything we can see and know in this universe.
Think, now, about all the technological devices you use on a regular basis. Computers, smartphones, global positioningsystems (GPSs), MP3 players, and satellite radio might come to mind. Then, think about the most exciting moderntechnologies you have heard about in the news, such as trains that levitate above tracks, “invisibility cloaks” that bend lightaround them, and microscopic robots that fight cancer cells in our bodies. All these groundbreaking advances, commonplaceor unbelievable, rely on the principles of physics. Aside from playing a significant role in technology, professionals suchas engineers, pilots, physicians, physical therapists, electricians, and computer programmers apply physics concepts in theirdaily work. For example, a pilot must understand how wind forces affect a flight path; a physical therapist must understandhow the muscles in the body experience forces as they move and bend. As you will learn in this text, the principles ofphysics are propelling new, exciting technologies, and these principles are applied in a wide range of careers.
The underlying order of nature makes science in general, and physics in particular, interesting and enjoyable to study. Forexample, what do a bag of chips and a car battery have in common? Both contain energy that can be converted to otherforms. The law of conservation of energy (which says that energy can change form but is never lost) ties together suchtopics as food calories, batteries, heat, light, and watch springs. Understanding this law makes it easier to learn about thevarious forms energy takes and how they relate to one another. Apparently unrelated topics are connected through broadlyapplicable physical laws, permitting an understanding beyond just the memorization of lists of facts.
Science consists of theories and laws that are the general truths of nature, as well as the body of knowledge they encompass.Scientists are continuously trying to expand this body of knowledge and to perfect the expression of the laws that describeit. Physics, which comes from the Greek phúsis, meaning “nature,” is concerned with describing the interactions ofenergy, matter, space, and time to uncover the fundamental mechanisms that underlie every phenomenon. This concern fordescribing the basic phenomena in nature essentially defines the scope of physics.
Physics aims to understand the world around us at the most basic level. It emphasizes the use of a small number ofquantitative laws to do this, which can be useful to other fields pushing the performance boundaries of existing technologies.Consider a smartphone (Figure 1.2). Physics describes how electricity interacts with the various circuits inside the device.This knowledge helps engineers select the appropriate materials and circuit layout when building a smartphone. Knowledge


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of the physics underlying these devices is required to shrink their size or increase their processing speed. Or, think about aGPS. Physics describes the relationship between the speed of an object, the distance over which it travels, and the time ittakes to travel that distance. When you use a GPS in a vehicle, it relies on physics equations to determine the travel timefrom one location to another.


Figure 1.2 The Apple iPhone is a common smartphone with aGPS function. Physics describes the way that electricity flowsthrough the circuits of this device. Engineers use theirknowledge of physics to construct an iPhone with features thatconsumers will enjoy. One specific feature of an iPhone is theGPS function. A GPS uses physics equations to determine thedrive time between two locations on a map.


Knowledge of physics is useful in everyday situations as well as in nonscientific professions. It can help you understandhow microwave ovens work, why metals should not be put into them, and why they might affect pacemakers. Physics allowsyou to understand the hazards of radiation and to evaluate these hazards rationally and more easily. Physics also explains thereason why a black car radiator helps remove heat in a car engine, and it explains why a white roof helps keep the inside ofa house cool. Similarly, the operation of a car’s ignition system as well as the transmission of electrical signals throughoutour body’s nervous system are much easier to understand when you think about them in terms of basic physics.
Physics is a key element of many important disciplines and contributes directly to others. Chemistry, for example—sinceit deals with the interactions of atoms and molecules—has close ties to atomic and molecular physics. Most branches ofengineering are concerned with designing new technologies, processes, or structures within the constraints set by the lawsof physics. In architecture, physics is at the heart of structural stability and is involved in the acoustics, heating, lighting, andcooling of buildings. Parts of geology rely heavily on physics, such as radioactive dating of rocks, earthquake analysis, andheat transfer within Earth. Some disciplines, such as biophysics and geophysics, are hybrids of physics and other disciplines.
Physics has many applications in the biological sciences. On the microscopic level, it helps describe the properties ofcells and their environments. On the macroscopic level, it explains the heat, work, and power associated with the humanbody and its various organ systems. Physics is involved in medical diagnostics, such as radiographs, magnetic resonanceimaging, and ultrasonic blood flow measurements. Medical therapy sometimes involves physics directly; for example,cancer radiotherapy uses ionizing radiation. Physics also explains sensory phenomena, such as how musical instrumentsmake sound, how the eye detects color, and how lasers transmit information.
It is not necessary to study all applications of physics formally. What is most useful is knowing the basic laws of physicsand developing skills in the analytical methods for applying them. The study of physics also can improve your problem-solving skills. Furthermore, physics retains the most basic aspects of science, so it is used by all the sciences, and the studyof physics makes other sciences easier to understand.


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The Scale of Physics
From the discussion so far, it should be clear that to accomplish your goals in any of the various fields within the naturalsciences and engineering, a thorough grounding in the laws of physics is necessary. The reason for this is simply that thelaws of physics govern everything in the observable universe at all measurable scales of length, mass, and time. Now, thatis easy enough to say, but to come to grips with what it really means, we need to get a little bit quantitative. So, beforesurveying the various scales that physics allows us to explore, let’s first look at the concept of “order of magnitude,” whichwe use to come to terms with the vast ranges of length, mass, and time that we consider in this text (Figure 1.3).


Figure 1.3 (a) Using a scanning tunneling microscope, scientists can see the individual atoms (diameters around 10–10 m) thatcompose this sheet of gold. (b) Tiny phytoplankton swim among crystals of ice in the Antarctic Sea. They range from a fewmicrometers (1 μm is 10–6 m) to as much as 2 mm (1 mm is 10–3 m) in length. (c) These two colliding galaxies, known as NGC4676A (right) and NGC 4676B (left), are nicknamed “The Mice” because of the tail of gas emanating from each one. They arelocated 300 million light-years from Earth in the constellation Coma Berenices. Eventually, these two galaxies will merge intoone. (credit a: modification of work by Erwinrossen; credit b: modification of work by Prof. Gordon T. Taylor, Stony BrookUniversity; NOAA Corps Collections; credit c: modification of work by NASA, H. Ford (JHU), G. Illingworth (UCSC/LO), M.Clampin (STScI), G. Hartig (STScI), the ACS Science Team, and ESA)
Order of magnitude
The order of magnitude of a number is the power of 10 that most closely approximates it. Thus, the order of magnituderefers to the scale (or size) of a value. Each power of 10 represents a different order of magnitude. For example,
101, 102, 103, and so forth, are all different orders of magnitude, as are 100 = 1, 10−1, 10−2, and 10−3. To find the
order of magnitude of a number, take the base-10 logarithm of the number and round it to the nearest integer, then the orderof magnitude of the number is simply the resulting power of 10. For example, the order of magnitude of 800 is 103 because
log10800 ≈ 2.903, which rounds to 3. Similarly, the order of magnitude of 450 is 103 because log10450 ≈ 2.653,
which rounds to 3 as well. Thus, we say the numbers 800 and 450 are of the same order of magnitude: 103. However, theorder of magnitude of 250 is 102 because log10250 ≈ 2.397, which rounds to 2.
An equivalent but quicker way to find the order of magnitude of a number is first to write it in scientific notation and then
check to see whether the first factor is greater than or less than 10 = 100.5 ≈ 3. The idea is that 10 = 100.5 is halfway
between 1 = 100 and 10 = 101 on a log base-10 scale. Thus, if the first factor is less than 10, then we round it down
to 1 and the order of magnitude is simply whatever power of 10 is required to write the number in scientific notation. On
the other hand, if the first factor is greater than 10, then we round it up to 10 and the order of magnitude is one power of
10 higher than the power needed to write the number in scientific notation. For example, the number 800 can be written in
scientific notation as 8 × 102. Because 8 is bigger than 10 ≈ 3, we say the order of magnitude of 800 is 102 + 1 = 103.
The number 450 can be written as 4.5 × 102, so its order of magnitude is also 103 because 4.5 is greater than 3. However,
250 written in scientific notation is 2.5 × 102 and 2.5 is less than 3, so its order of magnitude is 102.
The order of magnitude of a number is designed to be a ballpark estimate for the scale (or size) of its value. It is simply away of rounding numbers consistently to the nearest power of 10. This makes doing rough mental math with very big andvery small numbers easier. For example, the diameter of a hydrogen atom is on the order of 10−10 m, whereas the diameter
of the Sun is on the order of 109 m, so it would take roughly 109 /10−10 = 1019 hydrogen atoms to stretch across the


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diameter of the Sun. This is much easier to do in your head than using the more precise values of 1.06 × 10−10m for a
hydrogen atom diameter and 1.39 × 109m for the Sun’s diameter, to find that it would take 1.31 × 1019 hydrogen atoms
to stretch across the Sun’s diameter. In addition to being easier, the rough estimate is also nearly as informative as the precisecalculation.
Known ranges of length, mass, and time
The vastness of the universe and the breadth over which physics applies are illustrated by the wide range of examples ofknown lengths, masses, and times (given as orders of magnitude) in Figure 1.4. Examining this table will give you afeeling for the range of possible topics in physics and numerical values. A good way to appreciate the vastness of the rangesof values in Figure 1.4 is to try to answer some simple comparative questions, such as the following:


• How many hydrogen atoms does it take to stretch across the diameter of the Sun?(Answer: 109 m/10–10 m = 1019 hydrogen atoms)
• How many protons are there in a bacterium?(Answer: 10–15 kg/10–27 kg = 1012 protons)
• How many floating-point operations can a supercomputer do in 1 day?(Answer: 105 s/10–17 s = 1022 floating-point operations)


In studying Figure 1.4, take some time to come up with similar questions that interest you and then try answering them.Doing this can breathe some life into almost any table of numbers.


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Figure 1.4 This table shows the orders of magnitude of length, mass, and time.
Visit this site (https://openstaxcollege.org/l/21scaleuniv) to explore interactively the vast range of lengthscales in our universe. Scroll down and up the scale to view hundreds of organisms and objects, and click on theindividual objects to learn more about each one.


Building Models
How did we come to know the laws governing natural phenomena? What we refer to as the laws of nature are concisedescriptions of the universe around us. They are human statements of the underlying laws or rules that all natural processesfollow. Such laws are intrinsic to the universe; humans did not create them and cannot change them. We can only discoverand understand them. Their discovery is a very human endeavor, with all the elements of mystery, imagination, struggle,triumph, and disappointment inherent in any creative effort (Figure 1.5). The cornerstone of discovering natural laws isobservation; scientists must describe the universe as it is, not as we imagine it to be.


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Figure 1.5 (a) Enrico Fermi (1901–1954) was born in Italy. On accepting the Nobel Prize inStockholm in 1938 for his work on artificial radioactivity produced by neutrons, he took hisfamily to America rather than return home to the government in power at the time. He became anAmerican citizen and was a leading participant in the Manhattan Project. (b) Marie Curie(1867–1934) sacrificed monetary assets to help finance her early research and damaged herphysical well-being with radiation exposure. She is the only person to win Nobel prizes in bothphysics and chemistry. One of her daughters also won a Nobel Prize. (credit a: United StatesDepartment of Energy)


A model is a representation of something that is often too difficult (or impossible) to display directly. Although a modelis justified by experimental tests, it is only accurate in describing certain aspects of a physical system. An example is theBohr model of single-electron atoms, in which the electron is pictured as orbiting the nucleus, analogous to the way planetsorbit the Sun (Figure 1.6). We cannot observe electron orbits directly, but the mental image helps explain some of theobservations we can make, such as the emission of light from hot gases (atomic spectra). However, other observationsshow that the picture in the Bohr model is not really what atoms look like. The model is “wrong,” but is still useful forsome purposes. Physicists use models for a variety of purposes. For example, models can help physicists analyze a scenarioand perform a calculation or models can be used to represent a situation in the form of a computer simulation. Ultimately,however, the results of these calculations and simulations need to be double-checked by other means—namely, observationand experimentation.


Figure 1.6 What is a model? The Bohr model of a single-electron atom shows the electron orbiting the nucleus in one ofseveral possible circular orbits. Like all models, it capturessome, but not all, aspects of the physical system.


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The word theory means something different to scientists than what is often meant when the word is used in everydayconversation. In particular, to a scientist a theory is not the same as a “guess” or an “idea” or even a “hypothesis.” Thephrase “it’s just a theory” seems meaningless and silly to scientists because science is founded on the notion of theories. Toa scientist, a theory is a testable explanation for patterns in nature supported by scientific evidence and verified multipletimes by various groups of researchers. Some theories include models to help visualize phenomena whereas others do not.Newton’s theory of gravity, for example, does not require a model or mental image, because we can observe the objectsdirectly with our own senses. The kinetic theory of gases, on the other hand, is a model in which a gas is viewed as beingcomposed of atoms and molecules. Atoms and molecules are too small to be observed directly with our senses—thus, wepicture them mentally to understand what the instruments tell us about the behavior of gases. Although models are meantonly to describe certain aspects of a physical system accurately, a theory should describe all aspects of any system that fallswithin its domain of applicability. In particular, any experimentally testable implication of a theory should be verified. If anexperiment ever shows an implication of a theory to be false, then the theory is either thrown out or modified suitably (forexample, by limiting its domain of applicability).
A law uses concise language to describe a generalized pattern in nature supported by scientific evidence and repeatedexperiments. Often, a law can be expressed in the form of a single mathematical equation. Laws and theories are similarin that they are both scientific statements that result from a tested hypothesis and are supported by scientific evidence.However, the designation law is usually reserved for a concise and very general statement that describes phenomena innature, such as the law that energy is conserved during any process, or Newton’s second law of motion, which relates force(F), mass (m), and acceleration (a) by the simple equation F = ma. A theory, in contrast, is a less concise statement of
observed behavior. For example, the theory of evolution and the theory of relativity cannot be expressed concisely enough tobe considered laws. The biggest difference between a law and a theory is that a theory is much more complex and dynamic.A law describes a single action whereas a theory explains an entire group of related phenomena. Less broadly applicablestatements are usually called principles (such as Pascal’s principle, which is applicable only in fluids), but the distinctionbetween laws and principles often is not made carefully.
The models, theories, and laws we devise sometimes imply the existence of objects or phenomena that are as yetunobserved. These predictions are remarkable triumphs and tributes to the power of science. It is the underlying order inthe universe that enables scientists to make such spectacular predictions. However, if experimentation does not verify ourpredictions, then the theory or law is wrong, no matter how elegant or convenient it is. Laws can never be known withabsolute certainty because it is impossible to perform every imaginable experiment to confirm a law for every possiblescenario. Physicists operate under the assumption that all scientific laws and theories are valid until a counterexample isobserved. If a good-quality, verifiable experiment contradicts a well-established law or theory, then the law or theory mustbe modified or overthrown completely.
The study of science in general, and physics in particular, is an adventure much like the exploration of an uncharted ocean.Discoveries are made; models, theories, and laws are formulated; and the beauty of the physical universe is made moresublime for the insights gained.
1.2 | Units and Standards


Learning Objectives
By the end of this section, you will be able to:
• Describe how SI base units are defined.
• Describe how derived units are created from base units.
• Express quantities given in SI units using metric prefixes.


As we saw previously, the range of objects and phenomena studied in physics is immense. From the incredibly short lifetimeof a nucleus to the age of Earth, from the tiny sizes of subnuclear particles to the vast distance to the edges of the knownuniverse, from the force exerted by a jumping flea to the force between Earth and the Sun, there are enough factors of 10to challenge the imagination of even the most experienced scientist. Giving numerical values for physical quantities andequations for physical principles allows us to understand nature much more deeply than qualitative descriptions alone. Tocomprehend these vast ranges, we must also have accepted units in which to express them. We shall find that even in thepotentially mundane discussion of meters, kilograms, and seconds, a profound simplicity of nature appears: all physicalquantities can be expressed as combinations of only seven base physical quantities.


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We define a physical quantity either by specifying how it is measured or by stating how it is calculated from othermeasurements. For example, we might define distance and time by specifying methods for measuring them, such as usinga meter stick and a stopwatch. Then, we could define average speed by stating that it is calculated as the total distancetraveled divided by time of travel.
Measurements of physical quantities are expressed in terms of units, which are standardized values. For example, the lengthof a race, which is a physical quantity, can be expressed in units of meters (for sprinters) or kilometers (for distance runners).Without standardized units, it would be extremely difficult for scientists to express and compare measured values in ameaningful way (Figure 1.7).


Figure 1.7 Distances given in unknown units are maddeninglyuseless.


Two major systems of units are used in the world: SI units (for the French Système International d’Unités), also knownas the metric system, and English units (also known as the customary or imperial system). English units were historicallyused in nations once ruled by the British Empire and are still widely used in the United States. English units may also bereferred to as the foot–pound–second (fps) system, as opposed to the centimeter–gram–second (cgs) system. You may alsoencounter the term SAE units, named after the Society of Automotive Engineers. Products such as fasteners and automotivetools (for example, wrenches) that are measured in inches rather than metric units are referred to as SAE fasteners or SAEwrenches.
Virtually every other country in the world (except the United States) now uses SI units as the standard. The metric systemis also the standard system agreed on by scientists and mathematicians.
SI Units: Base and Derived Units
In any system of units, the units for some physical quantities must be defined through a measurement process. These arecalled the base quantities for that system and their units are the system’s base units. All other physical quantities canthen be expressed as algebraic combinations of the base quantities. Each of these physical quantities is then known as aderived quantity and each unit is called a derived unit. The choice of base quantities is somewhat arbitrary, as long as theyare independent of each other and all other quantities can be derived from them. Typically, the goal is to choose physicalquantities that can be measured accurately to a high precision as the base quantities. The reason for this is simple. Since thederived units can be expressed as algebraic combinations of the base units, they can only be as accurate and precise as thebase units from which they are derived.
Based on such considerations, the International Standards Organization recommends using seven base quantities, whichform the International System of Quantities (ISQ). These are the base quantities used to define the SI base units. Table 1.1lists these seven ISQ base quantities and the corresponding SI base units.


ISQ Base Quantity SI Base Unit
Length meter (m)
Mass kilogram (kg)
Time second (s)


Table 1.1 ISQ Base Quantities and Their SI Units


Chapter 1 | Units and Measurement 15




ISQ Base Quantity SI Base Unit
Electrical current ampere (A)
Thermodynamic temperature kelvin (K)
Amount of substance mole (mol)
Luminous intensity candela (cd)


Table 1.1 ISQ Base Quantities and Their SI Units
You are probably already familiar with some derived quantities that can be formed from the base quantities in Table 1.1.For example, the geometric concept of area is always calculated as the product of two lengths. Thus, area is a derived
quantity that can be expressed in terms of SI base units using square meters (m × m = m2). Similarly, volume is a derived
quantity that can be expressed in cubic meters (m3). Speed is length per time; so in terms of SI base units, we could
measure it in meters per second (m/s). Volume mass density (or just density) is mass per volume, which is expressed interms of SI base units such as kilograms per cubic meter (kg/m3). Angles can also be thought of as derived quantitiesbecause they can be defined as the ratio of the arc length subtended by two radii of a circle to the radius of the circle. Thisis how the radian is defined. Depending on your background and interests, you may be able to come up with other derivedquantities, such as the mass flow rate (kg/s) or volume flow rate (m3/s) of a fluid, electric charge (A · s), mass flux density
[kg/(m2 · s)], and so on. We will see many more examples throughout this text. For now, the point is that every physical
quantity can be derived from the seven base quantities in Table 1.1, and the units of every physical quantity can be derivedfrom the seven SI base units.
For the most part, we use SI units in this text. Non-SI units are used in a few applications in which they are in very commonuse, such as the measurement of temperature in degrees Celsius (°C), the measurement of fluid volume in liters (L), and
the measurement of energies of elementary particles in electron-volts (eV). Whenever non-SI units are discussed, they are
tied to SI units through conversions. For example, 1 L is 10−3 m3.


Check out a comprehensive source of information on SI units (https://openstaxcollege.org/l/21SIUnits) atthe National Institute of Standards and Technology (NIST) Reference on Constants, Units, and Uncertainty.


Units of Time, Length, and Mass: The Second, Meter, and Kilogram
The initial chapters in this textbook are concerned with mechanics, fluids, and waves. In these subjects all pertinent physicalquantities can be expressed in terms of the base units of length, mass, and time. Therefore, we now turn to a discussion ofthese three base units, leaving discussion of the others until they are needed later.
The second
The SI unit for time, the second (abbreviated s), has a long history. For many years it was defined as 1/86,400 of a meansolar day. More recently, a new standard was adopted to gain greater accuracy and to define the second in terms of anonvarying or constant physical phenomenon (because the solar day is getting longer as a result of the very gradual slowingof Earth’s rotation). Cesium atoms can be made to vibrate in a very steady way, and these vibrations can be readily observedand counted. In 1967, the second was redefined as the time required for 9,192,631,770 of these vibrations to occur (Figure1.8). Note that this may seem like more precision than you would ever need, but it isn’t—GPSs rely on the precision ofatomic clocks to be able to give you turn-by-turn directions on the surface of Earth, far from the satellites broadcasting theirlocation.


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Figure 1.8 An atomic clock such as this one uses thevibrations of cesium atoms to keep time to a precision of betterthan a microsecond per year. The fundamental unit of time, thesecond, is based on such clocks. This image looks down fromthe top of an atomic fountain nearly 30 feet tall. (credit: SteveJurvetson)
The meter
The SI unit for length is the meter (abbreviated m); its definition has also changed over time to become more precise. Themeter was first defined in 1791 as 1/10,000,000 of the distance from the equator to the North Pole. This measurement wasimproved in 1889 by redefining the meter to be the distance between two engraved lines on a platinum–iridium bar nowkept near Paris. By 1960, it had become possible to define the meter even more accurately in terms of the wavelength oflight, so it was again redefined as 1,650,763.73 wavelengths of orange light emitted by krypton atoms. In 1983, the meterwas given its current definition (in part for greater accuracy) as the distance light travels in a vacuum in 1/299,792,458 of asecond (Figure 1.9). This change came after knowing the speed of light to be exactly 299,792,458 m/s. The length of themeter will change if the speed of light is someday measured with greater accuracy.


Figure 1.9 The meter is defined to be the distance light travels in 1/299,792,458 of a second in a vacuum. Distancetraveled is speed multiplied by time.
The kilogram
The SI unit for mass is the kilogram (abbreviated kg); it is defined to be the mass of a platinum–iridium cylinderkept with the old meter standard at the International Bureau of Weights and Measures near Paris. Exact replicas of thestandard kilogram are also kept at the U.S. National Institute of Standards and Technology (NIST), located in Gaithersburg,Maryland, outside of Washington, DC, and at other locations around the world. Scientists at NIST are currently investigatingtwo complementary methods of redefining the kilogram (see Figure 1.10). The determination of all other masses can betraced ultimately to a comparison with the standard mass.


There is currently an effort to redefine the SI unit of mass in terms of more fundamental processes by 2018. Youcan explore the history of mass standards and the contenders in the quest to devise a new one at the website(https://openstaxcollege.org/l/21redefkilo) of the Physical Measurement Laboratory.


Chapter 1 | Units and Measurement 17




Figure 1.10 Redefining the SI unit of mass. Complementary methods are being investigated for use in an upcomingredefinition of the SI unit of mass. (a) The U.S. National Institute of Standards and Technology’s watt balance is a machine thatbalances the weight of a test mass against the current and voltage (the “watt”) produced by a strong system of magnets. (b) TheInternational Avogadro Project is working to redefine the kilogram based on the dimensions, mass, and other known properties ofa silicon sphere. (credit a and credit b: National Institute of Standards and Technology)
Metric Prefixes
SI units are part of the metric system, which is convenient for scientific and engineering calculations because the units arecategorized by factors of 10. Table 1.2 lists the metric prefixes and symbols used to denote various factors of 10 in SI units.For example, a centimeter is one-hundredth of a meter (in symbols, 1 cm = 10–2 m) and a kilometer is a thousand meters (1km = 103 m). Similarly, a megagram is a million grams (1 Mg = 106 g), a nanosecond is a billionth of a second (1 ns = 10–9s), and a terameter is a trillion meters (1 Tm = 1012 m).


Prefix Symbol Meaning Prefix Symbol Meaning
yotta- Y 1024 yocto- y 10–24
zetta- Z 1021 zepto- z 10–21
exa- E 1018 atto- a 10–18
peta- P 1015 femto- f 10–15
tera- T 1012 pico- p 10–12
giga- G 109 nano- n 10–9
mega- M 106 micro- µ 10–6
kilo- k 103 milli- m 10–3
hecto- h 102 centi- c 10–2
deka- da 101 deci- d 10–1


Table 1.2 Metric Prefixes for Powers of 10 and Their Symbols
The only rule when using metric prefixes is that you cannot “double them up.” For example, if you have measurements in
petameters (1 Pm = 1015 m), it is not proper to talk about megagigameters, although 106 × 109 = 1015. In practice, the


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1.1


only time this becomes a bit confusing is when discussing masses. As we have seen, the base SI unit of mass is the kilogram(kg), but metric prefixes need to be applied to the gram (g), because we are not allowed to “double-up” prefixes. Thus, athousand kilograms (103 kg) is written as a megagram (1 Mg) since
103kg = 103 × 103g = 106g = 1Mg.


Incidentally, 103 kg is also called a metric ton, abbreviated t. This is one of the units outside the SI system consideredacceptable for use with SI units.
As we see in the next section, metric systems have the advantage that conversions of units involve only powers of 10. Thereare 100 cm in 1 m, 1000 m in 1 km, and so on. In nonmetric systems, such as the English system of units, the relationshipsare not as simple—there are 12 in. in 1 ft, 5280 ft in 1 mi, and so on.
Another advantage of metric systems is that the same unit can be used over extremely large ranges of values simply byscaling it with an appropriate metric prefix. The prefix is chosen by the order of magnitude of physical quantities commonlyfound in the task at hand. For example, distances in meters are suitable in construction, whereas distances in kilometers areappropriate for air travel, and nanometers are convenient in optical design. With the metric system there is no need to inventnew units for particular applications. Instead, we rescale the units with which we are already familiar.
Example 1.1


Using Metric Prefixes
Restate the mass 1.93 × 1013kg using a metric prefix such that the resulting numerical value is bigger than one
but less than 1000.
Strategy
Since we are not allowed to “double-up” prefixes, we first need to restate the mass in grams by replacing theprefix symbol k with a factor of 103 (see Table 1.2). Then, we should see which two prefixes in Table 1.2 areclosest to the resulting power of 10 when the number is written in scientific notation. We use whichever of thesetwo prefixes gives us a number between one and 1000.
Solution
Replacing the k in kilogram with a factor of 103, we find that


1.93 × 1013kg = 1.93 × 1013 × 103g = 1.93 × 1016g.


From Table 1.2, we see that 1016 is between “peta-” (1015) and “exa-” (1018). If we use the “peta-” prefix, then
we find that 1.93 × 1016g = 1.93 × 101Pg, since 16 = 1 + 15. Alternatively, if we use the “exa-” prefix we
find that 1.93 × 1016g = 1.93 × 10−2Eg, since 16 = −2 + 18. Because the problem asks for the numerical
value between one and 1000, we use the “peta-” prefix and the answer is 19.3 Pg.
Significance
It is easy to make silly arithmetic errors when switching from one prefix to another, so it is always a good idea tocheck that our final answer matches the number we started with. An easy way to do this is to put both numbers inscientific notation and count powers of 10, including the ones hidden in prefixes. If we did not make a mistake,
the powers of 10 should match up. In this problem, we started with 1.93 × 1013kg, so we have 13 + 3 = 16
powers of 10. Our final answer in scientific notation is 1.93 × 101 Pg, so we have 1 + 15 = 16 powers of 10. So,
everything checks out.
If this mass arose from a calculation, we would also want to check to determine whether a mass this large makesany sense in the context of the problem. For this, Figure 1.4 might be helpful.


Check Your Understanding Restate 4.79 × 105kg using a metric prefix such that the resulting number
is bigger than one but less than 1000.


Chapter 1 | Units and Measurement 19




1.3 | Unit Conversion
Learning Objectives


By the end of this section, you will be able to:
• Use conversion factors to express the value of a given quantity in different units.


It is often necessary to convert from one unit to another. For example, if you are reading a European cookbook, somequantities may be expressed in units of liters and you need to convert them to cups. Or perhaps you are reading walkingdirections from one location to another and you are interested in how many miles you will be walking. In this case, you mayneed to convert units of feet or meters to miles.
Let’s consider a simple example of how to convert units. Suppose we want to convert 80 m to kilometers. The first thing todo is to list the units you have and the units to which you want to convert. In this case, we have units in meters and we wantto convert to kilometers. Next, we need to determine a conversion factor relating meters to kilometers. A conversion factoris a ratio that expresses how many of one unit are equal to another unit. For example, there are 12 in. in 1 ft, 1609 m in 1mi, 100 cm in 1 m, 60 s in 1 min, and so on. Refer to Appendix B for a more complete list of conversion factors. In thiscase, we know that there are 1000 m in 1 km. Now we can set up our unit conversion. We write the units we have and thenmultiply them by the conversion factor so the units cancel out, as shown:


80 m × 1 km
1000 m


= 0.080 km.


Note that the unwanted meter unit cancels, leaving only the desired kilometer unit. You can use this method to convertbetween any type of unit. Now, the conversion of 80 m to kilometers is simply the use of a metric prefix, as we saw in thepreceding section, so we can get the same answer just as easily by noting that
80 m = 8.0 × 101m = 8.0 × 10−2km = 0.080 km,


since “kilo-” means 103 (see Table 1.2) and 1 = −2 + 3. However, using conversion factors is handy when converting
between units that are not metric or when converting between derived units, as the following examples illustrate.
Example 1.2


Converting Nonmetric Units to Metric
The distance from the university to home is 10 mi and it usually takes 20 min to drive this distance. Calculate theaverage speed in meters per second (m/s). (Note: Average speed is distance traveled divided by time of travel.)
Strategy
First we calculate the average speed using the given units, then we can get the average speed into the desiredunits by picking the correct conversion factors and multiplying by them. The correct conversion factors are thosethat cancel the unwanted units and leave the desired units in their place. In this case, we want to convert miles tometers, so we need to know the fact that there are 1609 m in 1 mi. We also want to convert minutes to seconds,so we use the conversion of 60 s in 1 min.
Solution


1. Calculate average speed. Average speed is distance traveled divided by time of travel. (Take this definitionas a given for now. Average speed and other motion concepts are covered in later chapters.) In equationform,
Average speed = Distance


Time
.


2. Substitute the given values for distance and time:
Average speed = 10 mi


20 min
= 0.50 mi


min
.


3. Convert miles per minute to meters per second by multiplying by the conversion factor that cancels milesand leave meters, and also by the conversion factor that cancels minutes and leave seconds:


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1.2


1.3


0.50 mile
min


× 1609 m
1 mile


× 1 min
60 s


= (0.50)(1609)
60


m/s = 13 m/s.


Significance
Check the answer in the following ways:


1. Be sure the units in the unit conversion cancel correctly. If the unit conversion factor was written upsidedown, the units do not cancel correctly in the equation. We see the “miles” in the numerator in 0.50mi/min cancels the “mile” in the denominator in the first conversion factor. Also, the “min” in thedenominator in 0.50 mi/min cancels the “min” in the numerator in the second conversion factor.
2. Check that the units of the final answer are the desired units. The problem asked us to solve for averagespeed in units of meters per second and, after the cancellations, the only units left are a meter (m) in thenumerator and a second (s) in the denominator, so we have indeed obtained these units.


Check Your Understanding Light travels about 9 Pm in a year. Given that a year is about 3 × 107 s,
what is the speed of light in meters per second?


Example 1.3
Converting between Metric Units
The density of iron is 7.86 g/cm3 under standard conditions. Convert this to kg/m3.
Strategy
We need to convert grams to kilograms and cubic centimeters to cubic meters. The conversion factors we need are
1 kg = 103g and 1 cm = 10−2m. However, we are dealing with cubic centimeters (cm3 = cm × cm × cm),
so we have to use the second conversion factor three times (that is, we need to cube it). The idea is still to multiplyby the conversion factors in such a way that they cancel the units we want to get rid of and introduce the units wewant to keep.
Solution


7.86
g


cm 3
×


kg


103 g
×



cm
10−2m





3


= 7.86
(103)(10−6)


kg/m3 = 7.86 × 103 kg/m3


Significance
Remember, it’s always important to check the answer.


1. Be sure to cancel the units in the unit conversion correctly. We see that the gram (“g”) in the numeratorin 7.86 g/cm3 cancels the “g” in the denominator in the first conversion factor. Also, the three factors of“cm” in the denominator in 7.86 g/cm3 cancel with the three factors of “cm” in the numerator that we getby cubing the second conversion factor.
2. Check that the units of the final answer are the desired units. The problem asked for us to convert tokilograms per cubic meter. After the cancellations just described, we see the only units we have left are“kg” in the numerator and three factors of “m” in the denominator (that is, one factor of “m” cubed, or“m3”). Therefore, the units on the final answer are correct.


Check Your Understanding We know from Figure 1.4 that the diameter of Earth is on the order of 107m, so the order of magnitude of its surface area is 1014m2. What is that in square kilometers (that is, km2)? (Trydoing this both by converting 107 m to km and then squaring it and then by converting 1014 m2 directly tosquare kilometers. You should get the same answer both ways.)
Unit conversions may not seem very interesting, but not doing them can be costly. One famous example of this situation was


Chapter 1 | Units and Measurement 21




1.4


seen with the Mars Climate Orbiter. This probe was launched by NASA on December 11, 1998. On September 23, 1999,while attempting to guide the probe into its planned orbit around Mars, NASA lost contact with it. Subsequent investigationsshowed a piece of software called SM_FORCES (or “small forces”) was recording thruster performance data in the Englishunits of pound-seconds (lb-s). However, other pieces of software that used these values for course corrections expected themto be recorded in the SI units of newton-seconds (N-s), as dictated in the software interface protocols. This error caused theprobe to follow a very different trajectory from what NASA thought it was following, which most likely caused the probeeither to burn up in the Martian atmosphere or to shoot out into space. This failure to pay attention to unit conversions costhundreds of millions of dollars, not to mention all the time invested by the scientists and engineers who worked on theproject.
Check Your Understanding Given that 1 lb (pound) is 4.45 N, were the numbers being output bySM_FORCES too big or too small?


1.4 | Dimensional Analysis
Learning Objectives


By the end of this section, you will be able to:
• Find the dimensions of a mathematical expression involving physical quantities.
• Determine whether an equation involving physical quantities is dimensionally consistent.


The dimension of any physical quantity expresses its dependence on the base quantities as a product of symbols (or powersof symbols) representing the base quantities. Table 1.3 lists the base quantities and the symbols used for their dimension.For example, a measurement of length is said to have dimension L or L1, a measurement of mass has dimension M orM1, and a measurement of time has dimension T or T1. Like units, dimensions obey the rules of algebra. Thus, area isthe product of two lengths and so has dimension L2, or length squared. Similarly, volume is the product of three lengthsand has dimension L3, or length cubed. Speed has dimension length over time, L/T or LT–1. Volumetric mass density hasdimension M/L3 or ML–3, or mass over length cubed. In general, the dimension of any physical quantity can be written as
LaMbTc IdΘeN f Jg for some powers a, b, c, d, e, f , and g. We can write the dimensions of a length in this form with
a = 1 and the remaining six powers all set equal to zero: L1 = L1M0T0 I0Θ0N0 J0. Any quantity with a dimension that
can be written so that all seven powers are zero (that is, its dimension is L0M0T0 I0Θ0N0 J0 ) is called dimensionless
(or sometimes “of dimension 1,” because anything raised to the zero power is one). Physicists often call dimensionlessquantities pure numbers.


Base Quantity Symbol for Dimension
Length L
Mass M
Time T
Current I
Thermodynamic temperature Θ
Amount of substance N
Luminous intensity J


Table 1.3 Base Quantities and Their Dimensions
Physicists often use square brackets around the symbol for a physical quantity to represent the dimensions of that quantity.For example, if r is the radius of a cylinder and h is its height, then we write [r] = L and [h] = L to indicate the
dimensions of the radius and height are both those of length, or L. Similarly, if we use the symbol A for the surface area of
a cylinder and V for its volume, then [A] = L2 and [V] = L3. If we use the symbol m for the mass of the cylinder and ρ


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for the density of the material from which the cylinder is made, then [m] = M and [ρ] = ML−3.
The importance of the concept of dimension arises from the fact that any mathematical equation relating physical quantitiesmust be dimensionally consistent, which means the equation must obey the following rules:


• Every term in an expression must have the same dimensions; it does not make sense to add or subtract quantities ofdiffering dimension (think of the old saying: “You can’t add apples and oranges”). In particular, the expressions oneach side of the equality in an equation must have the same dimensions.
• The arguments of any of the standard mathematical functions such as trigonometric functions (such as sine andcosine), logarithms, or exponential functions that appear in the equation must be dimensionless. These functionsrequire pure numbers as inputs and give pure numbers as outputs.


If either of these rules is violated, an equation is not dimensionally consistent and cannot possibly be a correct statementof physical law. This simple fact can be used to check for typos or algebra mistakes, to help remember the various laws ofphysics, and even to suggest the form that new laws of physics might take. This last use of dimensions is beyond the scopeof this text, but is something you will undoubtedly learn later in your academic career.
Example 1.4


Using Dimensions to Remember an Equation
Suppose we need the formula for the area of a circle for some computation. Like many people who learnedgeometry too long ago to recall with any certainty, two expressions may pop into our mind when we think of
circles: πr2 and 2πr. One expression is the circumference of a circle of radius r and the other is its area. But
which is which?
Strategy
One natural strategy is to look it up, but this could take time to find information from a reputable source. Besides,even if we think the source is reputable, we shouldn’t trust everything we read. It is nice to have a way to double-check just by thinking about it. Also, we might be in a situation in which we cannot look things up (such as duringa test). Thus, the strategy is to find the dimensions of both expressions by making use of the fact that dimensionsfollow the rules of algebra. If either expression does not have the same dimensions as area, then it cannot possiblybe the correct equation for the area of a circle.
Solution
We know the dimension of area is L2. Now, the dimension of the expression πr2 is


[πr2] = [π] · [r]2 = 1 · L2 = L2,


since the constant π is a pure number and the radius r is a length. Therefore, πr2 has the dimension of area.
Similarly, the dimension of the expression 2πr is


[2πr] = [2] · [π] · [r] = 1 · 1 · L = L,


since the constants 2 and π are both dimensionless and the radius r is a length. We see that 2πr has the
dimension of length, which means it cannot possibly be an area.
We rule out 2πr because it is not dimensionally consistent with being an area. We see that πr2 is dimensionally
consistent with being an area, so if we have to choose between these two expressions, πr2 is the one to choose.
Significance
This may seem like kind of a silly example, but the ideas are very general. As long as we know the dimensionsof the individual physical quantities that appear in an equation, we can check to see whether the equation isdimensionally consistent. On the other hand, knowing that true equations are dimensionally consistent, we canmatch expressions from our imperfect memories to the quantities for which they might be expressions. Doing thiswill not help us remember dimensionless factors that appear in the equations (for example, if you had accidentally
conflated the two expressions from the example into 2πr2, then dimensional analysis is no help), but it does
help us remember the correct basic form of equations.


Chapter 1 | Units and Measurement 23




1.5 Check Your Understanding Suppose we want the formula for the volume of a sphere. The two
expressions commonly mentioned in elementary discussions of spheres are 4πr2 and 4πr3 /3. One is the
volume of a sphere of radius r and the other is its surface area. Which one is the volume?


Example 1.5
Checking Equations for Dimensional Consistency
Consider the physical quantities s, v, a, and t with dimensions [s] = L, [v] = LT−1, [a] = LT−2,
and [t] = T. Determine whether each of the following equations is dimensionally consistent: (a)
s = vt + 0.5at2; (b) s = vt2 + 0.5at; and (c) v = sin(at2 /s).
Strategy
By the definition of dimensional consistency, we need to check that each term in a given equation has the samedimensions as the other terms in that equation and that the arguments of any standard mathematical functions aredimensionless.
Solution


a. There are no trigonometric, logarithmic, or exponential functions to worry about in this equation, so weneed only look at the dimensions of each term appearing in the equation. There are three terms, one in theleft expression and two in the expression on the right, so we look at each in turn:
[s] = L
[vt] = [v] · [t] = LT−1 · T = LT 0 = L


[0.5at2] = [a] · [t]2 = LT−2 · T2 = LT 0 = L.


All three terms have the same dimension, so this equation is dimensionally consistent.
b. Again, there are no trigonometric, exponential, or logarithmic functions, so we only need to look at thedimensions of each of the three terms appearing in the equation:


[s] = L
[vt2] = [v] · [t]2 = LT−1 · T2 = LT


[at] = [a] · [t] = LT−2 · T = LT−1.


None of the three terms has the same dimension as any other, so this is about as far from beingdimensionally consistent as you can get. The technical term for an equation like this is nonsense.
c. This equation has a trigonometric function in it, so first we should check that the argument of the sinefunction is dimensionless:




at2
s

⎦ =


[a] · [t]2


[s]
= LT


−2 · T2
L


= L
L


= 1.


The argument is dimensionless. So far, so good. Now we need to check the dimensions of each of the twoterms (that is, the left expression and the right expression) in the equation:
[v] = LT−1



⎣sin


at2
s



⎦ = 1.


The two terms have different dimensions—meaning, the equation is not dimensionally consistent. This equationis another example of “nonsense.”
Significance
If we are trusting people, these types of dimensional checks might seem unnecessary. But, rest assured, any


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1.6


textbook on a quantitative subject such as physics (including this one) almost certainly contains some equationswith typos. Checking equations routinely by dimensional analysis save us the embarrassment of using an incorrectequation. Also, checking the dimensions of an equation we obtain through algebraic manipulation is a great wayto make sure we did not make a mistake (or to spot a mistake, if we made one).
Check Your Understanding Is the equation v = at dimensionally consistent?


One further point that needs to be mentioned is the effect of the operations of calculus on dimensions. We have seen thatdimensions obey the rules of algebra, just like units, but what happens when we take the derivative of one physical quantitywith respect to another or integrate a physical quantity over another? The derivative of a function is just the slope of the linetangent to its graph and slopes are ratios, so for physical quantities v and t, we have that the dimension of the derivative ofv with respect to t is just the ratio of the dimension of v over that of t:


dv
dt

⎦ =


[v]
[t]


.


Similarly, since integrals are just sums of products, the dimension of the integral of v with respect to t is simply thedimension of v times the dimension of t:

⎣∫ vdt



⎦ = [v] · [t].


By the same reasoning, analogous rules hold for the units of physical quantities derived from other quantities by integrationor differentiation.
1.5 | Estimates and Fermi Calculations


Learning Objectives
By the end of this section, you will be able to:
• Estimate the values of physical quantities.


On many occasions, physicists, other scientists, and engineers need to make estimates for a particular quantity. Otherterms sometimes used are guesstimates, order-of-magnitude approximations, back-of-the-envelope calculations, or Fermicalculations. (The physicist Enrico Fermi mentioned earlier was famous for his ability to estimate various kinds of data withsurprising precision.) Will that piece of equipment fit in the back of the car or do we need to rent a truck? How long willthis download take? About how large a current will there be in this circuit when it is turned on? How many houses coulda proposed power plant actually power if it is built? Note that estimating does not mean guessing a number or a formulaat random. Rather, estimation means using prior experience and sound physical reasoning to arrive at a rough idea of aquantity’s value. Because the process of determining a reliable approximation usually involves the identification of correctphysical principles and a good guess about the relevant variables, estimating is very useful in developing physical intuition.Estimates also allow us perform “sanity checks” on calculations or policy proposals by helping us rule out certain scenariosor unrealistic numbers. They allow us to challenge others (as well as ourselves) in our efforts to learn truths about the world.
Many estimates are based on formulas in which the input quantities are known only to a limited precision. As you developphysics problem-solving skills (which are applicable to a wide variety of fields), you also will develop skills at estimating.You develop these skills by thinking more quantitatively and by being willing to take risks. As with any skill, experiencehelps. Familiarity with dimensions (see Table 1.3) and units (see Table 1.1 and Table 1.2), and the scales of basequantities (see Figure 1.4) also helps.
To make some progress in estimating, you need to have some definite ideas about how variables may be related. Thefollowing strategies may help you in practicing the art of estimation:


• Get big lengths from smaller lengths. When estimating lengths, remember that anything can be a ruler. Thus,imagine breaking a big thing into smaller things, estimate the length of one of the smaller things, and multiply toget the length of the big thing. For example, to estimate the height of a building, first count how many floors ithas. Then, estimate how big a single floor is by imagining how many people would have to stand on each other’s


Chapter 1 | Units and Measurement 25




shoulders to reach the ceiling. Last, estimate the height of a person. The product of these three estimates is yourestimate of the height of the building. It helps to have memorized a few length scales relevant to the sorts ofproblems you find yourself solving. For example, knowing some of the length scales in Figure 1.4 might comein handy. Sometimes it also helps to do this in reverse—that is, to estimate the length of a small thing, imagine abunch of them making up a bigger thing. For example, to estimate the thickness of a sheet of paper, estimate thethickness of a stack of paper and then divide by the number of pages in the stack. These same strategies of breakingbig things into smaller things or aggregating smaller things into a bigger thing can sometimes be used to estimateother physical quantities, such as masses and times.
• Get areas and volumes from lengths.When dealing with an area or a volume of a complex object, introduce a simplemodel of the object such as a sphere or a box. Then, estimate the linear dimensions (such as the radius of the sphereor the length, width, and height of the box) first, and use your estimates to obtain the volume or area from standardgeometric formulas. If you happen to have an estimate of an object’s area or volume, you can also do the reverse;that is, use standard geometric formulas to get an estimate of its linear dimensions.
• Get masses from volumes and densities.When estimating masses of objects, it can help first to estimate its volumeand then to estimate its mass from a rough estimate of its average density (recall, density has dimension mass overlength cubed, so mass is density times volume). For this, it helps to remember that the density of air is around 1 kg/m3, the density of water is 103 kg/m3, and the densest everyday solids max out at around 104 kg/m3. Asking yourselfwhether an object floats or sinks in either air or water gets you a ballpark estimate of its density. You can also dothis the other way around; if you have an estimate of an object’s mass and its density, you can use them to get anestimate of its volume.
• If all else fails, bound it. For physical quantities for which you do not have a lot of intuition, sometimes the bestyou can do is think something like: Well, it must be bigger than this and smaller than that. For example, supposeyou need to estimate the mass of a moose. Maybe you have a lot of experience with moose and know their averagemass offhand. If so, great. But for most people, the best they can do is to think something like: It must be biggerthan a person (of order 102 kg) and less than a car (of order 103 kg). If you need a single number for a subsequentcalculation, you can take the geometric mean of the upper and lower bound—that is, you multiply them togetherand then take the square root. For the moose mass example, this would be



⎝10


2 × 103⎞⎠
0.5


= 102.5 = 100.5 × 102 ≈ 3 × 102kg.


The tighter the bounds, the better. Also, no rules are unbreakable when it comes to estimation. If you think the valueof the quantity is likely to be closer to the upper bound than the lower bound, then you may want to bump up yourestimate from the geometric mean by an order or two of magnitude.
• One “sig. fig.” is fine. There is no need to go beyond one significant figure when doing calculations to obtain anestimate. In most cases, the order of magnitude is good enough. The goal is just to get in the ballpark figure, so keepthe arithmetic as simple as possible.
• Ask yourself: Does this make any sense? Last, check to see whether your answer is reasonable. How does it comparewith the values of other quantities with the same dimensions that you already know or can look up easily? If youget some wacky answer (for example, if you estimate the mass of the Atlantic Ocean to be bigger than the mass ofEarth, or some time span to be longer than the age of the universe), first check to see whether your units are correct.Then, check for arithmetic errors. Then, rethink the logic you used to arrive at your answer. If everything checksout, you may have just proved that some slick new idea is actually bogus.
Example 1.6


Mass of Earth’s Oceans
Estimate the total mass of the oceans on Earth.
Strategy
We know the density of water is about 103 kg/m3, so we start with the advice to “get masses from densitiesand volumes.” Thus, we need to estimate the volume of the planet’s oceans. Using the advice to “get areas andvolumes from lengths,” we can estimate the volume of the oceans as surface area times average depth, or V = AD.We know the diameter of Earth from Figure 1.4 and we know that most of Earth’s surface is covered in water,so we can estimate the surface area of the oceans as being roughly equal to the surface area of the planet. By


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1.7


following the advice to “get areas and volumes from lengths” again, we can approximate Earth as a sphere and
use the formula for the surface area of a sphere of diameter d—that is, A = πd2, to estimate the surface area of
the oceans. Now we just need to estimate the average depth of the oceans. For this, we use the advice: “If all elsefails, bound it.” We happen to know the deepest points in the ocean are around 10 km and that it is not uncommon
for the ocean to be deeper than 1 km, so we take the average depth to be around (103 × 104)0.5 ≈ 3 × 103m.
Now we just need to put it all together, heeding the advice that “one ‘sig. fig.’ is fine.”
Solution
We estimate the surface area of Earth (and hence the surface area of Earth’s oceans) to be roughly


A = πd2 = π(107m)2 ≈ 3 × 1014m2.


Next, using our average depth estimate of D = 3 × 103m, which was obtained by bounding, we estimate the
volume of Earth’s oceans to be


V = AD = (3 × 1014m2)(3 × 103m) = 9 × 1017m3.


Last, we estimate the mass of the world’s oceans to be
M = ρV = (103 kg/m3)(9 × 1017m3) = 9 × 1020kg.


Thus, we estimate that the order of magnitude of the mass of the planet’s oceans is 1021 kg.
Significance
To verify our answer to the best of our ability, we first need to answer the question: Does this make any sense?From Figure 1.4, we see the mass of Earth’s atmosphere is on the order of 1019 kg and the mass of Earth is onthe order of 1025 kg. It is reassuring that our estimate of 1021 kg for the mass of Earth’s oceans falls somewherebetween these two. So, yes, it does seem to make sense. It just so happens that we did a search on the Web for
“mass of oceans” and the top search results all said 1.4 × 1021kg, which is the same order of magnitude as our
estimate. Now, rather than having to trust blindly whoever first put that number up on a website (most of the othersites probably just copied it from them, after all), we can have a little more confidence in it.


Check Your Understanding Figure 1.4 says the mass of the atmosphere is 1019 kg. Assuming thedensity of the atmosphere is 1 kg/m3, estimate the height of Earth’s atmosphere. Do you think your answer is anunderestimate or an overestimate? Explain why.
How many piano tuners are there in New York City? How many leaves are on that tree? If you are studying photosynthesisor thinking of writing a smartphone app for piano tuners, then the answers to these questions might be of great interest toyou. Otherwise, you probably couldn’t care less what the answers are. However, these are exactly the sorts of estimationproblems that people in various tech industries have been asking potential employees to evaluate their quantitative reasoningskills. If building physical intuition and evaluating quantitative claims do not seem like sufficient reasons for you to practiceestimation problems, how about the fact that being good at them just might land you a high-paying job?


For practice estimating relative lengths, areas, and volumes, check out this PhET(https://openstaxcollege.org/l/21lengthgame) simulation, titled “Estimation.”


Chapter 1 | Units and Measurement 27




1.6 | Significant Figures
Learning Objectives


By the end of this section, you will be able to:
• Determine the correct number of significant figures for the result of a computation.
• Describe the relationship between the concepts of accuracy, precision, uncertainty, anddiscrepancy.
• Calculate the percent uncertainty of a measurement, given its value and its uncertainty.
• Determine the uncertainty of the result of a computation involving quantities with givenuncertainties.


Figure 1.11 shows two instruments used to measure the mass of an object. The digital scale has mostly replaced thedouble-pan balance in physics labs because it gives more accurate and precise measurements. But what exactly do wemean by accurate and precise? Aren’t they the same thing? In this section we examine in detail the process of making andreporting a measurement.


Figure 1.11 (a) A double-pan mechanical balance is used to compare different masses. Usually an object with unknown massis placed in one pan and objects of known mass are placed in the other pan. When the bar that connects the two pans ishorizontal, then the masses in both pans are equal. The “known masses” are typically metal cylinders of standard mass such as 1g, 10 g, and 100 g. (b) Many mechanical balances, such as double-pan balances, have been replaced by digital scales, which cantypically measure the mass of an object more precisely. A mechanical balance may read only the mass of an object to the nearesttenth of a gram, but many digital scales can measure the mass of an object up to the nearest thousandth of a gram. (credit a:modification of work by Serge Melki; credit b: modification of work by Karel Jakubec)
Accuracy and Precision of a Measurement
Science is based on observation and experiment—that is, on measurements. Accuracy is how close a measurement is to theaccepted reference value for that measurement. For example, let’s say we want to measure the length of standard printerpaper. The packaging in which we purchased the paper states that it is 11.0 in. long. We then measure the length of the paperthree times and obtain the following measurements: 11.1 in., 11.2 in., and 10.9 in. These measurements are quite accuratebecause they are very close to the reference value of 11.0 in. In contrast, if we had obtained a measurement of 12 in., ourmeasurement would not be very accurate. Notice that the concept of accuracy requires that an accepted reference value begiven.
The precision of measurements refers to how close the agreement is between repeated independent measurements (whichare repeated under the same conditions). Consider the example of the paper measurements. The precision of themeasurements refers to the spread of the measured values. One way to analyze the precision of the measurements is todetermine the range, or difference, between the lowest and the highest measured values. In this case, the lowest value was10.9 in. and the highest value was 11.2 in. Thus, the measured values deviated from each other by, at most, 0.3 in. Thesemeasurements were relatively precise because they did not vary too much in value. However, if the measured values hadbeen 10.9 in., 11.1 in., and 11.9 in., then the measurements would not be very precise because there would be significant


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variation from one measurement to another. Notice that the concept of precision depends only on the actual measurementsacquired and does not depend on an accepted reference value.
The measurements in the paper example are both accurate and precise, but in some cases, measurements are accurate butnot precise, or they are precise but not accurate. Let’s consider an example of a GPS attempting to locate the positionof a restaurant in a city. Think of the restaurant location as existing at the center of a bull’s-eye target and think of eachGPS attempt to locate the restaurant as a black dot. In Figure 1.12(a), we see the GPS measurements are spread out farapart from each other, but they are all relatively close to the actual location of the restaurant at the center of the target.This indicates a low-precision, high-accuracy measuring system. However, in Figure 1.12(b), the GPS measurements areconcentrated quite closely to one another, but they are far away from the target location. This indicates a high-precision,low-accuracy measuring system.


Figure 1.12 A GPS attempts to locate a restaurant at the center of the bull’s-eye. Theblack dots represent each attempt to pinpoint the location of the restaurant. (a) The dotsare spread out quite far apart from one another, indicating low precision, but they are eachrather close to the actual location of the restaurant, indicating high accuracy. (b) The dotsare concentrated rather closely to one another, indicating high precision, but they arerather far away from the actual location of the restaurant, indicating low accuracy. (credita and credit b: modification of works by Dark Evil)
Accuracy, Precision, Uncertainty, and Discrepancy
The precision of a measuring system is related to the uncertainty in the measurements whereas the accuracy is relatedto the discrepancy from the accepted reference value. Uncertainty is a quantitative measure of how much your measuredvalues deviate from one another. There are many different methods of calculating uncertainty, each of which is appropriateto different situations. Some examples include taking the range (that is, the biggest less the smallest) or finding the standarddeviation of the measurements. Discrepancy (or “measurement error”) is the difference between the measured value and agiven standard or expected value. If the measurements are not very precise, then the uncertainty of the values is high. If themeasurements are not very accurate, then the discrepancy of the values is high.
Recall our example of measuring paper length; we obtained measurements of 11.1 in., 11.2 in., and 10.9 in., and the acceptedvalue was 11.0 in. We might average the three measurements to say our best guess is 11.1 in.; in this case, our discrepancyis 11.1 – 11.0 = 0.1 in., which provides a quantitative measure of accuracy. We might calculate the uncertainty in our bestguess by using the range of our measured values: 0.3 in. Then we would say the length of the paper is 11.1 in. plus or minus0.3 in. The uncertainty in a measurement, A, is often denoted as δA (read “delta A”), so the measurement result would berecorded as A ± δA. Returning to our paper example, the measured length of the paper could be expressed as 11.1 ± 0.3in. Since the discrepancy of 0.1 in. is less than the uncertainty of 0.3 in., we might say the measured value agrees with theaccepted reference value to within experimental uncertainty.
Some factors that contribute to uncertainty in a measurement include the following:


• Limitations of the measuring device
• The skill of the person taking the measurement
• Irregularities in the object being measured
• Any other factors that affect the outcome (highly dependent on the situation)


In our example, such factors contributing to the uncertainty could be the smallest division on the ruler is 1/16 in., the personusing the ruler has bad eyesight, the ruler is worn down on one end, or one side of the paper is slightly longer than the other.


Chapter 1 | Units and Measurement 29




1.8


At any rate, the uncertainty in a measurement must be calculated to quantify its precision. If a reference value is known, itmakes sense to calculate the discrepancy as well to quantify its accuracy.
Percent uncertainty
Another method of expressing uncertainty is as a percent of the measured value. If a measurement A is expressed withuncertainty δA, the percent uncertainty is defined as


Percent uncertainty = δA
A


× 100%.


Example 1.7
Calculating Percent Uncertainty: A Bag of Apples
A grocery store sells 5-lb bags of apples. Let’s say we purchase four bags during the course of a month and weighthe bags each time. We obtain the following measurements:


• Week 1 weight: 4.8 lb
• Week 2 weight: 5.3 lb
• Week 3 weight: 4.9 lb
• Week 4 weight: 5.4 lb


We then determine the average weight of the 5-lb bag of apples is 5.1 ± 0.2 lb. What is the percent uncertainty ofthe bag’s weight?
Strategy
First, observe that the average value of the bag’s weight, A, is 5.1 lb. The uncertainty in this value, δA, is 0.2 lb.
We can use the following equation to determine the percent uncertainty of the weight:


(1.1)Percent uncertainty = δA
A


× 100%.


Solution
Substitute the values into the equation:


Percent uncertainty = δA
A


× 100% = 0.2 lb
5.1 lb


× 100% = 3.9% ≈ 4%.


Significance
We can conclude the average weight of a bag of apples from this store is 5.1 lb ± 4%. Notice the percentuncertainty is dimensionless because the units of weight in δA = 0.2 lb canceled those inn A = 5.1 lb when we
took the ratio.


Check Your Understanding A high school track coach has just purchased a new stopwatch. Thestopwatch manual states the stopwatch has an uncertainty of ±0.05 s. Runners on the track coach’s teamregularly clock 100-m sprints of 11.49 s to 15.01 s. At the school’s last track meet, the first-place sprinter camein at 12.04 s and the second-place sprinter came in at 12.07 s. Will the coach’s new stopwatch be helpful intiming the sprint team? Why or why not?


Uncertainties in calculations
Uncertainty exists in anything calculated from measured quantities. For example, the area of a floor calculated frommeasurements of its length and width has an uncertainty because the length and width have uncertainties. How big is theuncertainty in something you calculate by multiplication or division? If the measurements going into the calculation havesmall uncertainties (a few percent or less), then the method of adding percents can be used for multiplication or division.This method states the percent uncertainty in a quantity calculated by multiplication or division is the sum of the percent


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uncertainties in the items used to make the calculation. For example, if a floor has a length of 4.00 m and a width of 3.00 m,with uncertainties of 2% and 1%, respectively, then the area of the floor is 12.0 m2 and has an uncertainty of 3%. (Expressed
as an area, this is 0.36 m2 [ 12.0 m2 × 0.03 ], which we round to 0.4 m2 since the area of the floor is given to a tenth of a
square meter.)
Precision of Measuring Tools and Significant Figures
An important factor in the precision of measurements involves the precision of the measuring tool. In general, a precisemeasuring tool is one that can measure values in very small increments. For example, a standard ruler can measure length tothe nearest millimeter whereas a caliper can measure length to the nearest 0.01 mm. The caliper is a more precise measuringtool because it can measure extremely small differences in length. The more precise the measuring tool, the more precisethe measurements.
When we express measured values, we can only list as many digits as we measured initially with our measuring tool. Forexample, if we use a standard ruler to measure the length of a stick, we may measure it to be 36.7 cm. We can’t express thisvalue as 36.71 cm because our measuring tool is not precise enough to measure a hundredth of a centimeter. It should benoted that the last digit in a measured value has been estimated in some way by the person performing the measurement. Forexample, the person measuring the length of a stick with a ruler notices the stick length seems to be somewhere in between36.6 cm and 36.7 cm, and he or she must estimate the value of the last digit. Using the method of significant figures, therule is that the last digit written down in a measurement is the first digit with some uncertainty. To determine the numberof significant digits in a value, start with the first measured value at the left and count the number of digits through the lastdigit written on the right. For example, the measured value 36.7 cm has three digits, or three significant figures. Significantfigures indicate the precision of the measuring tool used to measure a value.
Zeros
Special consideration is given to zeros when counting significant figures. The zeros in 0.053 are not significant becausethey are placeholders that locate the decimal point. There are two significant figures in 0.053. The zeros in 10.053 arenot placeholders; they are significant. This number has five significant figures. The zeros in 1300 may or may not besignificant, depending on the style of writing numbers. They could mean the number is known to the last digit or theycould be placeholders. So 1300 could have two, three, or four significant figures. To avoid this ambiguity, we should write
1300 in scientific notation as 1.3 × 103, 1.30 × 103, or 1.300 × 103, depending on whether it has two, three, or four
significant figures. Zeros are significant except when they serve only as placeholders.
Significant figures in calculations
When combining measurements with different degrees of precision, the number of significant digits in the final answer canbe no greater than the number of significant digits in the least-precise measured value. There are two different rules, onefor multiplication and division and the other for addition and subtraction.


1. For multiplication and division, the result should have the same number of significant figures as the quantitywith the least number of significant figures entering into the calculation. For example, the area of a circle can becalculated from its radius using A = πr2. Let’s see how many significant figures the area has if the radius has onlytwo—say, r = 1.2 m. Using a calculator with an eight-digit output, we would calculate
A = πr2 = (3.1415927…) × (1.2 m)2 = 4.5238934 m2.


But because the radius has only two significant figures, it limits the calculated quantity to two significant figures, or
A = 4.5 m2,


although π is good to at least eight digits.2. For addition and subtraction, the answer can contain no more decimal places than the least-precise measurement.Suppose we buy 7.56 kg of potatoes in a grocery store as measured with a scale with precision 0.01 kg, then we dropoff 6.052 kg of potatoes at your laboratory as measured by a scale with precision 0.001 kg. Then, we go home andadd 13.7 kg of potatoes as measured by a bathroom scale with precision 0.1 kg. How many kilograms of potatoes dowe now have and how many significant figures are appropriate in the answer? The mass is found by simple additionand subtraction:


Chapter 1 | Units and Measurement 31




7.56 kg


−6.052 kg


+13.7 kg
15.208 kg


= 15.2 kg.


Next, we identify the least-precise measurement: 13.7 kg. This measurement is expressed to the 0.1 decimal place,so our final answer must also be expressed to the 0.1 decimal place. Thus, the answer is rounded to the tenths place,giving us 15.2 kg.
Significant figures in this text
In this text, most numbers are assumed to have three significant figures. Furthermore, consistent numbers of significantfigures are used in all worked examples. An answer given to three digits is based on input good to at least three digits,for example. If the input has fewer significant figures, the answer will also have fewer significant figures. Care is alsotaken that the number of significant figures is reasonable for the situation posed. In some topics, particularly in optics, moreaccurate numbers are needed and we use more than three significant figures. Finally, if a number is exact, such as the twoin the formula for the circumference of a circle, C = 2πr, it does not affect the number of significant figures in a calculation.Likewise, conversion factors such as 100 cm/1 m are considered exact and do not affect the number of significant figures ina calculation.
1.7 | Solving Problems in Physics


Learning Objectives
By the end of this section, you will be able to:
• Describe the process for developing a problem-solving strategy.
• Explain how to find the numerical solution to a problem.
• Summarize the process for assessing the significance of the numerical solution to a problem.


Figure 1.13 Problem-solving skills are essential to your success in physics. (credit:“scui3asteveo”/Flickr)


Problem-solving skills are clearly essential to success in a quantitative course in physics. More important, the ability


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to apply broad physical principles—usually represented by equations—to specific situations is a very powerful form ofknowledge. It is much more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can beapplied to new situations whereas a list of facts cannot be made long enough to contain every possible circumstance. Suchanalytical skills are useful both for solving problems in this text and for applying physics in everyday life.
As you are probably well aware, a certain amount of creativity and insight is required to solve problems. No rigid procedureworks every time. Creativity and insight grow with experience. With practice, the basics of problem solving become almostautomatic. One way to get practice is to work out the text’s examples for yourself as you read. Another is to work asmany end-of-section problems as possible, starting with the easiest to build confidence and then progressing to the moredifficult. After you become involved in physics, you will see it all around you, and you can begin to apply it to situationsyou encounter outside the classroom, just as is done in many of the applications in this text.
Although there is no simple step-by-step method that works for every problem, the following three-stage process facilitatesproblem solving and makes it more meaningful. The three stages are strategy, solution, and significance. This process isused in examples throughout the book. Here, we look at each stage of the process in turn.
Strategy
Strategy is the beginning stage of solving a problem. The idea is to figure out exactly what the problem is and then developa strategy for solving it. Some general advice for this stage is as follows:


• Examine the situation to determine which physical principles are involved. It often helps to draw a simple sketchat the outset. You often need to decide which direction is positive and note that on your sketch. When you haveidentified the physical principles, it is much easier to find and apply the equations representing those principles.Although finding the correct equation is essential, keep in mind that equations represent physical principles, laws ofnature, and relationships among physical quantities. Without a conceptual understanding of a problem, a numericalsolution is meaningless.
• Make a list of what is given or can be inferred from the problem as stated (identify the “knowns”). Many problemsare stated very succinctly and require some inspection to determine what is known. Drawing a sketch be very usefulat this point as well. Formally identifying the knowns is of particular importance in applying physics to real-worldsituations. For example, the word stopped means the velocity is zero at that instant. Also, we can often take initialtime and position as zero by the appropriate choice of coordinate system.
• Identify exactly what needs to be determined in the problem (identify the unknowns). In complex problems,especially, it is not always obvious what needs to be found or in what sequence. Making a list can help identify theunknowns.
• Determine which physical principles can help you solve the problem. Since physical principles tend to be expressedin the form of mathematical equations, a list of knowns and unknowns can help here. It is easiest if you can findequations that contain only one unknown—that is, all the other variables are known—so you can solve for theunknown easily. If the equation contains more than one unknown, then additional equations are needed to solve theproblem. In some problems, several unknowns must be determined to get at the one needed most. In such problemsit is especially important to keep physical principles in mind to avoid going astray in a sea of equations. You mayhave to use two (or more) different equations to get the final answer.


Solution
The solution stage is when you do the math. Substitute the knowns (along with their units) into the appropriate equationand obtain numerical solutions complete with units. That is, do the algebra, calculus, geometry, or arithmetic necessary tofind the unknown from the knowns, being sure to carry the units through the calculations. This step is clearly importantbecause it produces the numerical answer, along with its units. Notice, however, that this stage is only one-third of theoverall problem-solving process.
Significance
After having done the math in the solution stage of problem solving, it is tempting to think you are done. But, alwaysremember that physics is not math. Rather, in doing physics, we use mathematics as a tool to help us understand nature. So,after you obtain a numerical answer, you should always assess its significance:


• Check your units. If the units of the answer are incorrect, then an error has been made and you should go back overyour previous steps to find it. One way to find the mistake is to check all the equations you derived for dimensionalconsistency. However, be warned that correct units do not guarantee the numerical part of the answer is also correct.


Chapter 1 | Units and Measurement 33




• Check the answer to see whether it is reasonable. Does it make sense? This step is extremely important: –the goalof physics is to describe nature accurately. To determine whether the answer is reasonable, check both its magnitudeand its sign, in addition to its units. The magnitude should be consistent with a rough estimate of what it shouldbe. It should also compare reasonably with magnitudes of other quantities of the same type. The sign usually tellsyou about direction and should be consistent with your prior expectations. Your judgment will improve as you solvemore physics problems, and it will become possible for you to make finer judgments regarding whether nature isdescribed adequately by the answer to a problem. This step brings the problem back to its conceptual meaning. Ifyou can judge whether the answer is reasonable, you have a deeper understanding of physics than just being able tosolve a problem mechanically.
• Check to see whether the answer tells you something interesting. What does it mean? This is the flip side of thequestion: Does it make sense? Ultimately, physics is about understanding nature, and we solve physics problems tolearn a little something about how nature operates. Therefore, assuming the answer does make sense, you shouldalways take a moment to see if it tells you something about the world that you find interesting. Even if the answer tothis particular problem is not very interesting to you, what about the method you used to solve it? Could the methodbe adapted to answer a question that you do find interesting? In many ways, it is in answering questions such asthese science that progresses.


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accuracy
base quantity
base unit
conversion factor
derived quantity
derived units
dimension


dimensionally consistent
dimensionless


discrepancy
English units
estimation


kilogram
law
meter
method of adding percents
metric system
model
order of magnitude
percent uncertainty
physical quantity
physics
precision
second
SI units
significant figures


CHAPTER 1 REVIEW
KEY TERMS


the degree to which a measured value agrees with an accepted reference value for that measurement
physical quantity chosen by convention and practical considerations such that all other physical quantitiescan be expressed as algebraic combinations of them


standard for expressing the measurement of a base quantity within a particular system of units; defined by aparticular procedure used to measure the corresponding base quantity
a ratio that expresses how many of one unit are equal to another unit


physical quantity defined using algebraic combinations of base quantities
units that can be calculated using algebraic combinations of the fundamental units


expression of the dependence of a physical quantity on the base quantities as a product of powers of symbols
representing the base quantities; in general, the dimension of a quantity has the form LaMbTc IdΘeNf Jg for some
powers a, b, c, d, e, f, and g.


equation in which every term has the same dimensions and the arguments of anymathematical functions appearing in the equation are dimensionless
quantity with a dimension of L0M0T0 I0Θ0N0 J0 = 1; also called quantity of dimension 1 or a pure


number
the difference between the measured value and a given standard or expected value
system of measurement used in the United States; includes units of measure such as feet, gallons, andpounds


using prior experience and sound physical reasoning to arrive at a rough idea of a quantity’s value; sometimescalled an “order-of-magnitude approximation,” a “guesstimate,” a “back-of-the-envelope calculation”, or a “Fermicalculation”
SI unit for mass, abbreviated kg


description, using concise language or a mathematical formula, of a generalized pattern in nature supported byscientific evidence and repeated experiments
SI unit for length, abbreviated m


the percent uncertainty in a quantity calculated by multiplication or division is the sum ofthe percent uncertainties in the items used to make the calculation.
system in which values can be calculated in factors of 10


representation of something often too difficult (or impossible) to display directly
the size of a quantity as it relates to a power of 10
the ratio of the uncertainty of a measurement to the measured value, expressed as a percentage


characteristic or property of an object that can be measured or calculated from other measurements
science concerned with describing the interactions of energy, matter, space, and time; especially interested inwhat fundamental mechanisms underlie every phenomenon
the degree to which repeated measurements agree with each other


the SI unit for time, abbreviated s
the international system of units that scientists in most countries have agreed to use; includes units such asmeters, liters, and grams


used to express the precision of a measuring tool used to measure a value


Chapter 1 | Units and Measurement 35




theory
uncertainty
units


testable explanation for patterns in nature supported by scientific evidence and verified multiple times by variousgroups of researchers
a quantitative measure of how much measured values deviate from one another


standards used for expressing and comparing measurements


KEY EQUATIONS
Percent uncertainty Percent uncertainty = δAA × 100%


SUMMARY
1.1 The Scope and Scale of Physics


• Physics is about trying to find the simple laws that describe all natural phenomena.
• Physics operates on a vast range of scales of length, mass, and time. Scientists use the concept of the order ofmagnitude of a number to track which phenomena occur on which scales. They also use orders of magnitude tocompare the various scales.
• Scientists attempt to describe the world by formulating models, theories, and laws.


1.2 Units and Standards
• Systems of units are built up from a small number of base units, which are defined by accurate and precisemeasurements of conventionally chosen base quantities. Other units are then derived as algebraic combinations ofthe base units.
• Two commonly used systems of units are English units and SI units. All scientists and most of the other people inthe world use SI, whereas nonscientists in the United States still tend to use English units.
• The SI base units of length, mass, and time are the meter (m), kilogram (kg), and second (s), respectively.
• SI units are a metric system of units, meaning values can be calculated by factors of 10. Metric prefixes may beused with metric units to scale the base units to sizes appropriate for almost any application.


1.3 Unit Conversion
• To convert a quantity from one unit to another, multiply by conversions factors in such a way that you cancel theunits you want to get rid of and introduce the units you want to end up with.
• Be careful with areas and volumes. Units obey the rules of algebra so, for example, if a unit is squared we need twofactors to cancel it.


1.4 Dimensional Analysis
• The dimension of a physical quantity is just an expression of the base quantities from which it is derived.
• All equations expressing physical laws or principles must be dimensionally consistent. This fact can be used as anaid in remembering physical laws, as a way to check whether claimed relationships between physical quantities arepossible, and even to derive new physical laws.


1.5 Estimates and Fermi Calculations
• An estimate is a rough educated guess at the value of a physical quantity based on prior experience and soundphysical reasoning. Some strategies that may help when making an estimate are as follows:


◦ Get big lengths from smaller lengths.
◦ Get areas and volumes from lengths.
◦ Get masses from volumes and densities.
◦ If all else fails, bound it.


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◦ One “sig. fig.” is fine.
◦ Ask yourself: Does this make any sense?


1.6 Significant Figures
• Accuracy of a measured value refers to how close a measurement is to an accepted reference value. The discrepancyin a measurement is the amount by which the measurement result differs from this value.
• Precision of measured values refers to how close the agreement is between repeated measurements. The uncertaintyof a measurement is a quantification of this.
• The precision of a measuring tool is related to the size of its measurement increments. The smaller the measurementincrement, the more precise the tool.
• Significant figures express the precision of a measuring tool.
• When multiplying or dividing measured values, the final answer can contain only as many significant figures as theleast-precise value.
• When adding or subtracting measured values, the final answer cannot contain more decimal places than the least-precise value.


1.7 Solving Problems in Physics
The three stages of the process for solving physics problems used in this book are as follows:


• Strategy: Determine which physical principles are involved and develop a strategy for using them to solve theproblem.
• Solution: Do the math necessary to obtain a numerical solution complete with units.
• Significance: Check the solution to make sure it makes sense (correct units, reasonable magnitude and sign) andassess its significance.


CONCEPTUAL QUESTIONS
1.1 The Scope and Scale of Physics
1. What is physics?
2. Some have described physics as a “search forsimplicity.” Explain why this might be an appropriatedescription.
3. If two different theories describe experimentalobservations equally well, can one be said to be more validthan the other (assuming both use accepted rules of logic)?
4. What determines the validity of a theory?
5. Certain criteria must be satisfied if a measurement orobservation is to be believed. Will the criteria necessarilybe as strict for an expected result as for an unexpectedresult?
6. Can the validity of a model be limited or must it beuniversally valid? How does this compare with the requiredvalidity of a theory or a law?


1.2 Units and Standards
7. Identify some advantages of metric units.
8. What are the SI base units of length, mass, and time?
9. What is the difference between a base unit and a derivedunit? (b) What is the difference between a base quantity anda derived quantity? (c) What is the difference between abase quantity and a base unit?
10. For each of the following scenarios, refer to Figure1.4 and Table 1.2 to determine which metric prefix onthe meter is most appropriate for each of the followingscenarios. (a) You want to tabulate the mean distance fromthe Sun for each planet in the solar system. (b) You wantto compare the sizes of some common viruses to design amechanical filter capable of blocking the pathogenic ones.(c) You want to list the diameters of all the elements on theperiodic table. (d) You want to list the distances to all thestars that have now received any radio broadcasts sent fromEarth 10 years ago.


Chapter 1 | Units and Measurement 37




1.6 Significant Figures
11. (a) What is the relationship between the precisionand the uncertainty of a measurement? (b) What is therelationship between the accuracy and the discrepancy of ameasurement?


1.7 Solving Problems in Physics
12. What information do you need to choose whichequation or equations to use to solve a problem?
13. What should you do after obtaining a numericalanswer when solving a problem?


PROBLEMS
1.1 The Scope and Scale of Physics
14. Find the order of magnitude of the following physicalquantities. (a) The mass of Earth’s atmosphere:
5.1 × 1018kg; (b) The mass of the Moon’s atmosphere:
25,000 kg; (c) The mass of Earth’s hydrosphere:
1.4 × 1021kg; (d) The mass of Earth: 5.97 × 1024kg;
(e) The mass of the Moon: 7.34 × 1022kg; (f) The
Earth–Moon distance (semimajor axis): 3.84 × 108m; (g)
The mean Earth–Sun distance: 1.5 × 1011m; (h) The
equatorial radius of Earth: 6.38 × 106m; (i) The mass of
an electron: 9.11 × 10−31kg; (j) The mass of a proton:
1.67 × 10−27kg; (k) The mass of the Sun:
1.99 × 1030kg.


15. Use the orders of magnitude you found in the previousproblem to answer the following questions to within anorder of magnitude. (a) How many electrons would it taketo equal the mass of a proton? (b) How many Earths wouldit take to equal the mass of the Sun? (c) How manyEarth–Moon distances would it take to cover the distancefrom Earth to the Sun? (d) How many Moon atmosphereswould it take to equal the mass of Earth’s atmosphere? (e)How many moons would it take to equal the mass of Earth?(f) How many protons would it take to equal the mass ofthe Sun?
For the remaining questions, you need to use Figure 1.4to obtain the necessary orders of magnitude of lengths,masses, and times.
16. Roughly how many heartbeats are there in a lifetime?
17. A generation is about one-third of a lifetime.Approximately how many generations have passed sincethe year 0 AD?
18. Roughly how many times longer than the mean life ofan extremely unstable atomic nucleus is the lifetime of ahuman?


19. Calculate the approximate number of atoms in abacterium. Assume the average mass of an atom in thebacterium is 10 times the mass of a proton.
20. (a) Calculate the number of cells in a hummingbirdassuming the mass of an average cell is 10 times the massof a bacterium. (b) Making the same assumption, howmany cells are there in a human?
21. Assuming one nerve impulse must end before anothercan begin, what is the maximum firing rate of a nerve inimpulses per second?
22. About how many floating-point operations can asupercomputer perform each year?
23. Roughly how many floating-point operations can asupercomputer perform in a human lifetime?


1.2 Units and Standards
24. The following times are given using metric prefixeson the base SI unit of time: the second. Rewrite them inscientific notation without the prefix. For example, 47 Ts
would be rewritten as 4.7 × 1013 s. (a) 980 Ps; (b) 980 fs;
(c) 17 ns; (d) 577 µs.
25. The following times are given in seconds. Use metricprefixes to rewrite them so the numerical value is greater
than one but less than 1000. For example, 7.9 × 10−2 s
could be written as either 7.9 cs or 79 ms. (a) 9.57 × 105 s;
(b) 0.045 s; (c) 5.5 × 10−7 s; (d) 3.16 × 107 s.
26. The following lengths are given using metric prefixeson the base SI unit of length: the meter. Rewrite them inscientific notation without the prefix. For example, 4.2 Pm
would be rewritten as 4.2 × 1015m. (a) 89 Tm; (b) 89 pm;
(c) 711 mm; (d) 0.45 µm.
27. The following lengths are given in meters. Use metricprefixes to rewrite them so the numerical value is bigger


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than one but less than 1000. For example, 7.9 × 10−2m
could be written either as 7.9 cm or 79 mm. (a)
7.59 × 107m; (b) 0.0074 m; (c) 8.8 × 10−11m; (d)
1.63 × 1013m.


28. The following masses are written using metricprefixes on the gram. Rewrite them in scientific notationin terms of the SI base unit of mass: the kilogram. For
example, 40 Mg would be written as 4 × 104kg. (a) 23
mg; (b) 320 Tg; (c) 42 ng; (d) 7 g; (e) 9 Pg.
29. The following masses are given in kilograms. Usemetric prefixes on the gram to rewrite them so thenumerical value is bigger than one but less than 1000.
For example, 7 × 10−4kg could be written as 70 cg or
700 mg. (a) 3.8 × 10−5kg; (b) 2.3 × 1017kg; (c)
2.4 × 10−11kg; (d) 8 × 1015kg; (e) 4.2 × 10−3kg.


1.3 Unit Conversion
30. The volume of Earth is on the order of 1021 m3.(a) What is this in cubic kilometers (km3)? (b) What is itin cubic miles (mi3)? (c) What is it in cubic centimeters(cm3)?
31. The speed limit on some interstate highways isroughly 100 km/h. (a) What is this in meters per second?(b) How many miles per hour is this?
32. A car is traveling at a speed of 33 m/s. (a) What is itsspeed in kilometers per hour? (b) Is it exceeding the 90 km/h speed limit?
33. In SI units, speeds are measured in meters per second(m/s). But, depending on where you live, you’re probablymore comfortable of thinking of speeds in terms of eitherkilometers per hour (km/h) or miles per hour (mi/h). Inthis problem, you will see that 1 m/s is roughly 4 km/hor 2 mi/h, which is handy to use when developing yourphysical intuition. More precisely, show that (a)
1.0 m/s = 3.6 km/h and (b) 1.0 m/s = 2.2 mi/h.
34. American football is played on a 100-yd-long field,excluding the end zones. How long is the field in meters?(Assume that 1 m = 3.281 ft.)
35. Soccer fields vary in size. A large soccer field is 115m long and 85.0 m wide. What is its area in square feet?(Assume that 1 m = 3.281 ft.)
36. What is the height in meters of a person who is 6 ft 1.0in. tall?


37. Mount Everest, at 29,028 ft, is the tallest mountain onEarth. What is its height in kilometers? (Assume that 1 m =3.281 ft.)
38. The speed of sound is measured to be 342 m/s ona certain day. What is this measurement in kilometers perhour?
39. Tectonic plates are large segments of Earth’s crust thatmove slowly. Suppose one such plate has an average speedof 4.0 cm/yr. (a) What distance does it move in 1.0 s atthis speed? (b) What is its speed in kilometers per millionyears?
40. The average distance between Earth and the Sun is
1.5 × 1011m. (a) Calculate the average speed of Earth in
its orbit (assumed to be circular) in meters per second. (b)What is this speed in miles per hour?
41. The density of nuclear matter is about 1018 kg/m3.Given that 1 mL is equal in volume to cm3, what is thedensity of nuclear matter in megagrams per microliter (thatis, Mg/µL )?
42. The density of aluminum is 2.7 g/cm3. What is thedensity in kilograms per cubic meter?
43. A commonly used unit of mass in the English systemis the pound-mass, abbreviated lbm, where 1 lbm = 0.454kg. What is the density of water in pound-mass per cubicfoot?
44. A furlong is 220 yd. A fortnight is 2 weeks. Converta speed of one furlong per fortnight to millimeters persecond.
45. It takes 2π radians (rad) to get around a circle, which
is the same as 360°. How many radians are in 1°?
46. Light travels a distance of about 3 × 108m/s. A
light-minute is the distance light travels in 1 min. If the Sun
is 1.5 × 1011m from Earth, how far away is it in light-
minutes?
47. A light-nanosecond is the distance light travels in 1 ns.Convert 1 ft to light-nanoseconds.
48. An electron has a mass of 9.11 × 10−31kg. A proton
has a mass of 1.67 × 10−27kg. What is the mass of a
proton in electron-masses?
49. A fluid ounce is about 30 mL. What is the volume of a12 fl-oz can of soda pop in cubic meters?


Chapter 1 | Units and Measurement 39




1.4 Dimensional Analysis
50. A student is trying to remember some formulas fromgeometry. In what follows, assume A is area, V is
volume, and all other variables are lengths. Determinewhich formulas are dimensionally consistent. (a)
V = πr2h; (b) A = 2πr2 + 2πrh; (c) V = 0.5bh; (d)
V = πd2; (e) V = πd3 /6.
51. Consider the physical quantities s, v, a, and t with
dimensions [s] = L, [v] = LT−1, [a] = LT−2, and
[t] = T. Determine whether each of the following
equations is dimensionally consistent. (a) v2 = 2as; (b)
s = vt2 + 0.5at2; (c) v = s/t; (d) a = v/t.
52. Consider the physical quantities m, s, v, a,
and t with dimensions [m] = M, [s] = L, [v] = LT–1,
[a] = LT–2, and [t] = T. Assuming each of the followingequations is dimensionally consistent, find the dimensionof the quantity on the left-hand side of the equation: (a) F =ma; (b) K = 0.5mv2; (c) p = mv; (d) W = mas; (e) L = mvr.
53. Suppose quantity s is a length and quantity t is a
time. Suppose the quantities v and a are defined by v
= ds/dt and a = dv/dt. (a) What is the dimension of v?(b) What is the dimension of the quantity a? What are the
dimensions of (c) ∫ vdt, (d) ∫ adt, and (e) da/dt?


54. Suppose [V] = L3, [ρ] = ML–3, and [t] = T. (a)
What is the dimension of ∫ ρdV ? (b) What is the
dimension of dV/dt? (c) What is the dimension of
ρ(dV /dt)?


55. The arc length formula says the length s of arc
subtended by angle Ɵ in a circle of radius r is given by
the equation s = r Ɵ . What are the dimensions of (a) s,
(b) r, and (c) Ɵ?


1.5 Estimates and Fermi Calculations
56. Assuming the human body is made primarily of water,estimate the volume of a person.
57. Assuming the human body is primarily made of water,estimate the number of molecules in it. (Note that water hasa molecular mass of 18 g/mol and there are roughly 1024atoms in a mole.)
58. Estimate the mass of air in a classroom.


59. Estimate the number of molecules that make up Earth,assuming an average molecular mass of 30 g/mol. (Notethere are on the order of 1024 objects per mole.)
60. Estimate the surface area of a person.
61. Roughly how many solar systems would it take to tilethe disk of the Milky Way?
62. (a) Estimate the density of the Moon. (b) Estimate thediameter of the Moon. (c) Given that the Moon subtendsat an angle of about half a degree in the sky, estimate itsdistance from Earth.
63. The average density of the Sun is on the order 103 kg/m3. (a) Estimate the diameter of the Sun. (b) Given that theSun subtends at an angle of about half a degree in the sky,estimate its distance from Earth.
64. Estimate the mass of a virus.
65. A floating-point operation is a single arithmeticoperation such as addition, subtraction, multiplication, ordivision. (a) Estimate the maximum number of floating-point operations a human being could possibly perform ina lifetime. (b) How long would it take a supercomputer toperform that many floating-point operations?


1.6 Significant Figures
66. Consider the equation 4000/400 = 10.0. Assuming thenumber of significant figures in the answer is correct, whatcan you say about the number of significant figures in 4000and 400?
67. Suppose your bathroom scale reads your mass as 65kg with a 3% uncertainty. What is the uncertainty in yourmass (in kilograms)?
68. A good-quality measuring tape can be off by 0.50 cmover a distance of 20 m. What is its percent uncertainty?
69. An infant’s pulse rate is measured to be 130 ± 5 beats/min. What is the percent uncertainty in this measurement?
70. (a) Suppose that a person has an average heart rate of72.0 beats/min. How many beats does he or she have in 2.0years? (b) In 2.00 years? (c) In 2.000 years?
71. A can contains 375 mL of soda. How much is left after308 mL is removed?
72. State how many significant figures are proper in theresults of the following calculations: (a)
(106.7)(98.2) / (46.210)(1.01); (b) (18.7)2; (c)


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⎝1.60 × 10


−19⎞
⎠(3712)


73. (a) How many significant figures are in the numbers99 and 100.? (b) If the uncertainty in each number is 1,what is the percent uncertainty in each? (c) Which is amore meaningful way to express the accuracy of these twonumbers: significant figures or percent uncertainties?
74. (a) If your speedometer has an uncertainty of 2.0 km/hat a speed of 90 km/h, what is the percent uncertainty? (b)If it has the same percent uncertainty when it reads 60 km/h, what is the range of speeds you could be going?
75. (a) A person’s blood pressure is measured to be
120 ± 2 mm Hg. What is its percent uncertainty? (b)
Assuming the same percent uncertainty, what is theuncertainty in a blood pressure measurement of 80 mmHg?
76. A person measures his or her heart rate by counting thenumber of beats in 30 s. If 40 ± 1 beats are counted in 30.0


± 0.5 s, what is the heart rate and its uncertainty in beats perminute?
77. What is the area of a circle 3.102 cm in diameter?
78. Determine the number of significant figures in thefollowing measurements: (a) 0.0009, (b) 15,450.0, (c)6×103, (d) 87.990, and (e) 30.42.
79. Perform the following calculations and express youranswer using the correct number of significant digits. (a)A woman has two bags weighing 13.5 lb and one bagwith a weight of 10.2 lb. What is the total weight of thebags? (b) The force F on an object is equal to its mass mmultiplied by its acceleration a. If a wagon with mass 55kg accelerates at a rate of 0.0255 m/s2, what is the force onthe wagon? (The unit of force is called the newton and it isexpressed with the symbol N.)


ADDITIONAL PROBLEMS
80. Consider the equation y = mt +b, where the dimensionof y is length and the dimension of t is time, and m and bare constants. What are the dimensions and SI units of (a)m and (b) b?
81. Consider the equation
s = s0 + v0 t + a0 t


2 /2 + j0 t
3 /6 + S0 t


4 /24 + ct5 /120,


where s is a length and t is a time. What are the dimensionsand SI units of (a) s0, (b) v0, (c) a0, (d) j0, (e) S0,
and (f) c?
82. (a) A car speedometer has a 5% uncertainty. What isthe range of possible speeds when it reads 90 km/h? (b)Convert this range to miles per hour. Note 1 km = 0.6214mi.
83. A marathon runner completes a 42.188-km course in 2h, 30 min, and 12 s. There is an uncertainty of 25 m in thedistance traveled and an uncertainty of 1 s in the elapsedtime. (a) Calculate the percent uncertainty in the distance.(b) Calculate the percent uncertainty in the elapsed time. (c)What is the average speed in meters per second? (d) Whatis the uncertainty in the average speed?


84. The sides of a small rectangular box are measured tobe 1.80 ± 0.1 cm, 2.05 ± 0.02 cm, and 3.1 ± 0.1 cm long.Calculate its volume and uncertainty in cubic centimeters.
85. When nonmetric units were used in the UnitedKingdom, a unit of mass called the pound-mass (lbm) wasused, where 1 lbm = 0.4539 kg. (a) If there is an uncertaintyof 0.0001 kg in the pound-mass unit, what is its percentuncertainty? (b) Based on that percent uncertainty, whatmass in pound-mass has an uncertainty of 1 kg whenconverted to kilograms?
86. The length and width of a rectangular room aremeasured to be 3.955 ± 0.005 m and 3.050 ± 0.005 m.Calculate the area of the room and its uncertainty in squaremeters.
87. A car engine moves a piston with a circular cross-section of 7.500 ± 0.002 cm in diameter a distance of3.250 ± 0.001 cm to compress the gas in the cylinder. (a)By what amount is the gas decreased in volume in cubiccentimeters? (b) Find the uncertainty in this volume.


CHALLENGE PROBLEMS
88. The first atomic bomb was detonated on July 16, 1945,at the Trinity test site about 200 mi south of Los Alamos. In 1947, the U.S. government declassified a film reel of theexplosion. From this film reel, British physicist G. I. Taylor


Chapter 1 | Units and Measurement 41




was able to determine the rate at which the radius of thefireball from the blast grew. Using dimensional analysis, hewas then able to deduce the amount of energy released inthe explosion, which was a closely guarded secret at thetime. Because of this, Taylor did not publish his results until1950. This problem challenges you to recreate this famouscalculation. (a) Using keen physical insight developed fromyears of experience, Taylor decided the radius r of thefireball should depend only on time since the explosion,t, the density of the air, ρ, and the energy of the initial
explosion, E. Thus, he made the educated guess that
r = kEa ρb tc for some dimensionless constant k and some
unknown exponents a, b, and c. Given that [E] = ML2T–2,determine the values of the exponents necessary to makethis equation dimensionally consistent. (Hint: Notice the
equation implies that k = rE−a ρ−b t−c and that [k] = 1.
) (b) By analyzing data from high-energy conventionalexplosives, Taylor found the formula he derived seemed tobe valid as long as the constant k had the value 1.03. Fromthe film reel, he was able to determine many values of rand the corresponding values of t. For example, he foundthat after 25.0 ms, the fireball had a radius of 130.0 m.Use these values, along with an average air density of 1.25kg/m3, to calculate the initial energy release of the Trinitydetonation in joules (J). (Hint: To get energy in joules,you need to make sure all the numbers you substitute inare expressed in terms of SI base units.) (c) The energyreleased in large explosions is often cited in units of “tonsof TNT” (abbreviated “t TNT”), where 1 t TNT is about4.2 GJ. Convert your answer to (b) into kilotons of TNT


(that is, kt TNT). Compare your answer with the quick-and-dirty estimate of 10 kt TNT made by physicist EnricoFermi shortly after witnessing the explosion from what wasthought to be a safe distance. (Reportedly, Fermi made hisestimate by dropping some shredded bits of paper rightbefore the remnants of the shock wave hit him and lookedto see how far they were carried by it.)
89. The purpose of this problem is to show the entireconcept of dimensional consistency can be summarizedby the old saying “You can’t add apples and oranges.” Ifyou have studied power series expansions in a calculuscourse, you know the standard mathematical functions suchas trigonometric functions, logarithms, and exponentialfunctions can be expressed as infinite sums of the form

n = 0




an x
n = a0 + a1 x + a2 x


2 + a3 x
3 + ⋯ , where


the an are dimensionless constants for all
n = 0, 1, 2, ⋯ and x is the argument of the function.
(If you have not studied power series in calculus yet, justtrust us.) Use this fact to explain why the requirementthat all terms in an equation have the same dimensions issufficient as a definition of dimensional consistency. Thatis, it actually implies the arguments of standardmathematical functions must be dimensionless, so it is notreally necessary to make this latter condition a separaterequirement of the definition of dimensional consistency aswe have done in this section.


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2 | VECTORS


Figure 2.1 A signpost gives information about distances and directions to towns or to other locations relative to the location ofthe signpost. Distance is a scalar quantity. Knowing the distance alone is not enough to get to the town; we must also know thedirection from the signpost to the town. The direction, together with the distance, is a vector quantity commonly called thedisplacement vector. A signpost, therefore, gives information about displacement vectors from the signpost to towns. (credit:modification of work by “studio tdes”/Flickr)


Chapter Outline
2.1 Scalars and Vectors
2.2 Coordinate Systems and Components of a Vector
2.3 Algebra of Vectors
2.4 Products of Vectors


Introduction
Vectors are essential to physics and engineering. Many fundamental physical quantities are vectors, including displacement,velocity, force, and electric and magnetic vector fields. Scalar products of vectors define other fundamental scalar physicalquantities, such as energy. Vector products of vectors define still other fundamental vector physical quantities, such as torqueand angular momentum. In other words, vectors are a component part of physics in much the same way as sentences are acomponent part of literature.
In introductory physics, vectors are Euclidean quantities that have geometric representations as arrows in one dimension (ina line), in two dimensions (in a plane), or in three dimensions (in space). They can be added, subtracted, or multiplied. Inthis chapter, we explore elements of vector algebra for applications in mechanics and in electricity and magnetism. Vectoroperations also have numerous generalizations in other branches of physics.


Chapter 2 | Vectors 43




2.1 | Scalars and Vectors
Learning Objectives


By the end of this section, you will be able to:
• Describe the difference between vector and scalar quantities.
• Identify the magnitude and direction of a vector.
• Explain the effect of multiplying a vector quantity by a scalar.
• Describe how one-dimensional vector quantities are added or subtracted.
• Explain the geometric construction for the addition or subtraction of vectors in a plane.
• Distinguish between a vector equation and a scalar equation.


Many familiar physical quantities can be specified completely by giving a single number and the appropriate unit. Forexample, “a class period lasts 50 min” or “the gas tank in my car holds 65 L” or “the distance between two posts is 100m.” A physical quantity that can be specified completely in this manner is called a scalar quantity. Scalar is a synonym of“number.” Time, mass, distance, length, volume, temperature, and energy are examples of scalar quantities.
Scalar quantities that have the same physical units can be added or subtracted according to the usual rules of algebra fornumbers. For example, a class ending 10 min earlier than 50 min lasts 50 min − 10 min = 40 min . Similarly, a 60-cal
serving of corn followed by a 200-cal serving of donuts gives 60 cal + 200 cal = 260 cal of energy. When we multiply
a scalar quantity by a number, we obtain the same scalar quantity but with a larger (or smaller) value. For example, ifyesterday’s breakfast had 200 cal of energy and today’s breakfast has four times as much energy as it had yesterday, thentoday’s breakfast has 4(200 cal) = 800 cal of energy. Two scalar quantities can also be multiplied or divided by each other
to form a derived scalar quantity. For example, if a train covers a distance of 100 km in 1.0 h, its speed is 100.0 km/1.0 h =27.8 m/s, where the speed is a derived scalar quantity obtained by dividing distance by time.
Many physical quantities, however, cannot be described completely by just a single number of physical units. For example,when the U.S. Coast Guard dispatches a ship or a helicopter for a rescue mission, the rescue team must know not only thedistance to the distress signal, but also the direction from which the signal is coming so they can get to its origin as quicklyas possible. Physical quantities specified completely by giving a number of units (magnitude) and a direction are calledvector quantities. Examples of vector quantities include displacement, velocity, position, force, and torque. In the languageof mathematics, physical vector quantities are represented by mathematical objects called vectors (Figure 2.2). We canadd or subtract two vectors, and we can multiply a vector by a scalar or by another vector, but we cannot divide by a vector.The operation of division by a vector is not defined.


Figure 2.2 We draw a vector from the initial point or origin(called the “tail” of a vector) to the end or terminal point (calledthe “head” of a vector), marked by an arrowhead. Magnitude isthe length of a vector and is always a positive scalar quantity.(credit: modification of work by Cate Sevilla)


Let’s examine vector algebra using a graphical method to be aware of basic terms and to develop a qualitativeunderstanding. In practice, however, when it comes to solving physics problems, we use analytical methods, which we’llsee in the next section. Analytical methods are more simple computationally and more accurate than graphical methods.From now on, to distinguish between a vector and a scalar quantity, we adopt the common convention that a letter in bold


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type with an arrow above it denotes a vector, and a letter without an arrow denotes a scalar. For example, a distance of 2.0km, which is a scalar quantity, is denoted by d = 2.0 km, whereas a displacement of 2.0 km in some direction, which is a
vector quantity, is denoted by d→ .
Suppose you tell a friend on a camping trip that you have discovered a terrific fishing hole 6 km from your tent. It is unlikelyyour friend would be able to find the hole easily unless you also communicate the direction in which it can be found withrespect to your campsite. You may say, for example, “Walk about 6 km northeast from my tent.” The key concept here isthat you have to give not one but two pieces of information—namely, the distance or magnitude (6 km) and the direction(northeast).
Displacement is a general term used to describe a change in position, such as during a trip from the tent to the fishing hole.Displacement is an example of a vector quantity. If you walk from the tent (location A) to the hole (location B), as shown
in Figure 2.3, the vector D→ , representing your displacement, is drawn as the arrow that originates at point A and ends
at point B. The arrowhead marks the end of the vector. The direction of the displacement vector D→ is the direction of the
arrow. The length of the arrow represents themagnitude D of vector D→ . Here, D = 6 km. Since the magnitude of a vector
is its length, which is a positive number, the magnitude is also indicated by placing the absolute value notation around the
symbol that denotes the vector; so, we can write equivalently that D ≡ | D→ | . To solve a vector problem graphically, we
need to draw the vector D→ to scale. For example, if we assume 1 unit of distance (1 km) is represented in the drawing
by a line segment of length u = 2 cm, then the total displacement in this example is represented by a vector of length
d = 6u = 6(2 cm) = 12 cm , as shown in Figure 2.4. Notice that here, to avoid confusion, we used D = 6 km to denote
the magnitude of the actual displacement and d = 12 cm to denote the length of its representation in the drawing.


Figure 2.3 The displacement vector from point A (the initialposition at the campsite) to point B (the final position at thefishing hole) is indicated by an arrow with origin at point A andend at point B. The displacement is the same for any of theactual paths (dashed curves) that may be taken between points Aand B.


Chapter 2 | Vectors 45




Figure 2.4 A displacement D→ of magnitude 6 km is drawn
to scale as a vector of length 12 cm when the length of 2 cmrepresents 1 unit of displacement (which in this case is 1 km).


Suppose your friend walks from the campsite at A to the fishing pond at B and then walks back: from the fishing pond at
B to the campsite at A. The magnitude of the displacement vector D→ AB from A to B is the same as the magnitude of the
displacement vector D→ BA from B to A (it equals 6 km in both cases), so we can write DAB = DBA . However, vector
D


AB is not equal to vector D→ BA because these two vectors have different directions: D→ AB ≠ D→ BA . In Figure
2.3, vector D→ BA would be represented by a vector with an origin at point B and an end at point A, indicating vector
D


BA points to the southwest, which is exactly 180° opposite to the direction of vector D→ AB . We say that vector
D


BA is antiparallel to vector D→ AB and write D→ AB = − D→ BA , where the minus sign indicates the antiparallel
direction.
Two vectors that have identical directions are said to be parallel vectors—meaning, they are parallel to each other. Two
parallel vectors A→ and B→ are equal, denoted by A→ = B→ , if and only if they have equal magnitudes | A→ | = | B→ | .
Two vectors with directions perpendicular to each other are said to be orthogonal vectors. These relations between vectorsare illustrated in Figure 2.5.


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2.1


Figure 2.5 Various relations between two vectors A→ and B→ . (a)
A


≠ B
→ because A ≠ B . (b) A→ ≠ B→ because they are not


parallel and A ≠ B . (c) A→ ≠ − A→ because they have different
directions (even though | A→ | = | − A→ | = A) . (d) A→ = B→
because they are parallel and have identical magnitudes A = B. (e)
A


≠ B
→ because they have different directions (are not parallel);


here, their directions differ by 90°—meaning, they are orthogonal.
Check Your Understanding Two motorboats named Alice and Bob are moving on a lake. Given theinformation about their velocity vectors in each of the following situations, indicate whether their velocityvectors are equal or otherwise. (a) Alicemoves north at 6 knots and Bobmoves west at 6 knots. (b) Alicemoveswest at 6 knots and Bob moves west at 3 knots. (c) Alice moves northeast at 6 knots and Bob moves south at 3knots. (d) Alice moves northeast at 6 knots and Bob moves southwest at 6 knots. (e) Alice moves northeast at 2knots and Bob moves closer to the shore northeast at 2 knots.


Algebra of Vectors in One Dimension
Vectors can be multiplied by scalars, added to other vectors, or subtracted from other vectors. We can illustrate these vectorconcepts using an example of the fishing trip seen in Figure 2.6.


Chapter 2 | Vectors 47




Figure 2.6 Displacement vectors for a fishing trip. (a) Stopping to rest at point C while walking from camp (point A) to thepond (point B). (b) Going back for the dropped tackle box (point D). (c) Finishing up at the fishing pond.


Suppose your friend departs from point A (the campsite) and walks in the direction to point B (the fishing pond), but,along the way, stops to rest at some point C located three-quarters of the distance between A and B, beginning from
point A (Figure 2.6(a)). What is his displacement vector D→ AC when he reaches point C? We know that if he walks
all the way to B, his displacement vector relative to A is D→ AB , which has magnitude DAB = 6 km and a direction
of northeast. If he walks only a 0.75 fraction of the total distance, maintaining the northeasterly direction, at point C hemust be 0.75DAB = 4.5 km away from the campsite at A. So, his displacement vector at the rest point C has magnitude
DAC = 4.5 km = 0.75DAB and is parallel to the displacement vector D→ AB . All of this can be stated succinctly in the
form of the following vector equation:


D


AC = 0.75 D


AB.


In a vector equation, both sides of the equation are vectors. The previous equation is an example of a vector multiplied by a
positive scalar (number) α = 0.75 . The result, D→ AC , of such a multiplication is a new vector with a direction parallel to
the direction of the original vector D→ AB .
In general, when a vector A→ is multiplied by a positive scalar α , the result is a new vector B→ that is parallel to A→ :


(2.1)B→ = α A→ .


The magnitude | B→ | of this new vector is obtained by multiplying the magnitude | A→ | of the original vector, as expressed
by the scalar equation:


(2.2)B = |α|A.


In a scalar equation, both sides of the equation are numbers. Equation 2.2 is a scalar equation because the magnitudesof vectors are scalar quantities (and positive numbers). If the scalar α is negative in the vector equation Equation 2.1,
then the magnitude | B→ | of the new vector is still given by Equation 2.2, but the direction of the new vector B→ is


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antiparallel to the direction of A→ . These principles are illustrated in Figure 2.7(a) by two examples where the length
of vector A→ is 1.5 units. When α = 2 , the new vector B→ = 2 A→ has length B = 2A = 3.0 units (twice as long
as the original vector) and is parallel to the original vector. When α = −2 , the new vector C→ = −2 A→ has length
C = | − 2|A = 3.0 units (twice as long as the original vector) and is antiparallel to the original vector.


Figure 2.7 Algebra of vectors in one dimension. (a) Multiplication by a
scalar. (b) Addition of two vectors ( R→ is called the resultant of vectors
A
→ and B→ ) . (c) Subtraction of two vectors ( D→ is the difference of
vectors A→ and B→ ) .


Now suppose your fishing buddy departs from point A (the campsite), walking in the direction to point B (the fishinghole), but he realizes he lost his tackle box when he stopped to rest at point C (located three-quarters of the distancebetween A and B, beginning from point A). So, he turns back and retraces his steps in the direction toward the campsiteand finds the box lying on the path at some point D only 1.2 km away from point C (see Figure 2.6(b)). What is his
displacement vector D→ AD when he finds the box at point D? What is his displacement vector D→ DB from point D to the
hole? We have already established that at rest point C his displacement vector is D→ AC = 0.75 D→ AB . Starting at point
C, he walks southwest (toward the campsite), which means his new displacement vector D→ CD from point C to point
D is antiparallel to D→ AB . Its magnitude | D→ CD| is DCD = 1.2 km = 0.2DAB , so his second displacement vector is
D


CD = −0.2 D


AB . His total displacement D→ AD relative to the campsite is the vector sum of the two displacement
vectors: vector D→ AC (from the campsite to the rest point) and vector D→ CD (from the rest point to the point where he
finds his box):


(2.3)D→ AD = D→ AC + D→ CD.


The vector sum of two (or more) vectors is called the resultant vector or, for short, the resultant. When the vectors on the
right-hand-side of Equation 2.3 are known, we can find the resultant D→ AD as follows:


(2.4)D→ AD = D→ AC + D→ CD = 0.75 D→ AB − 0.2 D→ AB = (0.75 − 0.2) D→ AB = 0.55 D→ AB.
When your friend finally reaches the pond at B, his displacement vector D→ AB from point A is the vector sum of his


Chapter 2 | Vectors 49




displacement vector D→ AD from point A to point D and his displacement vector D→ DB from point D to the fishing hole:
D


AB = D


AD + D


DB (see Figure 2.6(c)). This means his displacement vector D→ DB is the difference of two
vectors:


(2.5)D→ DB = D→ AB − D→ AD = D→ AB + (− D→ AD).
Notice that a difference of two vectors is nothing more than a vector sum of two vectors because the second term in
Equation 2.5 is vector − D→ AD (which is antiparallel to D→ AD) . When we substitute Equation 2.4 into Equation
2.5, we obtain the second displacement vector:


(2.6)D→ DB = D→ AB − D→ AD = D→ AB − 0.55 D→ AB = (1.0 − 0.55) D→ AB = 0.45 D→ AB.
This result means your friend walked DDB = 0.45DAB = 0.45(6.0 km) = 2.7 km from the point where he finds his tackle
box to the fishing hole.
When vectors A→ and B→ lie along a line (that is, in one dimension), such as in the camping example, their resultant
R


= A


+ B
→ and their difference D→ = A→ − B→ both lie along the same direction. We can illustrate the addition


or subtraction of vectors by drawing the corresponding vectors to scale in one dimension, as shown in Figure 2.7.
To illustrate the resultant when A→ and B→ are two parallel vectors, we draw them along one line by placing the origin
of one vector at the end of the other vector in head-to-tail fashion (see Figure 2.7(b)). The magnitude of this resultant
is the sum of their magnitudes: R = A + B. The direction of the resultant is parallel to both vectors. When vector A→
is antiparallel to vector B→ , we draw them along one line in either head-to-head fashion (Figure 2.7(c)) or tail-to-
tail fashion. The magnitude of the vector difference, then, is the absolute value D = |A − B| of the difference of their
magnitudes. The direction of the difference vector D→ is parallel to the direction of the longer vector.
In general, in one dimension—as well as in higher dimensions, such as in a plane or in space—we can add any number ofvectors and we can do so in any order because the addition of vectors is commutative,


(2.7)A→ + B→ = B→ + A→ ,


and associative,


(2.8)( A→ + B→ ) + C→ = A→ + ( B→ + C→ ).


Moreover, multiplication by a scalar is distributive:


(2.9)α1 A→ + α2 A→ = (α1 + α2) A→ .


We used the distributive property in Equation 2.4 and Equation 2.6.
When adding many vectors in one dimension, it is convenient to use the concept of a unit vector. A unit vector, which
is denoted by a letter symbol with a hat, such as û , has a magnitude of one and does not have any physical unit so that
| û | ≡ u = 1 . The only role of a unit vector is to specify direction. For example, instead of saying vector D→ AB has a


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magnitude of 6.0 km and a direction of northeast, we can introduce a unit vector û that points to the northeast and say
succinctly that D→ AB = (6.0 km) û . Then the southwesterly direction is simply given by the unit vector − û . In this way,
the displacement of 6.0 km in the southwesterly direction is expressed by the vector


D


BA = (−6.0 km) û .


Example 2.1
A Ladybug Walker
A long measuring stick rests against a wall in a physics laboratory with its 200-cm end at the floor. A ladybuglands on the 100-cm mark and crawls randomly along the stick. It first walks 15 cm toward the floor, then it walks56 cm toward the wall, then it walks 3 cm toward the floor again. Then, after a brief stop, it continues for 25 cmtoward the floor and then, again, it crawls up 19 cm toward the wall before coming to a complete rest (Figure2.8). Find the vector of its total displacement and its final resting position on the stick.
Strategy
If we choose the direction along the stick toward the floor as the direction of unit vector û , then the direction
toward the floor is + û and the direction toward the wall is − û . The ladybug makes a total of five
displacements:


D


1 = (15 cm)( + û ),


D


2 = (56 cm)(− û ),


D


3 = (3 cm)( + û ),


D


4 = (25 cm)( + û ), and


D


5 = (19 cm)(− û ).


The total displacement D→ is the resultant of all its displacement vectors.


Figure 2.8 Five displacements of the ladybug. Note that in this schematic drawing,magnitudes of displacements are not drawn to scale. (credit: modification of work by“Persian Poet Gal”/Wikimedia Commons)


Chapter 2 | Vectors 51




2.2


Solution
The resultant of all the displacement vectors is


D


= D


1 + D


2 + D


3 + D


4 + D


5


= (15 cm)( + û ) + (56 cm)(− û ) + (3 cm)( + û ) + (25 cm)( + û ) + (19 cm)(− û )


= (15 − 56 + 3 + 25 − 19)cm û


= −32 cm û .


In this calculation, we use the distributive law given by Equation 2.9. The result reads that the totaldisplacement vector points away from the 100-cm mark (initial landing site) toward the end of the meter stickthat touches the wall. The end that touches the wall is marked 0 cm, so the final position of the ladybug is at the(100 – 32)cm = 68-cm mark.
Check Your Understanding A cave diver enters a long underwater tunnel. When her displacement withrespect to the entry point is 20 m, she accidentally drops her camera, but she doesn’t notice it missing until sheis some 6 m farther into the tunnel. She swims back 10 m but cannot find the camera, so she decides to end thedive. How far from the entry point is she? Taking the positive direction out of the tunnel, what is herdisplacement vector relative to the entry point?


Algebra of Vectors in Two Dimensions
When vectors lie in a plane—that is, when they are in two dimensions—they can be multiplied by scalars, added to othervectors, or subtracted from other vectors in accordance with the general laws expressed by Equation 2.1, Equation 2.2,Equation 2.7, and Equation 2.8. However, the addition rule for two vectors in a plane becomes more complicated thanthe rule for vector addition in one dimension. We have to use the laws of geometry to construct resultant vectors, followedby trigonometry to find vector magnitudes and directions. This geometric approach is commonly used in navigation(Figure 2.9). In this section, we need to have at hand two rulers, a triangle, a protractor, a pencil, and an eraser for drawingvectors to scale by geometric constructions.


Figure 2.9 In navigation, the laws of geometry are used to draw resultant displacements onnautical maps.


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For a geometric construction of the sum of two vectors in a plane, we follow the parallelogram rule. Suppose two vectors
A
→ and B→ are at the arbitrary positions shown in Figure 2.10. Translate either one of them in parallel to the beginning
of the other vector, so that after the translation, both vectors have their origins at the same point. Now, at the end of vector
A
→ we draw a line parallel to vector B→ and at the end of vector B→ we draw a line parallel to vector A→ (the dashed
lines in Figure 2.10). In this way, we obtain a parallelogram. From the origin of the two vectors we draw a diagonal that is
the resultant R→ of the two vectors: R→ = A→ + B→ (Figure 2.10(a)). The other diagonal of this parallelogram is the
vector difference of the two vectors D→ = A→ − B→ , as shown in Figure 2.10(b). Notice that the end of the difference
vector is placed at the end of vector A→ .


Figure 2.10 The parallelogram rule for the addition of two vectors. Make the parallel translation of each vector to a pointwhere their origins (marked by the dot) coincide and construct a parallelogram with two sides on the vectors and the other
two sides (indicated by dashed lines) parallel to the vectors. (a) Draw the resultant vector R→ along the diagonal of the
parallelogram from the common point to the opposite corner. Length R of the resultant vector is not equal to the sum of the
magnitudes of the two vectors. (b) Draw the difference vector D→ = A→ − B→ along the diagonal connecting the ends of
the vectors. Place the origin of vector D→ at the end of vector B→ and the end (arrowhead) of vector D→ at the end of
vector A→ . Length D of the difference vector is not equal to the difference of magnitudes of the two vectors.


It follows from the parallelogram rule that neither the magnitude of the resultant vector nor the magnitude of the differencevector can be expressed as a simple sum or difference of magnitudes A and B, because the length of a diagonal cannot be
expressed as a simple sum of side lengths. When using a geometric construction to find magnitudes | R→ | and | D→ | , we
have to use trigonometry laws for triangles, which may lead to complicated algebra. There are two ways to circumvent thisalgebraic complexity. One way is to use the method of components, which we examine in the next section. The other way isto draw the vectors to scale, as is done in navigation, and read approximate vector lengths and angles (directions) from thegraphs. In this section we examine the second approach.
If we need to add three or more vectors, we repeat the parallelogram rule for the pairs of vectors until we find the resultantof all of the resultants. For three vectors, for example, we first find the resultant of vector 1 and vector 2, and then wefind the resultant of this resultant and vector 3. The order in which we select the pairs of vectors does not matter becausethe operation of vector addition is commutative and associative (see Equation 2.7 and Equation 2.8). Before we state ageneral rule that follows from repetitive applications of the parallelogram rule, let’s look at the following example.
Suppose you plan a vacation trip in Florida. Departing from Tallahassee, the state capital, you plan to visit your uncleJoe in Jacksonville, see your cousin Vinny in Daytona Beach, stop for a little fun in Orlando, see a circus performancein Tampa, and visit the University of Florida in Gainesville. Your route may be represented by five displacement vectors
A


, B
→ , C→ , D→ , and E→ , which are indicated by the red vectors in Figure 2.11. What is your total displacement


when you reach Gainesville? The total displacement is the vector sum of all five displacement vectors, which may befound by using the parallelogram rule four times. Alternatively, recall that the displacement vector has its beginning atthe initial position (Tallahassee) and its end at the final position (Gainesville), so the total displacement vector can bedrawn directly as an arrow connecting Tallahassee with Gainesville (see the green vector in Figure 2.11). When we use
the parallelogram rule four times, the resultant R→ we obtain is exactly this green vector connecting Tallahassee with
Gainesville: R→ = A→ + B→ + C→ + D→ + E→ .


Chapter 2 | Vectors 53




Figure 2.11 When we use the parallelogram rule four times, we obtain the resultant vector
R


= A


+ B


+ C


+ D


+ E
→ , which is the green vector connecting Tallahassee with Gainesville.


Drawing the resultant vector of many vectors can be generalized by using the following tail-to-head geometric
construction. Suppose we want to draw the resultant vector R→ of four vectors A→ , B→ , C→ , and D→ (Figure
2.12(a)). We select any one of the vectors as the first vector and make a parallel translation of a second vector to a positionwhere the origin (“tail”) of the second vector coincides with the end (“head”) of the first vector. Then, we select a thirdvector and make a parallel translation of the third vector to a position where the origin of the third vector coincides withthe end of the second vector. We repeat this procedure until all the vectors are in a head-to-tail arrangement like the one
shown in Figure 2.12. We draw the resultant vector R→ by connecting the origin (“tail”) of the first vector with the end
(“head”) of the last vector. The end of the resultant vector is at the end of the last vector. Because the addition of vectors isassociative and commutative, we obtain the same resultant vector regardless of which vector we choose to be first, second,third, or fourth in this construction.


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Figure 2.12 Tail-to-head method for drawing the resultant vector
R


= A


+ B


+ C


+ D
→ . (a) Four vectors of different magnitudes and


directions. (b) Vectors in (a) are translated to new positions where the origin (“tail”) ofone vector is at the end (“head”) of another vector. The resultant vector is drawn fromthe origin (“tail”) of the first vector to the end (“head”) of the last vector in thisarrangement.


Example 2.2
Geometric Construction of the Resultant
The three displacement vectors A→ , B→ , and C→ in Figure 2.13 are specified by their magnitudes A = 10.0,
B = 7.0, and C = 8.0, respectively, and by their respective direction angles with the horizontal direction α = 35° ,
β = −110° , and γ = 30° . The physical units of the magnitudes are centimeters. Choose a convenient scale and
use a ruler and a protractor to find the following vector sums: (a) R→ = A→ + B→ , (b) D→ = A→ − B→ , and
(c) S→ = A→ − 3 B→ + C→ .


Figure 2.13 Vectors used in Example 2.2 and in the Check Your Understanding feature that follows.


Strategy
In geometric construction, to find a vector means to find its magnitude and its direction angle with the horizontaldirection. The strategy is to draw to scale the vectors that appear on the right-hand side of the equation andconstruct the resultant vector. Then, use a ruler and a protractor to read the magnitude of the resultant and thedirection angle. For parts (a) and (b) we use the parallelogram rule. For (c) we use the tail-to-head method.
Solution
For parts (a) and (b), we attach the origin of vector B→ to the origin of vector A→ , as shown in Figure 2.14,
and construct a parallelogram. The shorter diagonal of this parallelogram is the sum A→ + B→ . The longer of


Chapter 2 | Vectors 55




the diagonals is the difference A→ − B→ . We use a ruler to measure the lengths of the diagonals, and a protractor
to measure the angles with the horizontal. For the resultant R→ , we obtain R = 5.8 cm and θR ≈ 0° . For the
difference D→ , we obtain D = 16.2 cm and θD = 49.3° , which are shown in Figure 2.14.


Figure 2.14 Using the parallelogram rule to solve (a) (finding the resultant, red) and (b) (findingthe difference, blue).


For (c), we can start with vector −3 B→ and draw the remaining vectors tail-to-head as shown in Figure 2.15.
In vector addition, the order in which we draw the vectors is unimportant, but drawing the vectors to scale is very
important. Next, we draw vector S→ from the origin of the first vector to the end of the last vector and place the
arrowhead at the end of S→ . We use a ruler to measure the length of S→ , and find that its magnitude is
S = 36.9 cm. We use a protractor and find that its direction angle is θS = 52.9° . This solution is shown in Figure
2.15.


Figure 2.15 Using the tail-to-head method to solve (c)
(finding vector S→ , green).


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2.3 Check Your Understanding Using the three displacement vectors A→ , B→ , and F→ in Figure
2.13, choose a convenient scale, and use a ruler and a protractor to find vector G→ given by the vector
equation G→ = A→ + 2 B→ − F→ .


Observe the addition of vectors in a plane by visiting this vector calculator (https://openstaxcollege.org/l/21compveccalc) and this Phet simulation (https://openstaxcollege.org/l/21phetvecaddsim) .


2.2 | Coordinate Systems and Components of a Vector
Learning Objectives


By the end of this section, you will be able to:
• Describe vectors in two and three dimensions in terms of their components, using unit vectorsalong the axes.
• Distinguish between the vector components of a vector and the scalar components of a vector.
• Explain how the magnitude of a vector is defined in terms of the components of a vector.
• Identify the direction angle of a vector in a plane.
• Explain the connection between polar coordinates and Cartesian coordinates in a plane.


Vectors are usually described in terms of their components in a coordinate system. Even in everyday life we naturally invokethe concept of orthogonal projections in a rectangular coordinate system. For example, if you ask someone for directions toa particular location, you will more likely be told to go 40 km east and 30 km north than 50 km in the direction 37° north
of east.
In a rectangular (Cartesian) xy-coordinate system in a plane, a point in a plane is described by a pair of coordinates (x, y).
In a similar fashion, a vector A→ in a plane is described by a pair of its vector coordinates. The x-coordinate of vector
A
→ is called its x-component and the y-coordinate of vector A→ is called its y-component. The vector x-component is
a vector denoted by A→ x . The vector y-component is a vector denoted by A→ y . In the Cartesian system, the x and y
vector components of a vector are the orthogonal projections of this vector onto the x- and y-axes, respectively. In this way,following the parallelogram rule for vector addition, each vector on a Cartesian plane can be expressed as the vector sum ofits vector components:


(2.10)A→ = A→ x + A→ y.
As illustrated in Figure 2.16, vector A→ is the diagonal of the rectangle where the x-component A→ x is the side parallel
to the x-axis and the y-component A→ y is the side parallel to the y-axis. Vector component A→ x is orthogonal to vector
component A→ y .


Chapter 2 | Vectors 57




Figure 2.16 Vector A→ in a plane in the Cartesian coordinate
system is the vector sum of its vector x- and y-components. The
x-vector component A→ x is the orthogonal projection of vector
A
→ onto the x-axis. The y-vector component A→ y is the
orthogonal projection of vector A→ onto the y-axis. The numbers
Ax and Ay that multiply the unit vectors are the scalar components
of the vector.


It is customary to denote the positive direction on the x-axis by the unit vector i^ and the positive direction on the y-axis
by the unit vector j^ . Unit vectors of the axes, i^ and j^ , define two orthogonal directions in the plane. As shown in
Figure 2.16, the x- and y- components of a vector can now be written in terms of the unit vectors of the axes:


(2.11)⎧

⎨ A



x = Ax i


^


A


y = Ay j
^
.


The vectors A→ x and A→ y defined by Equation 2.11 are the vector components of vector A→ . The numbers Ax and
Ay that define the vector components in Equation 2.11 are the scalar components of vector A→ . Combining Equation
2.10 with Equation 2.11, we obtain the component form of a vector:


(2.12)
A


= Ax i
^


+ Ay j
^
.


If we know the coordinates b(xb, yb) of the origin point of a vector (where b stands for “beginning”) and the coordinates
e(xe, ye) of the end point of a vector (where e stands for “end”), we can obtain the scalar components of a vector simply
by subtracting the origin point coordinates from the end point coordinates:


(2.13)⎧


Ax = xe − xb
Ay = ye − yb.


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Example 2.3
Displacement of a Mouse Pointer
A mouse pointer on the display monitor of a computer at its initial position is at point (6.0 cm, 1.6 cm) withrespect to the lower left-side corner. If you move the pointer to an icon located at point (2.0 cm, 4.5 cm), what isthe displacement vector of the pointer?
Strategy
The origin of the xy-coordinate system is the lower left-side corner of the computer monitor. Therefore, the unit
vector i^ on the x-axis points horizontally to the right and the unit vector j^ on the y-axis points vertically
upward. The origin of the displacement vector is located at point b(6.0, 1.6) and the end of the displacement vectoris located at point e(2.0, 4.5). Substitute the coordinates of these points into Equation 2.13 to find the scalar
components Dx and Dy of the displacement vector D→ . Finally, substitute the coordinates into Equation
2.12 to write the displacement vector in the vector component form.
Solution
We identify xb = 6.0 , xe = 2.0 , yb = 1.6 , and ye = 4.5 , where the physical unit is 1 cm. The scalar x- and
y-components of the displacement vector are


Dx = xe − xb = (2.0 − 6.0)cm = −4.0 cm,


Dy = ye − yb = (4.5 − 1.6)cm = + 2.9 cm.


The vector component form of the displacement vector is
(2.14)


D


= Dx i
^


+ Dy j
^


= (−4.0 cm) i
^


+ (2.9 cm) j
^


= (−4.0 i
^


+ 2.9 j
^
)cm.


This solution is shown in Figure 2.17.


Figure 2.17 The graph of the displacement vector. The vector points fromthe origin point at b to the end point at e.


Significance
Notice that the physical unit—here, 1 cm—can be placed either with each component immediately before the unitvector or globally for both components, as in Equation 2.14. Often, the latter way is more convenient becauseit is simpler.


Chapter 2 | Vectors 59




2.4


The vector x-component D→ x = −4.0 i^ = 4.0(− i^ ) of the displacement vector has the magnitude
| D→ x| = | − 4.0|| i^ | = 4.0 because the magnitude of the unit vector is | i^ | = 1 . Notice, too, that the direction
of the x-component is − i^ , which is antiparallel to the direction of the +x-axis; hence, the x-component vector
D


x points to the left, as shown in Figure 2.17. The scalar x-component of vector D→ is Dx = −4.0 .
Similarly, the vector y-component D→ y = + 2.9 j^ of the displacement vector has magnitude
| D→ y| = |2.9|| j^ | = 2.9 because the magnitude of the unit vector is | j^ | = 1 . The direction of the y-component
is + j^ , which is parallel to the direction of the +y-axis. Therefore, the y-component vector D→ y points up, as
seen in Figure 2.17. The scalar y-component of vector D→ is Dy = + 2.9 . The displacement vector D→ is
the resultant of its two vector components.
The vector component form of the displacement vector Equation 2.14 tells us that the mouse pointer has beenmoved on the monitor 4.0 cm to the left and 2.9 cm upward from its initial position.


Check Your Understanding A blue fly lands on a sheet of graph paper at a point located 10.0 cm to theright of its left edge and 8.0 cm above its bottom edge and walks slowly to a point located 5.0 cm from the leftedge and 5.0 cm from the bottom edge. Choose the rectangular coordinate system with the origin at the lowerleft-side corner of the paper and find the displacement vector of the fly. Illustrate your solution by graphing.


When we know the scalar components Ax and Ay of a vector A→ , we can find its magnitude A and its direction angle
θA . The direction angle—or direction, for short—is the angle the vector forms with the positive direction on the x-axis.
The angle θA is measured in the counterclockwise direction from the +x-axis to the vector (Figure 2.18). Because the
lengths A, Ax , and Ay form a right triangle, they are related by the Pythagorean theorem:


(2.15)A2 = Ax2 + Ay2 ⇔ A = Ax2 + Ay2.


This equation works even if the scalar components of a vector are negative. The direction angle θA of a vector is defined
via the tangent function of angle θA in the triangle shown in Figure 2.18:


(2.16)
tan θA =


Ay
Ax


⇒ θA = tan
−1⎛

Ay
Ax

⎠.


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Figure 2.18 For vector A→ , its magnitude A and its direction
angle θA are related to the magnitudes of its scalar components
because A, Ax , and Ay form a right triangle.


When the vector lies either in the first quadrant or in the fourth quadrant, where component Ax is positive (Figure 2.19),
the angle θ in Equation 2.16) is identical to the direction angle θA . For vectors in the fourth quadrant, angle θ is
negative, which means that for these vectors, direction angle θA is measured clockwise from the positive x-axis. Similarly,
for vectors in the second quadrant, angle θ is negative. When the vector lies in either the second or third quadrant, where
component Ax is negative, the direction angle is θA = θ + 180° (Figure 2.19).


Figure 2.19 Scalar components of a vector may be positive or negative.Vectors in the first quadrant (I)have both scalar components positive andvectors in the third quadrant have both scalar components negative. Forvectors in quadrants II and III, the direction angle of a vector is
θA = θ + 180° .


Chapter 2 | Vectors 61




2.5


Example 2.4
Magnitude and Direction of the Displacement Vector
You move a mouse pointer on the display monitor from its initial position at point (6.0 cm, 1.6 cm) to an iconlocated at point (2.0 cm, 4.5 cm). What is the magnitude and direction of the displacement vector of the pointer?
Strategy
In Example 2.3, we found the displacement vector D→ of the mouse pointer (see Equation 2.14). We identify
its scalar components Dx = −4.0 cm and Dy = + 2.9 cm and substitute into Equation 2.15 and Equation
2.16 to find the magnitude D and direction θD , respectively.
Solution
The magnitude of vector D→ is


D = Dx
2 + Dy


2 = (−4.0 cm)2 + (2.9 cm)2 = (4.0)2 + (2.9)2 cm = 4.9 cm.


The direction angle is
tan θ =


Dy
Dx


= +2.9 cm
−4.0 cm


= −0.725 ⇒ θ = tan−1 (−0.725) = −35.9°.


Vector D→ lies in the second quadrant, so its direction angle is
θD = θ + 180° = −35.9° + 180° = 144.1°.


Check Your Understanding If the displacement vector of a blue fly walking on a sheet of graph paper is
D


= (−5.00 i
^


− 3.00 j
^
)cm , find its magnitude and direction.


In many applications, the magnitudes and directions of vector quantities are known and we need to find the resultant ofmany vectors. For example, imagine 400 cars moving on the Golden Gate Bridge in San Francisco in a strong wind. Eachcar gives the bridge a different push in various directions and we would like to know how big the resultant push can possiblybe. We have already gained some experience with the geometric construction of vector sums, so we know the task of findingthe resultant by drawing the vectors and measuring their lengths and angles may become intractable pretty quickly, leadingto huge errors. Worries like this do not appear when we use analytical methods. The very first step in an analytical approachis to find vector components when the direction and magnitude of a vector are known.
Let us return to the right triangle in Figure 2.18. The quotient of the adjacent side Ax to the hypotenuse A is the cosine
function of direction angle θA , Ax/A = cos θA , and the quotient of the opposite side Ay to the hypotenuse A is the sine
function of θA , Ay/A = sin θA . When magnitude A and direction θA are known, we can solve these relations for the
scalar components:


(2.17)⎧


Ax = A cos θA
Ay = A sin θA


.


When calculating vector components with Equation 2.17, care must be taken with the angle. The direction angle θA
of a vector is the angle measured counterclockwise from the positive direction on the x-axis to the vector. The clockwisemeasurement gives a negative angle.


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Example 2.5
Components of Displacement Vectors
A rescue party for a missing child follows a search dog named Trooper. Trooper wanders a lot and makes manytrial sniffs along many different paths. Trooper eventually finds the child and the story has a happy ending, but hisdisplacements on various legs seem to be truly convoluted. On one of the legs he walks 200.0 m southeast, thenhe runs north some 300.0 m. On the third leg, he examines the scents carefully for 50.0 m in the direction 30°
west of north. On the fourth leg, Trooper goes directly south for 80.0 m, picks up a fresh scent and turns 23°
west of south for 150.0 m. Find the scalar components of Trooper’s displacement vectors and his displacementvectors in vector component form for each leg.
Strategy
Let’s adopt a rectangular coordinate system with the positive x-axis in the direction of geographic east, with the
positive y-direction pointed to geographic north. Explicitly, the unit vector i^ of the x-axis points east and the
unit vector j^ of the y-axis points north. Trooper makes five legs, so there are five displacement vectors. We start
by identifying their magnitudes and direction angles, then we use Equation 2.17 to find the scalar componentsof the displacements and Equation 2.12 for the displacement vectors.
Solution
On the first leg, the displacement magnitude is L1 = 200.0 m and the direction is southeast. For direction
angle θ1 we can take either 45° measured clockwise from the east direction or 45° + 270° measured
counterclockwise from the east direction. With the first choice, θ1 = −45° . With the second choice,
θ1 = + 315° . We can use either one of these two angles. The components are


L1x = L1 cos θ1 = (200.0 m) cos 315° = 141.4 m,


L1y = L1 sin θ1 = (200.0 m) sin 315° = −141.4 m.


The displacement vector of the first leg is
L


1 = L1x i
^


+ L1y j
^


= (141.4 i
^


− 141.4 j
^
) m.


On the second leg of Trooper’s wanderings, the magnitude of the displacement is L2 = 300.0 m and the
direction is north. The direction angle is θ2 = + 90° . We obtain the following results:


L2x = L2 cos θ2 = (300.0 m) cos 90° = 0.0 ,


L2y = L2 sin θ2 = (300.0 m) sin 90° = 300.0 m,


L


2 = L2x i
^


+ L2y j
^


= (300.0 m) j
^
.


On the third leg, the displacement magnitude is L3 = 50.0 m and the direction is 30° west of north. The
direction angle measured counterclockwise from the eastern direction is θ3 = 30° + 90° = + 120° . This gives
the following answers:


L3x = L3 cos θ3 = (50.0 m) cos 120° = −25.0 m,


L3y = L3 sin θ3 = (50.0 m) sin 120° = + 43.3 m,


L


3 = L3x i
^


+ L3y j
^


= (−25.0 i
^


+ 43.3 j
^
)m.


On the fourth leg of the excursion, the displacement magnitude is L4 = 80.0 m and the direction is south. The
direction angle can be taken as either θ4 = −90° or θ4 = + 270° . We obtain


Chapter 2 | Vectors 63




2.6


L4x = L4 cos θ4 = (80.0 m) cos (−90°) = 0 ,


L4y = L4 sin θ4 = (80.0 m) sin (−90°) = −80.0 m,


L


4 = L4x i
^


+ L4y j
^


= (−80.0 m) j
^
.


On the last leg, the magnitude is L5 = 150.0 m and the angle is θ5 = −23° + 270° = + 247° (23° west of
south), which gives


L5x = L5 cos θ5 = (150.0 m) cos 247° = −58.6 m,


L5y = L5 sin θ5 = (150.0 m) sin 247° = −138.1 m,


L


5 = L5x i
^


+ L5y j
^


= (−58.6 i
^


− 138.1 j
^
)m.


Check Your Understanding If Trooper runs 20 m west before taking a rest, what is his displacementvector?


Polar Coordinates
To describe locations of points or vectors in a plane, we need two orthogonal directions. In the Cartesian coordinate system
these directions are given by unit vectors i^ and j^ along the x-axis and the y-axis, respectively. The Cartesian coordinate
system is very convenient to use in describing displacements and velocities of objects and the forces acting on them.However, it becomes cumbersome when we need to describe the rotation of objects. When describing rotation, we usuallywork in the polar coordinate system.
In the polar coordinate system, the location of point P in a plane is given by two polar coordinates (Figure 2.20). The firstpolar coordinate is the radial coordinate r, which is the distance of point P from the origin. The second polar coordinate isan angle φ that the radial vector makes with some chosen direction, usually the positive x-direction. In polar coordinates,
angles are measured in radians, or rads. The radial vector is attached at the origin and points away from the origin to point
P. This radial direction is described by a unit radial vector r̂ . The second unit vector t^ is a vector orthogonal to the
radial direction r̂ . The positive + t^ direction indicates how the angle φ changes in the counterclockwise direction. In
this way, a point P that has coordinates (x, y) in the rectangular system can be described equivalently in the polar coordinatesystem by the two polar coordinates (r, φ) . Equation 2.17 is valid for any vector, so we can use it to express the x-
and y-coordinates of vector r→ . In this way, we obtain the connection between the polar coordinates and rectangular
coordinates of point P:


(2.18)⎧


x = r cos φ
y = r sin φ


.


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Figure 2.20 Using polar coordinates, the unit vector r̂ defines
the positive direction along the radius r (radial direction) and,
orthogonal to it, the unit vector t^ defines the positive direction of
rotation by the angle φ .


Example 2.6
Polar Coordinates
A treasure hunter finds one silver coin at a location 20.0 m away from a dry well in the direction 20° north of
east and finds one gold coin at a location 10.0 m away from the well in the direction 20° north of west. What are
the polar and rectangular coordinates of these findings with respect to the well?
Strategy
The well marks the origin of the coordinate system and east is the +x-direction. We identify radial distances fromthe locations to the origin, which are rS = 20.0 m (for the silver coin) and rG = 10.0 m (for the gold coin). To
find the angular coordinates, we convert 20° to radians: 20° = π20/180 = π/9 . We use Equation 2.18 to find
the x- and y-coordinates of the coins.
Solution
The angular coordinate of the silver coin is φS = π/9 , whereas the angular coordinate of the gold coin is
φG = π − π/9 = 8π/9 . Hence, the polar coordinates of the silver coin are (rS, φS) = (20.0 m, π/9) and those
of the gold coin are (rG, φG) = (10.0 m, 8π/9) . We substitute these coordinates into Equation 2.18 to obtain
rectangular coordinates. For the gold coin, the coordinates are






xG = rG cos φG = (10.0 m) cos 8π/9 = −9.4 m


yG = rG sin φG = (10.0 m) sin 8π/9 = 3.4 m
⇒ (xG, yG) = (−9.4 m, 3.4 m).


For the silver coin, the coordinates are




xS = rS cos φS = (20.0 m) cos π/9 = 18.9 m


yS = rS sin φS = (20.0 m) sin π/9 = 6.8 m
⇒ (xS, yS) = (18.9 m, 6.8 m).


Vectors in Three Dimensions
To specify the location of a point in space, we need three coordinates (x, y, z), where coordinates x and y specify locations ina plane, and coordinate z gives a vertical positions above or below the plane. Three-dimensional space has three orthogonal


Chapter 2 | Vectors 65




directions, so we need not two but three unit vectors to define a three-dimensional coordinate system. In the Cartesian
coordinate system, the first two unit vectors are the unit vector of the x-axis i^ and the unit vector of the y-axis j^ . The
third unit vector k^ is the direction of the z-axis (Figure 2.21). The order in which the axes are labeled, which is the
order in which the three unit vectors appear, is important because it defines the orientation of the coordinate system. The
order x-y-z, which is equivalent to the order i^ - j^ - k^ , defines the standard right-handed coordinate system (positive
orientation).


Figure 2.21 Three unit vectors define a Cartesian system inthree-dimensional space. The order in which these unit vectorsappear defines the orientation of the coordinate system. Theorder shown here defines the right-handed orientation.


In three-dimensional space, vector A→ has three vector components: the x-component A→ x = Ax i^ , which is the part
of vector A→ along the x-axis; the y-component A→ y = Ay j^ , which is the part of A→ along the y-axis; and the
z-component A→ z = Az k^ , which is the part of the vector along the z-axis. A vector in three-dimensional space is the
vector sum of its three vector components (Figure 2.22):


(2.19)
A


= Ax i
^


+ Ay j
^


+ Az k
^
.


If we know the coordinates of its origin b(xb, yb, zb) and of its end e(xe, ye, ze) , its scalar components are obtained by
taking their differences: Ax and Ay are given by Equation 2.13 and the z-component is given by


(2.20)Az = ze − zb.


Magnitude A is obtained by generalizing Equation 2.15 to three dimensions:


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(2.21)A = Ax2 + Ay2 + Az2.


This expression for the vector magnitude comes from applying the Pythagorean theorem twice. As seen in Figure 2.22,
the diagonal in the xy-plane has length Ax2 + Ay2 and its square adds to the square Az2 to give A2 . Note that when the
z-component is zero, the vector lies entirely in the xy-plane and its description is reduced to two dimensions.


Figure 2.22 A vector in three-dimensional space is the vectorsum of its three vector components.


Example 2.7
Takeoff of a Drone
During a takeoff of IAI Heron (Figure 2.23), its position with respect to a control tower is 100 m above theground, 300 m to the east, and 200 m to the north. One minute later, its position is 250 m above the ground, 1200m to the east, and 2100 m to the north. What is the drone’s displacement vector with respect to the control tower?What is the magnitude of its displacement vector?


Figure 2.23 The drone IAI Heron in flight. (credit: SSgtReynaldo Ramon, USAF)


Strategy
We take the origin of the Cartesian coordinate system as the control tower. The direction of the +x-axis is given


Chapter 2 | Vectors 67




2.7


by unit vector i^ to the east, the direction of the +y-axis is given by unit vector j^ to the north, and the direction
of the +z-axis is given by unit vector k^ , which points up from the ground. The drone’s first position is the origin
(or, equivalently, the beginning) of the displacement vector and its second position is the end of the displacementvector.
Solution
We identify b(300.0 m, 200.0 m, 100.0 m) and e(480.0 m, 370.0 m, 250.0m), and use Equation 2.13 andEquation 2.20 to find the scalar components of the drone’s displacement vector:






Dx = xe − xb = 1200.0 m − 300.0 m = 900.0 m,


Dy = ye − yb = 2100.0 m − 200.0 m = 1900.0 m,


Dz = ze − zb = 250.0 m − 100.0 m = 150.0 m.


We substitute these components into Equation 2.19 to find the displacement vector:
D


= Dx i
^


+ Dy j
^


+ Dz k
^


= 900.0 m i
^


+ 1900.0 m j
^


+ 150.0 mk
^


= (0.90 i
^


+ 1.90 j
^


+ 0.15k
^
) km.


We substitute into Equation 2.21 to find the magnitude of the displacement”
D = Dx


2 + Dy
2 + Dz


2 = (0.90 km)2 + (1.90 km)2 + (0.15 km)2 = 4.44 km.


Check Your Understanding If the average velocity vector of the drone in the displacement in Example
2.7 is u→ = (15.0 i^ + 31.7 j^ + 2.5k^ )m/s , what is the magnitude of the drone’s velocity vector?


2.3 | Algebra of Vectors
Learning Objectives


By the end of this section, you will be able to:
• Apply analytical methods of vector algebra to find resultant vectors and to solve vectorequations for unknown vectors.
• Interpret physical situations in terms of vector expressions.


Vectors can be added together and multiplied by scalars. Vector addition is associative (Equation 2.8) and commutative(Equation 2.7), and vector multiplication by a sum of scalars is distributive (Equation 2.9). Also, scalar multiplicationby a sum of vectors is distributive:


(2.22)α( A→ + B→ ) = α A→ + α B→ .


In this equation, α is any number (a scalar). For example, a vector antiparallel to vector A→ = Ax i^ + Ay j^ + Az k^ can
be expressed simply by multiplying A→ by the scalar α = −1 :


(2.23)
− A


= −Ax i
^


− Ay j
^


− Az k
^
.


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Example 2.8
Direction of Motion
In a Cartesian coordinate system where i^ denotes geographic east, j^ denotes geographic north, and k^
denotes altitude above sea level, a military convoy advances its position through unknown territory with velocity
v→ = (4.0 i


^
+ 3.0 j


^
+ 0.1k


^
)km/h . If the convoy had to retreat, in what geographic direction would it be


moving?
Solution
The velocity vector has the third component v→ z = ( + 0.1km/h)k^ , which says the convoy is climbing at a rate
of 100 m/h through mountainous terrain. At the same time, its velocity is 4.0 km/h to the east and 3.0 km/h to
the north, so it moves on the ground in direction tan−1(3 /4) ≈ 37° north of east. If the convoy had to retreat,
its new velocity vector u→ would have to be antiparallel to v→ and be in the form u→ = −α v→ , where
α is a positive number. Thus, the velocity of the retreat would be u→ = α(−4.0 i^ − 3.0 j^ − 0.1k^ )km/h . The
negative sign of the third component indicates the convoy would be descending. The direction angle of the retreat
velocity is tan−1(−3α/ − 4α) ≈ 37° south of west. Therefore, the convoy would be moving on the ground in
direction 37° south of west while descending on its way back.


The generalization of the number zero to vector algebra is called the null vector, denoted by 0→ . All components of the
null vector are zero, 0→ = 0 i^ + 0 j^ + 0k^ , so the null vector has no length and no direction.
Two vectors A→ and B→ are equal vectors if and only if their difference is the null vector:


0


= A


− B


= (Ax i
^


+ Ay j
^


+ Az k
^
) − (Bx i


^
+ By j


^
+ Bz k


^
) = (Ax − Bx) i


^
+ (Ay − By) j


^
+ (Az − Bz)k


^
.


This vector equation means we must have simultaneously Ax − Bx = 0 , Ay − By = 0 , and Az − Bz = 0 . Hence, we can
write A→ = B→ if and only if the corresponding components of vectors A→ and B→ are equal:


(2.24)
A


= B







Ax = Bx
Ay = By
Az = Bz


.


Two vectors are equal when their corresponding scalar components are equal.
Resolving vectors into their scalar components (i.e., finding their scalar components) and expressing them analytically invector component form (given by Equation 2.19) allows us to use vector algebra to find sums or differences of many
vectors analytically (i.e., without using graphical methods). For example, to find the resultant of two vectors A→ and B→
, we simply add them component by component, as follows:


R


= A


+ B


= (Ax i
^


+ Ay j
^


+ Az k
^
) + (Bx i


^
+ By j


^
+ Bz k


^
) = (Ax + Bx) i


^
+ (Ay + By) j


^
+ (Az + Bz)k


^
.


In this way, using Equation 2.24, scalar components of the resultant vector R→ = Rx i^ + Ry j^ + Rz k^ are the sums of
corresponding scalar components of vectors A→ and B→ :


Chapter 2 | Vectors 69






Rx = Ax + Bx,
Ry = Ay + By,


Rz = Az + Bz.


Analytical methods can be used to find components of a resultant of many vectors. For example, if we are to sum up
N vectors F→ 1, F→ 2, F→ 3, … , F→ N , where each vector is F→ k = Fkx i^ + Fky j^ + Fkz k^ , the resultant vector
F


R is
F


R = F


1 + F


2 + F


3 +… + F


N = ∑
k = 1


N


F


k = ∑
k = 1


N ⎛
⎝Fkx i


^
+ Fky j


^
+ Fkz k


^⎞


=



⎜∑
k = 1


N


Fkx



⎟ i
^


+



⎜∑
k = 1


N


Fky



⎟ j
^


+



⎜∑
k = 1


N


Fkz



⎟k
^
.


Therefore, scalar components of the resultant vector are


(2.25)⎧













FRx = ∑
k = 1


N


Fkx = F1x + F2x +… + FNx


FRy = ∑
k = 1


N


Fky = F1y + F2y +… + FNy


FRz = ∑
k = 1


N


Fkz = F1z + F2z +… + FNz.


Having found the scalar components, we can write the resultant in vector component form:
F


R = FRx i
^


+ FRy j
^


+ FRz k
^
.


Analytical methods for finding the resultant and, in general, for solving vector equations are very important in physicsbecause many physical quantities are vectors. For example, we use this method in kinematics to find resultant displacementvectors and resultant velocity vectors, in mechanics to find resultant force vectors and the resultants of many derived vectorquantities, and in electricity and magnetism to find resultant electric or magnetic vector fields.
Example 2.9


Analytical Computation of a Resultant
Three displacement vectors A→ , B→ , and C→ in a plane (Figure 2.13) are specified by their magnitudes
A = 10.0, B = 7.0, and C = 8.0, respectively, and by their respective direction angles with the horizontaldirection α = 35°, β = −110° , and γ = 30° . The physical units of the magnitudes are centimeters. Resolve
the vectors to their scalar components and find the following vector sums: (a) R→ = A→ + B→ + C→ , (b)
D


= A


− B
→ , and (c) S→ = A→ − 3 B→ + C→ .


Strategy
First, we use Equation 2.17 to find the scalar components of each vector and then we express each vector in itsvector component form given by Equation 2.12. Then, we use analytical methods of vector algebra to find theresultants.
Solution
We resolve the given vectors to their scalar components:


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Ax = A cos α = (10.0 cm) cos 35° = 8.19 cm
Ay = A sin α = (10.0 cm) sin 35° = 5.73 cm






Bx = B cos β = (7.0 cm) cos (−110°) = −2.39 cm


By = B sin β = (7.0 cm) sin (−110°) = −6.58 cm






Cx = C cos γ = (8.0 cm) cos 30° = 6.93 cm
Cy = C sin γ = (8.0 cm) sin 30° = 4.00 cm


.


For (a) we may substitute directly into Equation 2.24 to find the scalar components of the resultant:




Rx = Ax + Bx + Cx = 8.19 cm − 2.39 cm + 6.93 cm = 12.73 cm
Ry = Ay + By + Cy = 5.73 cm − 6.58 cm + 4.00 cm = 3.15 cm


.


Therefore, the resultant vector is R→ = Rx i^ + Ry j^ = (12.7 i^ + 3.1 j^ )cm .
For (b), we may want to write the vector difference as


D


= A


− B


= (Ax i
^


+ Ay j
^
) − (Bx i


^
+ By j


^
) = (Ax − Bx) i


^
+ (Ay − By) j


^
.


Then, the scalar components of the vector difference are




Dx = Ax − Bx = 8.19 cm − (−2.39 cm) = 10.58 cm
Dy = Ay − By = 5.73 cm − (−6.58 cm) = 12.31 cm


.


Hence, the difference vector is D→ = Dx i^ + Dy j^ = (10.6 i^ + 12.3 j^ )cm .
For (c), we can write vector S→ in the following explicit form:


S


= A


− 3 B


+ C


= (Ax i
^


+ Ay j
^
) − 3(Bx i


^
+ By j


^
) + (Cx i


^
+ Cy j


^
)


= (Ax − 3Bx + Cx) i
^


+ (Ay − 3By + Cy) j
^
.


Then, the scalar components of S→ are




Sx = Ax − 3Bx + Cx = 8.19 cm − 3(−2.39 cm) + 6.93 cm = 22.29 cm
Sy = Ay − 3By + Cy = 5.73 cm − 3(−6.58 cm) + 4.00 cm = 29.47 cm


.


The vector is S→ = Sx i^ + Sy j^ = (22.3 i^ + 29.5 j^ )cm .
Significance
Having found the vector components, we can illustrate the vectors by graphing or we can compute magnitudesand direction angles, as shown in Figure 2.24. Results for the magnitudes in (b) and (c) can be compared withresults for the same problems obtained with the graphical method, shown in Figure 2.14 and Figure 2.15.Notice that the analytical method produces exact results and its accuracy is not limited by the resolution of a ruleror a protractor, as it was with the graphical method used in Example 2.2 for finding this same resultant.


Chapter 2 | Vectors 71




2.8
Figure 2.24 Graphical illustration of the solutions obtained analytically in Example 2.9.


Check Your Understanding Three displacement vectors A→ , B→ , and F→ (Figure 2.13) are
specified by their magnitudes A = 10.00, B = 7.00, and F = 20.00, respectively, and by their respective directionangles with the horizontal direction α = 35° , β = −110° , and φ = 110° . The physical units of the
magnitudes are centimeters. Use the analytical method to find vector G→ = A→ + 2 B→ − F→ . Verify that
G = 28.15 cm and that θG = −68.65° .


Example 2.10
The Tug-of-War Game
Four dogs named Ang, Bing, Chang, and Dong play a tug-of-war game with a toy (Figure 2.25). Ang pulls onthe toy in direction α = 55° south of east, Bing pulls in direction β = 60° east of north, and Chang pulls in
direction γ = 55° west of north. Ang pulls strongly with 160.0 units of force (N), which we abbreviate as A =
160.0 N. Bing pulls even stronger than Ang with a force of magnitude B = 200.0 N, and Chang pulls with a forceof magnitude C = 140.0 N. When Dong pulls on the toy in such a way that his force balances out the resultant ofthe other three forces, the toy does not move in any direction. With how big a force and in what direction mustDong pull on the toy for this to happen?


Figure 2.25 Four dogs play a tug-of-war game with a toy.


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2.9


Strategy
We assume that east is the direction of the positive x-axis and north is the direction of the positive y-axis. As
in Example 2.9, we have to resolve the three given forces— A→ (the pull from Ang), B→ (the pull from
Bing), and C→ (the pull from Chang)—into their scalar components and then find the scalar components of the
resultant vector R→ = A→ + B→ + C→ . When the pulling force D→ from Dong balances out this resultant,
the sum of D→ and R→ must give the null vector D→ + R→ = 0→ . This means that D→ = − R→ , so the
pull from Dong must be antiparallel to R→ .
Solution
The direction angles are θA = −α = −55° , θB = 90° − β = 30° , and θC = 90° + γ = 145° , and substituting
them into Equation 2.17 gives the scalar components of the three given forces:






Ax = A cos θA = (160.0 N) cos (−55°) = + 91.8 N


Ay = A sin θA = (160.0 N) sin (−55°) = −131.1 N






Bx = B cos θB = (200.0 N) cos 30° = + 173.2 N


By = B sin θB = (200.0 N) sin 30° = + 100.0 N






Cx = C cos θC = (140.0 N) cos 145° = −114.7 N


Cy = C sin θC = (140.0 N) sin 145° = + 80.3 N


.


Now we compute scalar components of the resultant vector R→ = A→ + B→ + C→ :




Rx = Ax + Bx + Cx = + 91.8 N + 173.2 N − 114.7 N = + 150.3 N
Ry = Ay + By + Cy = −131.1 N + 100.0 N + 80.3 N = + 49.2 N


.


The antiparallel vector to the resultant R→ is
D


= − R


= −Rx i
^


− Ry j
^


= (−150.3 i
^


− 49.2 j
^
) N.


The magnitude of Dong’s pulling force is
D = Dx


2 + Dy
2 = (−150.3)2 + (−49.2)2 N = 158.1 N.


The direction of Dong’s pulling force is
θ = tan−1




Dy
Dx

⎠ = tan


−1 ⎛

−49.2 N
−150.3 N



⎠ = tan


−1 ⎛

49.2
150.3



⎠ = 18.1°.


Dong pulls in the direction 18.1° south of west because both components are negative, which means the pull
vector lies in the third quadrant (Figure 2.19).


Check Your Understanding Suppose that Bing in Example 2.10 leaves the game to attend to moreimportant matters, but Ang, Chang, and Dong continue playing. Ang and Chang’s pull on the toy does notchange, but Dong runs around and bites on the toy in a different place. With how big a force and in whatdirection must Dong pull on the toy now to balance out the combined pulls from Chang and Ang? Illustrate thissituation by drawing a vector diagram indicating all forces involved.


Chapter 2 | Vectors 73




Example 2.11
Vector Algebra
Find the magnitude of the vector C→ that satisfies the equation 2 A→ − 6 B→ + 3 C→ = 2 j^ , where
A


= i
^


− 2k
^ and B→ = − j^ + k^ /2 .


Strategy
We first solve the given equation for the unknown vector C→ . Then we substitute A→ and B→ ; group the
terms along each of the three directions i^ , j^ , and k^ ; and identify the scalar components Cx , Cy , and Cz .
Finally, we substitute into Equation 2.21 to find magnitude C.
Solution


2 A


− 6 B


+ 3 C


= 2 j
^


3 C


= 2 j
^


− 2 A


+ 6 B


C


= 2
3
j
^


− 2
3
A


+ 2 B


= 2
3
j
^


− 2
3
( i
^


− 2k
^
) + 2



⎜− j


^
+ k


^


2





⎟ = 2


3
j
^


− 2
3
i
^


+ 4
3
k
^


− 2 j
^


+ k
^


= −2
3
i
^


+ ⎛⎝
2
3
− 2⎞⎠ j


^
+ ⎛⎝


4
3
+ 1⎞⎠k


^


= −2
3
i
^


− 4
3
j
^


+ 7
3
k
^
.


The components are Cx = −2 /3 , Cy = −4/3 , and Cz = 7 /3 , and substituting into Equation 2.21 gives
C = Cx


2 + Cy
2 + Cz


2 = (−2/3)2 + (−4 /3)2 + (7 /3)2 = 23/3.


Example 2.12
Displacement of a Skier
Starting at a ski lodge, a cross-country skier goes 5.0 km north, then 3.0 km west, and finally 4.0 km southwestbefore taking a rest. Find his total displacement vector relative to the lodge when he is at the rest point. How farand in what direction must he ski from the rest point to return directly to the lodge?
Strategy
We assume a rectangular coordinate system with the origin at the ski lodge and with the unit vector i^ pointing
east and the unit vector j^ pointing north. There are three displacements: D→ 1 , D→ 2 , and D→ 3 . We
identify their magnitudes as D1 = 5.0 km , D2 = 3.0 km , and D3 = 4.0 km . We identify their directions are
the angles θ1 = 90° , θ2 = 180° , and θ3 = 180° + 45° = 225° . We resolve each displacement vector to its
scalar components and substitute the components into Equation 2.24 to obtain the scalar components of the
resultant displacement D→ from the lodge to the rest point. On the way back from the rest point to the lodge, the
displacement is B→ = − D→ . Finally, we find the magnitude and direction of B→ .
Solution
Scalar components of the displacement vectors are


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D1x = D1 cos θ1 = (5.0 km) cos 90° = 0


D1y = D1 sin θ1 = (5.0 km) sin 90° = 5.0 km






D2x = D2 cos θ2 = (3.0 km) cos 180° = −3.0 km


D2y = D2 sin θ2 = (3.0 km) sin 180° = 0






D3x = D3 cos θ3 = (4.0 km) cos 225° = −2.8 km


D3y = D3 sin θ3 = (4.0 km) sin 225° = −2.8 km


.


Scalar components of the net displacement vector are




Dx = D1x + D2x + D3x = (0 − 3.0 − 2.8)km = −5.8 km


Dy = D1y + D2y + D3y = (5.0 + 0 − 2.8)km = + 2.2 km
.


Hence, the skier’s net displacement vector is D→ = Dx i^ + Dy j^ = (−5.8 i^ + 2.2 j^ )km . On the way back
to the lodge, his displacement is B→ = − D→ = −(−5.8 i^ + 2.2 j^ )km = (5.8 i^ − 2.2 j^ )km . Its magnitude is
B = Bx


2 + By
2 = (5.8)2 + (−2.2)2 km = 6.2 km and its direction angle is θ = tan−1(−2.2/5.8) = −20.8° .


Therefore, to return to the lodge, he must go 6.2 km in a direction about 21° south of east.
Significance
Notice that no figure is needed to solve this problem by the analytical method. Figures are required when using agraphical method; however, we can check if our solution makes sense by sketching it, which is a useful final stepin solving any vector problem.


Example 2.13
Displacement of a Jogger
A jogger runs up a flight of 200 identical steps to the top of a hill and then runs along the top of the hill 50.0 mbefore he stops at a drinking fountain (Figure 2.26). His displacement vector from point A at the bottom of the
steps to point B at the fountain is D→ AB = (−90.0 i^ + 30.0 j^ )m . What is the height and width of each step in
the flight? What is the actual distance the jogger covers? If he makes a loop and returns to point A, what is his netdisplacement vector?


Figure 2.26 A jogger runs up a flight of steps.


Chapter 2 | Vectors 75




Strategy
The displacement vector D→ AB is the vector sum of the jogger’s displacement vector D→ AT along the stairs
(from point A at the bottom of the stairs to point T at the top of the stairs) and his displacement vector D→ TB on
the top of the hill (from point T at the top of the stairs to the fountain at point B). We must find the horizontal and
the vertical components of D→ TB . If each step has width w and height h, the horizontal component of D→ TB
must have a length of 200w and the vertical component must have a length of 200h. The actual distance the joggercovers is the sum of the distance he runs up the stairs and the distance of 50.0 m that he runs along the top of thehill.
Solution
In the coordinate system indicated in Figure 2.26, the jogger’s displacement vector on the top of the hill is
D


TB = (−50.0 m) i
^ . His net displacement vector is


D


AB = D


AT + D


TB.


Therefore, his displacement vector D→ TB along the stairs is
D


AT = D


AB − D


TB = (−90.0 i
^


+ 30.0 j
^
)m − (−50.0 m) i


^
= [(−90.0 + 50.0) i


^
+ 30.0 j


^
)]m


= (−40.0 i
^


+ 30.0 j
^
)m.


Its scalar components are DATx = −40.0 m and DATy = 30.0 m . Therefore, we must have
200w = | − 40.0|m and 200h = 30.0 m.


Hence, the step width is w = 40.0 m/200 = 0.2 m = 20 cm, and the step height is w = 30.0 m/200 = 0.15 m = 15cm. The distance that the jogger covers along the stairs is
DAT = DATx


2 + DATy
2 = (−40.0)2 + (30.0)2 m = 50.0 m.


Thus, the actual distance he runs is DAT + DTB = 50.0 m + 50.0 m = 100.0 m . When he makes a loop and
comes back from the fountain to his initial position at point A, the total distance he covers is twice this distance,or 200.0 m. However, his net displacement vector is zero, because when his final position is the same as his initialposition, the scalar components of his net displacement vector are zero (Equation 2.13).


In many physical situations, we often need to know the direction of a vector. For example, we may want to know thedirection of a magnetic field vector at some point or the direction of motion of an object. We have already said direction isgiven by a unit vector, which is a dimensionless entity—that is, it has no physical units associated with it. When the vectorin question lies along one of the axes in a Cartesian system of coordinates, the answer is simple, because then its unit vectorof direction is either parallel or antiparallel to the direction of the unit vector of an axis. For example, the direction of vector
d


= −5m i
^ is unit vector d^ = − i^ . The general rule of finding the unit vector V^ of direction for any vector V→ is


to divide it by its magnitude V:


(2.26)
V
^


= V


V
.


We see from this expression that the unit vector of direction is indeed dimensionless because the numerator and thedenominator in Equation 2.26 have the same physical unit. In this way, Equation 2.26 allows us to express the unitvector of direction in terms of unit vectors of the axes. The following example illustrates this principle.


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2.10


Example 2.14
The Unit Vector of Direction
If the velocity vector of the military convoy in Example 2.8 is v→ = (4.000 i^ + 3.000 j^ + 0.100k^ )km/h ,
what is the unit vector of its direction of motion?
Strategy
The unit vector of the convoy’s direction of motion is the unit vector v̂ that is parallel to the velocity vector. The
unit vector is obtained by dividing a vector by its magnitude, in accordance with Equation 2.26.
Solution
The magnitude of the vector v→ is


v = vx
2 + vy


2 + vz
2 = 4.0002 + 3.0002 + 0.1002km/h = 5.001km/h.


To obtain the unit vector v̂ , divide v→ by its magnitude:
v̂ = v



v =


(4.000 i
^


+ 3.000 j
^


+ 0.100k
^
)km/h


5.001km/h


=
(4.000 i


^
+ 3.000 j


^
+ 0.100k


^
)


5.001


= 4.000
5.001


i
^


+ 3.000
5.001


j
^


+ 0.100
5.001


k
^


= (79.98 i
^


+ 59.99 j
^


+ 2.00k
^
) × 10−2.


Significance
Note that when using the analytical method with a calculator, it is advisable to carry out your calculations to atleast three decimal places and then round off the final answer to the required number of significant figures, whichis the way we performed calculations in this example. If you round off your partial answer too early, you risk yourfinal answer having a huge numerical error, and it may be far off from the exact answer or from a value measuredin an experiment.


Check Your Understanding Verify that vector v̂ obtained in Example 2.14 is indeed a unit vector
by computing its magnitude. If the convoy in Example 2.8 was moving across a desert flatland—that is, if thethird component of its velocity was zero—what is the unit vector of its direction of motion? Which geographicdirection does it represent?


2.4 | Products of Vectors
Learning Objectives


By the end of this section, you will be able to:
• Explain the difference between the scalar product and the vector product of two vectors.
• Determine the scalar product of two vectors.
• Determine the vector product of two vectors.
• Describe how the products of vectors are used in physics.


A vector can be multiplied by another vector but may not be divided by another vector. There are two kinds of productsof vectors used broadly in physics and engineering. One kind of multiplication is a scalar multiplication of two vectors.


Chapter 2 | Vectors 77




Taking a scalar product of two vectors results in a number (a scalar), as its name indicates. Scalar products are used todefine work and energy relations. For example, the work that a force (a vector) performs on an object while causing itsdisplacement (a vector) is defined as a scalar product of the force vector with the displacement vector. A quite different kindof multiplication is a vector multiplication of vectors. Taking a vector product of two vectors returns as a result a vector, asits name suggests. Vector products are used to define other derived vector quantities. For example, in describing rotations,a vector quantity called torque is defined as a vector product of an applied force (a vector) and its distance from pivot toforce (a vector). It is important to distinguish between these two kinds of vector multiplications because the scalar productis a scalar quantity and a vector product is a vector quantity.
The Scalar Product of Two Vectors (the Dot Product)
Scalar multiplication of two vectors yields a scalar product.


Scalar Product (Dot Product)
The scalar product A→ · B→ of two vectors A→ and B→ is a number defined by the equation


(2.27)A→ · B→ = AB cos φ,
where φ is the angle between the vectors (shown in Figure 2.27). The scalar product is also called the dot product
because of the dot notation that indicates it.


In the definition of the dot product, the direction of angle φ does not matter, and φ can be measured from either of
the two vectors to the other because cos φ = cos (−φ) = cos (2π − φ) . The dot product is a negative number when
90° < φ ≤ 180° and is a positive number when 0° ≤ φ < 90° . Moreover, the dot product of two parallel vectors is
A


· B


= AB cos 0° = AB , and the dot product of two antiparallel vectors is A→ · B→ = AB cos 180° = −AB . The
scalar product of two orthogonal vectors vanishes: A→ · B→ = AB cos 90° = 0 . The scalar product of a vector with itself
is the square of its magnitude:


(2.28)A→ 2 ≡ A→ · A→ = AA cos 0° = A2.


Figure 2.27 The scalar product of two vectors. (a) The angle between the two vectors. (b) The orthogonal projection A⊥ of
vector A→ onto the direction of vector B→ . (c) The orthogonal projection B⊥ of vector B→ onto the direction of vector
A
→ .


Example 2.15
The Scalar Product
For the vectors shown in Figure 2.13, find the scalar product A→ · F→ .


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2.11


Strategy
From Figure 2.13, the magnitudes of vectors A→ and F→ are A = 10.0 and F = 20.0. Angle θ , between
them, is the difference: θ = φ − α = 110° − 35° = 75° . Substituting these values into Equation 2.27 gives
the scalar product.
Solution
A straightforward calculation gives us


A


· F


= AF cos θ = (10.0)(20.0) cos 75° = 51.76.


Check Your Understanding For the vectors given in Figure 2.13, find the scalar products A→ · B→
and F→ · C→ .


In the Cartesian coordinate system, scalar products of the unit vector of an axis with other unit vectors of axes always vanishbecause these unit vectors are orthogonal:
(2.29)


i
^
· j
^


= | i^ || j^ | cos 90° = (1)(1)(0) = 0,
i
^
· k
^


= | i^ ||k^ | cos 90° = (1)(1)(0) = 0,
k
^
· j
^


= |k^ || j^ | cos 90° = (1)(1)(0) = 0.
In these equations, we use the fact that the magnitudes of all unit vectors are one: | i^ | = | j^ | = |k^ | = 1 . For unit vectors of
the axes, Equation 2.28 gives the following identities:


(2.30)
i
^
· i
^


= i2 = j
^
· j
^


= j2 = k
^
· k
^


= k2 = 1.


The scalar product A→ · B→ can also be interpreted as either the product of B with the orthogonal projection A⊥ of
vector A→ onto the direction of vector B→ (Figure 2.27(b)) or the product of A with the orthogonal projection B⊥ of
vector B→ onto the direction of vector A→ (Figure 2.27(c)):


A


· B


= AB cos φ
= B(A cos φ) = BA⊥
= A(B cos φ) = AB⊥ .


For example, in the rectangular coordinate system in a plane, the scalar x-component of a vector is its dot product with the
unit vector i^ , and the scalar y-component of a vector is its dot product with the unit vector j^ :











A


· i
^


= | A→ || i^ | cos θA = A cos θA = Ax
A


· j
^


= | A→ || j^ | cos (90° − θA) = A sin θA = Ay
.


Scalar multiplication of vectors is commutative,


(2.31)A→ · B→ = B→ · A→ ,


Chapter 2 | Vectors 79




2.12


and obeys the distributive law:


(2.32)A→ · ( B→ + C→ ) = A→ · B→ + A→ · C→ .


We can use the commutative and distributive laws to derive various relations for vectors, such as expressing the dot productof two vectors in terms of their scalar components.
Check Your Understanding For vector A→ = Ax i^ + Ay j^ + Az k^ in a rectangular coordinate


system, use Equation 2.29 through Equation 2.32 to show that A→ · i^ = Ax A→ · j^ = Ay and
A


· k
^


= Az .


When the vectors in Equation 2.27 are given in their vector component forms,
A


= Ax i
^


+ Ay j
^


+ Az k
^


and B


= Bx i
^


+ By j
^


+ Bz k
^
,


we can compute their scalar product as follows:
A


· B


= (Ax i
^


+ Ay j
^


+ Az k
^
) · (Bx i


^
+ By j


^
+ Bz k


^
)


= AxBx i
^
· i
^


+ AxBy i
^
· j
^


+ AxBz i
^
· k
^


+AyBx j
^
· i
^


+ AyBy j
^
· j
^


+ AyBz j
^
· k
^


+AzBx k
^
· i
^


+ AzBy k
^
· j
^


+ AzBz k
^
· k
^
.


Since scalar products of two different unit vectors of axes give zero, and scalar products of unit vectors with themselvesgive one (see Equation 2.29 and Equation 2.30), there are only three nonzero terms in this expression. Thus, the scalarproduct simplifies to


(2.33)A→ · B→ = AxBx + AyBy + AzBz.


We can use Equation 2.33 for the scalar product in terms of scalar components of vectors to find the angle between twovectors. When we divide Equation 2.27 by AB, we obtain the equation for cos φ , into which we substitute Equation
2.33:


(2.34)
cos φ = A



· B


AB
=


AxBx + AyBy + AzBz
AB


.


Angle φ between vectors A→ and B→ is obtained by taking the inverse cosine of the expression in Equation 2.34.
Example 2.16


Angle between Two Forces
Three dogs are pulling on a stick in different directions, as shown in Figure 2.28. The first dog pulls with


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force F→ 1 = (10.0 i^ − 20.4 j^ + 2.0k^ )N , the second dog pulls with force F→ 2 = (−15.0 i^ − 6.2k^ )N , and
the third dog pulls with force F→ 3 = (5.0 i^ + 12.5 j^ )N . What is the angle between forces F→ 1 and F→ 2 ?


Figure 2.28 Three dogs are playing with a stick.


Strategy
The components of force vector F→ 1 are F1x = 10.0 N , F1y = −20.4 N , and F1z = 2.0 N , whereas those
of force vector F→ 2 are F2x = −15.0 N , F2y = 0.0 N , and F2z = −6.2 N . Computing the scalar product of
these vectors and their magnitudes, and substituting into Equation 2.34 gives the angle of interest.
Solution
The magnitudes of forces F→ 1 and F→ 2 are


F1 = F1x
2 + F1y


2 + F1z
2 = 10.02 + 20.42 + 2.02 N = 22.8 N


and
F2 = F2x


2 + F2y
2 + F2z


2 = 15.02 + 6.22 N = 16.2 N.


Substituting the scalar components into Equation 2.33 yields the scalar product
F


1 · F


2 = F1xF2x + F1yF2y + F1zF2z
= (10.0 N)(−15.0 N) + (−20.4 N)(0.0 N) + (2.0 N)(−6.2 N)


= −162.4 N2.


Finally, substituting everything into Equation 2.34 gives the angle
cos φ =


F


1 · F


2
F1F2


= −162.4N
2


(22.8 N)(16.2 N)
= −0.439 ⇒ φ = cos−1(−0.439) = 116.0°.


Significance
Notice that when vectors are given in terms of the unit vectors of axes, we can find the angle between themwithout knowing the specifics about the geographic directions the unit vectors represent. Here, for example, the+x-direction might be to the east and the +y-direction might be to the north. But, the angle between the forces inthe problem is the same if the +x-direction is to the west and the +y-direction is to the south.


Chapter 2 | Vectors 81




2.13


2.14


Check Your Understanding Find the angle between forces F→ 1 and F→ 3 in Example 2.16.


Example 2.17
The Work of a Force
When force F→ pulls on an object and when it causes its displacement D→ , we say the force performs work.
The amount of work the force does is the scalar product F→ · D→ . If the stick in Example 2.16 moves
momentarily and gets displaced by vector D→ = (−7.9 j^ − 4.2k^ ) cm , how much work is done by the third dog
in Example 2.16?
Strategy
We compute the scalar product of displacement vector D→ with force vector F→ 3 = (5.0 i^ + 12.5 j^ )N , which
is the pull from the third dog. Let’s use W3 to denote the work done by force F→ 3 on displacement D→ .
Solution
Calculating the work is a straightforward application of the dot product:


W3 = F


3 · D


= F3xDx + F3yDy + F3zDz


= (5.0 N)(0.0 cm) + (12.5 N)(−7.9 cm) + (0.0 N)(−4.2 cm)
= −98.7 N · cm.


Significance
The SI unit of work is called the joule (J) , where 1 J = 1 N ·m . The unit cm ·N can be written as
10−2m ·N = 10−2 J , so the answer can be expressed as W3 = −0.9875 J ≈ −1.0 J .


Check Your Understanding How much work is done by the first dog and by the second dog inExample 2.16 on the displacement in Example 2.17?


The Vector Product of Two Vectors (the Cross Product)
Vector multiplication of two vectors yields a vector product.


Vector Product (Cross Product)
The vector product of two vectors A→ and B→ is denoted by A→ × B→ and is often referred to as a cross
product. The vector product is a vector that has its direction perpendicular to both vectors A→ and B→ . In other
words, vector A→ × B→ is perpendicular to the plane that contains vectors A→ and B→ , as shown in Figure 2.29.
The magnitude of the vector product is defined as


(2.35)| A→ × B→ | = AB sin φ,
where angle φ , between the two vectors, is measured from vector A→ (first vector in the product) to vector B→
(second vector in the product), as indicated in Figure 2.29, and is between 0° and 180° .


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According to Equation 2.35, the vector product vanishes for pairs of vectors that are either parallel ⎛⎝φ = 0°⎞⎠ or
antiparallel ⎛⎝φ = 180°⎞⎠ because sin 0° = sin 180° = 0 .


Figure 2.29 The vector product of two vectors is drawn in three-
dimensional space. (a) The vector product A→ × B→ is a vector
perpendicular to the plane that contains vectors A→ and B→ . Small
squares drawn in perspective mark right angles between A→ and C→ ,
and between B→ and C→ so that if A→ and B→ lie on the floor,
vector C→ points vertically upward to the ceiling. (b) The vector
product B→ × A→ is a vector antiparallel to vector A→ × B→ .


On the line perpendicular to the plane that contains vectors A→ and B→ there are two alternative directions—either up or
down, as shown in Figure 2.29—and the direction of the vector product may be either one of them. In the standard right-
handed orientation, where the angle between vectors is measured counterclockwise from the first vector, vector A→ × B→
points upward, as seen in Figure 2.29(a). If we reverse the order of multiplication, so that now B→ comes first in the
product, then vector B→ × A→ must point downward, as seen in Figure 2.29(b). This means that vectors A→ × B→
and B→ × A→ are antiparallel to each other and that vector multiplication is not commutative but anticommutative. The
anticommutative property means the vector product reverses the sign when the order of multiplication is reversed:


(2.36)A→ × B→ = − B→ × A→ .


The corkscrew right-hand rule is a common mnemonic used to determine the direction of the vector product. As shown
in Figure 2.30, a corkscrew is placed in a direction perpendicular to the plane that contains vectors A→ and B→ , and
its handle is turned in the direction from the first to the second vector in the product. The direction of the cross product isgiven by the progression of the corkscrew.


Chapter 2 | Vectors 83




Figure 2.30 The corkscrew right-hand rule can be used to determine
the direction of the cross product A→ × B→ . Place a corkscrew in the
direction perpendicular to the plane that contains vectors A→ and B→
, and turn it in the direction from the first to the second vector in theproduct. The direction of the cross product is given by the progression ofthe corkscrew. (a) Upward movement means the cross-product vectorpoints up. (b) Downward movement means the cross-product vectorpoints downward.


Example 2.18
The Torque of a Force
The mechanical advantage that a familiar tool called a wrench provides (Figure 2.31) depends on magnitude Fof the applied force, on its direction with respect to the wrench handle, and on how far from the nut this force is
applied. The distance R from the nut to the point where force vector F→ is attached and is represented by the
radial vector R→ . The physical vector quantity that makes the nut turn is called torque (denoted by τ→ ) , and it
is the vector product of the distance between the pivot to force with the force: τ→ = R→ × F→ .
To loosen a rusty nut, a 20.00-N force is applied to the wrench handle at angle φ = 40° and at a distance of 0.25
m from the nut, as shown in Figure 2.31(a). Find the magnitude and direction of the torque applied to the nut.What would the magnitude and direction of the torque be if the force were applied at angle φ = 45° , as shown
in Figure 2.31(b)? For what value of angle φ does the torque have the largest magnitude?


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Figure 2.31 A wrench provides grip and mechanical advantage in applying torque to turn a nut. (a) Turncounterclockwise to loosen the nut. (b) Turn clockwise to tighten the nut.


Strategy
We adopt the frame of reference shown in Figure 2.31, where vectors R→ and F→ lie in the xy-plane and
the origin is at the position of the nut. The radial direction along vector R→ (pointing away from the origin)
is the reference direction for measuring the angle φ because R→ is the first vector in the vector product
τ→ = R



× F


→ . Vector τ→ must lie along the z-axis because this is the axis that is perpendicular to the
xy-plane, where both R→ and F→ lie. To compute the magnitude τ , we use Equation 2.35. To find the
direction of τ→ , we use the corkscrew right-hand rule (Figure 2.30).
Solution
For the situation in (a), the corkscrew rule gives the direction of R→ × F→ in the positive direction of the z-axis.
Physically, it means the torque vector τ→ points out of the page, perpendicular to the wrench handle. We identify
F = 20.00 N and R = 0.25 m, and compute the magnitude using Equation 2.11:


τ = | R→ × F→ | = RF sin φ = (0.25 m)(20.00 N) sin 40° = 3.21 N ·m.
For the situation in (b), the corkscrew rule gives the direction of R→ × F→ in the negative direction of
the z-axis. Physically, it means the vector τ→ points into the page, perpendicular to the wrench handle. The
magnitude of this torque is


τ = | R→ × F→ | = RF sin φ = (0.25 m)(20.00 N) sin 45° = 3.53 N ·m.
The torque has the largest value when sin φ = 1 , which happens when φ = 90° . Physically, it means the
wrench is most effective—giving us the best mechanical advantage—when we apply the force perpendicularto the wrench handle. For the situation in this example, this best-torque value is
τbest = RF = (0.25 m)(20.00 N) = 5.00 N ·m .


Chapter 2 | Vectors 85




2.15


Significance
When solving mechanics problems, we often do not need to use the corkscrew rule at all, as we’ll see now in the
following equivalent solution. Notice that once we have identified that vector R→ × F→ lies along the z-axis,
we can write this vector in terms of the unit vector k^ of the z-axis:


R


× F


= RF sin φk
^
.


In this equation, the number that multiplies k^ is the scalar z-component of the vector R→ × F→ . In the
computation of this component, care must be taken that the angle φ is measured counterclockwise from
R
→ (first vector) to F→ (second vector). Following this principle for the angles, we obtain
RF sin ( + 40°) = + 3.2 N ·m for the situation in (a), and we obtain RF sin (−45°) = −3.5 N ·m for the
situation in (b). In the latter case, the angle is negative because the graph in Figure 2.31 indicates theangle is measured clockwise; but, the same result is obtained when this angle is measured counterclockwisebecause +(360° − 45°) = + 315° and sin ( + 315°) = sin (−45°) . In this way, we obtain the solution without
reference to the corkscrew rule. For the situation in (a), the solution is R→ × F→ = + 3.2 N ·mk^ ; for the
situation in (b), the solution is R→ × F→ = −3.5 N ·mk^ .


Check Your Understanding For the vectors given in Figure 2.13, find the vector products
A


× B
→ and C→ × F→ .


Similar to the dot product (Equation 2.31), the cross product has the following distributive property:


(2.37)A→ × ( B→ + C→ ) = A→ × B→ + A→ × C→ .


The distributive property is applied frequently when vectors are expressed in their component forms, in terms of unit vectorsof Cartesian axes.
When we apply the definition of the cross product, Equation 2.35, to unit vectors i^ , j^ , and k^ that define the positive
x-, y-, and z-directions in space, we find that


(2.38)
i
^


× i
^


= j
^


× j
^


= k
^


× k
^


= 0.


All other cross products of these three unit vectors must be vectors of unit magnitudes because i^ , j^ , and k^ are
orthogonal. For example, for the pair i^ and j^ , the magnitude is | i^ × j^ | = i j sin 90° = (1)(1)(1) = 1 . The direction
of the vector product i^ × j^ must be orthogonal to the xy-plane, which means it must be along the z-axis. The only unit
vectors along the z-axis are −k^ or + k^ . By the corkscrew rule, the direction of vector i^ × j^ must be parallel to the
positive z-axis. Therefore, the result of the multiplication i^ × j^ is identical to + k^ . We can repeat similar reasoning for
the remaining pairs of unit vectors. The results of these multiplications are


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(2.39)⎧









i
^


× j
^


= + k
^
,


j
^


× k
^


= + i
^
,


k
^


× i
^


= + j
^
.


Notice that in Equation 2.39, the three unit vectors i^ , j^ , and k^ appear in the cyclic order shown in a diagram in
Figure 2.32(a). The cyclic order means that in the product formula, i^ follows k^ and comes before j^ , or k^ follows
j
^ and comes before i^ , or j^ follows i^ and comes before k^ . The cross product of two different unit vectors is always
a third unit vector. When two unit vectors in the cross product appear in the cyclic order, the result of such a multiplicationis the remaining unit vector, as illustrated in Figure 2.32(b). When unit vectors in the cross product appear in a differentorder, the result is a unit vector that is antiparallel to the remaining unit vector (i.e., the result is with the minus sign, asshown by the examples in Figure 2.32(c) and Figure 2.32(d). In practice, when the task is to find cross products ofvectors that are given in vector component form, this rule for the cross-multiplication of unit vectors is very useful.


Figure 2.32 (a) The diagram of the cyclic order of the unit vectors of theaxes. (b) The only cross products where the unit vectors appear in the cyclicorder. These products have the positive sign. (c, d) Two examples of crossproducts where the unit vectors do not appear in the cyclic order. Theseproducts have the negative sign.


Suppose we want to find the cross product A→ × B→ for vectors A→ = Ax i^ + Ay j^ + Az k^ and
B


= Bx i
^


+ By j
^


+ Bz k
^ . We can use the distributive property (Equation 2.37), the anticommutative property


(Equation 2.36), and the results in Equation 2.38 and Equation 2.39 for unit vectors to perform the following algebra:


Chapter 2 | Vectors 87




A


× B


= (Ax i
^


+ Ay j
^


+ Az k
^
) × (Bx i


^
+ By j


^
+ Bz k


^
)


= Ax i
^


× (Bx i
^


+ By j
^


+ Bz k
^
) + Ay j


^
× (Bx i


^
+ By j


^
+ Bz k


^
) + Az k


^
× (Bx i


^
+ By j


^
+ Bz k


^
)


= AxBx i
^


× i
^


+ AxBy i
^


× j
^


+ AxBz i
^


× k
^


+AyBx j
^


× i
^


+ AyBy j
^


× j
^


+ AyBz j
^


× k
^


+AzBx k
^


× i
^


+ AzBy k
^


× j
^


+ AzBz k
^


× k
^


= AxBx(0) + AxBy( + k
^
) + AxBz(− j


^
)


+AyBx(−k
^
) + AyBy(0) + AyBz( + i


^
)


+AzBx( + j
^
) + AzBy(− i


^
) + AzBz(0).


When performing algebraic operations involving the cross product, be very careful about keeping the correct order ofmultiplication because the cross product is anticommutative. The last two steps that we still have to do to complete our taskare, first, grouping the terms that contain a common unit vector and, second, factoring. In this way we obtain the followingvery useful expression for the computation of the cross product:


(2.40)
C


= A


× B


= (AyBz − AzBy) i
^


+ (AzBx − AxBz) j
^


+ (AxBy − AyBx)k
^
.


In this expression, the scalar components of the cross-product vector are
(2.41)⎧









Cx = AyBz − AzBy,


Cy = AzBx − AxBz,


Cz = AxBy − AyBx.


When finding the cross product, in practice, we can use either Equation 2.35 or Equation 2.40, depending on which oneof them seems to be less complex computationally. They both lead to the same final result. One way to make sure if thefinal result is correct is to use them both.
Example 2.19


A Particle in a Magnetic Field
When moving in a magnetic field, some particles may experience a magnetic force. Without going into details—a
detailed study of magnetic phenomena comes in later chapters—let’s acknowledge that the magnetic field B→
is a vector, the magnetic force F→ is a vector, and the velocity u→ of the particle is a vector. The magnetic
force vector is proportional to the vector product of the velocity vector with the magnetic field vector, which we
express as F→ = ζ u→ × B→ . In this equation, a constant ζ takes care of the consistency in physical units, so
we can omit physical units on vectors u→ and B→ . In this example, let’s assume the constant ζ is positive.
A particle moving in space with velocity vector u→ = −5.0 i^ − 2.0 j^ + 3.5k^ enters a region with a magnetic
field and experiences a magnetic force. Find the magnetic force F→ on this particle at the entry point to the
region where the magnetic field vector is (a) B→ = 7.2 i^ − j^ − 2.4k^ and (b) B→ = 4.5k^ . In each case, find
magnitude F of the magnetic force and angle θ the force vector F→ makes with the given magnetic field vector
B
→ .


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Strategy
First, we want to find the vector product u→ × B→ , because then we can determine the magnetic force
using F→ = ζ u→ × B→ . Magnitude F can be found either by using components, F = Fx2 + Fy2 + Fz2 , or by
computing the magnitude | u→ × B→ | directly using Equation 2.35. In the latter approach, we would have to
find the angle between vectors u→ and B→ . When we have F→ , the general method for finding the direction
angle θ involves the computation of the scalar product F→ · B→ and substitution into Equation 2.34. To
compute the vector product we can either use Equation 2.40 or compute the product directly, whichever way issimpler.
Solution
The components of the velocity vector are ux = −5.0 , uy = −2.0 , and uz = 3.5 .
(a) The components of the magnetic field vector are Bx = 7.2 , By = −1.0 , and Bz = −2.4 . Substituting them
into Equation 2.41 gives the scalar components of vector F→ = ζ u→ × B→ :











Fx = ζ(uyBz − uzBy) = ζ[(−2.0)(−2.4) − (3.5)(−1.0)] = 8.3ζ


Fy = ζ(uzBx − uxBz) = ζ[(3.5)(7.2) − (−5.0)(−2.4)] = 13.2ζ


Fz = ζ(uxBy − uyBx) = ζ[(−5.0)(−1.0) − (−2.0)(7.2)] = 19.4ζ


.


Thus, the magnetic force is F→ = ζ(8.3 i^ + 13.2 j^ + 19.4k^ ) and its magnitude is
F = Fx


2 + Fy
2 + Fz


2 = ζ (8.3)2 + (13.2)2 + (19.4)2 = 24.9ζ.


To compute angle θ , we may need to find the magnitude of the magnetic field vector,
B = Bx


2 + By
2 + Bz


2 = (7.2)2 + (−1.0)2 + (−2.4)2 = 7.6,


and the scalar product F→ · B→ :
F


· B


= FxBx + FyBy + FzBz = (8.3ζ)(7.2) + (13.2ζ)(−1.0) + (19.4ζ)(−2.4) = 0.


Now, substituting into Equation 2.34 gives angle θ :
cos θ = F



· B


FB
= 0


(18.2ζ)(7.6)
= 0 ⇒ θ = 90°.


Hence, the magnetic force vector is perpendicular to the magnetic field vector. (We could have saved some timeif we had computed the scalar product earlier.)
(b) Because vector B→ = 4.5k^ has only one component, we can perform the algebra quickly and find the vector
product directly:


F


= ζ u→ × B


= ζ(−5.0 i
^


− 2.0 j
^


+ 3.5k
^
) × (4.5k


^
)


= ζ[(−5.0)(4.5) i
^


× k
^


+ (−2.0)(4.5) j
^


× k
^


+ (3.5)(4.5)k
^


× k
^
]


= ζ[−22.5(− j
^
) − 9.0( + i


^
) + 0] = ζ(−9.0 i


^
+ 22.5 j


^
).


The magnitude of the magnetic force is
F = Fx


2 + Fy
2 + Fz


2 = ζ (−9.0)2 + (22.5)2 + (0.0)2 = 24.2ζ.


Chapter 2 | Vectors 89




2.16


Because the scalar product is
F


· B


= FxBx + FyBy + FzBz = (−9.0ζ)(0) + (22.5ζ)(0) + (0)(4.5) = 0,


the magnetic force vector F→ is perpendicular to the magnetic field vector B→ .
Significance
Even without actually computing the scalar product, we can predict that the magnetic force vector must alwaysbe perpendicular to the magnetic field vector because of the way this vector is constructed. Namely, the magnetic
force vector is the vector product F→ = ζ u→ × B→ and, by the definition of the vector product (see Figure
2.29), vector F→ must be perpendicular to both vectors u→ and B→ .


Check Your Understanding Given two vectors A→ = − i^ + j^ and B→ = 3 i^ − j^ , find (a)
A


× B
→ , (b) | A→ × B→ | , (c) the angle between A→ and B→ , and (d) the angle between A→ × B→ and


vector C→ = i^ + k^ .


In conclusion to this section, we want to stress that “dot product” and “cross product” are entirely different mathematicalobjects that have different meanings. The dot product is a scalar; the cross product is a vector. Later chapters use theterms dot product and scalar product interchangeably. Similarly, the terms cross product and vector product are usedinterchangeably.


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anticommutative property
antiparallel vectors
associative
commutative
component form of a vector
corkscrew right-hand rule
cross product
difference of two vectors
direction angle
displacement
distributive
dot product
equal vectors
magnitude
null vector
orthogonal vectors
parallel vectors
parallelogram rule
polar coordinate system
polar coordinates
radial coordinate
resultant vector
scalar
scalar component
scalar equation
scalar product
scalar quantity
tail-to-head geometric construction
unit vector
unit vectors of the axes
vector
vector components


CHAPTER 2 REVIEW
KEY TERMS


change in the order of operation introduces the minus sign
two vectors with directions that differ by 180°


terms can be grouped in any fashion
operations can be performed in any order


a vector written as the vector sum of its components in terms of unit vectors
a rule used to determine the direction of the vector product


the result of the vector multiplication of vectors is a vector called a cross product; also called a vectorproduct
vector sum of the first vector with the vector antiparallel to the second


in a plane, an angle between the positive direction of the x-axis and the vector, measuredcounterclockwise from the axis to the vector
change in position


multiplication can be distributed over terms in summation
the result of the scalar multiplication of two vectors is a scalar called a dot product; also called a scalarproduct
two vectors are equal if and only if all their corresponding components are equal; alternately, two parallelvectors of equal magnitudes


length of a vector
a vector with all its components equal to zero


two vectors with directions that differ by exactly 90° , synonymous with perpendicular vectors
two vectors with exactly the same direction angles


geometric construction of the vector sum in a plane
an orthogonal coordinate system where location in a plane is given by polar coordinates


a radial coordinate and an angle
distance to the origin in a polar coordinate system
vector sum of two (or more) vectors


a number, synonymous with a scalar quantity in physics
a number that multiplies a unit vector in a vector component of a vector


equation in which the left-hand and right-hand sides are numbers
the result of the scalar multiplication of two vectors is a scalar called a scalar product; also called a dotproduct
quantity that can be specified completely by a single number with an appropriate physical unit


geometric construction for drawing the resultant vector of many vectors
vector of a unit magnitude that specifies direction; has no physical unit


unit vectors that define orthogonal directions in a plane or in space
mathematical object with magnitude and direction


orthogonal components of a vector; a vector is the vector sum of its vector components.


Chapter 2 | Vectors 91




vector equation
vector product
vector quantity
vector sum


equation in which the left-hand and right-hand sides are vectors
the result of the vector multiplication of vectors is a vector called a vector product; also called a crossproduct
physical quantity described by a mathematical vector—that is, by specifying both its magnitude and itsdirection; synonymous with a vector in physics


resultant of the combination of two (or more) vectors


KEY EQUATIONS
Multiplication by a scalar(vector equation) B→ = α A→
Multiplication by a scalar(scalar equation formagnitudes) B = |α|A


Resultant of two vectors D→ AD = D→ AC + D→ CD
Commutative law A→ + B→ = B→ + A→
Associative law ( A→ + B→ ) + C→ = A→ + ( B→ + C→ )
Distributive law α1 A→ + α2 A→ = (α1 + α2) A→
The component form of avector in two dimensions A→ = Ax i^ + Ay j^


Scalar components of a vectorin two dimensions




Ax = xe − xb
Ay = ye − yb


Magnitude of a vector in aplane A = Ax2 + Ay2
The direction angle of a vectorin a plane θA = tan−1




Ay
Ax



Scalar components of a vectorin a plane




Ax = A cos θA
Ay = A sin θA


Polar coordinates in a plane ⎧⎩⎨
x = r cos φ
y = r sin φ


The component form of avector in three dimensions A→ = Ax i^ + Ay j^ + Az k^
The scalar z-component of avector in three dimensions Az = ze − zb
Magnitude of a vector in threedimensions A = Ax2 + Ay2 + Az2
Distributive property α( A→ + B→ ) = α A→ + α B→


Antiparallel vector to A→ − A→ = −Ax i^ − Ay j^ − Az k^


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Equal vectors A→ = B→ ⇔




Ax = Bx
Ay = By
Az = Bz


Components of the resultant ofN vectors















FRx = ∑
k = 1


N


Fkx = F1x + F2x +… + FNx


FRy = ∑
k = 1


N


Fky = F1y + F2y +… + FNy


FRz = ∑
k = 1


N


Fkz = F1z + F2z +… + FNz


General unit vector V^ = V→
V


Definition of the scalar product A→ · B→ = AB cos φ
Commutative property of thescalar product A→ · B→ = B→ · A→
Distributive property of thescalar product A→ · ( B→ + C→ ) = A→ · B→ + A→ · C→
Scalar product in terms ofscalar components of vectors A→ · B→ = AxBx + AyBy + AzBz
Cosine of the angle betweentwo vectors cos φ = A



· B


AB


Dot products of unit vectors i^ · j^ = j^ · k^ = k^ · i^ = 0
Magnitude of the vector product(definition) | A



× B


→ | = AB sin φ
Anticommutative property of thevector product A→ × B→ = − B→ × A→
Distributive property of thevector product A→ × ( B→ + C→ ) = A→ × B→ + A→ × C→


Cross products of unit vectors









i
^


× j
^


= + k
^
,


j
^


× k
^


= + i
^
,


k
^


× i
^


= + j
^
.


The cross product in terms ofscalarcomponents of vectors A


× B


= (AyBz − AzBy) i
^


+ (AzBx − AxBz) j
^


+ (AxBy − AyBx)k
^


SUMMARY
2.1 Scalars and Vectors


• A vector quantity is any quantity that has magnitude and direction, such as displacement or velocity. Vectorquantities are represented by mathematical objects called vectors.
• Geometrically, vectors are represented by arrows, with the end marked by an arrowhead. The length of the vector isits magnitude, which is a positive scalar. On a plane, the direction of a vector is given by the angle the vector makes


Chapter 2 | Vectors 93




with a reference direction, often an angle with the horizontal. The direction angle of a vector is a scalar.
• Two vectors are equal if and only if they have the same magnitudes and directions. Parallel vectors have the samedirection angles but may have different magnitudes. Antiparallel vectors have direction angles that differ by 180° .
Orthogonal vectors have direction angles that differ by 90° .


• When a vector is multiplied by a scalar, the result is another vector of a different length than the length of the originalvector. Multiplication by a positive scalar does not change the original direction; only the magnitude is affected.Multiplication by a negative scalar reverses the original direction. The resulting vector is antiparallel to the originalvector. Multiplication by a scalar is distributive. Vectors can be divided by nonzero scalars but cannot be divided byvectors.
• Two or more vectors can be added to form another vector. The vector sum is called the resultant vector. We can addvectors to vectors or scalars to scalars, but we cannot add scalars to vectors. Vector addition is commutative andassociative.
• To construct a resultant vector of two vectors in a plane geometrically, we use the parallelogram rule. To constructa resultant vector of many vectors in a plane geometrically, we use the tail-to-head method.


2.2 Coordinate Systems and Components of a Vector
• Vectors are described in terms of their components in a coordinate system. In two dimensions (in a plane), vectorshave two components. In three dimensions (in space), vectors have three components.
• A vector component of a vector is its part in an axis direction. The vector component is the product of the unit vectorof an axis with its scalar component along this axis. A vector is the resultant of its vector components.
• Scalar components of a vector are differences of coordinates, where coordinates of the origin are subtracted fromend point coordinates of a vector. In a rectangular system, the magnitude of a vector is the square root of the sum ofthe squares of its components.
• In a plane, the direction of a vector is given by an angle the vector has with the positive x-axis. This direction angleis measured counterclockwise. The scalar x-component of a vector can be expressed as the product of its magnitudewith the cosine of its direction angle, and the scalar y-component can be expressed as the product of its magnitudewith the sine of its direction angle.
• In a plane, there are two equivalent coordinate systems. The Cartesian coordinate system is defined by unit vectors


i
^ and j^ along the x-axis and the y-axis, respectively. The polar coordinate system is defined by the radial unit
vector r̂ , which gives the direction from the origin, and a unit vector t^ , which is perpendicular (orthogonal) to
the radial direction.


2.3 Algebra of Vectors
• Analytical methods of vector algebra allow us to find resultants of sums or differences of vectors without having todraw them. Analytical methods of vector addition are exact, contrary to graphical methods, which are approximate.
• Analytical methods of vector algebra are used routinely in mechanics, electricity, and magnetism. They areimportant mathematical tools of physics.


2.4 Products of Vectors
• There are two kinds of multiplication for vectors. One kind of multiplication is the scalar product, also known asthe dot product. The other kind of multiplication is the vector product, also known as the cross product. The scalarproduct of vectors is a number (scalar). The vector product of vectors is a vector.
• Both kinds of multiplication have the distributive property, but only the scalar product has the commutativeproperty. The vector product has the anticommutative property, which means that when we change the order inwhich two vectors are multiplied, the result acquires a minus sign.
• The scalar product of two vectors is obtained by multiplying their magnitudes with the cosine of the angle betweenthem. The scalar product of orthogonal vectors vanishes; the scalar product of antiparallel vectors is negative.
• The vector product of two vectors is a vector perpendicular to both of them. Its magnitude is obtained by multiplying


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their magnitudes by the sine of the angle between them. The direction of the vector product can be determined by thecorkscrew right-hand rule. The vector product of two either parallel or antiparallel vectors vanishes. The magnitudeof the vector product is largest for orthogonal vectors.
• The scalar product of vectors is used to find angles between vectors and in the definitions of derived scalar physicalquantities such as work or energy.
• The cross product of vectors is used in definitions of derived vector physical quantities such as torque or magneticforce, and in describing rotations.


CONCEPTUAL QUESTIONS
2.1 Scalars and Vectors
1. A weather forecast states the temperature is predicted tobe −5 °C the following day. Is this temperature a vector or
a scalar quantity? Explain.
2. Which of the following is a vector: a person’s height,the altitude on Mt. Everest, the velocity of a fly, the age ofEarth, the boiling point of water, the cost of a book, Earth’spopulation, or the acceleration of gravity?
3. Give a specific example of a vector, stating itsmagnitude, units, and direction.
4. What do vectors and scalars have in common? How dothey differ?
5. Suppose you add two vectors A→ and B→ . What
relative direction between them produces the resultant withthe greatest magnitude? What is the maximum magnitude?What relative direction between them produces theresultant with the smallest magnitude? What is theminimum magnitude?
6. Is it possible to add a scalar quantity to a vectorquantity?
7. Is it possible for two vectors of different magnitudesto add to zero? Is it possible for three vectors of differentmagnitudes to add to zero? Explain.
8. Does the odometer in an automobile indicate a scalar ora vector quantity?
9. When a 10,000-m runner competing on a 400-m trackcrosses the finish line, what is the runner’s netdisplacement? Can this displacement be zero? Explain.
10. A vector has zero magnitude. Is it necessary to specifyits direction? Explain.
11. Can a magnitude of a vector be negative?


12. Can the magnitude of a particle’s displacement begreater that the distance traveled?
13. If two vectors are equal, what can you say about theircomponents? What can you say about their magnitudes?What can you say about their directions?
14. If three vectors sum up to zero, what geometriccondition do they satisfy?


2.2 Coordinate Systems and Components of a
Vector
15. Give an example of a nonzero vector that has acomponent of zero.
16. Explain why a vector cannot have a component greaterthan its own magnitude.
17. If two vectors are equal, what can you say about theircomponents?
18. If vectors A→ and B→ are orthogonal, what is the
component of B→ along the direction of A→ ? What is
the component of A→ along the direction of B→ ?
19. If one of the two components of a vector is not zero,can the magnitude of the other vector component of thisvector be zero?
20. If two vectors have the same magnitude, do theircomponents have to be the same?


2.4 Products of Vectors
21. What is wrong with the following expressions? How
can you correct them? (a) C = A→ B→ , (b)
C


= A


B
→ , (c) C = A→ × B→ , (d) C = A B→ , (e)


C + 2 A


= B , (f) C→ = A × B→ , (g)
A


· B


= A


× B
→ , (h) C→ = 2 A→ · B→ , (i)


Chapter 2 | Vectors 95




C = A


/ B
→ , and (j) C = A→ /B .


22. If the cross product of two vectors vanishes, what canyou say about their directions?


23. If the dot product of two vectors vanishes, what canyou say about their directions?
24. What is the dot product of a vector with the crossproduct that this vector has with another vector?


PROBLEMS
2.1 Scalars and Vectors
25. A scuba diver makes a slow descent into the depthsof the ocean. His vertical position with respect to a boaton the surface changes several times. He makes the firststop 9.0 m from the boat but has a problem with equalizingthe pressure, so he ascends 3.0 m and then continuesdescending for another 12.0 m to the second stop. Fromthere, he ascends 4 m and then descends for 18.0 m,ascends again for 7 m and descends again for 24.0 m,where he makes a stop, waiting for his buddy. Assuming thepositive direction up to the surface, express his net verticaldisplacement vector in terms of the unit vector. What is hisdistance to the boat?
26. In a tug-of-war game on one campus, 15 students pullon a rope at both ends in an effort to displace the centralknot to one side or the other. Two students pull with force196 N each to the right, four students pull with force 98N each to the left, five students pull with force 62 N eachto the left, three students pull with force 150 N each to theright, and one student pulls with force 250 N to the left.Assuming the positive direction to the right, express the netpull on the knot in terms of the unit vector. How big is thenet pull on the knot? In what direction?
27. Suppose you walk 18.0 m straight west and then 25.0m straight north. How far are you from your starting pointand what is the compass direction of a line connectingyour starting point to your final position? Use a graphicalmethod.
28. For the vectors given in the following figure, usea graphical method to find the following resultants: (a)
A


+ B
→ , (b) C→ + B→ , (c) D→ + F→ , (d)


A


− B
→ , (e) D→ − F→ , (f) A→ + 2 F→ , (g); and (h)


A


− 4 D


+ 2 F
→ .


29. A delivery man starts at the post office, drives 40 kmnorth, then 20 km west, then 60 km northeast, and finally50 km north to stop for lunch. Use a graphical method tofind his net displacement vector.
30. An adventurous dog strays from home, runs threeblocks east, two blocks north, one block east, one blocknorth, and two blocks west. Assuming that each block isabout 100 m, how far from home and in what direction isthe dog? Use a graphical method.
31. In an attempt to escape a desert island, a castawaybuilds a raft and sets out to sea. The wind shifts a greatdeal during the day and he is blown along the followingdirections: 2.50 km and 45.0° north of west, then 4.70 km
and 60.0° south of east, then 1.30 km and 25.0° south of
west, then 5.10 km straight east, then 1.70 km and 5.00°
east of north, then 7.20 km and 55.0° south of west, and
finally 2.80 km and 10.0° north of east. Use a graphical
method to find the castaway’s final position relative to theisland.
32. A small plane flies 40.0 km in a direction 60° north
of east and then flies 30.0 km in a direction 15° north of


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east. Use a graphical method to find the total distance theplane covers from the starting point and the direction of thepath to the final position.
33. A trapper walks a 5.0-km straight-line distance fromhis cabin to the lake, as shown in the following figure. Usea graphical method (the parallelogram rule) to determinethe trapper’s displacement directly to the east anddisplacement directly to the north that sum up to hisresultant displacement vector. If the trapper walked only indirections east and north, zigzagging his way to the lake,how many kilometers would he have to walk to get to thelake?


34. A surveyor measures the distance across a river thatflows straight north by the following method. Startingdirectly across from a tree on the opposite bank, thesurveyor walks 100 m along the river to establish abaseline. She then sights across to the tree and reads thatthe angle from the baseline to the tree is 35° . How wide is
the river?
35. A pedestrian walks 6.0 km east and then 13.0 kmnorth. Use a graphical method to find the pedestrian’sresultant displacement and geographic direction.
36. The magnitudes of two displacement vectors are A =20 m and B = 6 m. What are the largest and the smallestvalues of the magnitude of the resultant
R


= A


+ B


?


2.2 Coordinate Systems and Components of a
Vector
37. Assuming the +x-axis is horizontal and points to theright, resolve the vectors given in the following figure totheir scalar components and express them in vectorcomponent form.
38. Suppose you walk 18.0 m straight west and then 25.0m straight north. How far are you from your starting point?What is your displacement vector? What is the direction ofyour displacement? Assume the +x-axis is horizontal to the


right.
39. You drive 7.50 km in a straight line in a direction 15°
east of north. (a) Find the distances you would have to drivestraight east and then straight north to arrive at the samepoint. (b) Show that you still arrive at the same point ifthe east and north legs are reversed in order. Assume the+x-axis is to the east.
40. A sledge is being pulled by two horses on a flatterrain. The net force on the sledge can be expressed inthe Cartesian coordinate system as vector
F


= (−2980.0 i
^


+ 8200.0 j
^
)N , where i^ and j^


denote directions to the east and north, respectively. Findthe magnitude and direction of the pull.
41. A trapper walks a 5.0-km straight-line distance fromher cabin to the lake, as shown in the following figure.Determine the east and north components of herdisplacement vector. How many more kilometers wouldshe have to walk if she walked along the componentdisplacements? What is her displacement vector?


42. The polar coordinates of a point are 4π/3 and 5.50 m.
What are its Cartesian coordinates?
43. Two points in a plane have polar coordinates
P1(2.500 m, π/6) and P2(3.800 m, 2π/3) . Determine
their Cartesian coordinates and the distance between themin the Cartesian coordinate system. Round the distance to anearest centimeter.
44. A chameleon is resting quietly on a lanai screen,waiting for an insect to come by. Assume the origin of aCartesian coordinate system at the lower left-hand cornerof the screen and the horizontal direction to the right as the+x-direction. If its coordinates are (2.000 m, 1.000 m), (a)how far is it from the corner of the screen? (b) What is itslocation in polar coordinates?
45. Two points in the Cartesian plane are A(2.00 m, −4.00m) and B(−3.00 m, 3.00 m). Find the distance betweenthem and their polar coordinates.


Chapter 2 | Vectors 97




46. A fly enters through an open window and zoomsaround the room. In a Cartesian coordinate system withthree axes along three edges of the room, the fly changesits position from point b(4.0 m, 1.5 m, 2.5 m) to point e(1.0m, 4.5 m, 0.5 m). Find the scalar components of the fly’sdisplacement vector and express its displacement vector invector component form. What is its magnitude?


2.3 Algebra of Vectors
47. For vectors B→ = − i^ − 4 j^ and
A


= −3 i
^


− 2 j
^ , calculate (a) A→ + B→ and its


magnitude and direction angle, and (b) A→ − B→ and its
magnitude and direction angle.
48. A particle undergoes three consecutive displacements
given by vectors D→ 1 = (3.0 i^ − 4.0 j^ − 2.0k^ )mm ,
D


2 = (1.0 i
^


− 7.0 j
^


+ 4.0k
^
)mm , and


D


3 = (−7.0 i
^


+ 4.0 j
^


+ 1.0k
^
)mm . (a) Find the


resultant displacement vector of the particle. (b) What isthe magnitude of the resultant displacement? (c) If alldisplacements were along one line, how far would theparticle travel?
49. Given two displacement vectors
A


= (3.00 i
^


− 4.00 j
^


+ 4.00k
^
)m and


B


= (2.00 i
^


+ 3.00 j
^


− 7.00k
^
)m , find the


displacements and their magnitudes for (a)
C


= A


+ B
→ and (b) D→ = 2 A→ − B→ .


50. A small plane flies 40.0 km in a direction 60° north
of east and then flies 30.0 km in a direction 15° north
of east. Use the analytical method to find the total distancethe plane covers from the starting point, and the geographicdirection of its displacement vector. What is itsdisplacement vector?
51. In an attempt to escape a desert island, a castawaybuilds a raft and sets out to sea. The wind shifts a greatdeal during the day, and she is blown along the followingstraight lines: 2.50 km and 45.0° north of west, then 4.70
km and 60.0° south of east, then 1.30 km and 25.0° south
of west, then 5.10 km due east, then 1.70 km and 5.00°
east of north, then 7.20 km and 55.0° south of west, and
finally 2.80 km and 10.0° north of east. Use the analytical
method to find the resultant vector of all her displacementvectors. What is its magnitude and direction?


52. Assuming the +x-axis is horizontal to the right forthe vectors given in the following figure, use the analytical
method to find the following resultants: (a) A→ + B→ ,
(b) C→ + B→ , (c) D→ + F→ , (d) A→ − B→ , (e)
D


− F
→ , (f) A→ + 2 F→ , (g) C→ − 2 D→ + 3 F→ ,


and (h) A→ − 4 D→ + 2 F→ .


Figure 2.33
53. Given the vectors in the preceding figure, find vector
R
→ that solves equations (a) D→ + R→ = F→ and (b)
C


− 2 D


+ 5 R


= 3 F
→ . Assume the +x-axis is


horizontal to the right.
54. A delivery man starts at the post office, drives 40 kmnorth, then 20 km west, then 60 km northeast, and finally50 km north to stop for lunch. Use the analytical methodto determine the following: (a) Find his net displacementvector. (b) How far is the restaurant from the post office?(c) If he returns directly from the restaurant to the postoffice, what is his displacement vector on the return trip?(d) What is his compass heading on the return trip? Assumethe +x-axis is to the east.
55. An adventurous dog strays from home, runs threeblocks east, two blocks north, and one block east, one blocknorth, and two blocks west. Assuming that each block isabout a 100 yd, use the analytical method to find the dog’snet displacement vector, its magnitude, and its direction.Assume the +x-axis is to the east. How would your answerbe affected if each block was about 100 m?


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56. If D→ = (6.00 i^ − 8.00 j^ )m ,
B


= (−8.00 i
^


+ 3.00 j
^
)m , and


A


= (26.0 i
^


+ 19.0 j
^
)m , find the unknown constants a


and b such that a D→ + b B→ + A→ = 0→ .


57. Given the displacement vector D→ = (3 i^ − 4 j^ )m,
find the displacement vector R→ so that
D


+ R


= −4D j
^ .


58. Find the unit vector of direction for the following
vector quantities: (a) Force F→ = (3.0 i^ − 2.0 j^ )N , (b)
displacement D→ = (−3.0 i^ − 4.0 j^ )m , and (c) velocity
v→ = (−5.00 i


^
+ 4.00 j


^
)m/s .


59. At one point in space, the direction of the electric fieldvector is given in the Cartesian system by the unit vector
E
^


= 1 / 5 i
^


− 2 / 5 j
^ . If the magnitude of the electric


field vector is E = 400.0 V/m, what are the scalarcomponents Ex , Ey , and Ez of the electric field vector
E
→ at this point? What is the direction angle θE of the
electric field vector at this point?
60. A barge is pulled by the two tugboats shown in thefollowing figure. One tugboat pulls on the barge with aforce of magnitude 4000 units of force at 15° above the
line AB (see the figure and the other tugboat pulls on thebarge with a force of magnitude 5000 units of force at 12°
below the line AB. Resolve the pulling forces to their scalarcomponents and find the components of the resultant forcepulling on the barge. What is the magnitude of the resultantpull? What is its direction relative to the line AB?


Figure 2.34
61. In the control tower at a regional airport, an air trafficcontroller monitors two aircraft as their positions changewith respect to the control tower. One plane is a cargocarrier Boeing 747 and the other plane is a Douglas DC-3.The Boeing is at an altitude of 2500 m, climbing at 10°
above the horizontal, and moving 30° north of west. The
DC-3 is at an altitude of 3000 m, climbing at 5° above the
horizontal, and cruising directly west. (a) Find the positionvectors of the planes relative to the control tower. (b) Whatis the distance between the planes at the moment the airtraffic controller makes a note about their positions?


2.4 Products of Vectors
62. Assuming the +x-axis is horizontal to the right for thevectors in the following figure, find the following scalar
products: (a) A→ · C→ , (b) A→ · F→ , (c) D→ · C→ , (d)
A


· ( F


+ 2 C


) , (e) i^ · B→ , (f) j^ · B→ , (g)


Chapter 2 | Vectors 99




(3 i
^


− j
^
) · B


→ , and (h) B^ · B→ .


63. Assuming the +x-axis is horizontal to the right for thevectors in the preceding figure, find (a) the component of
vector A→ along vector C→ , (b) the component of vector
C
→ along vector A→ , (c) the component of vector i^
along vector F→ , and (d) the component of vector F→
along vector i^ .
64. Find the angle between vectors for (a)
D


= (−3.0 i
^


− 4.0 j
^
)m and


A


= (−3.0 i
^


+ 4.0 j
^
)m and (b)


D


= (2.0 i
^


− 4.0 j
^


+ k
^
)m and


B


= (−2.0 i
^


+ 3.0 j
^


+ 2.0k
^
)m .


65. Find the angles that vector


D


= (2.0 i
^


− 4.0 j
^


+ k
^
)m makes with the x-, y-, and z-


axes.
66. Show that the force vector
D


= (2.0 i
^


− 4.0 j
^


+ k
^
)N is orthogonal to the force


vector G→ = (3.0 i^ + 4.0 j^ + 10.0k^ )N .
67. Assuming the +x-axis is horizontal to the right forthe vectors in the previous figure, find the following vector
products: (a) A→ × C→ , (b) A→ × F→ , (c) D→ × C→ ,
(d) A→ × ( F→ + 2 C→ ) , (e) i^ × B→ , (f) j^ × B→ , (g)
(3 i
^


− j
^
) × B


→ , and (h) B^ × B→ .


68. Find the cross product A→ × C→ for (a)
A


= 2.0 i
^


− 4.0 j
^


+ k
^ and


C


= 3.0 i
^


+ 4.0 j
^


+ 10.0k
^ , (b)


A


= 3.0 i
^


+ 4.0 j
^


+ 10.0k
^ and


C


= 2.0 i
^


− 4.0 j
^


+ k
^ , (c) A→ = −3.0 i^ − 4.0 j^


and C→ = −3.0 i^ + 4.0 j^ , and (d)
C


= −2.0 i
^


+ 3.0 j
^


+ 2.0k
^ and A→ = −9.0 j^ .


69. For the vectors in the earlier figure, find (a)
( A


× F


) · D
→ , (b) ( A→ × F→ ) · ( D→ × B→ ) , and


(c) ( A→ · F→ )( D→ × B→ ) .


70. (a) If A→ × F→ = B→ × F→ , can we conclude
A


= B
→ ? (b) If A→ · F→ = B→ · F→ , can we


conclude A→ = B→ ? (c) If F A→ = B→ F , can we
conclude A→ = B→ ? Why or why not?


ADDITIONAL PROBLEMS
71. You fly 32.0 km in a straight line in still air in the
direction 35.0° south of west. (a) Find the distances you
would have to fly due south and then due west to arriveat the same point. (b) Find the distances you would haveto fly first in a direction 45.0° south of west and then
in a direction 45.0° west of north. Note these are the
components of the displacement along a different set of


axes—namely, the one rotated by 45° with respect to the
axes in (a).
72. Rectangular coordinates of a point are given by (2, y)and its polar coordinates are given by (r, π/6) . Find y and
r.


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73. If the polar coordinates of a point are (r, φ) and
its rectangular coordinates are (x, y) , determine the polar
coordinates of the following points: (a) (−x, y), (b) (−2x,−2y), and (c) (3x, −3y).
74. Vectors A→ and B→ have identical magnitudes
of 5.0 units. Find the angle between them if
A


+ B


= 5 2 j
^ .


75. Starting at the island of Moi in an unknownarchipelago, a fishing boat makes a round trip with twostops at the islands of Noi and Poi. It sails from Moi for4.76 nautical miles (nmi) in a direction 37° north of east
to Noi. From Noi, it sails 69° west of north to Poi. On
its return leg from Poi, it sails 28° east of south. What
distance does the boat sail between Noi and Poi? Whatdistance does it sail between Moi and Poi? Express youranswer both in nautical miles and in kilometers. Note: 1nmi = 1852 m.
76. An air traffic controller notices two signals from twoplanes on the radar monitor. One plane is at altitude 800m and in a 19.2-km horizontal distance to the tower in adirection 25° south of west. The second plane is at altitude
1100 m and its horizontal distance is 17.6 km and 20°
south of west. What is the distance between these planes?
77. Show that when A→ + B→ = C→ , then
C2 = A2 + B2 + 2AB cos φ , where φ is the angle
between vectors A→ and B→ .
78. Four force vectors each have the same magnitude f.What is the largest magnitude the resultant force vectormay have when these forces are added? What is thesmallest magnitude of the resultant? Make a graph of bothsituations.
79. A skater glides along a circular path of radius 5.00m in clockwise direction. When he coasts around one-half of the circle, starting from the west point, find (a)the magnitude of his displacement vector and (b) how farhe actually skated. (c) What is the magnitude of hisdisplacement vector when he skates all the way around thecircle and comes back to the west point?
80. A stubborn dog is being walked on a leash by itsowner. At one point, the dog encounters an interesting scentat some spot on the ground and wants to explore it in detail,but the owner gets impatient and pulls on the leash with
force F→ = (98.0 i^ + 132.0 j^ + 32.0k^ )N along the
leash. (a) What is the magnitude of the pulling force? (b)


What angle does the leash make with the vertical?
81. If the velocity vector of a polar bear is
u→ = (−18.0 i


^
− 13.0 j


^
)km/h , how fast and in what


geographic direction is it heading? Here, i^ and j^ are
directions to geographic east and north, respectively.
82. Find the scalar components of three-dimensional
vectors G→ and H→ in the following figure and write
the vectors in vector component form in terms of the unitvectors of the axes.


83. A diver explores a shallow reef off the coast of Belize.She initially swims 90.0 m north, makes a turn to the eastand continues for 200.0 m, then follows a big grouper for80.0 m in the direction 30° north of east. In the meantime,
a local current displaces her by 150.0 m south. Assuming


Chapter 2 | Vectors 101




the current is no longer present, in what direction and howfar should she now swim to come back to the point whereshe started?
84. A force vector A→ has x- and y-components,
respectively, of −8.80 units of force and 15.00 units of
force. The x- and y-components of force vector B→ are,
respectively, 13.20 units of force and −6.60 units of force.
Find the components of force vector C→ that satisfies the
vector equation A→ − B→ + 3 C→ = 0 .


85. Vectors A→ and B→ are two orthogonal vectors
in the xy-plane and they have identical magnitudes. If
A


= 3.0 i
^


+ 4.0 j
^ , find B→ .


86. For the three-dimensional vectors in the following
figure, find (a) G→ × H→ , (b) | G→ × H→ | , and (c)
G


· H
→ .


87. Show that ( B→ × C→ ) · A→ is the volume of the
parallelepiped, with edges formed by the three vectors inthe following figure.


CHALLENGE PROBLEMS
88. Vector B→ is 5.0 cm long and vector A→ is 4.0 cm long. Find the angle between these two vectors when


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| A→ + B→ | = 3.0 cm and | A→ − B→ | = 3.0 cm .
89. What is the component of the force vector
G


= (3.0 i
^


+ 4.0 j
^


+ 10.0k
^
)N along the force vector


H


= (1.0 i
^


+ 4.0 j
^
)N ?


90. The following figure shows a triangle formed by the
three vectors A→ , B→ , and C→ . If vector C→ ′ is
drawn between the midpoints of vectors A→ and B→ ,
show that C→ ′ = C→ /2 .


91. Distances between points in a plane do not changewhen a coordinate system is rotated. In other words, themagnitude of a vector is invariant under rotations of thecoordinate system. Suppose a coordinate system S isrotated about its origin by angle φ to become a new
coordinate system S′ , as shown in the following figure. A
point in a plane has coordinates (x, y) in S and coordinates

⎝x′, y′⎞⎠ in S′ .


(a) Show that, during the transformation of rotation, thecoordinates in S′ are expressed in terms of the coordinates
in S by the following relations:




x′ = x cos φ + y sin φ
y′ = −x sin φ + y cos φ


.


(b) Show that the distance of point P to the origin isinvariant under rotations of the coordinate system. Here,you have to show that
x2 + y2 = x′2 + y′2.


(c) Show that the distance between points P and Q isinvariant under rotations of the coordinate system. Here,you have to show that
(xP − xQ)


2 + (yP − yQ)
2 = (x′P − x′Q)


2 + (y′P − y′Q)
2.


Chapter 2 | Vectors 103




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3 | MOTION ALONG ASTRAIGHT LINE


Figure 3.1 A JR Central L0 series five-car maglev (magnetic levitation) train undergoing a test run on the Yamanashi TestTrack. The maglev train’s motion can be described using kinematics, the subject of this chapter. (credit: modification of work by“Maryland GovPics”/Flickr)


Chapter Outline
3.1 Position, Displacement, and Average Velocity
3.2 Instantaneous Velocity and Speed
3.3 Average and Instantaneous Acceleration
3.4 Motion with Constant Acceleration
3.5 Free Fall
3.6 Finding Velocity and Displacement from Acceleration


Introduction
Our universe is full of objects in motion. From the stars, planets, and galaxies; to the motion of people and animals; downto the microscopic scale of atoms and molecules—everything in our universe is in motion. We can describe motion usingthe two disciplines of kinematics and dynamics. We study dynamics, which is concerned with the causes of motion, inNewton’s Laws of Motion; but, there is much to be learned about motion without referring to what causes it, and thisis the study of kinematics. Kinematics involves describing motion through properties such as position, time, velocity, andacceleration.
A full treatment of kinematics considers motion in two and three dimensions. For now, we discuss motion in one dimension,which provides us with the tools necessary to study multidimensional motion. A good example of an object undergoing one-dimensional motion is the maglev (magnetic levitation) train depicted at the beginning of this chapter. As it travels, say, fromTokyo to Kyoto, it is at different positions along the track at various times in its journey, and therefore has displacements,or changes in position. It also has a variety of velocities along its path and it undergoes accelerations (changes in velocity).With the skills learned in this chapter we can calculate these quantities and average velocity. All these quantities can bedescribed using kinematics, without knowing the train’s mass or the forces involved.


Chapter 3 | Motion Along a Straight Line 105




3.1 | Position, Displacement, and Average Velocity
Learning Objectives


By the end of this section, you will be able to:
• Define position, displacement, and distance traveled.
• Calculate the total displacement given the position as a function of time.
• Determine the total distance traveled.
• Calculate the average velocity given the displacement and elapsed time.


When you’re in motion, the basic questions to ask are: Where are you? Where are you going? How fast are you gettingthere? The answers to these questions require that you specify your position, your displacement, and your averagevelocity—the terms we define in this section.
Position
To describe the motion of an object, you must first be able to describe its position (x): where it is at any particular time.More precisely, we need to specify its position relative to a convenient frame of reference. A frame of reference is anarbitrary set of axes from which the position and motion of an object are described. Earth is often used as a frame ofreference, and we often describe the position of an object as it relates to stationary objects on Earth. For example, a rocketlaunch could be described in terms of the position of the rocket with respect to Earth as a whole, whereas a cyclist’s positioncould be described in terms of where she is in relation to the buildings she passes Figure 3.2. In other cases, we usereference frames that are not stationary but are in motion relative to Earth. To describe the position of a person in an airplane,for example, we use the airplane, not Earth, as the reference frame. To describe the position of an object undergoing one-dimensional motion, we often use the variable x. Later in the chapter, during the discussion of free fall, we use the variabley.


Figure 3.2 These cyclists in Vietnam can be described bytheir position relative to buildings or a canal. Their motion canbe described by their change in position, or displacement, in aframe of reference. (credit: Suzan Black)
Displacement
If an object moves relative to a frame of reference—for example, if a professor moves to the right relative to a whiteboardFigure 3.3—then the object’s position changes. This change in position is called displacement. The word displacementimplies that an object has moved, or has been displaced. Although position is the numerical value of x along a straight linewhere an object might be located, displacement gives the change in position along this line. Since displacement indicatesdirection, it is a vector and can be either positive or negative, depending on the choice of positive direction. Also, an analysisof motion can have many displacements embedded in it. If right is positive and an object moves 2 m to the right, then 4 mto the left, the individual displacements are 2 m and −4 m, respectively.


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Figure 3.3 A professor paces left and right while lecturing. Her position relative to Earthis given by x. The +2.0-m displacement of the professor relative to Earth is represented byan arrow pointing to the right.


Displacement
Displacement Δx is the change in position of an object:


(3.1)Δx = xf − x0,
where Δx is displacement, xf is the final position, and x0 is the initial position.


We use the uppercase Greek letter delta (Δ) to mean “change in” whatever quantity follows it; thus, Δx means change in
position (final position less initial position). We always solve for displacement by subtracting initial position x0 from final
position xf . Note that the SI unit for displacement is the meter, but sometimes we use kilometers or other units of length.
Keep in mind that when units other than meters are used in a problem, you may need to convert them to meters to completethe calculation (see Appendix B).
Objects in motion can also have a series of displacements. In the previous example of the pacing professor, the individualdisplacements are 2 m and −4 m, giving a total displacement of −2 m. We define total displacement ΔxTotal , as the sum
of the individual displacements, and express this mathematically with the equation


(3.2)ΔxTotal = ∑ Δxi,


where Δxi are the individual displacements. In the earlier example,
Δx1 = x1 − x0 = 2 − 0 = 2m.


Similarly,
Δx2 = x2 − x1 = −2 − (2) = −4m.


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Thus,
ΔxTotal = Δx1 + Δx2 = 2 − 4 = −2m.


The total displacement is 2 − 4 = −2 m to the left, or in the negative direction. It is also useful to calculate the magnitudeof the displacement, or its size. The magnitude of the displacement is always positive. This is the absolute value ofthe displacement, because displacement is a vector and cannot have a negative value of magnitude. In our example, themagnitude of the total displacement is 2 m, whereas the magnitudes of the individual displacements are 2 m and 4 m.
The magnitude of the total displacement should not be confused with the distance traveled. Distance traveled xTotal , is the
total length of the path traveled between two positions. In the previous problem, the distance traveled is the sum of themagnitudes of the individual displacements:


xTotal = |Δx1| + |Δx2| = 2 + 4 = 6m.Average Velocity
To calculate the other physical quantities in kinematics we must introduce the time variable. The time variable allows us notonly to state where the object is (its position) during its motion, but also how fast it is moving. How fast an object is movingis given by the rate at which the position changes with time.
For each position xi , we assign a particular time ti . If the details of the motion at each instant are not important, the rate
is usually expressed as the average velocity v– . This vector quantity is simply the total displacement between two points
divided by the time taken to travel between them. The time taken to travel between two points is called the elapsed time
Δt .


Average Velocity
If x1 and x2 are the positions of an object at times t1 and t2 , respectively, then


(3.3)
Average velocity = v– =


Displacement between two points
Elapsed time between two points


v– = Δx
Δt


=
x2 − x1
t2 − t1


.


It is important to note that the average velocity is a vector and can be negative, depending on positions x1 and x2 .
Example 3.1


Delivering Flyers
Jill sets out from her home to deliver flyers for her yard sale, traveling due east along her street lined with houses.At 0.5 km and 9 minutes later she runs out of flyers and has to retrace her steps back to her house to get more.
This takes an additional 9 minutes. After picking up more flyers, she sets out again on the same path, continuingwhere she left off, and ends up 1.0 km from her house. This third leg of her trip takes 15 minutes. At this point
she turns back toward her house, heading west. After 1.75 km and 25 minutes she stops to rest.


a. What is Jill’s total displacement to the point where she stops to rest?
b. What is the magnitude of the final displacement?
c. What is the average velocity during her entire trip?
d. What is the total distance traveled?
e. Make a graph of position versus time.


A sketch of Jill’s movements is shown in Figure 3.4.


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Figure 3.4 Timeline of Jill’s movements.


Strategy
The problem contains data on the various legs of Jill’s trip, so it would be useful to make a table of the physicalquantities. We are given position and time in the wording of the problem so we can calculate the displacementsand the elapsed time. We take east to be the positive direction. From this information we can find the totaldisplacement and average velocity. Jill’s home is the starting point x0 . The following table gives Jill’s time and
position in the first two columns, and the displacements are calculated in the third column.


Time ti (min) Position xi (km) Displacement Δxi (km)
t0 = 0 x0 = 0 Δx0 = 0


t1 = 9 x1 = 0.5 Δx1 = x1 − x0 = 0.5


t2 = 18 x2 = 0 Δx2 = x2 − x1 = −0.5


t3 = 33 x3 = 1.0 Δx3 = x3 − x2 = 1.0


t4 = 58 x4 = −0.75 Δx4 = x4 − x3 = −1.75


Solutiona. From the above table, the total displacement is
∑ Δxi = 0.5 − 0.5 + 1.0 − 1.75 km = −0.75 km.


b. The magnitude of the total displacement is |−0.75| km = 0.75 km .
c. Average velocity = Total displacement


Elapsed time
= v– = −0.75 km


58 min
= −0.013 km/min


d. The total distance traveled (sum of magnitudes of individual displacements) is
xTotal = ∑ |Δxi| = 0.5 + 0.5 + 1.0 + 1.75 km = 3.75 km .


e. We can graph Jill’s position versus time as a useful aid to see the motion; the graph is shown in Figure3.5.


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3.1


Figure 3.5 This graph depicts Jill’s position versus time. Theaverage velocity is the slope of a line connecting the initial andfinal points.
Significance
Jill’s total displacement is −0.75 km, which means at the end of her trip she ends up 0.75 km due west of her
home. The average velocity means if someone was to walk due west at 0.013 km/min starting at the same time
Jill left her home, they both would arrive at the final stopping point at the same time. Note that if Jill were toend her trip at her house, her total displacement would be zero, as well as her average velocity. The total distancetraveled during the 58 minutes of elapsed time for her trip is 3.75 km.


Check Your Understanding A cyclist rides 3 km west and then turns around and rides 2 km east. (a)What is his displacement? (b) What is the distance traveled? (c) What is the magnitude of his displacement?


3.2 | Instantaneous Velocity and Speed
Learning Objectives


By the end of this section, you will be able to:
• Explain the difference between average velocity and instantaneous velocity.
• Describe the difference between velocity and speed.
• Calculate the instantaneous velocity given the mathematical equation for the velocity.
• Calculate the speed given the instantaneous velocity.


We have now seen how to calculate the average velocity between two positions. However, since objects in the real worldmove continuously through space and time, we would like to find the velocity of an object at any single point. We can findthe velocity of the object anywhere along its path by using some fundamental principles of calculus. This section gives us


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better insight into the physics of motion and will be useful in later chapters.
Instantaneous Velocity
The quantity that tells us how fast an object is moving anywhere along its path is the instantaneous velocity, usuallycalled simply velocity. It is the average velocity between two points on the path in the limit that the time (and therefore thedisplacement) between the two points approaches zero. To illustrate this idea mathematically, we need to express position xas a continuous function of t denoted by x(t). The expression for the average velocity between two points using this notation
is v– = x(t2) − x(t1)t2 − t1 . To find the instantaneous velocity at any position, we let t1 = t and t2 = t + Δt . After inserting
these expressions into the equation for the average velocity and taking the limit as Δt → 0 , we find the expression for the
instantaneous velocity:


v(t) = lim
Δt → 0


x(t + Δt) − x(t)
Δt


= dx(t)
dt


.


Instantaneous Velocity
The instantaneous velocity of an object is the limit of the average velocity as the elapsed time approaches zero, or thederivative of x with respect to t:


(3.4)v(t) = d
dt
x(t).


Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity ata specific time point t0 is the rate of change of the position function, which is the slope of the position function x(t) at
t0 . Figure 3.6 shows how the average velocity v– = ΔxΔt between two times approaches the instantaneous velocity at t0.
The instantaneous velocity is shown at time t0 , which happens to be at the maximum of the position function. The slope
of the position graph is zero at this point, and thus the instantaneous velocity is zero. At other times, t1, t2 , and so on,
the instantaneous velocity is not zero because the slope of the position graph would be positive or negative. If the positionfunction had a minimum, the slope of the position graph would also be zero, giving an instantaneous velocity of zero thereas well. Thus, the zeros of the velocity function give the minimum and maximum of the position function.


Figure 3.6 In a graph of position versus time, theinstantaneous velocity is the slope of the tangent line at a given
point. The average velocities v– = Δx


Δt
=


xf − xi
tf − ti


between
times Δt = t6 − t1, Δt = t5 − t2, and Δt = t4 − t3 are
shown. When Δt → 0 , the average velocity approaches the
instantaneous velocity at t = t0 .


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Example 3.2
Finding Velocity from a Position-Versus-Time Graph
Given the position-versus-time graph of Figure 3.7, find the velocity-versus-time graph.


Figure 3.7 The object starts out in the positive direction, stopsfor a short time, and then reverses direction, heading backtoward the origin. Notice that the object comes to restinstantaneously, which would require an infinite force. Thus, thegraph is an approximation of motion in the real world. (Theconcept of force is discussed in Newton’s Laws of Motion.)


Strategy
The graph contains three straight lines during three time intervals. We find the velocity during each time intervalby taking the slope of the line using the grid.
Solution
Time interval 0 s to 0.5 s: v– = Δx


Δt
= 0.5 m − 0.0 m


0.5 s − 0.0 s
= 1.0 m/s


Time interval 0.5 s to 1.0 s: v– = Δx
Δt


= 0.0 m − 0.0 m
1.0 s − 0.5 s


= 0.0 m/s


Time interval 1.0 s to 2.0 s: v– = Δx
Δt


= 0.0 m − 0.5 m
2.0 s − 1.0 s


= −0.5 m/s


The graph of these values of velocity versus time is shown in Figure 3.8.


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Figure 3.8 The velocity is positive for the first part of the trip,zero when the object is stopped, and negative when the objectreverses direction.


Significance
During the time interval between 0 s and 0.5 s, the object’s position is moving away from the origin and theposition-versus-time curve has a positive slope. At any point along the curve during this time interval, we canfind the instantaneous velocity by taking its slope, which is +1 m/s, as shown in Figure 3.8. In the subsequenttime interval, between 0.5 s and 1.0 s, the position doesn’t change and we see the slope is zero. From 1.0 s to 2.0s, the object is moving back toward the origin and the slope is −0.5 m/s. The object has reversed direction and hasa negative velocity.


Speed
In everyday language, most people use the terms speed and velocity interchangeably. In physics, however, they do not havethe same meaning and are distinct concepts. One major difference is that speed has no direction; that is, speed is a scalar.
We can calculate the average speed by finding the total distance traveled divided by the elapsed time:


(3.5)Average speed = s– = Total distance
Elapsed time


.


Average speed is not necessarily the same as the magnitude of the average velocity, which is found by dividing themagnitude of the total displacement by the elapsed time. For example, if a trip starts and ends at the same location, the totaldisplacement is zero, and therefore the average velocity is zero. The average speed, however, is not zero, because the totaldistance traveled is greater than zero. If we take a road trip of 300 km and need to be at our destination at a certain time,then we would be interested in our average speed.
However, we can calculate the instantaneous speed from the magnitude of the instantaneous velocity:


(3.6)Instantaneous speed = |v(t)|.


If a particle is moving along the x-axis at +7.0 m/s and another particle is moving along the same axis at −7.0 m/s, they havedifferent velocities, but both have the same speed of 7.0 m/s. Some typical speeds are shown in the following table.


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Speed m/s mi/h
Continental drift 10−7 2 × 10−7
Brisk walk 1.7 3.9
Cyclist 4.4 10
Sprint runner 12.2 27
Rural speed limit 24.6 56
Official land speed record 341.1 763
Speed of sound at sea level 343 768
Space shuttle on reentry 7800 17,500
Escape velocity of Earth* 11,200 25,000
Orbital speed of Earth around the Sun 29,783 66,623
Speed of light in a vacuum 299,792,458 670,616,629


Table 3.1 Speeds of Various Objects *Escape velocity is the velocity atwhich an object must be launched so that it overcomes Earth’s gravity and isnot pulled back toward Earth.
Calculating Instantaneous Velocity
When calculating instantaneous velocity, we need to specify the explicit form of the position function x(t). For the moment,let’s use polynomials x(t) = Atn , because they are easily differentiated using the power rule of calculus:


(3.7)dx(t)
dt


= nAtn − 1.


The following example illustrates the use of Equation 3.7.
Example 3.3


Instantaneous Velocity Versus Average Velocity
The position of a particle is given by x(t) = 3.0t + 0.5t3 m .


a. Using Equation 3.4 and Equation 3.7, find the instantaneous velocity at t = 2.0 s.
b. Calculate the average velocity between 1.0 s and 3.0 s.


Strategy
Equation 3.4 give the instantaneous velocity of the particle as the derivative of the position function. Lookingat the form of the position function given, we see that it is a polynomial in t. Therefore, we can use Equation3.7, the power rule from calculus, to find the solution. We use Equation 3.6 to calculate the average velocity ofthe particle.
Solution


a. v(t) = dx(t)
dt


= 3.0 + 1.5t2 m/s .
Substituting t = 2.0 s into this equation gives v(2.0 s) = [3.0 + 1.5(2.0)2] m/s = 9.0 m/s .


b. To determine the average velocity of the particle between 1.0 s and 3.0 s, we calculate the values of x(1.0s) and x(3.0 s):


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x(1.0 s) = ⎡⎣(3.0)(1.0) + 0.5(1.0)
3⎤
⎦m = 3.5 m


x(3.0 s) = ⎡⎣(3.0)(3.0) + 0.5(3.0)
3⎤
⎦m = 22.5 m.


Then the average velocity is
v– = x(3.0 s) − x(1.0 s)


t(3.0 s) − t(1.0 s)
= 22.5 − 3.5 m


3.0 − 1.0 s
= 9.5 m/s.


Significance
In the limit that the time interval used to calculate v− goes to zero, the value obtained for v− converges to the
value of v.


Example 3.4
Instantaneous Velocity Versus Speed
Consider the motion of a particle in which the position is x(t) = 3.0t − 3t2 m .


a. What is the instantaneous velocity at t = 0.25 s, t = 0.50 s, and t = 1.0 s?
b. What is the speed of the particle at these times?


Strategy
The instantaneous velocity is the derivative of the position function and the speed is the magnitude of theinstantaneous velocity. We use Equation 3.4 and Equation 3.7 to solve for instantaneous velocity.
Solution


a. v(t) = dx(t)
dt


= 3.0 − 6.0tm/s


b. v(0.25 s) = 1.50 m/s, v(0.5 s) = 0 m/s, v(1.0 s) = −3.0 m/s
c. Speed = |v(t)| = 1.50 m/s, 0.0 m/s, and 3.0 m/s


Significance
The velocity of the particle gives us direction information, indicating the particle is moving to the left (west)or right (east). The speed gives the magnitude of the velocity. By graphing the position, velocity, and speed asfunctions of time, we can understand these concepts visually Figure 3.9. In (a), the graph shows the particlemoving in the positive direction until t = 0.5 s, when it reverses direction. The reversal of direction can also beseen in (b) at 0.5 s where the velocity is zero and then turns negative. At 1.0 s it is back at the origin where itstarted. The particle’s velocity at 1.0 s in (b) is negative, because it is traveling in the negative direction. But in(c), however, its speed is positive and remains positive throughout the travel time. We can also interpret velocityas the slope of the position-versus-time graph. The slope of x(t) is decreasing toward zero, becoming zero at 0.5 sand increasingly negative thereafter. This analysis of comparing the graphs of position, velocity, and speed helpscatch errors in calculations. The graphs must be consistent with each other and help interpret the calculations.


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3.2


Figure 3.9 (a) Position: x(t) versus time. (b) Velocity: v(t) versus time. The slope of the position graph is thevelocity. A rough comparison of the slopes of the tangent lines in (a) at 0.25 s, 0.5 s, and 1.0 s with the values forvelocity at the corresponding times indicates they are the same values. (c) Speed: |v(t)| versus time. Speed is always a
positive number.


Check Your Understanding The position of an object as a function of time is x(t) = −3t2 m . (a) What
is the velocity of the object as a function of time? (b) Is the velocity ever positive? (c) What are the velocity andspeed at t = 1.0 s?


3.3 | Average and Instantaneous Acceleration
Learning Objectives


By the end of this section, you will be able to:
• Calculate the average acceleration between two points in time.
• Calculate the instantaneous acceleration given the functional form of velocity.
• Explain the vector nature of instantaneous acceleration and velocity.
• Explain the difference between average acceleration and instantaneous acceleration.
• Find instantaneous acceleration at a specified time on a graph of velocity versus time.


The importance of understanding acceleration spans our day-to-day experience, as well as the vast reaches of outer spaceand the tiny world of subatomic physics. In everyday conversation, to accelerate means to speed up; applying the brakepedal causes a vehicle to slow down. We are familiar with the acceleration of our car, for example. The greater theacceleration, the greater the change in velocity over a given time. Acceleration is widely seen in experimental physics. Inlinear particle accelerator experiments, for example, subatomic particles are accelerated to very high velocities in collisionexperiments, which tell us information about the structure of the subatomic world as well as the origin of the universe.In space, cosmic rays are subatomic particles that have been accelerated to very high energies in supernovas (explodingmassive stars) and active galactic nuclei. It is important to understand the processes that accelerate cosmic rays becausethese rays contain highly penetrating radiation that can damage electronics flown on spacecraft, for example.
Average Acceleration
The formal definition of acceleration is consistent with these notions just described, but is more inclusive.


Average Acceleration
Average acceleration is the rate at which velocity changes:


(3.8)a– = Δv
Δt


=
vf − v0
tf − t0


,


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where a− is average acceleration, v is velocity, and t is time. (The bar over the a means average acceleration.)


Because acceleration is velocity in meters divided by time in seconds, the SI units for acceleration are often abbreviated m/s2—that is, meters per second squared or meters per second per second. This literally means by how many meters per secondthe velocity changes every second. Recall that velocity is a vector—it has both magnitude and direction—which means thata change in velocity can be a change in magnitude (or speed), but it can also be a change in direction. For example, if arunner traveling at 10 km/h due east slows to a stop, reverses direction, continues her run at 10 km/h due west, her velocityhas changed as a result of the change in direction, although the magnitude of the velocity is the same in both directions.Thus, acceleration occurs when velocity changes in magnitude (an increase or decrease in speed) or in direction, or both.
Acceleration as a Vector
Acceleration is a vector in the same direction as the change in velocity, Δv . Since velocity is a vector, it can change
in magnitude or in direction, or both. Acceleration is, therefore, a change in speed or direction, or both.


Keep in mind that although acceleration is in the direction of the change in velocity, it is not always in the direction ofmotion. When an object slows down, its acceleration is opposite to the direction of its motion. Although this is commonlyreferred to as deceleration Figure 3.10, we say the train is accelerating in a direction opposite to its direction of motion.


Figure 3.10 A subway train in Sao Paulo, Brazil, deceleratesas it comes into a station. It is accelerating in a directionopposite to its direction of motion. (credit: Yusuke Kawasaki)


The term deceleration can cause confusion in our analysis because it is not a vector and it does not point to a specificdirection with respect to a coordinate system, so we do not use it. Acceleration is a vector, so we must choose the appropriatesign for it in our chosen coordinate system. In the case of the train in Figure 3.10, acceleration is in the negative directionin the chosen coordinate system, so we say the train is undergoing negative acceleration.
If an object in motion has a velocity in the positive direction with respect to a chosen origin and it acquires a constantnegative acceleration, the object eventually comes to a rest and reverses direction. If we wait long enough, the object passesthrough the origin going in the opposite direction. This is illustrated in Figure 3.11.


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Figure 3.11 An object in motion with a velocity vector toward the east undernegative acceleration comes to a rest and reverses direction. It passes the origin goingin the opposite direction after a long enough time.


Example 3.5
Calculating Average Acceleration: A Racehorse Leaves the Gate
A racehorse coming out of the gate accelerates from rest to a velocity of 15.0 m/s due west in 1.80 s. What is itsaverage acceleration?


Figure 3.12 Racehorses accelerating out of the gate. (credit:Jon Sullivan)


Strategy
First we draw a sketch and assign a coordinate system to the problem Figure 3.13. This is a simple problem, butit always helps to visualize it. Notice that we assign east as positive and west as negative. Thus, in this case, wehave negative velocity.


Figure 3.13 Identify the coordinate system, the given information, and what you want todetermine.


We can solve this problem by identifying Δv and Δt from the given information, and then calculating the
average acceleration directly from the equation a– = Δv


Δt
=


vf − v0
tf − t0


.


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3.3


Solution
First, identify the knowns: v0 = 0, vf = −15.0 m/s (the negative sign indicates direction toward the west), Δt =
1.80 s.
Second, find the change in velocity. Since the horse is going from zero to –15.0 m/s, its change in velocity equalsits final velocity:


Δv = vf − v0 = vf = −15.0 m/s.


Last, substitute the known values (Δv and Δt ) and solve for the unknown a– :
a– = Δv


Δt
= −15.0 m/s


1.80 s
= −8.33m/s2.


Significance
The negative sign for acceleration indicates that acceleration is toward the west. An acceleration of 8.33 m/s2 duewest means the horse increases its velocity by 8.33 m/s due west each second; that is, 8.33 meters per second persecond, which we write as 8.33 m/s2. This is truly an average acceleration, because the ride is not smooth. Wesee later that an acceleration of this magnitude would require the rider to hang on with a force nearly equal to hisweight.


Check Your Understanding Protons in a linear accelerator are accelerated from rest to 2.0 × 107 m/s
in 10–4 s. What is the average acceleration of the protons?


Instantaneous Acceleration
Instantaneous acceleration a, or acceleration at a specific instant in time, is obtained using the same process discussedfor instantaneous velocity. That is, we calculate the average velocity between two points in time separated by Δt and let
Δt approach zero. The result is the derivative of the velocity function v(t), which is instantaneous acceleration and is
expressed mathematically as


(3.9)a(t) = d
dt
v(t).


Thus, similar to velocity being the derivative of the position function, instantaneous acceleration is the derivative of thevelocity function. We can show this graphically in the same way as instantaneous velocity. In Figure 3.14, instantaneousacceleration at time t0 is the slope of the tangent line to the velocity-versus-time graph at time t0. We see that average
acceleration a– = Δv


Δt
approaches instantaneous acceleration as Δt approaches zero. Also in part (a) of the figure, we see


that velocity has a maximum when its slope is zero. This time corresponds to the zero of the acceleration function. Inpart (b), instantaneous acceleration at the minimum velocity is shown, which is also zero, since the slope of the curve iszero there, too. Thus, for a given velocity function, the zeros of the acceleration function give either the minimum or themaximum velocity.


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Figure 3.14 In a graph of velocity versus time, instantaneous acceleration is the slope of the tangent line. (a)
Shown is average acceleration a– = Δv


Δt
=


vf − vi
tf − ti


between times Δt = t6 − t1, Δt = t5 − t2 , and
Δt = t4 − t3 . When Δt → 0 , the average acceleration approaches instantaneous acceleration at time t0. In view
(a), instantaneous acceleration is shown for the point on the velocity curve at maximum velocity. At this point,instantaneous acceleration is the slope of the tangent line, which is zero. At any other time, the slope of the tangentline—and thus instantaneous acceleration—would not be zero. (b) Same as (a) but shown for instantaneousacceleration at minimum velocity.


To illustrate this concept, let’s look at two examples. First, a simple example is shown using Figure 3.9(b), the velocity-versus-time graph of Example 3.3, to find acceleration graphically. This graph is depicted in Figure 3.15(a), which isa straight line. The corresponding graph of acceleration versus time is found from the slope of velocity and is shown inFigure 3.15(b). In this example, the velocity function is a straight line with a constant slope, thus acceleration is a constant.In the next example, the velocity function is has a more complicated functional dependence on time.


Figure 3.15 (a, b) The velocity-versus-time graph is linear and has a negative constant slope (a) that is equal toacceleration, shown in (b).


If we know the functional form of velocity, v(t), we can calculate instantaneous acceleration a(t) at any time point in themotion using Equation 3.9.


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Example 3.6
Calculating Instantaneous Acceleration
A particle is in motion and is accelerating. The functional form of the velocity is v(t) = 20t − 5t2 m/s .


a. Find the functional form of the acceleration.
b. Find the instantaneous velocity at t = 1, 2, 3, and 5 s.
c. Find the instantaneous acceleration at t = 1, 2, 3, and 5 s.
d. Interpret the results of (c) in terms of the directions of the acceleration and velocity vectors.


Strategy
We find the functional form of acceleration by taking the derivative of the velocity function. Then, we calculatethe values of instantaneous velocity and acceleration from the given functions for each. For part (d), we need tocompare the directions of velocity and acceleration at each time.
Solution


a. a(t) = dv(t)
dt


= 20 − 10tm/s2


b. v(1 s) = 15 m/s , v(2 s) = 20 m/s , v(3 s) = 15 m/s , v(5 s) = −25 m/s
c. a(1 s) = 10m/s2 , a(2 s) = 0m/s2 , a(3 s) = −10m/s2 , a(5 s) = −30m/s2
d. At t = 1 s, velocity v(1 s) = 15 m/s is positive and acceleration is positive, so both velocity and


acceleration are in the same direction. The particle is moving faster.
At t = 2 s, velocity has increased to v(2 s) = 20 m/s , where it is maximum, which corresponds to the time when
the acceleration is zero. We see that the maximum velocity occurs when the slope of the velocity function is zero,which is just the zero of the acceleration function.
At t = 3 s, velocity is v(3 s) = 15 m/s and acceleration is negative. The particle has reduced its velocity and the
acceleration vector is negative. The particle is slowing down.
At t = 5 s, velocity is v(5 s) = −25 m/s and acceleration is increasingly negative. Between the times t = 3 s and
t = 5 s the particle has decreased its velocity to zero and then become negative, thus reversing its direction. Theparticle is now speeding up again, but in the opposite direction.
We can see these results graphically in Figure 3.16.


Chapter 3 | Motion Along a Straight Line 121




Figure 3.16 (a) Velocity versus time. Tangent lines areindicated at times 1, 2, and 3 s. The slopes of the tangents linesare the accelerations. At t = 3 s, velocity is positive. At t = 5 s,velocity is negative, indicating the particle has reverseddirection. (b) Acceleration versus time. Comparing the values ofaccelerations given by the black dots with the correspondingslopes of the tangent lines (slopes of lines through black dots) in(a), we see they are identical.


Significance
By doing both a numerical and graphical analysis of velocity and acceleration of the particle, we can learnmuch about its motion. The numerical analysis complements the graphical analysis in giving a total view ofthe motion. The zero of the acceleration function corresponds to the maximum of the velocity in this example.Also in this example, when acceleration is positive and in the same direction as velocity, velocity increases. Asacceleration tends toward zero, eventually becoming negative, the velocity reaches a maximum, after which itstarts decreasing. If we wait long enough, velocity also becomes negative, indicating a reversal of direction. Areal-world example of this type of motion is a car with a velocity that is increasing to a maximum, after which itstarts slowing down, comes to a stop, then reverses direction.


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3.4 Check Your Understanding An airplane lands on a runway traveling east. Describe its acceleration.


Getting a Feel for Acceleration
You are probably used to experiencing acceleration when you step into an elevator, or step on the gas pedal in your car.However, acceleration is happening to many other objects in our universe with which we don’t have direct contact. Table3.2 presents the acceleration of various objects. We can see the magnitudes of the accelerations extend over many orders ofmagnitude.


Acceleration Value (m/s2)
High-speed train 0.25
Elevator 2
Cheetah 5
Object in a free fall without air resistance near the surface of Earth 9.8
Space shuttle maximum during launch 29
Parachutist peak during normal opening of parachute 59
F16 aircraft pulling out of a dive 79
Explosive seat ejection from aircraft 147
Sprint missile 982
Fastest rocket sled peak acceleration 1540
Jumping flea 3200
Baseball struck by a bat 30,000
Closing jaws of a trap-jaw ant 1,000,000
Proton in the large Hadron collider 1.9 × 109


Table 3.2 Typical Values of Acceleration (credit: Wikipedia: Orders of Magnitude(acceleration))
In this table, we see that typical accelerations vary widely with different objects and have nothing to do with object sizeor how massive it is. Acceleration can also vary widely with time during the motion of an object. A drag racer has a largeacceleration just after its start, but then it tapers off as the vehicle reaches a constant velocity. Its average acceleration can bequite different from its instantaneous acceleration at a particular time during its motion. Figure 3.17 compares graphicallyaverage acceleration with instantaneous acceleration for two very different motions.


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Figure 3.17 Graphs of instantaneous acceleration versus time for two different one-dimensional motions. (a) Accelerationvaries only slightly and is always in the same direction, since it is positive. The average over the interval is nearly the same asthe acceleration at any given time. (b) Acceleration varies greatly, perhaps representing a package on a post office conveyorbelt that is accelerated forward and backward as it bumps along. It is necessary to consider small time intervals (such as from0–1.0 s) with constant or nearly constant acceleration in such a situation.
Learn about position, velocity, and acceleration graphs. Move the little man back and forth with a mouse and plothis motion. Set the position, velocity, or acceleration and let the simulation move the man for you. Visit this link(https://openstaxcollege.org/l/21movmansimul) to use the moving man simulation.


3.4 | Motion with Constant Acceleration
Learning Objectives


By the end of this section, you will be able to:
• Identify which equations of motion are to be used to solve for unknowns.
• Use appropriate equations of motion to solve a two-body pursuit problem.


You might guess that the greater the acceleration of, say, a car moving away from a stop sign, the greater the car’sdisplacement in a given time. But, we have not developed a specific equation that relates acceleration and displacement. Inthis section, we look at some convenient equations for kinematic relationships, starting from the definitions of displacement,velocity, and acceleration. We first investigate a single object in motion, called single-body motion. Then we investigate themotion of two objects, called two-body pursuit problems.
Notation
First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch,is a great simplification. Since elapsed time is Δt = tf − t0 , taking t0 = 0 means thatΔt = tf , the final time on the
stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity.That is, x0 is the initial position and v0 is the initial velocity. We put no subscripts on the final values. That is, t is the
final time, x is the final position, and v is the final velocity. This gives a simpler expression for elapsed time, Δt = t . It also
simplifies the expression for x displacement, which is now Δx = x − x0 . Also, it simplifies the expression for change in
velocity, which is now Δv = v − v0 . To summarize, using the simplified notation, with the initial time taken to be zero,


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Δt = t
Δx = x − x0
Δv = v − v0,


where the subscript 0 denotes an initial value and the absence of a subscript denotes a final value in whatever motion isunder consideration.
We now make the important assumption that acceleration is constant. This assumption allows us to avoid using calculus tofind instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal—thatis,


a– = a = constant.


Thus, we can use the symbol a for acceleration at all times. Assuming acceleration to be constant does not seriously limitthe situations we can study nor does it degrade the accuracy of our treatment. For one thing, acceleration is constant ina great number of situations. Furthermore, in many other situations we can describe motion accurately by assuming aconstant acceleration equal to the average acceleration for that motion. Lastly, for motion during which acceleration changesdrastically, such as a car accelerating to top speed and then braking to a stop, motion can be considered in separate parts,each of which has its own constant acceleration.
Displacement and Position from Velocity
To get our first two equations, we start with the definition of average velocity:


v– = Δx
Δt


.


Substituting the simplified notation for Δx and Δt yields
v– =


x − x0
t .


Solving for x gives us


(3.10)x = x0 + v–t,


where the average velocity is


(3.11)v– = v0 + v
2


.


The equation v– = v0 + v
2


reflects the fact that when acceleration is constant, v is just the simple average of the initial and
final velocities. Figure 3.18 illustrates this concept graphically. In part (a) of the figure, acceleration is constant, withvelocity increasing at a constant rate. The average velocity during the 1-h interval from 40 km/h to 80 km/h is 60 km/h:


v– =
v0 + v


2
= 40 km/h + 80 km/h


2
= 60 km/h.


In part (b), acceleration is not constant. During the 1-h interval, velocity is closer to 80 km/h than 40 km/h. Thus, the averagevelocity is greater than in part (a).


Chapter 3 | Motion Along a Straight Line 125




Figure 3.18 (a) Velocity-versus-time graph with constant acceleration showing the initial and final velocities v0 and v .
The average velocity is 1


2
(v0 + v) = 60km/h . (b) Velocity-versus-time graph with an acceleration that changes with time.


The average velocity is not given by 1
2
(v0 + v) , but is greater than 60 km/h.


Solving for Final Velocity from Acceleration and Time
We can derive another useful equation by manipulating the definition of acceleration:


a = Δv
Δt


.


Substituting the simplified notation for Δv and Δt gives us
a =


v − v0
t (constant a).


Solving for v yields


(3.12)v = v0 + at (constant a).


Example 3.7
Calculating Final Velocity
An airplane lands with an initial velocity of 70.0 m/s and then decelerates at 1.50 m/s2 for 40.0 s. What is its finalvelocity?
Strategy
First, we identify the knowns: v0 = 70 m/s, a = −1.50 m/s2, t = 40 s .
Second, we identify the unknown; in this case, it is final velocity vf .
Last, we determine which equation to use. To do this we figure out which kinematic equation gives the unknownin terms of the knowns. We calculate the final velocity using Equation 3.12, v = v0 + at .
Solution
Substitute the known values and solve:


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v = v0 + at = 70.0 m/s +

⎝−1.50 m/s


2⎞
⎠(40.0 s) = 10.0 m/s.


Figure 3.19 is a sketch that shows the acceleration and velocity vectors.


Figure 3.19 The airplane lands with an initial velocity of 70.0 m/s and slows to a final velocity of 10.0 m/s beforeheading for the terminal. Note the acceleration is negative because its direction is opposite to its velocity, which ispositive.


Significance
The final velocity is much less than the initial velocity, as desired when slowing down, but is still positive (seefigure). With jet engines, reverse thrust can be maintained long enough to stop the plane and start moving itbackward, which is indicated by a negative final velocity, but is not the case here.


In addition to being useful in problem solving, the equation v = v0 + at gives us insight into the relationships among
velocity, acceleration, and time. We can see, for example, that


• Final velocity depends on how large the acceleration is and how long it lasts
• If the acceleration is zero, then the final velocity equals the initial velocity (v = v0), as expected (in other words,velocity is constant)
• If a is negative, then the final velocity is less than the initial velocity


All these observations fit our intuition. Note that it is always useful to examine basic equations in light of our intuition andexperience to check that they do indeed describe nature accurately.
Solving for Final Position with Constant Acceleration
We can combine the previous equations to find a third equation that allows us to calculate the final position of an objectexperiencing constant acceleration. We start with


v = v0 + at.


Adding v0 to each side of this equation and dividing by 2 gives
v0 + v


2
= v0 +


1
2
at.


Since v0 + v
2


= v– for constant acceleration, we have
v– = v0 +


1
2
at.


Now we substitute this expression for v– into the equation for displacement, x = x0 + v–t , yielding


(3.13)x = x0 + v0 t + 12at2 (constant a).


Chapter 3 | Motion Along a Straight Line 127




Example 3.8
Calculating Displacement of an Accelerating Object
Dragsters can achieve an average acceleration of 26.0 m/s2. Suppose a dragster accelerates from rest at this ratefor 5.56 s Figure 3.20. How far does it travel in this time?


Figure 3.20 U.S. Army Top Fuel pilot Tony “The Sarge”Schumacher begins a race with a controlled burnout. (credit: Lt.Col. William Thurmond. Photo Courtesy of U.S. Army.)


Strategy
First, let’s draw a sketch Figure 3.21. We are asked to find displacement, which is x if we take x0 to be zero.
(Think about x0 as the starting line of a race. It can be anywhere, but we call it zero and measure all other
positions relative to it.) We can use the equation x = x0 + v0 t + 12at2 when we identify v0 , a , and t from the
statement of the problem.


Figure 3.21 Sketch of an accelerating dragster.


Solution
First, we need to identify the knowns. Starting from rest means that v0 = 0 , a is given as 26.0 m/s2 and t is given
as 5.56 s.
Second, we substitute the known values into the equation to solve for the unknown:


x = x0 + v0 t +
1
2
at2.


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Since the initial position and velocity are both zero, this equation simplifies to
x = 1


2
at2.


Substituting the identified values of a and t gives
x = 1


2
(26.0 m/s2)(5.56 s)2 = 402 m.


Significance
If we convert 402 m to miles, we find that the distance covered is very close to one-quarter of a mile, the standarddistance for drag racing. So, our answer is reasonable. This is an impressive displacement to cover in only 5.56s, but top-notch dragsters can do a quarter mile in even less time than this. If the dragster were given an initialvelocity, this would add another term to the distance equation. If the same acceleration and time are used in theequation, the distance covered would be much greater.


What else can we learn by examining the equation x = x0 + v0 t + 12at2? We can see the following relationships:
• Displacement depends on the square of the elapsed time when acceleration is not zero. In Example 3.8, thedragster covers only one-fourth of the total distance in the first half of the elapsed time.
• If acceleration is zero, then initial velocity equals average velocity (v0 = v–) , and


x = x0 + v0 t +
1
2
at2 becomes x = x0 + v0 t.


Solving for Final Velocity from Distance and Acceleration
A fourth useful equation can be obtained from another algebraic manipulation of previous equations. If we solve
v = v0 + at for t, we get


t =
v − v0
a .


Substituting this and v– = v0 + v
2


into x = x0 + v–t , we get


(3.14)v2 = v02 + 2a(x − x0) (constant a).


Example 3.9
Calculating Final Velocity
Calculate the final velocity of the dragster in Example 3.8 without using information about time.
Strategy
The equation v2 = v02 + 2a(x − x0) is ideally suited to this task because it relates velocities, acceleration, and
displacement, and no time information is required.
Solution
First, we identify the known values. We know that v0 = 0, since the dragster starts from rest. We also know that x− x0 = 402 m (this was the answer in Example 3.8). The average acceleration was given by a = 26.0 m/s2.
Second, we substitute the knowns into the equation v2 = v02 + 2a(x − x0) and solve for v:


v2 = 0 + 2⎛⎝26.0 m/s
2⎞
⎠(402 m).


Chapter 3 | Motion Along a Straight Line 129




Thus,
v2 = 2.09 × 104 m /s2


v = 2.09 × 104 m2 /s2 = 145 m/s.


Significance
A velocity of 145 m/s is about 522 km/h, or about 324 mi/h, but even this breakneck speed is short of the recordfor the quarter mile. Also, note that a square root has two values; we took the positive value to indicate a velocityin the same direction as the acceleration.


An examination of the equation v2 = v02 + 2a(x − x0) can produce additional insights into the general relationships among
physical quantities:


• The final velocity depends on how large the acceleration is and the distance over which it acts.
• For a fixed acceleration, a car that is going twice as fast doesn’t simply stop in twice the distance. It takes muchfarther to stop. (This is why we have reduced speed zones near schools.)


Putting Equations Together
In the following examples, we continue to explore one-dimensional motion, but in situations requiring slightly morealgebraic manipulation. The examples also give insight into problem-solving techniques. The note that follows is providedfor easy reference to the equations needed. Be aware that these equations are not independent. In many situations we havetwo unknowns and need two equations from the set to solve for the unknowns. We need as many equations as there areunknowns to solve a given situation.


Summary of Kinematic Equations (constant a)
x = x0 + v


–t


v– =
v0 + v


2
v = v0 + at


x = x0 + v0 t +
1
2
at2


v2 = v0
2 + 2a(x − x0)


Before we get into the examples, let’s look at some of the equations more closely to see the behavior of acceleration atextreme values. Rearranging Equation 3.12, we have
a =


v − v0
t .


From this we see that, for a finite time, if the difference between the initial and final velocities is small, the acceleration issmall, approaching zero in the limit that the initial and final velocities are equal. On the contrary, in the limit t → 0 for a
finite difference between the initial and final velocities, acceleration becomes infinite.
Similarly, rearranging Equation 3.14, we can express acceleration in terms of velocities and displacement:


a =
v2 − v0


2


2(x − x0)
.


Thus, for a finite difference between the initial and final velocities acceleration becomes infinite in the limit thedisplacement approaches zero. Acceleration approaches zero in the limit the difference in initial and final velocitiesapproaches zero for a finite displacement.


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Example 3.10
How Far Does a Car Go?
On dry concrete, a car can decelerate at a rate of 7.00 m/s2, whereas on wet concrete it can decelerate at only 5.00m/s2. Find the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h) on (a) dry concrete and (b)wet concrete. (c) Repeat both calculations and find the displacement from the point where the driver sees a trafficlight turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.
Strategy
First, we need to draw a sketch Figure 3.22. To determine which equations are best to use, we need to list all theknown values and identify exactly what we need to solve for.


Figure 3.22 Sample sketch to visualize deceleration and stopping distance of a car.


Solutiona. First, we need to identify the knowns and what we want to solve for. We know that v0 = 30.0 m/s, v = 0,and a = −7.00 m/s2 (a is negative because it is in a direction opposite to velocity). We take x0 to be zero.We are looking for displacement Δx , or x − x0.Second, we identify the equation that will help us solve the problem. The best equation to use is
v2 = v0


2 + 2a(x − x0).


This equation is best because it includes only one unknown, x. We know the values of all the othervariables in this equation. (Other equations would allow us to solve for x, but they require us to know thestopping time, t, which we do not know. We could use them, but it would entail additional calculations.)Third, we rearrange the equation to solve for x:
x − x0 =


v2 − v0
2


2a


and substitute the known values:
x − 0 = 0


2 − (30.0 m/s)2


2(−7.00m/s2)
.


Thus,
x = 64.3 m on dry concrete.


b. This part can be solved in exactly the same manner as (a). The only difference is that the acceleration is−5.00 m/s2. The result is
xwet = 90.0 m on wet concrete.


c. When the driver reacts, the stopping distance is the same as it is in (a) and (b) for dry and wet concrete.So, to answer this question, we need to calculate how far the car travels during the reaction time, and then


Chapter 3 | Motion Along a Straight Line 131




add that to the stopping time. It is reasonable to assume the velocity remains constant during the driver’sreaction time.To do this, we, again, identify the knowns and what we want to solve for. We know that v– = 30.0 m/s ,
treaction = 0.500 s , and areaction = 0 . We take x0-reaction to be zero. We are looking for xreaction .
Second, as before, we identify the best equation to use. In this case, x = x0 + v–t works well because the
only unknown value is x, which is what we want to solve for.Third, we substitute the knowns to solve the equation:


x = 0 + (30.0 m/s)(0.500 s) = 15.0 m.


This means the car travels 15.0 m while the driver reacts, making the total displacements in the two casesof dry and wet concrete 15.0 m greater than if he reacted instantly.Last, we then add the displacement during the reaction time to the displacement when braking (Figure3.23),
xbraking + xreaction = xtotal,


and find (a) to be 64.3 m + 15.0 m = 79.3 m when dry and (b) to be 90.0 m + 15.0 m = 105 m when wet.


Figure 3.23 The distance necessary to stop a car varies greatly, depending on road conditions and driver reactiontime. Shown here are the braking distances for dry and wet pavement, as calculated in this example, for a cartraveling initially at 30.0 m/s. Also shown are the total distances traveled from the point when the driver first sees alight turn red, assuming a 0.500-s reaction time.


Significance
The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longerto stop a car on wet pavement than dry. It is interesting that reaction time adds significantly to the displacements,but more important is the general approach to solving problems. We identify the knowns and the quantities to bedetermined, then find an appropriate equation. If there is more than one unknown, we need as many independentequations as there are unknowns to solve. There is often more than one way to solve a problem. The various partsof this example can, in fact, be solved by other methods, but the solutions presented here are the shortest.


Example 3.11
Calculating Time
Suppose a car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is 10.0 m/s and it acceleratesat 2.00 m/s2, how long does it take the car to travel the 200 m up the ramp? (Such information might be useful toa traffic engineer.)


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3.5


Strategy
First, we draw a sketch Figure 3.24. We are asked to solve for time t. As before, we identify the known quantitiesto choose a convenient physical relationship (that is, an equation with one unknown, t.)


Figure 3.24 Sketch of a car accelerating on a freeway ramp.


Solution
Again, we identify the knowns and what we want to solve for. We know that x0 = 0,
v0 = 10 m/s, a = 2.00 m/s


2 , and x = 200 m.
We need to solve for t. The equation x = x0 + v0 t + 12at2 works best because the only unknown in the equation
is the variable t, for which we need to solve. From this insight we see that when we input the knowns into theequation, we end up with a quadratic equation.
We need to rearrange the equation to solve for t, then substituting the knowns into the equation:


200 m = 0m + (10.0 m/s)t + 1
2

⎝2.00 m/s


2⎞
⎠t
2.


We then simplify the equation. The units of meters cancel because they are in each term. We can get the units ofseconds to cancel by taking t = t s, where t is the magnitude of time and s is the unit. Doing so leaves
200 = 10t + t2.


We then use the quadratic formula to solve for t,
t2 + 10t − 200 = 0


t = −b ± b
2 − 4ac


2a
,


which yields two solutions: t = 10.0 and t = −20.0. A negative value for time is unreasonable, since it would meanthe event happened 20 s before the motion began. We can discard that solution. Thus,
t = 10.0 s.


Significance
Whenever an equation contains an unknown squared, there are two solutions. In some problems both solutions aremeaningful; in others, only one solution is reasonable. The 10.0-s answer seems reasonable for a typical freewayon-ramp.


Check Your Understanding A manned rocket accelerates at a rate of 20 m/s2 during launch. How longdoes it take the rocket to reach a velocity of 400 m/s?


Chapter 3 | Motion Along a Straight Line 133




Example 3.12
Acceleration of a Spaceship
A spaceship has left Earth’s orbit and is on its way to the Moon. It accelerates at 20 m/s2 for 2 min and covers adistance of 1000 km. What are the initial and final velocities of the spaceship?
Strategy
We are asked to find the initial and final velocities of the spaceship. Looking at the kinematic equations, we seethat one equation will not give the answer. We must use one kinematic equation to solve for one of the velocitiesand substitute it into another kinematic equation to get the second velocity. Thus, we solve two of the kinematicequations simultaneously.
Solution
First we solve for v0 using x = x0 + v0 t + 12at2 = 12at2 :


x − x0 = v0 t +
1
2
at


1.0 × 106 m = v0(120.0 s) +
1
2
(20.0m/s2)(120.0 s)2


v0 = 7133.3 m/s.


Then we substitute v0 into v = v0 + at to solve for the final velocity:
v = v0 + at = 7133.3 m/s + (20.0 m/s


2)(120.0 s) = 9533.3 m/s.


Significance
There are six variables in displacement, time, velocity, and acceleration that describe motion in one dimension.The initial conditions of a given problem can be many combinations of these variables. Because of this diversity,solutions may not be easy as simple substitutions into one of the equations. This example illustrates that solutionsto kinematics may require solving two simultaneous kinematic equations.


With the basics of kinematics established, we can go on to many other interesting examples and applications. In the processof developing kinematics, we have also glimpsed a general approach to problem solving that produces both correct answersand insights into physical relationships. The next level of complexity in our kinematics problems involves the motion oftwo interrelated bodies, called two-body pursuit problems.
Two-Body Pursuit Problems
Up until this point we have looked at examples of motion involving a single body. Even for the problem with two carsand the stopping distances on wet and dry roads, we divided this problem into two separate problems to find the answers.In a two-body pursuit problem, the motions of the objects are coupled—meaning, the unknown we seek depends on themotion of both objects. To solve these problems we write the equations of motion for each object and then solve themsimultaneously to find the unknown. This is illustrated in Figure 3.25.


Figure 3.25 A two-body pursuit scenario where car 2 has a constant velocity and car 1 isbehind with a constant acceleration. Car 1 catches up with car 2 at a later time.


The time and distance required for car 1 to catch car 2 depends on the initial distance car 1 is from car 2 as well as thevelocities of both cars and the acceleration of car 1. The kinematic equations describing the motion of both cars must besolved to find these unknowns.


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Consider the following example.
Example 3.13


Cheetah Catching a Gazelle
A cheetah waits in hiding behind a bush. The cheetah spots a gazelle running past at 10 m/s. At the instant thegazelle passes the cheetah, the cheetah accelerates from rest at 4 m/s2 to catch the gazelle. (a) How long does ittake the cheetah to catch the gazelle? (b) What is the displacement of the gazelle and cheetah?
Strategy
We use the set of equations for constant acceleration to solve this problem. Since there are two objects inmotion, we have separate equations of motion describing each animal. But what links the equations is a commonparameter that has the same value for each animal. If we look at the problem closely, it is clear the commonparameter to each animal is their position x at a later time t. Since they both start at x0 = 0 , their displacements
are the same at a later time t, when the cheetah catches up with the gazelle. If we pick the equation of motion thatsolves for the displacement for each animal, we can then set the equations equal to each other and solve for theunknown, which is time.
Solutiona. Equation for the gazelle: The gazelle has a constant velocity, which is its average velocity, since it is notaccelerating. Therefore, we use Equation 3.10 with x0 = 0 :


x = x0 + v
–t = v–t.


Equation for the cheetah: The cheetah is accelerating from rest, so we use Equation 3.13 with x0 = 0
and v0 = 0 :


x = x0 + v0 t +
1
2
at2 = 1


2
at2.


Now we have an equation of motion for each animal with a common parameter, which can be eliminatedto find the solution. In this case, we solve for t:
x = v–t = 1


2
at2


t = 2v


a .


The gazelle has a constant velocity of 10 m/s, which is its average velocity. The acceleration of thecheetah is 4 m/s2. Evaluating t, the time for the cheetah to reach the gazelle, we have
t = 2v




a =
2(10)
4


= 5 s.


b. To get the displacement, we use either the equation of motion for the cheetah or the gazelle, since theyshould both give the same answer.Displacement of the cheetah:
x = 1


2
at2 = 1


2
(4)(5)2 = 50 m.


Displacement of the gazelle:
x = v–t = 10(5) = 50 m.


We see that both displacements are equal, as expected.
Significance
It is important to analyze the motion of each object and to use the appropriate kinematic equations to describe the


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3.6


individual motion. It is also important to have a good visual perspective of the two-body pursuit problem to seethe common parameter that links the motion of both objects.
Check Your Understanding A bicycle has a constant velocity of 10 m/s. A person starts from rest andruns to catch up to the bicycle in 30 s. What is the acceleration of the person?


3.5 | Free Fall
Learning Objectives


By the end of this section, you will be able to:
• Use the kinematic equations with the variables y and g to analyze free-fall motion.
• Describe how the values of the position, velocity, and acceleration change during a free fall.
• Solve for the position, velocity, and acceleration as functions of time when an object is in a freefall.


An interesting application of Equation 3.4 through Equation 3.14 is called free fall, which describes the motion of anobject falling in a gravitational field, such as near the surface of Earth or other celestial objects of planetary size. Let’sassume the body is falling in a straight line perpendicular to the surface, so its motion is one-dimensional. For example, wecan estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. But“falling,” in the context of free fall, does not necessarily imply the body is moving from a greater height to a lesser height.If a ball is thrown upward, the equations of free fall apply equally to its ascent as well as its descent.
Gravity
The most remarkable and unexpected fact about falling objects is that if air resistance and friction are negligible, then ina given location all objects fall toward the center of Earth with the same constant acceleration, independent of their mass.This experimentally determined fact is unexpected because we are so accustomed to the effects of air resistance and frictionthat we expect light objects to fall slower than heavy ones. Until Galileo Galilei (1564–1642) proved otherwise, peoplebelieved that a heavier object has a greater acceleration in a free fall. We now know this is not the case. In the absence ofair resistance, heavy objects arrive at the ground at the same time as lighter objects when dropped from the same heightFigure 3.26.


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Figure 3.26 A hammer and a feather fall with the same constant acceleration if air resistance is negligible. This is a generalcharacteristic of gravity not unique to Earth, as astronaut David R. Scott demonstrated in 1971 on the Moon, where theacceleration from gravity is only 1.67 m/s2 and there is no atmosphere.


In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ballreaches the ground after a baseball dropped at the same time. (It might be difficult to observe the difference if the height isnot large.) Air resistance opposes the motion of an object through the air, and friction between objects—such as betweenclothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them.
For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be infree fall. The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objectsis therefore called acceleration due to gravity. Acceleration due to gravity is constant, which means we can apply thekinematic equations to any falling object where air resistance and friction are negligible. This opens to us a broad class ofinteresting situations.
Acceleration due to gravity is so important that its magnitude is given its own symbol, g. It is constant at any given locationon Earth and has the average value


g = 9.81 m/s2 (or 32.2 ft/s2).


Although g varies from 9.78 m/s2 to 9.83 m/s2, depending on latitude, altitude, underlying geological formations, andlocal topography, let’s use an average value of 9.8 m/s2 rounded to two significant figures in this text unless specifiedotherwise. Neglecting these effects on the value of g as a result of position on Earth’s surface, as well as effects resultingfrom Earth’s rotation, we take the direction of acceleration due to gravity to be downward (toward the center of Earth).In fact, its direction defines what we call vertical. Note that whether acceleration a in the kinematic equations has thevalue +g or −g depends on how we define our coordinate system. If we define the upward direction as positive, then
a = −g = −9.8 m/s2, and if we define the downward direction as positive, then a = g = 9.8 m/s2 .
One-Dimensional Motion Involving Gravity
The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progresstoward more complex ones. So, we start by considering straight up-and-down motion with no air resistance or friction.These assumptions mean the velocity (if there is any) is vertical. If an object is dropped, we know the initial velocity is zerowhen in free fall. When the object has left contact with whatever held or threw it, the object is in free fall. When the objectis thrown, it has the same initial speed in free fall as it did before it was released. When the object comes in contact with theground or any other object, it is no longer in free fall and its acceleration of g is no longer valid. Under these circumstances,the motion is one-dimensional and has constant acceleration of magnitude g. We represent vertical displacement with thesymbol y.


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Kinematic Equations for Objects in Free Fall
We assume here that acceleration equals −g (with the positive direction upward).


(3.15)v = v0 − gt
(3.16)y = y0 + v0 t − 12gt2
(3.17)v2 = v02 − 2g(y − y0)


Problem-Solving Strategy: Free Fall
1. Decide on the sign of the acceleration of gravity. In Equation 3.15 through Equation 3.17, accelerationg is negative, which says the positive direction is upward and the negative direction is downward. In someproblems, it may be useful to have acceleration g as positive, indicating the positive direction is downward.
2. Draw a sketch of the problem. This helps visualize the physics involved.
3. Record the knowns and unknowns from the problem description. This helps devise a strategy for selecting theappropriate equations to solve the problem.
4. Decide which of Equation 3.15 through Equation 3.17 are to be used to solve for the unknowns.


Example 3.14
Free Fall of a Ball
Figure 3.27 shows the positions of a ball, at 1-s intervals, with an initial velocity of 4.9 m/s downward, that isthrown from the top of a 98-m-high building. (a) How much time elapses before the ball reaches the ground? (b)What is the velocity when it arrives at the ground?


Figure 3.27 The positions and velocities at 1-s intervals of aball thrown downward from a tall building at 4.9 m/s.


Strategy
Choose the origin at the top of the building with the positive direction upward and the negative directiondownward. To find the time when the position is −98 m, we use Equation 3.16, with
y0 = 0, v0 = −4.9 m/s, and g = 9.8 m/s


2 .
Solutiona. Substitute the given values into the equation:


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y = y0 + v0 t −
1
2
gt2


−98.0 m = 0 − (4.9 m/s)t − 1
2
(9.8 m/s2)t2.


This simplifies to
t2 + t − 20 = 0.


This is a quadratic equation with roots t = −5.0s and t = 4.0s . The positive root is the one we are
interested in, since time t = 0 is the time when the ball is released at the top of the building. (The time
t = −5.0s represents the fact that a ball thrown upward from the ground would have been in the air for
5.0 s when it passed by the top of the building moving downward at 4.9 m/s.)


b. Using Equation 3.15, we have
v = v0 − gt = −4.9 m/s − (9.8m/s


2)(4.0 s) = −44.1 m/s.


Significance
For situations when two roots are obtained from a quadratic equation in the time variable, we must look at thephysical significance of both roots to determine which is correct. Since t = 0 corresponds to the time when
the ball was released, the negative root would correspond to a time before the ball was released, which is notphysically meaningful. When the ball hits the ground, its velocity is not immediately zero, but as soon as the ballinteracts with the ground, its acceleration is not g and it accelerates with a different value over a short time to zerovelocity. This problem shows how important it is to establish the correct coordinate system and to keep the signsof g in the kinematic equations consistent.


Example 3.15
Vertical Motion of a Baseball
A batter hits a baseball straight upward at home plate and the ball is caught 5.0 s after it is struck Figure 3.28.(a) What is the initial velocity of the ball? (b) What is the maximum height the ball reaches? (c) How long does ittake to reach the maximum height? (d) What is the acceleration at the top of its path? (e) What is the velocity ofthe ball when it is caught? Assume the ball is hit and caught at the same location.


Figure 3.28 A baseball hit straight up is caught by the catcher 5.0 s later.


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3.7


Strategy
Choose a coordinate system with a positive y-axis that is straight up and with an origin that is at the spot wherethe ball is hit and caught.
Solutiona. Equation 3.16 gives


y = y0 + v0 t −
1
2
gt2


0 = 0 + v0(5.0 s) −
1
2

⎝9.8 m/s


2⎞
⎠(5.0 s)


2,


which gives v0 = 24.5 m/sec .
b. At the maximum height, v = 0 . With v0 = 24.5 m/s , Equation 3.17 gives


v2 = v0
2 − 2g(y − y0)


0 = (24.5 m/s)2 − 2(9.8m/s2)(y − 0)


or
y = 30.6 m.


c. To find the time when v = 0 , we use Equation 3.15:
v = v0 − gt


0 = 24.5 m/s − (9.8m/s2)t.


This gives t = 2.5 s . Since the ball rises for 2.5 s, the time to fall is 2.5 s.
d. The acceleration is 9.8 m/s2 everywhere, even when the velocity is zero at the top of the path. Althoughthe velocity is zero at the top, it is changing at the rate of 9.8 m/s2 downward.
e. The velocity at t = 5.0s can be determined with Equation 3.15:


v = v0 − gt


= 24.5 m/s − 9.8m/s2(5.0 s)
= −24.5 m/s.


Significance
The ball returns with the speed it had when it left. This is a general property of free fall for any initial velocity. Weused a single equation to go from throw to catch, and did not have to break the motion into two segments, upwardand downward. We are used to thinking of the effect of gravity is to create free fall downward toward Earth. It isimportant to understand, as illustrated in this example, that objects moving upward away from Earth are also in astate of free fall.


Check Your Understanding A chunk of ice breaks off a glacier and falls 30.0 m before it hits the water.Assuming it falls freely (there is no air resistance), how long does it take to hit the water? Which quantityincreases faster, the speed of the ice chunk or its distance traveled?


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Example 3.16
Rocket Booster
A small rocket with a booster blasts off and heads straight upward. When at a height of 5.0 km and velocity of
200.0 m/s, it releases its booster. (a) What is the maximum height the booster attains? (b) What is the velocity ofthe booster at a height of 6.0 km? Neglect air resistance.


Figure 3.29 A rocket releases its booster at a given height andvelocity. How high and how fast does the booster go?


Strategy
We need to select the coordinate system for the acceleration of gravity, which we take as negative downward. Weare given the initial velocity of the booster and its height. We consider the point of release as the origin. We knowthe velocity is zero at the maximum position within the acceleration interval; thus, the velocity of the booster iszero at its maximum height, so we can use this information as well. From these observations, we use Equation3.17, which gives us the maximum height of the booster. We also use Equation 3.17 to give the velocity at 6.0km. The initial velocity of the booster is 200.0 m/s.
Solution


a. From Equation 3.17, v2 = v02 − 2g(y − y0) . With v = 0 and y0 = 0 , we can solve for y:
y =


v0
2


−2g
= (2.0 × 10


2m/s)2


−2(9.8 m/s2)
= 2040.8 m.


This solution gives the maximum height of the booster in our coordinate system, which has its origin atthe point of release, so the maximum height of the booster is roughly 7.0 km.
b. An altitude of 6.0 km corresponds to y = 1.0 × 103 m in the coordinate system we are using. The other


Chapter 3 | Motion Along a Straight Line 141




initial conditions are y0 = 0, and v0 = 200.0 m/s .
We have, from Equation 3.17,


v2 = (200.0 m/s)2 − 2(9.8 m/s2)(1.0 × 103 m) ⇒ v = ± 142.8 m/s.


Significance
We have both a positive and negative solution in (b). Since our coordinate system has the positive directionupward, the +142.8 m/s corresponds to a positive upward velocity at 6000 m during the upward leg of thetrajectory of the booster. The value v = −142.8 m/s corresponds to the velocity at 6000 m on the downward leg.This example is also important in that an object is given an initial velocity at the origin of our coordinate system,but the origin is at an altitude above the surface of Earth, which must be taken into account when forming thesolution.
Visit this site (https://openstaxcollege.org/l/21equatgraph) to learn about graphing polynomials. Theshape of the curve changes as the constants are adjusted. View the curves for the individual terms (for example, y= bx) to see how they add to generate the polynomial curve.


3.6 | Finding Velocity and Displacement from Acceleration
Learning Objectives


By the end of this section, you will be able to:
• Derive the kinematic equations for constant acceleration using integral calculus.
• Use the integral formulation of the kinematic equations in analyzing motion.
• Find the functional form of velocity versus time given the acceleration function.
• Find the functional form of position versus time given the velocity function.


This section assumes you have enough background in calculus to be familiar with integration. In Instantaneous Velocityand Speed and Average and Instantaneous Acceleration we introduced the kinematic functions of velocity andacceleration using the derivative. By taking the derivative of the position function we found the velocity function, andlikewise by taking the derivative of the velocity function we found the acceleration function. Using integral calculus, wecan work backward and calculate the velocity function from the acceleration function, and the position function from thevelocity function.
Kinematic Equations from Integral Calculus
Let’s begin with a particle with an acceleration a(t) is a known function of time. Since the time derivative of the velocityfunction is acceleration,


d
dt
v(t) = a(t),


we can take the indefinite integral of both sides, finding


d
dt
v(t)dt = ∫ a(t)dt + C1,


where C1 is a constant of integration. Since ⌠⌡ddtv(t)dt = v(t) , the velocity is given by


(3.18)v(t) = ∫ a(t)dt + C1.


Similarly, the time derivative of the position function is the velocity function,


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d
dt
x(t) = v(t).


Thus, we can use the same mathematical manipulations we just used and find


(3.19)x(t) = ∫ v(t)dt + C2,


where C2 is a second constant of integration.
We can derive the kinematic equations for a constant acceleration using these integrals. With a(t) = a a constant, and doingthe integration in Equation 3.18, we find


v(t) = ∫ adt + C1 = at + C1.


If the initial velocity is v(0) = v0, then
v0 = 0 + C1.


Then, C1 = v0 and
v(t) = v0 + at,


which is Equation 3.12. Substituting this expression into Equation 3.19 gives
x(t) = ∫ (v0 + at)dt + C2.


Doing the integration, we find
x(t) = v0 t +


1
2
at2 + C2.


If x(0) = x0, we have
x0 = 0 + 0 + C2;


so, C2 = x0. Substituting back into the equation for x(t), we finally have
x(t) = x0 + v0 t +


1
2
at2,


which is Equation 3.13.
Example 3.17


Motion of a Motorboat
A motorboat is traveling at a constant velocity of 5.0 m/s when it starts to decelerate to arrive at the dock. Its
acceleration is a(t) = − 1


4
tm/s2 . (a) What is the velocity function of the motorboat? (b) At what time does


the velocity reach zero? (c) What is the position function of the motorboat? (d) What is the displacement of themotorboat from the time it begins to decelerate to when the velocity is zero? (e) Graph the velocity and positionfunctions.
Strategy
(a) To get the velocity function we must integrate and use initial conditions to find the constant of integration.(b) We set the velocity function equal to zero and solve for t. (c) Similarly, we must integrate to find the positionfunction and use initial conditions to find the constant of integration. (d) Since the initial position is taken to bezero, we only have to evaluate the position function at t = 0 .
Solution
We take t = 0 to be the time when the boat starts to decelerate.


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a. From the functional form of the acceleration we can solve Equation 3.18 to get v(t):
v(t) = ⌠



a(t)dt + C1 =




−1
4
tdt + C1 = −


1
8
t2 + C1.


At t = 0 we have v(0) = 5.0 m/s = 0 + C1, so C1 = 5.0 m/s or v(t) = 5.0 m/s − 18t2 .
b. v(t) = 0 = 5.0 m/s − 1


8
t2 ⇒ t = 6.3 s


c. Solve Equation 3.19:
x(t) = ∫ v(t)dt + C2 = ⌠⌡(5.0 −


1
8
t2)dt + C2 = 5.0t −


1
24


t3 + C2.


At t = 0, we set x(0) = 0 = x0, since we are only interested in the displacement from when the boat startsto decelerate. We have
x(0) = 0 = C2.


Therefore, the equation for the position is
x(t) = 5.0t − 1


24
t3.


d. Since the initial position is taken to be zero, we only have to evaluate x(t) when the velocity is zero. Thisoccurs at t = 6.3 s. Therefore, the displacement is
x(6.3) = 5.0(6.3) − 1


24
(6.3)3 = 21.1 m.


Figure 3.30 (a) Velocity of the motorboat as a function of time. The motorboat decreases its velocity to zero in6.3 s. At times greater than this, velocity becomes negative—meaning, the boat is reversing direction. (b) Positionof the motorboat as a function of time. At t = 6.3 s, the velocity is zero and the boat has stopped. At times greaterthan this, the velocity becomes negative—meaning, if the boat continues to move with the same acceleration, itreverses direction and heads back toward where it originated.


Significance
The acceleration function is linear in time so the integration involves simple polynomials. In Figure 3.30, we


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3.8


see that if we extend the solution beyond the point when the velocity is zero, the velocity becomes negative andthe boat reverses direction. This tells us that solutions can give us information outside our immediate interest andwe should be careful when interpreting them.
Check Your Understanding A particle starts from rest and has an acceleration function 5 − 10tm/s2 .


(a) What is the velocity function? (b) What is the position function? (c) When is the velocity zero?


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acceleration due to gravity
average acceleration
average speed
average velocity
displacement
distance traveled
elapsed time
free fall
instantaneous acceleration
instantaneous speed
instantaneous velocity
kinematics
position
total displacement
two-body pursuit problem


CHAPTER 3 REVIEW
KEY TERMS


acceleration of an object as a result of gravity
the rate of change in velocity; the change in velocity over time


the total distance traveled divided by elapsed time
the displacement divided by the time over which displacement occurs


the change in position of an object
the total length of the path traveled between two positions


the difference between the ending time and the beginning time
the state of movement that results from gravitational force only


acceleration at a specific point in time
the absolute value of the instantaneous velocity
the velocity at a specific instant or time point


the description of motion through properties such as position, time, velocity, and acceleration
the location of an object at a particular time


the sum of individual displacements over a given time period
a kinematics problem in which the unknowns are calculated by solving the kinematicequations simultaneously for two moving objects


KEY EQUATIONS
Displacement Δx = xf − xi
Total displacement ΔxTotal = ∑ Δxi
Average velocity v– = ΔxΔt = x2 − x1t2 − t1
Instantaneous velocity v(t) = dx(t)dt
Average speed Average speed = s– = Total distanceElapsed time
Instantaneous speed Instantaneous speed = |v(t)|
Average acceleration a– = ΔvΔt =


v f − v0
t f − t0


Instantaneous acceleration a(t) = dv(t)dt
Position from average velocity x = x0 + v–t
Average velocity v– = v0 + v2
Velocity from acceleration v = v0 + at (constant a)


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Position from velocity and acceleration x = x0 + v0 t + 12at2 (constant a)
Velocity from distance v2 = v02 + 2a(x − x0) (constant a)
Velocity of free fall v = v0 − gt (positive upward)
Height of free fall y = y0 + v0 t − 12gt2
Velocity of free fall from height v2 = v02 − 2g(y − y0)
Velocity from acceleration v(t) = ∫ a(t)dt + C1
Position from velocity x(t) = ∫ v(t)dt + C2


SUMMARY
3.1 Position, Displacement, and Average Velocity


• Kinematics is the description of motion without considering its causes. In this chapter, it is limited to motion alonga straight line, called one-dimensional motion.
• Displacement is the change in position of an object. The SI unit for displacement is the meter. Displacement hasdirection as well as magnitude.
• Distance traveled is the total length of the path traveled between two positions.
• Time is measured in terms of change. The time between two position points x1 and x2 is Δt = t2 − t1 . Elapsed
time for an event is Δt = tf − t0 , where tf is the final time and t0 is the initial time. The initial time is often taken
to be zero.


• Average velocity v– is defined as displacement divided by elapsed time. If x1, t1 and x2, t2 are two position time
points, the average velocity between these points is


v– = Δx
Δt


=
x2 − x1
t2 − t1


.


3.2 Instantaneous Velocity and Speed
• Instantaneous velocity is a continuous function of time and gives the velocity at any point in time during a particle’smotion. We can calculate the instantaneous velocity at a specific time by taking the derivative of the positionfunction, which gives us the functional form of instantaneous velocity v(t).
• Instantaneous velocity is a vector and can be negative.
• Instantaneous speed is found by taking the absolute value of instantaneous velocity, and it is always positive.
• Average speed is total distance traveled divided by elapsed time.
• The slope of a position-versus-time graph at a specific time gives instantaneous velocity at that time.


3.3 Average and Instantaneous Acceleration
• Acceleration is the rate at which velocity changes. Acceleration is a vector; it has both a magnitude and direction.The SI unit for acceleration is meters per second squared.
• Acceleration can be caused by a change in the magnitude or the direction of the velocity, or both.
• Instantaneous acceleration a(t) is a continuous function of time and gives the acceleration at any specific time duringthe motion. It is calculated from the derivative of the velocity function. Instantaneous acceleration is the slope ofthe velocity-versus-time graph.


Chapter 3 | Motion Along a Straight Line 147




• Negative acceleration (sometimes called deceleration) is acceleration in the negative direction in the chosencoordinate system.
3.4 Motion with Constant Acceleration


• When analyzing one-dimensional motion with constant acceleration, identify the known quantities and choose theappropriate equations to solve for the unknowns. Either one or two of the kinematic equations are needed to solvefor the unknowns, depending on the known and unknown quantities.
• Two-body pursuit problems always require two equations to be solved simultaneously for the unknowns.


3.5 Free Fall
• An object in free fall experiences constant acceleration if air resistance is negligible.
• On Earth, all free-falling objects have an acceleration g due to gravity, which averages g = 9.81 m/s2 .
• For objects in free fall, the upward direction is normally taken as positive for displacement, velocity, andacceleration.


3.6 Finding Velocity and Displacement from Acceleration
• Integral calculus gives us a more complete formulation of kinematics.
• If acceleration a(t) is known, we can use integral calculus to derive expressions for velocity v(t) and position x(t).
• If acceleration is constant, the integral equations reduce to Equation 3.12 and Equation 3.13 for motion withconstant acceleration.


CONCEPTUAL QUESTIONS
3.1 Position, Displacement, and Average
Velocity
1. Give an example in which there are clear distinctionsamong distance traveled, displacement, and magnitude ofdisplacement. Identify each quantity in your examplespecifically.
2. Under what circumstances does distance traveled equalmagnitude of displacement? What is the only case in whichmagnitude of displacement and displacement are exactlythe same?
3. Bacteria move back and forth using their flagella(structures that look like little tails). Speeds of up to 50 μm/s (50 × 10−6 m/s) have been observed. The total distancetraveled by a bacterium is large for its size, whereas itsdisplacement is small. Why is this?
4. Give an example of a device used to measure time andidentify what change in that device indicates a change intime.
5. Does a car’s odometer measure distance traveled ordisplacement?
6. During a given time interval the average velocity ofan object is zero. What can you say conclude about its


displacement over the time interval?


3.2 Instantaneous Velocity and Speed
7. There is a distinction between average speed and themagnitude of average velocity. Give an example thatillustrates the difference between these two quantities.
8. Does the speedometer of a car measure speed orvelocity?
9. If you divide the total distance traveled on a car trip(as determined by the odometer) by the elapsed time ofthe trip, are you calculating average speed or magnitude ofaverage velocity? Under what circumstances are these twoquantities the same?
10. How are instantaneous velocity and instantaneousspeed related to one another? How do they differ?


3.3 Average and Instantaneous Acceleration
11. Is it possible for speed to be constant whileacceleration is not zero?
12. Is it possible for velocity to be constant whileacceleration is not zero? Explain.


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13. Give an example in which velocity is zero yetacceleration is not.
14. If a subway train is moving to the left (has a negativevelocity) and then comes to a stop, what is the direction ofits acceleration? Is the acceleration positive or negative?
15. Plus and minus signs are used in one-dimensionalmotion to indicate direction. What is the sign of anacceleration that reduces the magnitude of a negativevelocity? Of a positive velocity?


3.4 Motion with Constant Acceleration
16. When analyzing the motion of a single object, whatis the required number of known physical variables thatare needed to solve for the unknown quantities using thekinematic equations?
17. State two scenarios of the kinematics of single objectwhere three known quantities require two kinematicequations to solve for the unknowns.


3.5 Free Fall
18. What is the acceleration of a rock thrown straightupward on the way up? At the top of its flight? On the waydown? Assume there is no air resistance.
19. An object that is thrown straight up falls back to Earth.This is one-dimensional motion. (a) When is its velocity


zero? (b) Does its velocity change direction? (c) Does theacceleration have the same sign on the way up as on theway down?
20. Suppose you throw a rock nearly straight up at acoconut in a palm tree and the rock just misses the coconuton the way up but hits the coconut on the way down.Neglecting air resistance and the slight horizontal variationin motion to account for the hit and miss of the coconut,how does the speed of the rock when it hits the coconuton the way down compare with what it would have been ifit had hit the coconut on the way up? Is it more likely todislodge the coconut on the way up or down? Explain.
21. The severity of a fall depends on your speed whenyou strike the ground. All factors but the acceleration fromgravity being the same, how many times higher could a safefall on the Moon than on Earth (gravitational accelerationon the Moon is about one-sixth that of the Earth)?
22. How many times higher could an astronaut jump onthe Moon than on Earth if her takeoff speed is the samein both locations (gravitational acceleration on the Moon isabout on-sixth of that on Earth)?


3.6 Finding Velocity and Displacement from
Acceleration
23. When given the acceleration function, what additionalinformation is needed to find the velocity function andposition function?


PROBLEMS
3.1 Position, Displacement, and Average
Velocity
24. Consider a coordinate system in which the positive xaxis is directed upward vertically. What are the positions ofa particle (a) 5.0 m directly above the origin and (b) 2.0 mbelow the origin?
25. A car is 2.0 km west of a traffic light at t = 0 and 5.0km east of the light at t = 6.0 min. Assume the origin of thecoordinate system is the light and the positive x directionis eastward. (a) What are the car’s position vectors at thesetwo times? (b) What is the car’s displacement between 0min and 6.0 min?
26. The Shanghai maglev train connects Longyang Roadto Pudong International Airport, a distance of 30 km. Thejourney takes 8 minutes on average. What is the maglevtrain’s average velocity?


27. The position of a particle moving along the x-axis isgiven by x(t) = 4.0 − 2.0t m. (a) At what time does the
particle cross the origin? (b) What is the displacement ofthe particle between t = 3.0 s and t = 6.0 s?
28. A cyclist rides 8.0 km east for 20 minutes, then heturns and heads west for 8 minutes and 3.2 km. Finally,he rides east for 16 km, which takes 40 minutes. (a) Whatis the final displacement of the cyclist? (b) What is hisaverage velocity?
29. On February 15, 2013, a superbolide meteor (brighterthan the Sun) entered Earth’s atmosphere overChelyabinsk, Russia, and exploded at an altitude of 23.5km. Eyewitnesses could feel the intense heat from thefireball, and the blast wave from the explosion blew outwindows in buildings. The blast wave took approximately 2minutes 30 seconds to reach ground level. (a) What was theaverage velocity of the blast wave? b) Compare this withthe speed of sound, which is 343 m/s at sea level.


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3.2 Instantaneous Velocity and Speed
30. A woodchuck runs 20 m to the right in 5 s, then turnsand runs 10 m to the left in 3 s. (a) What is the averagevelocity of the woodchuck? (b) What is its average speed?
31. Sketch the velocity-versus-time graph from thefollowing position-versus-time graph.


32. Sketch the velocity-versus-time graph from thefollowing position-versus-time graph.


33. Given the following velocity-versus-time graph,sketch the position-versus-time graph.


34. An object has a position function x(t) = 5t m. (a) Whatis the velocity as a function of time? (b) Graph the positionfunction and the velocity function.
35. A particle moves along the x-axis according to
x(t) = 10t − 2t2 m . (a) What is the instantaneous velocity
at t = 2 s and t = 3 s? (b) What is the instantaneous speed atthese times? (c) What is the average velocity between t = 2s and t = 3 s?
36. Unreasonable results. A particle moves along the
x-axis according to x(t) = 3t3 + 5t . At what time is the
velocity of the particle equal to zero? Is this reasonable?


3.3 Average and Instantaneous Acceleration
37. A cheetah can accelerate from rest to a speed of 30.0m/s in 7.00 s. What is its acceleration?
38. Dr. John Paul Stapp was a U.S. Air Force officer whostudied the effects of extreme acceleration on the humanbody. On December 10, 1954, Stapp rode a rocket sled,accelerating from rest to a top speed of 282 m/s (1015km/h) in 5.00 s and was brought jarringly back to rest inonly 1.40 s. Calculate his (a) acceleration in his directionof motion and (b) acceleration opposite to his directionof motion. Express each in multiples of g (9.80 m/s2) bytaking its ratio to the acceleration of gravity.
39. Sketch the acceleration-versus-time graph from thefollowing velocity-versus-time graph.


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40. A commuter backs her car out of her garage with anacceleration of 1.40 m/s2. (a) How long does it take her toreach a speed of 2.00 m/s? (b) If she then brakes to a stopin 0.800 s, what is her acceleration?
41. Assume an intercontinental ballistic missile goes fromrest to a suborbital speed of 6.50 km/s in 60.0 s (the actualspeed and time are classified). What is its averageacceleration in meters per second and in multiples of g(9.80 m/s2)?
42. An airplane, starting from rest, moves down therunway at constant acceleration for 18 s and then takes offat a speed of 60 m/s. What is the average acceleration of theplane?


3.4 Motion with Constant Acceleration
43. A particle moves in a straight line at a constantvelocity of 30 m/s. What is its displacement between t = 0and t = 5.0 s?
44. A particle moves in a straight line with an initialvelocity of 30 m/s and a constant acceleration of 30 m/s2. Ifat t = 0, x = 0 and v = 0 , what is the particle’s position
at t = 5 s?
45. A particle moves in a straight line with an initialvelocity of 30 m/s and constant acceleration 30 m/s2. (a)What is its displacement at t = 5 s? (b) What is its velocityat this same time?
46. (a) Sketch a graph of velocity versus timecorresponding to the graph of displacement versus timegiven in the following figure. (b) Identify the time or times(ta, tb, tc, etc.) at which the instantaneous velocity has thegreatest positive value. (c) At which times is it zero? (d) At


which times is it negative?


47. (a) Sketch a graph of acceleration versus timecorresponding to the graph of velocity versus time given inthe following figure. (b) Identify the time or times (ta, tb,tc, etc.) at which the acceleration has the greatest positivevalue. (c) At which times is it zero? (d) At which times is itnegative?


48. A particle has a constant acceleration of 6.0 m/s2.(a) If its initial velocity is 2.0 m/s, at what time is itsdisplacement 5.0 m? (b) What is its velocity at that time?
49. At t = 10 s, a particle is moving from left to rightwith a speed of 5.0 m/s. At t = 20 s, the particle is movingright to left with a speed of 8.0 m/s. Assuming the particle’sacceleration is constant, determine (a) its acceleration, (b)its initial velocity, and (c) the instant when its velocity iszero.
50. A well-thrown ball is caught in a well-padded mitt. If
the acceleration of the ball is 2.10 × 104 m/s2 , and 1.85
ms (1 ms = 10−3 s) elapses from the time the ball first
touches the mitt until it stops, what is the initial velocity ofthe ball?
51. A bullet in a gun is accelerated from the firingchamber to the end of the barrel at an average rate of
6.20 × 105 m/s2 for 8.10 × 10−4 s . What is its muzzle
velocity (that is, its final velocity)?


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52. (a) A light-rail commuter train accelerates at a rate of1.35 m/s2. How long does it take to reach its top speed of80.0 km/h, starting from rest? (b) The same train ordinarilydecelerates at a rate of 1.65 m/s2. How long does it taketo come to a stop from its top speed? (c) In emergencies,the train can decelerate more rapidly, coming to rest from80.0 km/h in 8.30 s. What is its emergency acceleration inmeters per second squared?
53. While entering a freeway, a car accelerates from restat a rate of 2.04 m/s2 for 12.0 s. (a) Draw a sketch of thesituation. (b) List the knowns in this problem. (c) Howfar does the car travel in those 12.0 s? To solve this part,first identify the unknown, then indicate how you chosethe appropriate equation to solve for it. After choosingthe equation, show your steps in solving for the unknown,check your units, and discuss whether the answer isreasonable. (d) What is the car’s final velocity? Solve forthis unknown in the same manner as in (c), showing allsteps explicitly.
54. Unreasonable results At the end of a race, a runnerdecelerates from a velocity of 9.00 m/s at a rate of 2.00 m/s2. (a) How far does she travel in the next 5.00 s? (b) Whatis her final velocity? (c) Evaluate the result. Does it makesense?
55. Blood is accelerated from rest to 30.0 cm/s in adistance of 1.80 cm by the left ventricle of the heart. (a)Make a sketch of the situation. (b) List the knowns inthis problem. (c) How long does the acceleration take? Tosolve this part, first identify the unknown, then discuss howyou chose the appropriate equation to solve for it. Afterchoosing the equation, show your steps in solving for theunknown, checking your units. (d) Is the answer reasonablewhen compared with the time for a heartbeat?
56. During a slap shot, a hockey player accelerates thepuck from a velocity of 8.00 m/s to 40.0 m/s in the same
direction. If this shot takes 3.33 × 10−2 s , what is the
distance over which the puck accelerates?
57. A powerful motorcycle can accelerate from rest to26.8 m/s (100 km/h) in only 3.90 s. (a) What is its averageacceleration? (b) How far does it travel in that time?
58. Freight trains can produce only relatively smallaccelerations. (a) What is the final velocity of a freight train
that accelerates at a rate of 0.0500 m/s2 for 8.00 min,
starting with an initial velocity of 4.00 m/s? (b) If the train
can slow down at a rate of 0.550 m/s2 , how long will it
take to come to a stop from this velocity? (c) How far willit travel in each case?
59. A fireworks shell is accelerated from rest to a velocityof 65.0 m/s over a distance of 0.250 m. (a) Calculate the


acceleration. (b) How long did the acceleration last?
60. A swan on a lake gets airborne by flapping its wingsand running on top of the water. (a) If the swan must reacha velocity of 6.00 m/s to take off and it accelerates from
rest at an average rate of 0.35 m/s2 , how far will it travel
before becoming airborne? (b) How long does this take?
61. Awoodpecker’s brain is specially protected from largeaccelerations by tendon-like attachments inside the skull.While pecking on a tree, the woodpecker’s head comes toa stop from an initial velocity of 0.600 m/s in a distanceof only 2.00 mm. (a) Find the acceleration in meters persecond squared and in multiples of g, where g = 9.80 m/s2. (b) Calculate the stopping time. (c) The tendons cradlingthe brain stretch, making its stopping distance 4.50 mm(greater than the head and, hence, less acceleration of thebrain). What is the brain’s acceleration, expressed inmultiples of g?
62. An unwary football player collides with a paddedgoalpost while running at a velocity of 7.50 m/s and comesto a full stop after compressing the padding and his body0.350 m. (a) What is his acceleration? (b) How long doesthe collision last?
63. A care package is dropped out of a cargo plane andlands in the forest. If we assume the care package speed onimpact is 54 m/s (123 mph), then what is its acceleration?Assume the trees and snow stops it over a distance of 3.0m.
64. An express train passes through a station. It enterswith an initial velocity of 22.0 m/s and decelerates at a rate
of 0.150 m/s2 as it goes through. The station is 210.0 m
long. (a) How fast is it going when the nose leaves thestation? (b) How long is the nose of the train in the station?(c) If the train is 130 m long, what is the velocity of the endof the train as it leaves? (d) When does the end of the trainleave the station?
65. Unreasonable results Dragsters can actually reach atop speed of 145.0 m/s in only 4.45 s. (a) Calculate theaverage acceleration for such a dragster. (b) Find the finalvelocity of this dragster starting from rest and acceleratingat the rate found in (a) for 402.0 m (a quarter mile) withoutusing any information on time. (c) Why is the final velocitygreater than that used to find the average acceleration?(Hint: Consider whether the assumption of constantacceleration is valid for a dragster. If not, discuss whetherthe acceleration would be greater at the beginning or endof the run and what effect that would have on the finalvelocity.)


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3.5 Free Fall
66. Calculate the displacement and velocity at times of(a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ballthrown straight up with an initial velocity of 15.0 m/s. Takethe point of release to be y0 = 0 .
67. Calculate the displacement and velocity at times of (a)0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s fora rock thrown straight down with an initial velocity of 14.0m/s from the Verrazano Narrows Bridge in New York City.The roadway of this bridge is 70.0 m above the water.
68. A basketball referee tosses the ball straight up for thestarting tip-off. At what velocity must a basketball playerleave the ground to rise 1.25 m above the floor in anattempt to get the ball?
69. A rescue helicopter is hovering over a person whoseboat has sunk. One of the rescuers throws a life preserverstraight down to the victim with an initial velocity of 1.40m/s and observes that it takes 1.8 s to reach the water. (a)List the knowns in this problem. (b) How high above thewater was the preserver released? Note that the downdraftof the helicopter reduces the effects of air resistance on thefalling life preserver, so that an acceleration equal to that ofgravity is reasonable.
70. Unreasonable results A dolphin in an aquatic showjumps straight up out of the water at a velocity of 15.0m/s. (a) List the knowns in this problem. (b) How highdoes his body rise above the water? To solve this part, firstnote that the final velocity is now a known, and identifyits value. Then, identify the unknown and discuss howyou chose the appropriate equation to solve for it. Afterchoosing the equation, show your steps in solving for theunknown, checking units, and discuss whether the answeris reasonable. (c) How long a time is the dolphin in the air?Neglect any effects resulting from his size or orientation.
71. A diver bounces straight up from a diving board,avoiding the diving board on the way down, and falls feetfirst into a pool. She starts with a velocity of 4.00 m/s andher takeoff point is 1.80 m above the pool. (a) What is herhighest point above the board? (b) How long a time are herfeet in the air? (c) What is her velocity when her feet hit thewater?
72. (a) Calculate the height of a cliff if it takes 2.35 s fora rock to hit the ground when it is thrown straight up fromthe cliff with an initial velocity of 8.00 m/s. (b) How long atime would it take to reach the ground if it is thrown straightdown with the same speed?
73. A very strong, but inept, shot putter puts the shotstraight up vertically with an initial velocity of 11.0 m/s.


How long a time does he have to get out of the way if theshot was released at a height of 2.20 m and he is 1.80 mtall?
74. You throw a ball straight up with an initial velocity of15.0 m/s. It passes a tree branch on the way up at a heightof 7.0 m. How much additional time elapses before the ballpasses the tree branch on the way back down?
75. A kangaroo can jump over an object 2.50 m high. (a)Considering just its vertical motion, calculate its verticalspeed when it leaves the ground. (b) How long a time is itin the air?
76. Standing at the base of one of the cliffs of Mt. Arapilesin Victoria, Australia, a hiker hears a rock break loose froma height of 105.0 m. He can’t see the rock right away, butthen does, 1.50 s later. (a) How far above the hiker is therock when he can see it? (b) How much time does he haveto move before the rock hits his head?
77. There is a 250-m-high cliff at Half Dome in YosemiteNational Park in California. Suppose a boulder breaks loosefrom the top of this cliff. (a) How fast will it be going whenit strikes the ground? (b) Assuming a reaction time of 0.300s, how long a time will a tourist at the bottom have to getout of the way after hearing the sound of the rock breakingloose (neglecting the height of the tourist, which wouldbecome negligible anyway if hit)? The speed of sound is335.0 m/s on this day.


3.6 Finding Velocity and Displacement from
Acceleration
78. The acceleration of a particle varies with time
according to the equation a(t) = pt2 − qt3 . Initially, the
velocity and position are zero. (a) What is the velocity as afunction of time? (b) What is the position as a function oftime?
79. Between t = 0 and t = t0, a rocket moves straightupward with an acceleration given by a(t) = A − Bt1 /2 ,
where A and B are constants. (a) If x is in meters and t isin seconds, what are the units of A and B? (b) If the rocketstarts from rest, how does the velocity vary between t =0 and t = t0? (c) If its initial position is zero, what is therocket’s position as a function of time during this same timeinterval?
80. The velocity of a particle moving along the x-axis
varies with time according to v(t) = A + Bt−1 , where A =
2 m/s, B = 0.25 m, and 1.0 s ≤ t ≤ 8.0 s . Determine the
acceleration and position of the particle at t = 2.0 s and t =5.0 s. Assume that x(t = 1 s) = 0 .


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81. A particle at rest leaves the origin with its velocityincreasing with time according to v(t) = 3.2t m/s. At 5.0 s,the particle’s velocity starts decreasing according to [16.0– 1.5(t – 5.0)] m/s. This decrease continues until t = 11.0


s, after which the particle’s velocity remains constant at 7.0m/s. (a) What is the acceleration of the particle as a functionof time? (b) What is the position of the particle at t = 2.0 s,t = 7.0 s, and t = 12.0 s?


ADDITIONAL PROBLEMS
82. Professional baseball player Nolan Ryan could pitcha baseball at approximately 160.0 km/h. At that averagevelocity, how long did it take a ball thrown by Ryan toreach home plate, which is 18.4 m from the pitcher’smound? Compare this with the average reaction time of ahuman to a visual stimulus, which is 0.25 s.
83. An airplane leaves Chicago and makes the 3000-kmtrip to Los Angeles in 5.0 h. A second plane leaves Chicagoone-half hour later and arrives in Los Angeles at the sametime. Compare the average velocities of the two planes.Ignore the curvature of Earth and the difference in altitudebetween the two cities.
84. Unreasonable Results A cyclist rides 16.0 km east,then 8.0 km west, then 8.0 km east, then 32.0 km west,and finally 11.2 km east. If his average velocity is 24 km/h, how long did it take him to complete the trip? Is this areasonable time?
85. An object has an acceleration of +1.2 cm/s2 . At
t = 4.0 s , its velocity is −3.4 cm/s . Determine the
object’s velocities at t = 1.0 s and t = 6.0 s .
86. A particle moves along the x-axis according to the
equation x(t) = 2.0 − 4.0t2 m. What are the velocity and
acceleration at t = 2.0 s and t = 5.0 s?
87. A particle moving at constant acceleration hasvelocities of 2.0 m/s at t = 2.0 s and −7.6 m/s at
t = 5.2 s. What is the acceleration of the particle?
88. A train is moving up a steep grade at constant velocity(see following figure) when its caboose breaks loose andstarts rolling freely along the track. After 5.0 s, the cabooseis 30 m behind the train. What is the acceleration of thecaboose?


89. An electron is moving in a straight line with a velocity
of 4.0 × 105 m/s. It enters a region 5.0 cm long where it
undergoes an acceleration of 6.0 × 1012 m/s2 along the
same straight line. (a) What is the electron’s velocity whenit emerges from this region? b) How long does the electrontake to cross the region?
90. An ambulance driver is rushing a patient to thehospital. While traveling at 72 km/h, she notices the trafficlight at the upcoming intersections has turned amber. Toreach the intersection before the light turns red, she musttravel 50 m in 2.0 s. (a) What minimum acceleration mustthe ambulance have to reach the intersection before thelight turns red? (b) What is the speed of the ambulancewhen it reaches the intersection?
91. A motorcycle that is slowing down uniformly covers2.0 successive km in 80 s and 120 s, respectively. Calculate(a) the acceleration of the motorcycle and (b) its velocity atthe beginning and end of the 2-km trip.
92. A cyclist travels from point A to point B in 10 min.During the first 2.0 min of her trip, she maintains a uniform
acceleration of 0.090 m/s2 . She then travels at constant
velocity for the next 5.0 min. Next, she decelerates at aconstant rate so that she comes to a rest at point B 3.0 minlater. (a) Sketch the velocity-versus-time graph for the trip.(b) What is the acceleration during the last 3 min? (c) Howfar does the cyclist travel?
93. Two trains are moving at 30 m/s in opposite directionson the same track. The engineers see simultaneously thatthey are on a collision course and apply the brakes whenthey are 1000 m apart. Assuming both trains have the sameacceleration, what must this acceleration be if the trains areto stop just short of colliding?
94. A 10.0-m-long truck moving with a constant velocityof 97.0 km/h passes a 3.0-m-long car moving with aconstant velocity of 80.0 km/h. How much time elapsesbetween the moment the front of the truck is even with theback of the car and the moment the back of the truck is evenwith the front of the car?


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95. A police car waits in hiding slightly off the highway.A speeding car is spotted by the police car doing 40 m/s. Atthe instant the speeding car passes the police car, the policecar accelerates from rest at 4 m/s2 to catch the speeding car.How long does it take the police car to catch the speedingcar?
96. Pablo is running in a half marathon at a velocity of3 m/s. Another runner, Jacob, is 50 meters behind Pablowith the same velocity. Jacob begins to accelerate at 0.05m/s2. (a) How long does it take Jacob to catch Pablo? (b)What is the distance covered by Jacob? (c) What is the finalvelocity of the Jacob?
97. Unreasonable results A runner approaches the finishline and is 75 m away; her average speed at this position is8 m/s. She decelerates at this point at 0.5 m/s2. How longdoes it take her to cross the finish line from 75 m away? Isthis reasonable?
98. An airplane accelerates at 5.0 m/s2 for 30.0 s. Duringthis time, it covers a distance of 10.0 km. What are theinitial and final velocities of the airplane?
99. Compare the distance traveled of an object thatundergoes a change in velocity that is twice its initialvelocity with an object that changes its velocity by fourtimes its initial velocity over the same time period. Theaccelerations of both objects are constant.


100. An object is moving east with a constant velocity andis at position x0 at time t0 = 0 . (a) With what acceleration
must the object have for its total displacement to be zero ata later time t ? (b) What is the physical interpretation of thesolution in the case for t → ∞ ?
101. A ball is thrown straight up. It passes a 2.00-m-highwindow 7.50 m off the ground on its path up and takes1.30 s to go past the window. What was the ball’s initialvelocity?
102. A coin is dropped from a hot-air balloon that is 300m above the ground and rising at 10.0 m/s upward. For thecoin, find (a) the maximum height reached, (b) its positionand velocity 4.00 s after being released, and (c) the timebefore it hits the ground.
103. A soft tennis ball is dropped onto a hard floor froma height of 1.50 m and rebounds to a height of 1.10 m.(a) Calculate its velocity just before it strikes the floor. (b)Calculate its velocity just after it leaves the floor on its wayback up. (c) Calculate its acceleration during contact with
the floor if that contact lasts 3.50 ms (3.50 × 10−3 s) (d)
How much did the ball compress during its collision withthe floor, assuming the floor is absolutely rigid?
104. Unreasonable results. A raindrop falls from a cloud100 m above the ground. Neglect air resistance. What isthe speed of the raindrop when it hits the ground? Is this areasonable number?
105. Compare the time in the air of a basketball playerwho jumps 1.0 m vertically off the floor with that of aplayer who jumps 0.3 m vertically.
106. Suppose that a person takes 0.5 s to react and movehis hand to catch an object he has dropped. (a) How far
does the object fall on Earth, where g = 9.8 m/s2? (b)
How far does the object fall on the Moon, where theacceleration due to gravity is 1/6 of that on Earth?
107. A hot-air balloon rises from ground level at aconstant velocity of 3.0 m/s. One minute after liftoff, asandbag is dropped accidentally from the balloon.Calculate (a) the time it takes for the sandbag to reach theground and (b) the velocity of the sandbag when it hits theground.
108. (a) A world record was set for the men’s 100-m dashin the 2008 Olympic Games in Beijing by Usain Bolt ofJamaica. Bolt “coasted” across the finish line with a timeof 9.69 s. If we assume that Bolt accelerated for 3.00 s toreach his maximum speed, and maintained that speed forthe rest of the race, calculate his maximum speed and hisacceleration. (b) During the same Olympics, Bolt also set


Chapter 3 | Motion Along a Straight Line 155




the world record in the 200-m dash with a time of 19.30 s.Using the same assumptions as for the 100-m dash, whatwas his maximum speed for this race?
109. An object is dropped from a height of 75.0 m aboveground level. (a) Determine the distance traveled during thefirst second. (b) Determine the final velocity at which theobject hits the ground. (c) Determine the distance traveledduring the last second of motion before hitting the ground.
110. A steel ball is dropped onto a hard floor from a heightof 1.50 m and rebounds to a height of 1.45 m. (a) Calculate


its velocity just before it strikes the floor. (b) Calculate itsvelocity just after it leaves the floor on its way back up.(c) Calculate its acceleration during contact with the floor
if that contact lasts 0.0800 ms (8.00 × 10−5 s) (d) How
much did the ball compress during its collision with thefloor, assuming the floor is absolutely rigid?
111. An object is dropped from a roof of a building ofheight h. During the last second of its descent, it drops adistance h/3. Calculate the height of the building.


CHALLENGE PROBLEMS
112. In a 100-m race, the winner is timed at 11.2 s. Thesecond-place finisher’s time is 11.6 s. How far is thesecond-place finisher behind the winner when she crossesthe finish line? Assume the velocity of each runner isconstant throughout the race.
113. The position of a particle moving along the x-axis
varies with time according to x(t) = 5.0t2 − 4.0t3 m. Find
(a) the velocity and acceleration of the particle as functionsof time, (b) the velocity and acceleration at t = 2.0 s, (c) thetime at which the position is a maximum, (d) the time atwhich the velocity is zero, and (e) the maximum position.
114. A cyclist sprints at the end of a race to clinch avictory. She has an initial velocity of 11.5 m/s andaccelerates at a rate of 0.500 m/s2 for 7.00 s. (a) What is herfinal velocity? (b) The cyclist continues at this velocity to


the finish line. If she is 300 m from the finish line when shestarts to accelerate, how much time did she save? (c) Thesecond-place winner was 5.00 m ahead when the winnerstarted to accelerate, but he was unable to accelerate, andtraveled at 11.8 m/s until the finish line. What was thedifference in finish time in seconds between the winnerand runner-up? How far back was the runner-up when thewinner crossed the finish line?
115. In 1967, New Zealander Burt Munro set the worldrecord for an Indian motorcycle, on the Bonneville SaltFlats in Utah, of 295.38 km/h. The one-way course was8.00 km long. Acceleration rates are often described by thetime it takes to reach 96.0 km/h from rest. If this time was4.00 s and Burt accelerated at this rate until he reached hismaximum speed, how long did it take Burt to complete thecourse?


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4 | MOTION IN TWO ANDTHREE DIMENSIONS


Figure 4.1 The Red Arrows is the aerobatics display team of Britain’s Royal Air Force. Based in Lincolnshire, England, theyperform precision flying shows at high speeds, which requires accurate measurement of position, velocity, and acceleration inthree dimensions. (credit: modification of work by Phil Long)


Chapter Outline
4.1 Displacement and Velocity Vectors
4.2 Acceleration Vector
4.3 Projectile Motion
4.4 Uniform Circular Motion
4.5 Relative Motion in One and Two Dimensions


Introduction
To give a complete description of kinematics, we must explore motion in two and three dimensions. After all, most objectsin our universe do not move in straight lines; rather, they follow curved paths. From kicked footballs to the flight paths ofbirds to the orbital motions of celestial bodies and down to the flow of blood plasma in your veins, most motion followscurved trajectories.
Fortunately, the treatment of motion in one dimension in the previous chapter has given us a foundation on which to build,as the concepts of position, displacement, velocity, and acceleration defined in one dimension can be expanded to two andthree dimensions. Consider the Red Arrows, also known as the Royal Air Force Aerobatic team of the United Kingdom.Each jet follows a unique curved trajectory in three-dimensional airspace, as well as has a unique velocity and acceleration.Thus, to describe the motion of any of the jets accurately, we must assign to each jet a unique position vector in threedimensions as well as a unique velocity and acceleration vector. We can apply the same basic equations for displacement,velocity, and acceleration we derived inMotion Along a Straight Line to describe the motion of the jets in two and threedimensions, but with some modifications—in particular, the inclusion of vectors.
In this chapter we also explore two special types of motion in two dimensions: projectile motion and circular motion. Last,we conclude with a discussion of relative motion. In the chapter-opening picture, each jet has a relative motion with respect


Chapter 4 | Motion in Two and Three Dimensions 157




to any other jet in the group or to the people observing the air show on the ground.
4.1 | Displacement and Velocity Vectors


Learning Objectives
By the end of this section, you will be able to:
• Calculate position vectors in a multidimensional displacement problem.
• Solve for the displacement in two or three dimensions.
• Calculate the velocity vector given the position vector as a function of time.
• Calculate the average velocity in multiple dimensions.


Displacement and velocity in two or three dimensions are straightforward extensions of the one-dimensional definitions.However, now they are vector quantities, so calculations with them have to follow the rules of vector algebra, not scalaralgebra.
Displacement Vector
To describe motion in two and three dimensions, we must first establish a coordinate system and a convention for the axes.We generally use the coordinates x, y, and z to locate a particle at point P(x, y, z) in three dimensions. If the particle ismoving, the variables x, y, and z are functions of time (t):


(4.1)x = x(t) y = y(t) z = z(t).
The position vector from the origin of the coordinate system to point P is r→ (t). In unit vector notation, introduced in
Coordinate Systems and Components of a Vector, r→ (t) is


(4.2)
r→ (t) = x(t) i


^
+ y(t) j


^
+ z(t)k


^
.


Figure 4.2 shows the coordinate system and the vector to point P, where a particle could be located at a particular timet. Note the orientation of the x, y, and z axes. This orientation is called a right-handed coordinate system (CoordinateSystems and Components of a Vector) and it is used throughout the chapter.


Figure 4.2 A three-dimensional coordinate system with aparticle at position P(x(t), y(t), z(t)).


With our definition of the position of a particle in three-dimensional space, we can formulate the three-dimensional
displacement. Figure 4.3 shows a particle at time t1 located at P1 with position vector r→ (t1). At a later time t2, the


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particle is located at P2 with position vector r→ (t2) . The displacement vector Δ r→ is found by subtracting r→ (t1)
from r→ (t2)  :


(4.3)Δ r→ = r→ (t2) − r→ (t1).


Vector addition is discussed in Vectors. Note that this is the same operation we did in one dimension, but now the vectorsare in three-dimensional space.


Figure 4.3 The displacement Δ r→ = r→ (t2) − r→ (t1) is
the vector from P1 to P2 .


The following examples illustrate the concept of displacement in multiple dimensions.
Example 4.1


Polar Orbiting Satellite
A satellite is in a circular polar orbit around Earth at an altitude of 400 km—meaning, it passes directly overheadat the North and South Poles. What is the magnitude and direction of the displacement vector from when it isdirectly over the North Pole to when it is at −45° latitude?
Strategy
We make a picture of the problem to visualize the solution graphically. This will aid in our understanding of thedisplacement. We then use unit vectors to solve for the displacement.
Solution
Figure 4.4 shows the surface of Earth and a circle that represents the orbit of the satellite. Although satellites aremoving in three-dimensional space, they follow trajectories of ellipses, which can be graphed in two dimensions.The position vectors are drawn from the center of Earth, which we take to be the origin of the coordinate system,with the y-axis as north and the x-axis as east. The vector between them is the displacement of the satellite. Wetake the radius of Earth as 6370 km, so the length of each position vector is 6770 km.


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Figure 4.4 Two position vectors are drawn from the center of Earth,which is the origin of the coordinate system, with the y-axis as north and thex-axis as east. The vector between them is the displacement of the satellite.


In unit vector notation, the position vectors are
r→ (t1) = 6770. km j


^


r→ (t2) = 6770. km (cos 45°) i
^


+ 6770. km (sin(−45°)) j
^
.


Evaluating the sine and cosine, we have


r→ (t1) = 6770. j
^


r→ (t2) = 4787 i
^


− 4787 j
^
.


Now we can find Δ r→ , the displacement of the satellite:
Δ r→ = r→ (t2) − r


→ (t1) = 4787 i
^


− 11,557 j
^
.


The magnitude of the displacement is |Δ r→ | = (4787)2 + (−11,557)2 = 12,509 km. The angle the displacement makes
with the x-axis is θ = tan−1 ⎛⎝−11,5574787 ⎞⎠ = −67.5°.
Significance
Plotting the displacement gives information and meaning to the unit vector solution to the problem. When plotting thedisplacement, we need to include its components as well as its magnitude and the angle it makes with a chosen axis—in thiscase, the x-axis (Figure 4.5).


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Figure 4.5 Displacement vector with components, angle, and magnitude.


Note that the satellite took a curved path along its circular orbit to get from its initial position to its final position in thisexample. It also could have traveled 4787 km east, then 11,557 km south to arrive at the same location. Both of these pathsare longer than the length of the displacement vector. In fact, the displacement vector gives the shortest path between twopoints in one, two, or three dimensions.
Many applications in physics can have a series of displacements, as discussed in the previous chapter. The totaldisplacement is the sum of the individual displacements, only this time, we need to be careful, because we are addingvectors. We illustrate this concept with an example of Brownian motion.
Example 4.2


Brownian Motion
Brownian motion is a chaotic random motion of particles suspended in a fluid, resulting from collisions withthe molecules of the fluid. This motion is three-dimensional. The displacements in numerical order of a particleundergoing Brownian motion could look like the following, in micrometers (Figure 4.6):


Δ r→ 1 = 2.0 i
^


+ j
^


+ 3.0k
^


Δ r→ 2 = − i
^


+ 3.0k
^


Δ r→ 3 = 4.0 i
^


− 2.0 j
^


+ k
^


Δ r→ 4 = −3.0 i
^


+ j
^


+ 2.0k
^
.


What is the total displacement of the particle from the origin?


Chapter 4 | Motion in Two and Three Dimensions 161




Figure 4.6 Trajectory of a particle undergoing randomdisplacements of Brownian motion. The total displacement isshown in red.


Solution
We form the sum of the displacements and add them as vectors:


Δ r→ Total = ∑ Δ r→ i = Δ r→ 1 + Δ r→ 2 + Δ r→ 3 + Δ r→ 4


= (2.0 − 1.0 + 4.0 − 3.0) i
^


+ (1.0 + 0 − 2.0 + 1.0) j
^


+ (3.0 + 3.0 + 1.0 + 2.0)k
^


= 2.0 i
^


+ 0 j
^


+ 9.0k
^
µm.


To complete the solution, we express the displacement as a magnitude and direction,
|Δ r→ Total| = 2.02 + 02 + 9.02 = 9.2 µm, θ = tan−1 ⎛⎝92



⎠ = 77°,


with respect to the x-axis in the xz-plane.
Significance
From the figure we can see the magnitude of the total displacement is less than the sum of the magnitudes of theindividual displacements.


Velocity Vector
In the previous chapter we found the instantaneous velocity by calculating the derivative of the position function withrespect to time. We can do the same operation in two and three dimensions, but we use vectors. The instantaneous velocityvector is now


(4.4)
v→ (t) = lim


Δt → 0


r→ (t + Δt) − r→ (t)
Δt


= d r


dt
.


Let’s look at the relative orientation of the position vector and velocity vector graphically. In Figure 4.7 we show the
vectors r→ (t) and r→ (t + Δt), which give the position of a particle moving along a path represented by the gray line.
As Δt goes to zero, the velocity vector, given by Equation 4.4, becomes tangent to the path of the particle at time t.


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Figure 4.7 A particle moves along a path given by the grayline. In the limit as Δt approaches zero, the velocity vector
becomes tangent to the path of the particle.


Equation 4.4 can also be written in terms of the components of v→ (t). Since
r→ (t) = x(t) i


^
+ y(t) j


^
+ z(t)k


^
,


we can write


(4.5)
v→ (t) = vx(t) i


^
+ vy(t) j


^
+ vz(t)k


^


where


(4.6)vx(t) = dx(t)dt , vy(t) = dy(t)dt , vz(t) = dz(t)dt .


If only the average velocity is of concern, we have the vector equivalent of the one-dimensional average velocity for twoand three dimensions:


(4.7)
v→ avg =


r→ (t2) − r
→ (t1)


t2 − t1
.


Example 4.3
Calculating the Velocity Vector
The position function of a particle is r→ (t) = 2.0t2 i^ + (2.0 + 3.0t) j^ + 5.0tk^m. (a) What is the
instantaneous velocity and speed at t = 2.0 s? (b) What is the average velocity between 1.0 s and 3.0 s?
Solution
Using Equation 4.5 and Equation 4.6, and taking the derivative of the position function with respect to time,we find


Chapter 4 | Motion in Two and Three Dimensions 163




4.1


(a) v(t) = dr(t)
dt


= 4.0t i
^


+ 3.0 j
^


+ 5.0k
^
m/s


v→ (2.0s) = 8.0 i
^


+ 3.0 j
^


+ 5.0k
^
m/s


Speed | v→ (2.0 s)| = 82 + 32 + 52 = 9.9 m/s.
(b) From Equation 4.7,
v→ avg =


r→ (t2) − r
→ (t1)


t2 − t1
= r


→ (3.0 s) − r→ (1.0 s)
3.0 s − 1.0 s


=
(18 i


^
+ 11 j


^
+ 15k


^
) m − (2 i


^
+ 5 j


^
+ 5k


^
) m


2.0 s


=
(16 i


^
+ 6 j


^
+ 10k


^
) m


2.0 s
= 8.0 i


^
+ 3.0 j


^
+ 5.0k


^
m/s.


Significance
We see the average velocity is the same as the instantaneous velocity at t = 2.0 s, as a result of the velocity functionbeing linear. This need not be the case in general. In fact, most of the time, instantaneous and average velocitiesare not the same.


Check Your Understanding The position function of a particle is r→ (t) = 3.0t3 i^ + 4.0 j^ . (a) What is
the instantaneous velocity at t = 3 s? (b) Is the average velocity between 2 s and 4 s equal to the instantaneousvelocity at t = 3 s?


The Independence of Perpendicular Motions
When we look at the three-dimensional equations for position and velocity written in unit vector notation, Equation 4.2and Equation 4.5, we see the components of these equations are separate and unique functions of time that do not dependon one another. Motion along the x direction has no part of its motion along the y and z directions, and similarly for the othertwo coordinate axes. Thus, the motion of an object in two or three dimensions can be divided into separate, independentmotions along the perpendicular axes of the coordinate system in which the motion takes place.
To illustrate this concept with respect to displacement, consider a woman walking from point A to point B in a city withsquare blocks. The woman taking the path from A to B may walk east for so many blocks and then north (two perpendiculardirections) for another set of blocks to arrive at B. How far she walks east is affected only by her motion eastward. Similarly,how far she walks north is affected only by her motion northward.


Independence of Motion
In the kinematic description of motion, we are able to treat the horizontal and vertical components of motion separately.In many cases, motion in the horizontal direction does not affect motion in the vertical direction, and vice versa.


An example illustrating the independence of vertical and horizontal motions is given by two baseballs. One baseball isdropped from rest. At the same instant, another is thrown horizontally from the same height and it follows a curved path. Astroboscope captures the positions of the balls at fixed time intervals as they fall (Figure 4.8).


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Figure 4.8 A diagram of the motions of two identical balls:one falls from rest and the other has an initial horizontalvelocity. Each subsequent position is an equal time interval.Arrows represent the horizontal and vertical velocities at eachposition. The ball on the right has an initial horizontal velocitywhereas the ball on the left has no horizontal velocity. Despitethe difference in horizontal velocities, the vertical velocities andpositions are identical for both balls, which shows the verticaland horizontal motions are independent.


It is remarkable that for each flash of the strobe, the vertical positions of the two balls are the same. This similarity impliesvertical motion is independent of whether the ball is moving horizontally. (Assuming no air resistance, the vertical motionof a falling object is influenced by gravity only, not by any horizontal forces.) Careful examination of the ball thrownhorizontally shows it travels the same horizontal distance between flashes. This is because there are no additional forces onthe ball in the horizontal direction after it is thrown. This result means horizontal velocity is constant and is affected neitherby vertical motion nor by gravity (which is vertical). Note this case is true for ideal conditions only. In the real world, airresistance affects the speed of the balls in both directions.
The two-dimensional curved path of the horizontally thrown ball is composed of two independent one-dimensionalmotions (horizontal and vertical). The key to analyzing such motion, called projectile motion, is to resolve it into motionsalong perpendicular directions. Resolving two-dimensional motion into perpendicular components is possible because thecomponents are independent.


Chapter 4 | Motion in Two and Three Dimensions 165




4.2 | Acceleration Vector
Learning Objectives


By the end of this section, you will be able to:
• Calculate the acceleration vector given the velocity function in unit vector notation.
• Describe the motion of a particle with a constant acceleration in three dimensions.
• Use the one-dimensional motion equations along perpendicular axes to solve a problem in twoor three dimensions with a constant acceleration.
• Express the acceleration in unit vector notation.


Instantaneous Acceleration
In addition to obtaining the displacement and velocity vectors of an object in motion, we often want to know its accelerationvector at any point in time along its trajectory. This acceleration vector is the instantaneous acceleration and it can beobtained from the derivative with respect to time of the velocity function, as we have seen in a previous chapter. The onlydifference in two or three dimensions is that these are now vector quantities. Taking the derivative with respect to time
v→ (t), we find


(4.8)
a→ (t) = lim


t → 0


v→ (t + Δt) − v→ (t)
Δt


= d v
→ (t)
dt


.


The acceleration in terms of components is


(4.9)
a→ (t) = dvx(t)


dt
i
^


+
dvy(t)
dt


j
^


+
dvz(t)
dt


k
^
.


Also, since the velocity is the derivative of the position function, we can write the acceleration in terms of the secondderivative of the position function:


(4.10)
a→ (t) = d


2 x(t)
dt2


i
^


+
d2 y(t)


dt2
j
^


+ d
2 z(t)
dt2


k
^
.


Example 4.4
Finding an Acceleration Vector
A particle has a velocity of v→ (t) = 5.0t i^ + t2 j^ − 2.0t3 k^m/s. (a) What is the acceleration function? (b)
What is the acceleration vector at t = 2.0 s? Find its magnitude and direction.
Solution
(a) We take the first derivative with respect to time of the velocity function to find the acceleration. The derivativeis taken component by component:


a→ (t) = 5.0 i
^


+ 2.0t j
^


− 6.0t2 k
^
m/s2.


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(b) Evaluating a→ (2.0 s) = 5.0 i^ + 4.0 j^ − 24.0k^m/s2 gives us the direction in unit vector notation. The
magnitude of the acceleration is | a→ (2.0 s)| = 5.02 + 4.02 + (−24.0)2 = 24.8 m/s2.
Significance
In this example we find that acceleration has a time dependence and is changing throughout the motion. Let’sconsider a different velocity function for the particle.


Example 4.5
Finding a Particle Acceleration
A particle has a position function r→ (t) = (10t − t2) i^ + 5t j^ + 5tk^m. (a) What is the velocity? (b) What is
the acceleration? (c) Describe the motion from t = 0 s.
Strategy
We can gain some insight into the problem by looking at the position function. It is linear in y and z, so we knowthe acceleration in these directions is zero when we take the second derivative. Also, note that the position in thex direction is zero for t = 0 s and t = 10 s.
Solution
(a) Taking the derivative with respect to time of the position function, we find


v→ (t) = (10 − 2t) i
^


+ 5 j
^


+ 5k
^


m/s.


The velocity function is linear in time in the x direction and is constant in the y and z directions.
(b) Taking the derivative of the velocity function, we find


a→ (t) = −2 i
^


m/s2.


The acceleration vector is a constant in the negative x-direction.
(c) The trajectory of the particle can be seen in Figure 4.9. Let’s look in the y and z directions first. The particle’sposition increases steadily as a function of time with a constant velocity in these directions. In the x direction,however, the particle follows a path in positive x until t = 5 s, when it reverses direction. We know this fromlooking at the velocity function, which becomes zero at this time and negative thereafter. We also know thisbecause the acceleration is negative and constant—meaning, the particle is decelerating, or accelerating in thenegative direction. The particle’s position reaches 25 m, where it then reverses direction and begins to acceleratein the negative x direction. The position reaches zero at t = 10 s.


Chapter 4 | Motion in Two and Three Dimensions 167




4.2


Figure 4.9 The particle starts at point (x, y, z) = (0, 0, 0) with position vector
r→ = 0. The projection of the trajectory onto the xy-plane is shown. The


values of y and z increase linearly as a function of time, whereas x has a turningpoint at t = 5 s and 25 m, when it reverses direction. At this point, the xcomponent of the velocity becomes negative. At t = 10 s, the particle is back to 0m in the x direction.


Significance
By graphing the trajectory of the particle, we can better understand its motion, given by the numerical results ofthe kinematic equations.


Check Your Understanding Suppose the acceleration function has the form
a→ (t) = a i


^
+ b j


^
+ ck


^
m/s2, where a, b, and c are constants. What can be said about the functional form of


the velocity function?


Constant Acceleration
Multidimensional motion with constant acceleration can be treated the same way as shown in the previous chapter forone-dimensional motion. Earlier we showed that three-dimensional motion is equivalent to three one-dimensional motions,each along an axis perpendicular to the others. To develop the relevant equations in each direction, let’s consider the two-dimensional problem of a particle moving in the xy plane with constant acceleration, ignoring the z-component for themoment. The acceleration vector is


a→ = a0x i
^


+ a0y j
^
.


Each component of the motion has a separate set of equations similar to Equation 3.10–Equation 3.14 of the previouschapter on one-dimensional motion. We show only the equations for position and velocity in the x- and y-directions. Asimilar set of kinematic equations could be written for motion in the z-direction:


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(4.11)x(t) = x0 + (vx)avg t
(4.12)vx(t) = v0x + ax t
(4.13)x(t) = x0 + v0x t + 12ax t2
(4.14)vx2(t) = v0x2 + 2ax(x − x0)
(4.15)y(t) = y0 + (vy)avg t
(4.16)vy(t) = v0y + ay t
(4.17)y(t) = y0 + v0y t + 12ay t2
(4.18)vy2(t) = v0y2 + 2ay(y − y0).


Here the subscript 0 denotes the initial position or velocity. Equation 4.11 to Equation 4.18 can be substituted intoEquation 4.2 and Equation 4.5 without the z-component to obtain the position vector and velocity vector as a functionof time in two dimensions:
r→ (t) = x(t) i


^
+ y(t) j


^
and v→ (t) = vx(t) i


^
+ vy(t) j


^
.


The following example illustrates a practical use of the kinematic equations in two dimensions.
Example 4.6


A Skier
Figure 4.10 shows a skier moving with an acceleration of 2.1 m/s2 down a slope of 15° at t = 0. With the
origin of the coordinate system at the front of the lodge, her initial position and velocity are


r→ (0) = (75.0 i
^


− 50.0 j
^
) m


and
v→ (0) = (4.1 i


^
− 1.1 j


^
) m/s.


(a) What are the x- and y-components of the skier’s position and velocity as functions of time? (b) What are herposition and velocity at t = 10.0 s?


Chapter 4 | Motion in Two and Three Dimensions 169




Figure 4.10 A skier has an acceleration of 2.1 m/s2 down a slope of 15°. The origin of the coordinate system is at
the ski lodge.


Strategy
Since we are evaluating the components of the motion equations in the x and y directions, we need to find thecomponents of the acceleration and put them into the kinematic equations. The components of the accelerationare found by referring to the coordinate system in Figure 4.10. Then, by inserting the components of the initialposition and velocity into the motion equations, we can solve for her position and velocity at a later time t.
Solution
(a) The origin of the coordinate system is at the top of the hill with y-axis vertically upward and the x-axishorizontal. By looking at the trajectory of the skier, the x-component of the acceleration is positive and they-component is negative. Since the angle is 15° down the slope, we find


ax = (2.1 m/s
2) cos(15°) = 2.0 m/s2


ay = (−2.1 m/s
2) sin 15° = −0.54 m/s2.


Inserting the initial position and velocity into Equation 4.12 and Equation 4.13 for x, we have
x(t) = 75.0 m + (4.1 m/s)t + 1


2
(2.0 m/s2)t2


vx(t) = 4.1 m/s + (2.0 m/s
2)t.


For y, we have
y(t) = −50.0 m + (−1.1 m/s)t + 1


2
(−0.54 m/s2)t2


vy(t) = −1.1 m/s + (−0.54 m/s
2)t.


(b) Now that we have the equations of motion for x and y as functions of time, we can evaluate them at t = 10.0 s:
x(10.0 s) = 75.0 m + (4.1 m/s2)(10.0 s) + 1


2
(2.0 m/s2)(10.0 s)2 = 216.0 m


vx(10.0 s) = 4.1 m/s + (2.0 m/s
2)(10.0 s) = 24.1m/s


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y(10.0 s) = −50.0 m + (−1.1 m/s)(10.0 s) + 1
2
(−0.54 m/s2)(10.0 s)2 = −88.0 m


vy(10.0 s) = −1.1 m/s + (−0.54 m/s
2)(10.0 s) = −6.5 m/s.


The position and velocity at t = 10.0 s are, finally,
r→ (10.0 s) = (216.0 i


^
− 88.0 j


^
) m


v→ (10.0 s) = (24.1 i
^


− 6.5 j
^
)m/s.


The magnitude of the velocity of the skier at 10.0 s is 25 m/s, which is 60 mi/h.
Significance
It is useful to know that, given the initial conditions of position, velocity, and acceleration of an object, we canfind the position, velocity, and acceleration at any later time.


With Equation 4.8 through Equation 4.10 we have completed the set of expressions for the position, velocity, andacceleration of an object moving in two or three dimensions. If the trajectories of the objects look something like the “RedArrows” in the opening picture for the chapter, then the expressions for the position, velocity, and acceleration can be quitecomplicated. In the sections to follow we examine two special cases of motion in two and three dimensions by looking atprojectile motion and circular motion.
At this University of Colorado Boulder website (https://openstaxcollege.org/l/21phetmotladyb) ,you can explore the position velocity and acceleration of a ladybug with an interactive simulation that allows youto change these parameters.


4.3 | Projectile Motion
Learning Objectives


By the end of this section, you will be able to:
• Use one-dimensional motion in perpendicular directions to analyze projectile motion.
• Calculate the range, time of flight, and maximum height of a projectile that is launched andimpacts a flat, horizontal surface.
• Find the time of flight and impact velocity of a projectile that lands at a different height from thatof launch.
• Calculate the trajectory of a projectile.


Projectile motion is the motion of an object thrown or projected into the air, subject only to acceleration as a result ofgravity. The applications of projectile motion in physics and engineering are numerous. Some examples include meteorsas they enter Earth’s atmosphere, fireworks, and the motion of any ball in sports. Such objects are called projectiles andtheir path is called a trajectory. The motion of falling objects as discussed in Motion Along a Straight Line is a simpleone-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, and our treatment neglects the effects of air resistance.
The most important fact to remember here is that motions along perpendicular axes are independent and thus can beanalyzed separately. We discussed this fact in Displacement and Velocity Vectors, where we saw that vertical andhorizontal motions are independent. The key to analyzing two-dimensional projectile motion is to break it into two motions:one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible because accelerationresulting from gravity is vertical; thus, there is no acceleration along the horizontal axis when air resistance is negligible.)As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. It is not required that we use thischoice of axes; it is simply convenient in the case of gravitational acceleration. In other cases we may choose a different set
of axes. Figure 4.11 illustrates the notation for displacement, where we define s→ to be the total displacement, and x→
and y→ are its component vectors along the horizontal and vertical axes, respectively. The magnitudes of these vectors are


Chapter 4 | Motion in Two and Three Dimensions 171




s, x, and y.


Figure 4.11 The total displacement s of a soccer ball at a point along its path. The vector s→ has
components x→ and y→ along the horizontal and vertical axes. Its magnitude is s and it makes an angle
θ with the horizontal.


To describe projectile motion completely, we must include velocity and acceleration, as well as displacement. We must findtheir components along the x- and y-axes. Let’s assume all forces except gravity (such as air resistance and friction, forexample) are negligible. Defining the positive direction to be upward, the components of acceleration are then very simple:
ay = −g = −9.8 m/s


2 ( − 32 ft/s2).


Because gravity is vertical, ax = 0. If ax = 0, this means the initial velocity in the x direction is equal to the final velocity
in the x direction, or vx = v0x. With these conditions on acceleration and velocity, we can write the kinematic Equation
4.11 through Equation 4.18 for motion in a uniform gravitational field, including the rest of the kinematic equations fora constant acceleration from Motion with Constant Acceleration. The kinematic equations for motion in a uniformgravitational field become kinematic equations with ay = −g, ax = 0 :
Horizontal Motion


(4.19)v0x = vx, x = x0 + vx t
Vertical Motion


(4.20)y = y0 + 12(v0y + vy)t
(4.21)vy = v0y − gt
(4.22)y = y0 + v0y t − 12gt2
(4.23)vy2 = v0y2 − 2g(y − y0)


Using this set of equations, we can analyze projectile motion, keeping in mind some important points.
Problem-Solving Strategy: Projectile Motion


1. Resolve the motion into horizontal and vertical components along the x- and y-axes. The magnitudes of the
components of displacement s→ along these axes are x and y. The magnitudes of the components of velocity


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v→ are vx = vcosθ and vy = vsinθ, where v is the magnitude of the velocity and θ is its direction relative
to the horizontal, as shown in Figure 4.12.


2. Treat the motion as two independent one-dimensional motions: one horizontal and the other vertical. Use thekinematic equations for horizontal and vertical motion presented earlier.
3. Solve for the unknowns in the two separate motions: one horizontal and one vertical. Note that the onlycommon variable between the motions is time t. The problem-solving procedures here are the same as thosefor one-dimensional kinematics and are illustrated in the following solved examples.
4. Recombine quantities in the horizontal and vertical directions to find the total displacement s→ and velocity


v→ . Solve for the magnitude and direction of the displacement and velocity using
s = x2 + y2, θ = tan−1(y/x), v = vx


2 + vy
2,


where θ is the direction of the displacement s→ .


Figure 4.12 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motionsalong the vertical and horizontal axes. (b) The horizontal motion is simple, because ax = 0 and vx is a constant. (c) The
velocity in the vertical direction begins to decrease as the object rises. At its highest point, the vertical velocity is zero. As theobject falls toward Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initialvertical velocity. (d) The x and y motions are recombined to give the total velocity at any given point on the trajectory.


Chapter 4 | Motion in Two and Three Dimensions 173




Example 4.7
A Fireworks Projectile Explodes High and Away
During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0° above
the horizontal, as illustrated in Figure 4.13. The fuse is timed to ignite the shell just as it reaches its highest pointabove the ground. (a) Calculate the height at which the shell explodes. (b) How much time passes between thelaunch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? (d)What is the total displacement from the point of launch to the highest point?


Figure 4.13 The trajectory of a fireworks shell. The fuse is setto explode the shell at the highest point in its trajectory, which isfound to be at a height of 233 m and 125 m away horizontally.


Strategy
The motion can be broken into horizontal and vertical motions in which ax = 0 and ay = −g. We can then
define x0 and y0 to be zero and solve for the desired quantities.
Solution
(a) By “height” we mean the altitude or vertical position y above the starting point. The highest point in anytrajectory, called the apex, is reached when vy = 0. Since we know the initial and final velocities, as well as the
initial position, we use the following equation to find y:


vy
2 = v0y


2 − 2g(y − y0).


Because y0 and vy are both zero, the equation simplifies to
0 = v0y


2 − 2gy.


Solving for y gives
y =


v0y
2


2g
.


Now we must find v0y, the component of the initial velocity in the y direction. It is given by v0y = v0 sinθ0,
where v0 is the initial velocity of 70.0 m/s and θ0 = 75° is the initial angle. Thus,


v0y = v0 sinθ = (70.0 m/s)sin 75° = 67.6 m/s


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and y is
y = (67.6 m/s)


2


2(9.80 m/s2)
.


Thus, we have
y = 233 m.


Note that because up is positive, the initial vertical velocity is positive, as is the maximum height, but theacceleration resulting from gravity is negative. Note also that the maximum height depends only on the verticalcomponent of the initial velocity, so that any projectile with a 67.6-m/s initial vertical component of velocityreaches a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable forlarge fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance isnot completely negligible, so the initial velocity would have to be somewhat larger than that given to reach thesame height.
(b) As in many physics problems, there is more than one way to solve for the time the projectile reaches its highestpoint. In this case, the easiest method is to use vy = v0y − gt. Because vy = 0 at the apex, this equation reduces
to simply


0 = v0y − gt


or
t =


v0y
g =


67.6 m/s
9.80 m/s2


= 6.90s.


This time is also reasonable for large fireworks. If you are able to see the launch of fireworks, notice that several
seconds pass before the shell explodes. Another way of finding the time is by using y= y0 + 12(v0y + vy)t. This
is left for you as an exercise to complete.
(c) Because air resistance is negligible, ax = 0 and the horizontal velocity is constant, as discussed earlier. The
horizontal displacement is the horizontal velocity multiplied by time as given by x = x0 + vx t, where x0 is
equal to zero. Thus,


x = vx t,


where vx is the x-component of the velocity, which is given by
vx = v0 cosθ = (70.0 m/s)cos75° = 18.1 m/s.


Time t for both motions is the same, so x is
x = (18.1 m/s)6.90 s = 125 m.


Horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement foundhere could be useful in keeping the fireworks fragments from falling on spectators. When the shell explodes, airresistance has a major effect, and many fragments land directly below.
(d) The horizontal and vertical components of the displacement were just calculated, so all that is needed here isto find the magnitude and direction of the displacement at the highest point:


s→ = 125 i
^


+ 233 j
^


| s→ | = 1252 + 2332 = 264 m
θ = tan−1 ⎛⎝


233
125

⎠ = 61.8°.


Note that the angle for the displacement vector is less than the initial angle of launch. To see why this is, reviewFigure 4.11, which shows the curvature of the trajectory toward the ground level.


Chapter 4 | Motion in Two and Three Dimensions 175




4.3


When solving Example 4.7(a), the expression we found for y is valid for any projectile motion when air resistance isnegligible. Call the maximum height y = h. Then,
h =


v0y
2


2g
.


This equation defines the maximum height of a projectile above its launch position and it depends only on the verticalcomponent of the initial velocity.
Check Your Understanding A rock is thrown horizontally off a cliff 100.0 m high with a velocity of


15.0 m/s. (a) Define the origin of the coordinate system. (b) Which equation describes the horizontal motion?(c) Which equations describe the vertical motion? (d) What is the rock’s velocity at the point of impact?


Example 4.8
Calculating Projectile Motion: Tennis Player
A tennis player wins a match at Arthur Ashe stadium and hits a ball into the stands at 30 m/s and at an angle
45° above the horizontal (Figure 4.14). On its way down, the ball is caught by a spectator 10 m above the
point where the ball was hit. (a) Calculate the time it takes the tennis ball to reach the spectator. (b) What are themagnitude and direction of the ball’s velocity at impact?


Figure 4.14 The trajectory of a tennis ball hit into the stands.


Strategy
Again, resolving this two-dimensional motion into two independent one-dimensional motions allows us to solvefor the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. Thus, we solvefor t first. While the ball is rising and falling vertically, the horizontal motion continues at a constant velocity.
This example asks for the final velocity. Thus, we recombine the vertical and horizontal results to obtain v→ at
final time t, determined in the first part of the example.
Solution
(a) While the ball is in the air, it rises and then falls to a final position 10.0 m higher than its starting altitude. Wecan find the time for this by using Equation 4.22:


y= y0 + v0y t −
1
2
gt2.


If we take the initial position y0 to be zero, then the final position is y = 10 m. The initial vertical velocity is the
vertical component of the initial velocity:


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v0y = v0 sinθ0 = (30.0 m/s)sin 45° = 21.2 m/s.


Substituting into Equation 4.22 for y gives us
10.0 m = (21.2 m/s)t − (4.90 m/s2)t2.


Rearranging terms gives a quadratic equation in t:
(4.90 m/s2)t2 − (21.2 m/s)t + 10.0 m = 0.


Use of the quadratic formula yields t = 3.79 s and t = 0.54 s. Since the ball is at a height of 10 m at two timesduring its trajectory—once on the way up and once on the way down—we take the longer solution for the time ittakes the ball to reach the spectator:
t = 3.79 s.


The time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has aninitial vertical velocity of 21.2 m/s and lands 10.0 m below its starting altitude spends 3.79 s in the air.
(b) We can find the final horizontal and vertical velocities vx and vy with the use of the result from (a). Then,
we can combine them to find the magnitude of the total velocity vector v→ and the angle θ it makes with
the horizontal. Since vx is constant, we can solve for it at any horizontal location. We choose the starting point
because we know both the initial velocity and the initial angle. Therefore,


vx = v0 cosθ0 = (30 m/s)cos 45° = 21.2 m/s.


The final vertical velocity is given by Equation 4.21:
vy = v0y − gt.


Since v0y was found in part (a) to be 21.2 m/s, we have
vy = 21.2 m/s − 9.8 m/s


2(3.79 s) = −15.9 m/s.


The magnitude of the final velocity v→ is
v = vx


2 + vy
2 = (21.2 m/s)2 + (− 15.9 m/s)2 = 26.5 m/s.


The direction θv is found using the inverse tangent:
θv = tan


−1⎛

vy
vx

⎠ = tan


−1 ⎛

21.2
−15.9



⎠ = −53.1°.


Significance
(a) As mentioned earlier, the time for projectile motion is determined completely by the vertical motion. Thus,any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m below its starting altitude spends3.79 s in the air. (b) The negative angle means the velocity is 53.1° below the horizontal at the point of impact.
This result is consistent with the fact that the ball is impacting at a point on the other side of the apex of thetrajectory and therefore has a negative y component of the velocity. The magnitude of the velocity is less than themagnitude of the initial velocity we expect since it is impacting 10.0 m above the launch elevation.


Time of Flight, Trajectory, and Range
Of interest are the time of flight, trajectory, and range for a projectile launched on a flat horizontal surface and impactingon the same surface. In this case, kinematic equations give useful expressions for these quantities, which are derived in thefollowing sections.
Time of flight
We can solve for the time of flight of a projectile that is both launched and impacts on a flat horizontal surface by performingsome manipulations of the kinematic equations. We note the position and displacement in y must be zero at launch and atimpact on an even surface. Thus, we set the displacement in y equal to zero and find


Chapter 4 | Motion in Two and Three Dimensions 177




y − y0 = v0y t −
1
2
gt2 = (v0 sinθ0)t −


1
2
gt2 = 0.


Factoring, we have
t⎛⎝v0 sinθ0 −


gt
2

⎠ = 0.


Solving for t gives us


(4.24)
Ttof =


2(v0 sinθ0)
g .


This is the time of flight for a projectile both launched and impacting on a flat horizontal surface. Equation 4.24 does notapply when the projectile lands at a different elevation than it was launched, as we saw in Example 4.8 of the tennis playerhitting the ball into the stands. The other solution, t = 0, corresponds to the time at launch. The time of flight is linearlyproportional to the initial velocity in the y direction and inversely proportional to g. Thus, on the Moon, where gravity isone-sixth that of Earth, a projectile launched with the same velocity as on Earth would be airborne six times as long.
Trajectory
The trajectory of a projectile can be found by eliminating the time variable t from the kinematic equations for arbitrary t andsolving for y(x). We take x0 = y0 = 0 so the projectile is launched from the origin. The kinematic equation for x gives


x = v0x t ⇒ t =
x
v0x


= x
v0 cosθ0


.


Substituting the expression for t into the equation for the position y = (v0 sinθ0)t − 12gt2 gives


y = (v0 sinθ0)



x
v0 cosθ0



⎠−


1
2
g



x
v0 cosθ0





2


.


Rearranging terms, we have


(4.25)
y = (tanθ0)x −





⎢ g


2(v0 cosθ0)
2





⎥x2.


This trajectory equation is of the form y = ax + bx2, which is an equation of a parabola with coefficients
a = tanθ0, b = −


g


2(v0 cosθ0)
2
.


Range
From the trajectory equation we can also find the range, or the horizontal distance traveled by the projectile. FactoringEquation 4.25, we have


y = x



⎢tanθ0 −


g


2(v0 cosθ0)
2
x



⎥.


The position y is zero for both the launch point and the impact point, since we are again considering only a flat horizontalsurface. Setting y = 0 in this equation gives solutions x = 0, corresponding to the launch point, and
x =


2v0
2 sinθ0 cosθ0


g ,


corresponding to the impact point. Using the trigonometric identity 2sinθcosθ = sin2θ and setting x = R for range, we
find


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(4.26)
R =


v0
2 sin2θ0


g .


Note particularly that Equation 4.26 is valid only for launch and impact on a horizontal surface. We see the range isdirectly proportional to the square of the initial speed v0 and sin2θ0 , and it is inversely proportional to the acceleration of
gravity. Thus, on the Moon, the range would be six times greater than on Earth for the same initial velocity. Furthermore, wesee from the factor sin2θ0 that the range is maximum at 45°. These results are shown in Figure 4.15. In (a) we see that
the greater the initial velocity, the greater the range. In (b), we see that the range is maximum at 45°. This is true only for
conditions neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interestingthat the same range is found for two initial launch angles that sum to 90°. The projectile launched with the smaller angle
has a lower apex than the higher angle, but they both have the same range.


Figure 4.15 Trajectories of projectiles on level ground. (a) Thegreater the initial speed v0, the greater the range for a given
initial angle. (b) The effect of initial angle θ0 on the range of a
projectile with a given initial speed. Note that the range is the samefor initial angles of 15° and 75°, although the maximum
heights of those paths are different.


Example 4.9
Comparing Golf Shots
A golfer finds himself in two different situations on different holes. On the second hole he is 120 m from thegreen and wants to hit the ball 90 m and let it run onto the green. He angles the shot low to the ground at 30°
to the horizontal to let the ball roll after impact. On the fourth hole he is 90 m from the green and wants to let


Chapter 4 | Motion in Two and Three Dimensions 179




the ball drop with a minimum amount of rolling after impact. Here, he angles the shot at 70° to the horizontal to
minimize rolling after impact. Both shots are hit and impacted on a level surface.
(a) What is the initial speed of the ball at the second hole?
(b) What is the initial speed of the ball at the fourth hole?
(c) Write the trajectory equation for both cases.
(d) Graph the trajectories.
Strategy
We see that the range equation has the initial speed and angle, so we can solve for the initial speed for both (a)and (b). When we have the initial speed, we can use this value to write the trajectory equation.
Solution
(a) R = v02 sin2θ0g ⇒ v0 = Rgsin2θ0 =


90.0 m(9.8 m/s2)
sin(2(70°))


= 37.0 m/s


(b) R = v02 sin2θ0g ⇒ v0 = Rgsin2θ0 =
90.0 m(9.8 m/s2)


sin(2(30°))
= 31.9 m/s


(c)
y = x



⎢tanθ0 −


g


2(v0 cosθ0)
2
x





Second hole: y = x



⎢tan 70° − 9.8 m/s


2


2[(37.0 m/s)(cos 70°)]2
x



⎥ = 2.75x − 0.0306x2


Fourth hole: y = x



⎢tan 30° − 9.8 m/s


2


2[(31.9 m/s)(cos30°)]2
x



⎥ = 0.58x − 0.0064x2


(d) Using a graphing utility, we can compare the two trajectories, which are shown in Figure 4.16.


Figure 4.16 Two trajectories of a golf ball with a range of 90 m. Theimpact points of both are at the same level as the launch point.


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4.4


Significance
The initial speed for the shot at 70° is greater than the initial speed of the shot at 30°. Note from Figure 4.16
that two projectiles launched at the same speed but at different angles have the same range if the launch anglesadd to 90°. The launch angles in this example add to give a number greater than 90°. Thus, the shot at 70° has
to have a greater launch speed to reach 90 m, otherwise it would land at a shorter distance.


Check Your Understanding If the two golf shots in Example 4.9 were launched at the same speed,which shot would have the greatest range?


When we speak of the range of a projectile on level ground, we assume R is very small compared with the circumferenceof Earth. If, however, the range is large, Earth curves away below the projectile and the acceleration resulting from gravitychanges direction along the path. The range is larger than predicted by the range equation given earlier because the projectilehas farther to fall than it would on level ground, as shown in Figure 4.17, which is based on a drawing in Newton’sPrincipia . If the initial speed is great enough, the projectile goes into orbit. Earth’s surface drops 5 m every 8000 m. In 1 san object falls 5 m without air resistance. Thus, if an object is given a horizontal velocity of 8000 m/s (or 18,000mi/hr)
near Earth’s surface, it will go into orbit around the planet because the surface continuously falls away from the object. Thisis roughly the speed of the Space Shuttle in a low Earth orbit when it was operational, or any satellite in a low Earth orbit.These and other aspects of orbital motion, such as Earth’s rotation, are covered in greater depth in Gravitation.


Figure 4.17 Projectile to satellite. In each case shown here, aprojectile is launched from a very high tower to avoid airresistance. With increasing initial speed, the range increases andbecomes longer than it would be on level ground because Earthcurves away beneath its path. With a speed of 8000 m/s, orbit isachieved.
At PhET Explorations: Projectile Motion (https://openstaxcollege.org/l/21phetpromot) , learn aboutprojectile motion in terms of the launch angle and initial velocity.


Chapter 4 | Motion in Two and Three Dimensions 181




4.4 | Uniform Circular Motion
Learning Objectives


By the end of this section, you will be able to:
• Solve for the centripetal acceleration of an object moving on a circular path.
• Use the equations of circular motion to find the position, velocity, and acceleration of a particleexecuting circular motion.
• Explain the differences between centripetal acceleration and tangential acceleration resultingfrom nonuniform circular motion.
• Evaluate centripetal and tangential acceleration in nonuniform circular motion, and find the totalacceleration vector.


Uniform circular motion is a specific type of motion in which an object travels in a circle with a constant speed. Forexample, any point on a propeller spinning at a constant rate is executing uniform circular motion. Other examples are thesecond, minute, and hour hands of a watch. It is remarkable that points on these rotating objects are actually accelerating,although the rotation rate is a constant. To see this, we must analyze the motion in terms of vectors.
Centripetal Acceleration
In one-dimensional kinematics, objects with a constant speed have zero acceleration. However, in two- and three-dimensional kinematics, even if the speed is a constant, a particle can have acceleration if it moves along a curved trajectory
such as a circle. In this case the velocity vector is changing, or d v→ /dt ≠ 0. This is shown in Figure 4.18. As the particle
moves counterclockwise in time Δt on the circular path, its position vector moves from r→ (t) to r→ (t + Δt). The
velocity vector has constant magnitude and is tangent to the path as it changes from v→ (t) to v→ (t + Δt), changing its
direction only. Since the velocity vector v→ (t) is perpendicular to the position vector r→ (t), the triangles formed by the
position vectors and Δ r→ , and the velocity vectors and Δ v→ are similar. Furthermore, since | r→ (t)| = | r→ (t + Δt)|
and | v→ (t)| = | v→ (t + Δt)|, the two triangles are isosceles. From these facts we can make the assertion
Δv
v =


Δr
r or Δv = vrΔr.


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Figure 4.18 (a) A particle is moving in a circle at a constant speed, with position and velocity vectors at times t
and t + Δt. (b) Velocity vectors forming a triangle. The two triangles in the figure are similar. The vector Δ v→
points toward the center of the circle in the limit Δt → 0.


We can find the magnitude of the acceleration from
a = lim


Δt → 0




Δv
Δt

⎠ =


v
r

⎝ limΔt → 0


Δr
Δt

⎠ =


v2
r .


The direction of the acceleration can also be found by noting that as Δt and therefore Δθ approach zero, the vector Δ v→
approaches a direction perpendicular to v→ . In the limit Δt → 0, Δ v→ is perpendicular to v→ . Since v→ is tangent
to the circle, the acceleration d v→ /dt points toward the center of the circle. Summarizing, a particle moving in a circle at
a constant speed has an acceleration with magnitude


(4.27)
aC =


v2
r .


The direction of the acceleration vector is toward the center of the circle (Figure 4.19). This is a radial acceleration andis called the centripetal acceleration, which is why we give it the subscript c. The word centripetal comes from the Latinwords centrum (meaning “center”) and petere (meaning to seek”), and thus takes the meaning “center seeking.”


Chapter 4 | Motion in Two and Three Dimensions 183




4.5


Figure 4.19 The centripetal acceleration vector points towardthe center of the circular path of motion and is an acceleration inthe radial direction. The velocity vector is also shown and istangent to the circle.


Let’s investigate some examples that illustrate the relative magnitudes of the velocity, radius, and centripetal acceleration.
Example 4.10


Creating an Acceleration of 1 g
A jet is flying at 134.1 m/s along a straight line and makes a turn along a circular path level with the ground. Whatdoes the radius of the circle have to be to produce a centripetal acceleration of 1 g on the pilot and jet toward thecenter of the circular trajectory?
Strategy
Given the speed of the jet, we can solve for the radius of the circle in the expression for the centripetalacceleration.
Solution
Set the centripetal acceleration equal to the acceleration of gravity: 9.8 m/s2 = v2/r.
Solving for the radius, we find


r = (134.1 m/s)
2


9.8 m/s2
= 1835 m = 1.835 km.


Significance
To create a greater acceleration than g on the pilot, the jet would either have to decrease the radius of its circulartrajectory or increase its speed on its existing trajectory or both.


Check Your Understanding A flywheel has a radius of 20.0 cm. What is the speed of a point on the edge
of the flywheel if it experiences a centripetal acceleration of 900.0 cm/s2?


Centripetal acceleration can have a wide range of values, depending on the speed and radius of curvature of the circularpath. Typical centripetal accelerations are given in the following table.
Object Centripetal Acceleration (m/s2 or factorsof g)
Earth around the Sun 5.93 × 10−3


Table 4.1 Typical Centripetal Accelerations


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Object Centripetal Acceleration (m/s2 or factorsof g)
Moon around the Earth 2.73 × 10−3
Satellite in geosynchronous orbit 0.233
Outer edge of a CD when playing 5.78
Jet in a barrel roll (2–3 g)
Roller coaster (5 g)
Electron orbiting a proton in a simple Bohr model of theatom 9.0 × 1022


Table 4.1 Typical Centripetal Accelerations
Equations of Motion for Uniform Circular Motion
A particle executing circular motion can be described by its position vector r→ (t). Figure 4.20 shows a particle executing
circular motion in a counterclockwise direction. As the particle moves on the circle, its position vector sweeps out the angle
θ with the x-axis. Vector r→ (t) making an angle θ with the x-axis is shown with its components along the x- and y-axes.
The magnitude of the position vector is A = | r→ (t)| and is also the radius of the circle, so that in terms of its components,


(4.28)
r→ (t) = Acosωt i


^
+ Asinωt j


^
.


Here, ω is a constant called the angular frequency of the particle. The angular frequency has units of radians (rad) per
second and is simply the number of radians of angular measure through which the particle passes per second. The angle θ
that the position vector has at any particular time is ωt .
If T is the period of motion, or the time to complete one revolution ( 2π rad), then


ω = 2π
T


.


Figure 4.20 The position vector for a particle in circularmotion with its components along the x- and y-axes. The particlemoves counterclockwise. Angle θ is the angular frequency ω
in radians per second multiplied by t.


Velocity and acceleration can be obtained from the position function by differentiation:


Chapter 4 | Motion in Two and Three Dimensions 185




(4.29)
v→ (t) = d r


→ (t)
dt


= −Aωsinωt i
^


+ Aωcosωt j
^
.


It can be shown from Figure 4.20 that the velocity vector is tangential to the circle at the location of the particle, withmagnitude Aω. Similarly, the acceleration vector is found by differentiating the velocity:


(4.30)
a→ (t) = d v


→ (t)
dt


= −Aω2 cosωt i
^


− Aω2 sinωt j
^
.


From this equation we see that the acceleration vector has magnitude Aω2 and is directed opposite the position vector,
toward the origin, because a→ (t) = −ω2 r→ (t).
Example 4.11


Circular Motion of a Proton
A proton has speed 5 × 106 m/s and is moving in a circle in the xy plane of radius r = 0.175 m. What is its
position in the xy plane at time t = 2.0 × 10−7 s = 200 ns? At t = 0, the position of the proton is 0.175 m i^
and it circles counterclockwise. Sketch the trajectory.
Solution
From the given data, the proton has period and angular frequency:


T = 2πrv =
2π(0.175 m)


5.0 × 106 m/s
= 2.20 × 10−7 s


ω = 2π
T


= 2π
2.20 × 10−7 s


= 2.856 × 107 rad/s.


The position of the particle at t = 2.0 × 10−7 s with A = 0.175 m is
r→ (2.0 × 10−7 s) = A cosω(2.0 × 10−7 s) i


^
+ A sinω(2.0 × 10−7 s) j


^
m


= 0.175cos[(2.856 × 107 rad/s)(2.0 × 10−7 s)] i
^


+0.175sin[(2.856 × 107 rad/s)(2.0 × 10−7 s)] j
^


m


= 0.175cos(5.712 rad) i
^


+ 0.175sin(5.712 rad) j
^


= 0.147 i
^


− 0.095 j
^


m.


From this result we see that the proton is located slightly below the x-axis. This is shown in Figure 4.21.


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Figure 4.21 Position vector of the proton at
t = 2.0 × 10−7 s = 200 ns. The trajectory of the proton is
shown. The angle through which the proton travels along thecircle is 5.712 rad, which a little less than one completerevolution.


Significance
We picked the initial position of the particle to be on the x-axis. This was completely arbitrary. If a differentstarting position were given, we would have a different final position at t = 200 ns.


Nonuniform Circular Motion
Circular motion does not have to be at a constant speed. A particle can travel in a circle and speed up or slow down, showingan acceleration in the direction of the motion.
In uniform circular motion, the particle executing circular motion has a constant speed and the circle is at a fixed radius. Ifthe speed of the particle is changing as well, then we introduce an additional acceleration in the direction tangential to thecircle. Such accelerations occur at a point on a top that is changing its spin rate, or any accelerating rotor. In Displacementand Velocity Vectors we showed that centripetal acceleration is the time rate of change of the direction of the velocityvector. If the speed of the particle is changing, then it has a tangential acceleration that is the time rate of change of themagnitude of the velocity:


(4.31)
aT =


d| v→ |
dt


.


The direction of tangential acceleration is tangent to the circle whereas the direction of centripetal acceleration is radiallyinward toward the center of the circle. Thus, a particle in circular motion with a tangential acceleration has a totalacceleration that is the vector sum of the centripetal and tangential accelerations:


(4.32)a→ = a→ C + a→ T.


Chapter 4 | Motion in Two and Three Dimensions 187




The acceleration vectors are shown in Figure 4.22. Note that the two acceleration vectors a→ C and a→ T are
perpendicular to each other, with a→ C in the radial direction and a→ T in the tangential direction. The total acceleration
a→ points at an angle between a→ C and a→ T.


Figure 4.22 The centripetal acceleration points toward thecenter of the circle. The tangential acceleration is tangential tothe circle at the particle’s position. The total acceleration is thevector sum of the tangential and centripetal accelerations, whichare perpendicular.


Example 4.12
Total Acceleration during Circular Motion
A particle moves in a circle of radius r = 2.0 m. During the time interval from t = 1.5 s to t = 4.0 s its speed varieswith time according to


v(t) = c1 −
c2
t2


, c1 = 4.0 m/s, c2 = 6.0 m · s.


What is the total acceleration of the particle at t = 2.0 s?
Strategy
We are given the speed of the particle and the radius of the circle, so we can calculate centripetal accelerationeasily. The direction of the centripetal acceleration is toward the center of the circle. We find the magnitude of thetangential acceleration by taking the derivative with respect to time of |v(t)| using Equation 4.31 and evaluating
it at t = 2.0 s. We use this and the magnitude of the centripetal acceleration to find the total acceleration.
Solution
Centripetal acceleration is


v(2.0s) =



⎜4.0 − 6.0


(2.0)2





⎟m/s = 2.5 m/s


aC =
v2
r =


(2.5 m/s)2


2.0 m
= 3.1 m/s2


directed toward the center of the circle. Tangential acceleration is
aT = |d v→dt | = 2c2t3 = 12.0(2.0)3m/s2 = 1.5 m/s2.


Total acceleration is
| a→ | = 3.12 + 1.52m/s2 = 3.44 m/s2


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and θ = tan−1 3.1
1.5


= 64° from the tangent to the circle. See Figure 4.23.


Figure 4.23 The tangential and centripetal acceleration
vectors. The net acceleration a→ is the vector sum of the two
accelerations.


Significance
The directions of centripetal and tangential accelerations can be described more conveniently in terms of a polarcoordinate system, with unit vectors in the radial and tangential directions. This coordinate system, which is usedfor motion along curved paths, is discussed in detail later in the book.


4.5 | Relative Motion in One and Two Dimensions
Learning Objectives


By the end of this section, you will be able to:
• Explain the concept of reference frames.
• Write the position and velocity vector equations for relative motion.
• Draw the position and velocity vectors for relative motion.
• Analyze one-dimensional and two-dimensional relative motion problems using the position andvelocity vector equations.


Motion does not happen in isolation. If you’re riding in a train moving at 10 m/s east, this velocity is measured relative tothe ground on which you’re traveling. However, if another train passes you at 15 m/s east, your velocity relative to this othertrain is different from your velocity relative to the ground. Your velocity relative to the other train is 5 m/s west. To explorethis idea further, we first need to establish some terminology.
Reference Frames
To discuss relative motion in one or more dimensions, we first introduce the concept of reference frames. When we say anobject has a certain velocity, we must state it has a velocity with respect to a given reference frame. In most examples wehave examined so far, this reference frame has been Earth. If you say a person is sitting in a train moving at 10 m/s east,then you imply the person on the train is moving relative to the surface of Earth at this velocity, and Earth is the referenceframe. We can expand our view of the motion of the person on the train and say Earth is spinning in its orbit around theSun, in which case the motion becomes more complicated. In this case, the solar system is the reference frame. In summary,all discussion of relative motion must define the reference frames involved. We now develop a method to refer to referenceframes in relative motion.


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Relative Motion in One Dimension
We introduce relative motion in one dimension first, because the velocity vectors simplify to having only two possibledirections. Take the example of the person sitting in a train moving east. If we choose east as the positive direction and
Earth as the reference frame, then we can write the velocity of the train with respect to the Earth as v→ TE = 10 m/s i^
east, where the subscripts TE refer to train and Earth. Let’s now say the person gets up out of /her seat and walks towardthe back of the train at 2 m/s. This tells us she has a velocity relative to the reference frame of the train. Since the person is
walking west, in the negative direction, we write her velocity with respect to the train as v→ PT = −2m/s i^ . We can add
the two velocity vectors to find the velocity of the person with respect to Earth. This relative velocity is written as


(4.33)v→ PE = v→ PT + v→ TE.
Note the ordering of the subscripts for the various reference frames in Equation 4.33. The subscripts for the couplingreference frame, which is the train, appear consecutively in the right-hand side of the equation. Figure 4.24 shows thecorrect order of subscripts when forming the vector equation.


Figure 4.24 When constructing the vector equation, thesubscripts for the coupling reference frame appear consecutivelyon the inside. The subscripts on the left-hand side of theequation are the same as the two outside subscripts on the right-hand side of the equation.


Adding the vectors, we find v→ PE = 8m/s i^ , so the person is moving 8 m/s east with respect to Earth. Graphically, this
is shown in Figure 4.25.


Figure 4.25 Velocity vectors of the train with respect to Earth, person with respect to thetrain, and person with respect to Earth.
Relative Velocity in Two Dimensions
We can now apply these concepts to describing motion in two dimensions. Consider a particle P and reference frames S and
S′, as shown in Figure 4.26. The position of the origin of S′ as measured in S is r→ S′S, the position of P as measured
in S′ is r→ PS′, and the position of P as measured in S is r→ PS.


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Figure 4.26 The positions of particle P relative to frames S
and S′ are r→ PS and r→ PS′, respectively.


From Figure 4.26 we see that


(4.34)r→ PS = r→ PS′ + r→ S′S.


The relative velocities are the time derivatives of the position vectors. Therefore,


(4.35)v→ PS = v→ PS′ + v→ S′S.


The velocity of a particle relative to S is equal to its velocity relative to S′ plus the velocity of S′ relative to S.
We can extend Equation 4.35 to any number of reference frames. For particle P with velocities
v→ PA, v



PB , and v



PC in frames A, B, and C,


(4.36)v→ PC = v→ PA + v→ AB + v→ BC.


We can also see how the accelerations are related as observed in two reference frames by differentiating Equation 4.35:


(4.37)a→ PS = a→ PS′ + a→ S′S.


We see that if the velocity of S′ relative to S is a constant, then a→ S′S = 0 and
(4.38)a→ PS = a→ PS′.


This says the acceleration of a particle is the same as measured by two observers moving at a constant velocity relative toeach other.


Chapter 4 | Motion in Two and Three Dimensions 191




Example 4.13
Motion of a Car Relative to a Truck
A truck is traveling south at a speed of 70 km/h toward an intersection. A car is traveling east toward theintersection at a speed of 80 km/h (Figure 4.27). What is the velocity of the car relative to the truck?


Figure 4.27 A car travels east toward an intersection while atruck travels south toward the same intersection.


Strategy
First, we must establish the reference frame common to both vehicles, which is Earth. Then, we write thevelocities of each with respect to the reference frame of Earth, which enables us to form a vector equation thatlinks the car, the truck, and Earth to solve for the velocity of the car with respect to the truck.
Solution
The velocity of the car with respect to Earth is v→ CE = 80 km/h i^ . The velocity of the truck with respect to
Earth is v→ TE = −70 km/h j^ . Using the velocity addition rule, the relative motion equation we are seeking is


v→ CT = v


CE + v


ET.


Here, v→ CT is the velocity of the car with respect to the truck, and Earth is the connecting reference frame.
Since we have the velocity of the truck with respect to Earth, the negative of this vector is the velocity of Earth
with respect to the truck: v→ ET = − v→ TE. The vector diagram of this equation is shown in Figure 4.28.


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4.6


Figure 4.28 Vector diagram of the vector equation
v→ CT = v



CE + v



ET .


We can now solve for the velocity of the car with respect to the truck:
| v→ CT| = (80.0 km/h)2 + (70.0 km/h)2 = 106. km/h


and
θ = tan−1 ⎛⎝


70.0
80.0

⎠ = 41.2° north of east.


Significance
Drawing a vector diagram showing the velocity vectors can help in understanding the relative velocity of the twoobjects.


Check Your Understanding A boat heads north in still water at 4.5 m/s directly across a river that isrunning east at 3.0 m/s. What is the velocity of the boat with respect to Earth?


Example 4.14
Flying a Plane in a Wind
A pilot must fly his plane due north to reach his destination. The plane can fly at 300 km/h in still air. A windis blowing out of the northeast at 90 km/h. (a) What is the speed of the plane relative to the ground? (b) In whatdirection must the pilot head her plane to fly due north?
Strategy
The pilot must point her plane somewhat east of north to compensate for the wind velocity. We need to constructa vector equation that contains the velocity of the plane with respect to the ground, the velocity of the planewith respect to the air, and the velocity of the air with respect to the ground. Since these last two quantities areknown, we can solve for the velocity of the plane with respect to the ground. We can graph the vectors and usethis diagram to evaluate the magnitude of the plane’s velocity with respect to the ground. The diagram will alsotell us the angle the plane’s velocity makes with north with respect to the air, which is the direction the pilot musthead her plane.
Solution
The vector equation is v→ PG = v→ PA + v→ AG, where P = plane, A = air, and G = ground. From the
geometry in Figure 4.29, we can solve easily for the magnitude of the velocity of the plane with respect to theground and the angle of the plane’s heading, θ.


Chapter 4 | Motion in Two and Three Dimensions 193




Figure 4.29 Vector diagram for Equation 4.34 showing the
vectors v→ PA , v→ AG, and v→ PG.


(a) Known quantities:
| v→ PA| = 300 km/h
| v→ AG| = 90 km/h


Substituting into the equation of motion, we obtain | v→ PG| = 230 km/h.
(b) The angle θ = tan−1 63.64


300
= 12° east of north.


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acceleration vector
angular frequency
centripetal acceleration
displacement vector
position vector
projectile motion
range
reference frame
relative velocity
tangential acceleration
time of flight
total acceleration
trajectory
velocity vector


CHAPTER 4 REVIEW
KEY TERMS


instantaneous acceleration found by taking the derivative of the velocity function with respect totime in unit vector notation
ω, rate of change of an angle with which an object that is moving on a circular path


component of acceleration of an object moving in a circle that is directed radially inwardtoward the center of the circle
vector from the initial position to a final position on a trajectory of a particle


vector from the origin of a chosen coordinate system to the position of a particle in two- or three-dimensional space
motion of an object subject only to the acceleration of gravity


maximum horizontal distance a projectile travels
coordinate system in which the position, velocity, and acceleration of an object at rest or moving ismeasured
velocity of an object as observed from a particular reference frame, or the velocity of one referenceframe with respect to another reference frame


magnitude of which is the time rate of change of speed. Its direction is tangent to the circle.
elapsed time a projectile is in the air


vector sum of centripetal and tangential accelerations
path of a projectile through the air


vector that gives the instantaneous speed and direction of a particle; tangent to the trajectory


KEY EQUATIONS
Position vector r→ (t) = x(t) i^ + y(t) j^ + z(t)k^
Displacement vector Δ r→ = r→ (t2) − r→ (t1)
Velocity vector v→ (t) = lim


Δt → 0


r→ (t + Δt) − r→ (t)
Δt


= d r


dt


Velocity in terms of components v→ (t) = vx(t) i^ + vy(t) j^ + vz(t)k^


Velocity components vx(t) = dx(t)dt vy(t) = dy(t)dt vz(t) = dz(t)dt
Average velocity v→ avg = r→ (t2) − r→ (t1)t2 − t1
Instantaneous acceleration a→ (t) = lim


t → 0


v→ (t + Δt) − v→ (t)
Δt


= d v
→ (t)
dt


Instantaneous acceleration, component form a→ (t) = dvx(t)dt i^ + dvy(t)dt j^ + dvz(t)dt k^


Instantaneous acceleration as secondderivatives of position a→ (t) = d
2 x(t)
dt2


i
^


+
d2 y(t)


dt2
j
^


+ d
2 z(t)
dt2


k
^


Chapter 4 | Motion in Two and Three Dimensions 195




Time of flight Ttof = 2(v0 sinθ)g
Trajectory y = (tanθ0)x − ⎡



⎢ g


2(v0 cosθ0)
2





⎥x2


Range R = v02 sin 2θ0g
Centripetal acceleration aC = v2r
Position vector, uniform circular motion r→ (t) = Acosωt i^ + A sinωt j^


Velocity vector, uniform circular motion v→ (t) = d r→ (t)
dt


= −Aω sinωt i
^


+ Aω cosωt j
^


Acceleration vector, uniform circular motion a→ (t) = d v→ (t)
dt


= −Aω2 cosωt i
^


− Aω2 sinωt j
^


Tangential acceleration aT = d| v
→ |
dt


Total acceleration a→ = a→ C + a→ T
Position vector in frameS is the positionvector in frame S′ plus the vector from the
origin of S to the origin of S′


r→ PS = r


PS′ + r


S′S


Relative velocity equation connecting tworeference frames v→ PS = v→ PS′ + v→ S′S
Relative velocity equation connecting morethan two reference frames v→ PC = v→ PA + v→ AB + v→ BC
Relative acceleration equation a→ PS = a→ PS′ + a→ S′S


SUMMARY
4.1 Displacement and Velocity Vectors


• The position function r→ (t) gives the position as a function of time of a particle moving in two or three
dimensions. Graphically, it is a vector from the origin of a chosen coordinate system to the point where the particleis located at a specific time.


• The displacement vector Δ r→ gives the shortest distance between any two points on the trajectory of a particle in
two or three dimensions.


• Instantaneous velocity gives the speed and direction of a particle at a specific time on its trajectory in two or threedimensions, and is a vector in two and three dimensions.
• The velocity vector is tangent to the trajectory of the particle.
• Displacement r→ (t) can be written as a vector sum of the one-dimensional displacements x→ (t), y→ (t), z→ (t)
along the x, y, and z directions.


• Velocity v→ (t) can be written as a vector sum of the one-dimensional velocities vx(t), vy(t), vz(t) along the x, y,
and z directions.


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• Motion in any given direction is independent of motion in a perpendicular direction.
4.2 Acceleration Vector


• In two and three dimensions, the acceleration vector can have an arbitrary direction and does not necessarily pointalong a given component of the velocity.
• The instantaneous acceleration is produced by a change in velocity taken over a very short (infinitesimal) timeperiod. Instantaneous acceleration is a vector in two or three dimensions. It is found by taking the derivative of thevelocity function with respect to time.
• In three dimensions, acceleration a→ (t) can be written as a vector sum of the one-dimensional accelerations


ax(t), ay(t), and az(t) along the x-, y-, and z-axes.
• The kinematic equations for constant acceleration can be written as the vector sum of the constant accelerationequations in the x, y, and z directions.


4.3 Projectile Motion
• Projectile motion is the motion of an object subject only to the acceleration of gravity, where the acceleration isconstant, as near the surface of Earth.
• To solve projectile motion problems, we analyze the motion of the projectile in the horizontal and vertical directionsusing the one-dimensional kinematic equations for x and y.
• The time of flight of a projectile launched with initial vertical velocity v0y on an even surface is given by


Tto f =
2(v0 sinθ)


g .


This equation is valid only when the projectile lands at the same elevation from which it was launched.
• The maximum horizontal distance traveled by a projectile is called the range. Again, the equation for range is validonly when the projectile lands at the same elevation from which it was launched.


4.4 Uniform Circular Motion
• Uniform circular motion is motion in a circle at constant speed.
• Centripetal acceleration a→ C is the acceleration a particle must have to follow a circular path. Centripetal
acceleration always points toward the center of rotation and has magnitude aC = v2/r.


• Nonuniform circular motion occurs when there is tangential acceleration of an object executing circular motion such
that the speed of the object is changing. This acceleration is called tangential acceleration a→ T. The magnitude of
tangential acceleration is the time rate of change of the magnitude of the velocity. The tangential acceleration vectoris tangential to the circle, whereas the centripetal acceleration vector points radially inward toward the center of thecircle. The total acceleration is the vector sum of tangential and centripetal accelerations.


• An object executing uniform circular motion can be described with equations of motion. The position vector of the
object is r→ (t) = A cosωt i^ + A sinωt j^ , where A is the magnitude | r→ (t)|, which is also the radius of the
circle, and ω is the angular frequency.


4.5 Relative Motion in One and Two Dimensions
• When analyzing motion of an object, the reference frame in terms of position, velocity, and acceleration needs to bespecified.
• Relative velocity is the velocity of an object as observed from a particular reference frame, and it varies with thechoice of reference frame.


Chapter 4 | Motion in Two and Three Dimensions 197




• If S and S′ are two reference frames moving relative to each other at a constant velocity, then the velocity of an
object relative to S is equal to its velocity relative to S′ plus the velocity of S′ relative to S.


• If two reference frames are moving relative to each other at a constant velocity, then the accelerations of an objectas observed in both reference frames are equal.


CONCEPTUAL QUESTIONS
4.1 Displacement and Velocity Vectors
1. What form does the trajectory of a particle have ifthe distance from any point A to point B is equal to themagnitude of the displacement from A to B?
2. Give an example of a trajectory in two or threedimensions caused by independent perpendicular motions.
3. If the instantaneous velocity is zero, what can be saidabout the slope of the position function?


4.2 Acceleration Vector
4. If the position function of a particle is a linear functionof time, what can be said about its acceleration?
5. If an object has a constant x-component of the velocityand suddenly experiences an acceleration in the y direction,does the x-component of its velocity change?
6. If an object has a constant x-component of velocity andsuddenly experiences an acceleration at an angle of 70° in
the x direction, does the x-component of velocity change?


4.3 Projectile Motion
7. Answer the following questions for projectile motionon level ground assuming negligible air resistance, withthe initial angle being neither 0° nor 90° : (a) Is the
velocity ever zero? (b) When is the velocity a minimum?A maximum? (c) Can the velocity ever be the same as theinitial velocity at a time other than at t = 0? (d) Can thespeed ever be the same as the initial speed at a time otherthan at t = 0?
8. Answer the following questions for projectile motionon level ground assuming negligible air resistance, withthe initial angle being neither 0° nor 90° : (a) Is the
acceleration ever zero? (b) Is the acceleration ever in thesame direction as a component of velocity? (c) Is theacceleration ever opposite in direction to a component ofvelocity?


9. A dime is placed at the edge of a table so it hangs overslightly. A quarter is slid horizontally on the table surfaceperpendicular to the edge and hits the dime head on. Whichcoin hits the ground first?


4.4 Uniform Circular Motion
10. Can centripetal acceleration change the speed of aparticle undergoing circular motion?
11. Can tangential acceleration change the speed of aparticle undergoing circular motion?


4.5 Relative Motion in One and Two
Dimensions
12. What frame or frames of reference do you useinstinctively when driving a car? When flying in acommercial jet?
13. A basketball player dribbling down the court usuallykeeps his eyes fixed on the players around him. He ismoving fast. Why doesn’t he need to keep his eyes on theball?
14. If someone is riding in the back of a pickup truck andthrows a softball straight backward, is it possible for theball to fall straight down as viewed by a person standingat the side of the road? Under what condition would thisoccur? How would the motion of the ball appear to theperson who threw it?
15. The hat of a jogger running at constant velocity fallsoff the back of his head. Draw a sketch showing the path ofthe hat in the jogger’s frame of reference. Draw its path asviewed by a stationary observer. Neglect air resistance.
16. A clod of dirt falls from the bed of a moving truck. Itstrikes the ground directly below the end of the truck. (a)What is the direction of its velocity relative to the truck justbefore it hits? (b) Is this the same as the direction of itsvelocity relative to ground just before it hits? Explain youranswers.


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PROBLEMS
4.1 Displacement and Velocity Vectors
17. The coordinates of a particle in a rectangularcoordinate system are (1.0, –4.0, 6.0). What is the positionvector of the particle?
18. The position of a particle changes from
r→ 1 = (2.0 i


^
+ 3.0 j


^
)cm to


r→ 2 = (−4.0 i
^


+ 3.0 j
^
) cm. What is the particle’s


displacement?
19. The 18th hole at Pebble Beach Golf Course is a doglegto the left of length 496.0 m. The fairway off the tee istaken to be the x direction. A golfer hits his tee shot adistance of 300.0 m, corresponding to a displacement
Δ r→ 1 = 300.0 m i


^
, and hits his second shot 189.0 m


with a displacement Δ r→ 2 = 172.0 m i^ + 80.3 m j^ .
What is the final displacement of the golf ball from the tee?
20. A bird flies straight northeast a distance of 95.0 kmfor 3.0 h. With the x-axis due east and the y-axis due north,what is the displacement in unit vector notation for thebird? What is the average velocity for the trip?
21. A cyclist rides 5.0 km due east, then 10.0 km 20°
west of north. From this point she rides 8.0 km due west.What is the final displacement from where the cycliststarted?
22. New York Rangers defenseman Daniel Girardi standsat the goal and passes a hockey puck 20 m and 45° from
straight down the ice to left wing Chris Kreider waiting atthe blue line. Kreider waits for Girardi to reach the blueline and passes the puck directly across the ice to him 10 maway. What is the final displacement of the puck? See thefollowing figure.


23. The position of a particle is
r→ (t) = 4.0t2 i


^
− 3.0 j


^
+ 2.0t3 k


^
m. (a) What is the


velocity of the particle at 0 s and at 1.0 s? (b) What is the
average velocity between 0 s and 1.0 s?
24. Clay Matthews, a linebacker for the Green BayPackers, can reach a speed of 10.0 m/s. At the start of aplay, Matthews runs downfield at 45° with respect to the
50-yard line and covers 8.0 m in 1 s. He then runs straightdown the field at 90° with respect to the 50-yard line for
12 m, with an elapsed time of 1.2 s. (a) What is Matthews’final displacement from the start of the play? (b) What ishis average velocity?
25. The F-35B Lighting II is a short-takeoff and verticallanding fighter jet. If it does a vertical takeoff to 20.00-mheight above the ground and then follows a flight pathangled at 30° with respect to the ground for 20.00 km,
what is the final displacement?


4.2 Acceleration Vector
26. The position of a particle is
r→ (t) = (3.0t2 i


^
+ 5.0 j


^
− 6.0tk


^
) m. (a) Determine its


velocity and acceleration as functions of time. (b) What areits velocity and acceleration at time t = 0?


27. A particle’s acceleration is (4.0 i^ + 3.0 j^ )m/s2. At
t = 0, its position and velocity are zero. (a) What are theparticle’s position and velocity as functions of time? (b)Find the equation of the path of the particle. Draw the x-and y-axes and sketch the trajectory of the particle.
28. A boat leaves the dock at t = 0 and heads out into
a lake with an acceleration of 2.0 m/s2 i^ . A strong wind
is pushing the boat, giving it an additional velocity of
2.0 m/s i


^
+ 1.0 m/s j


^
. (a) What is the velocity of the boat


at t = 10 s? (b) What is the position of the boat at t = 10s?Draw a sketch of the boat’s trajectory and position at t = 10s, showing the x- and y-axes.
29. The position of a particle for t > 0 is given by
r→ (t) = (3.0t2 i


^
− 7.0t3 j


^
− 5.0t−2 k


^
) m. (a) What is


the velocity as a function of time? (b) What is theacceleration as a function of time? (c) What is the particle’svelocity at t = 2.0 s? (d) What is its speed at t = 1.0 s and t= 3.0 s? (e) What is the average velocity between t = 1.0 sand t = 2.0 s?


Chapter 4 | Motion in Two and Three Dimensions 199




30. The acceleration of a particle is a constant. At t =
0 the velocity of the particle is (10 i^ + 20 j^ )m/s. At t
= 4 s the velocity is 10 j^m/s. (a) What is the particle’s
acceleration? (b) How do the position and velocity varywith time? Assume the particle is initially at the origin.
31. A particle has a position function
r→ (t) = cos(1.0t) i


^
+ sin(1.0t) j


^
+ tk


^
, where the


arguments of the cosine and sine functions are in radians.(a) What is the velocity vector? (b) What is the accelerationvector?
32. A Lockheed Martin F-35 II Lighting jet takes off froman aircraft carrier with a runway length of 90 m and atakeoff speed 70 m/s at the end of the runway. Jets arecatapulted into airspace from the deck of an aircraft carrierwith two sources of propulsion: the jet propulsion and thecatapult. At the point of leaving the deck of the aircraftcarrier, the F-35’s acceleration decreases to a constant
acceleration of 5.0 m/s2 at 30° with respect to the
horizontal. (a) What is the initial acceleration of the F-35 onthe deck of the aircraft carrier to make it airborne? (b) Writethe position and velocity of the F-35 in unit vector notationfrom the point it leaves the deck of the aircraft carrier. (c)At what altitude is the fighter 5.0 s after it leaves the deckof the aircraft carrier? (d) What is its velocity and speed atthis time? (e) How far has it traveled horizontally?


4.3 Projectile Motion
33. A bullet is shot horizontally from shoulder height (1.5m) with and initial speed 200 m/s. (a) How much timeelapses before the bullet hits the ground? (b) How far doesthe bullet travel horizontally?
34. A marble rolls off a tabletop 1.0 m high and hits thefloor at a point 3.0 m away from the table’s edge in thehorizontal direction. (a) How long is the marble in the air?(b) What is the speed of the marble when it leaves thetable’s edge? (c) What is its speed when it hits the floor?
35. A dart is thrown horizontally at a speed of 10 m/s at the bull’s-eye of a dartboard 2.4 m away, as in thefollowing figure. (a) How far below the intended targetdoes the dart hit? (b) What does your answer tell you abouthow proficient dart players throw their darts?
36. An airplane flying horizontally with a speed of 500km/h at a height of 800 m drops a crate of supplies (see thefollowing figure). If the parachute fails to open, how far infront of the release point does the crate hit the ground?


37. Suppose the airplane in the preceding problem fires aprojectile horizontally in its direction of motion at a speedof 300 m/s relative to the plane. (a) How far in front of therelease point does the projectile hit the ground? (b) What isits speed when it hits the ground?
38. A fastball pitcher can throw a baseball at a speed of 40m/s (90 mi/h). (a) Assuming the pitcher can release the ball16.7 m from home plate so the ball is moving horizontally,how long does it take the ball to reach home plate? (b) Howfar does the ball drop between the pitcher’s hand and homeplate?
39. A projectile is launched at an angle of 30° and lands
20 s later at the same height as it was launched. (a) What isthe initial speed of the projectile? (b) What is the maximumaltitude? (c) What is the range? (d) Calculate thedisplacement from the point of launch to the position on itstrajectory at 15 s.
40. A basketball player shoots toward a basket 6.1 m awayand 3.0 m above the floor. If the ball is released 1.8 m abovethe floor at an angle of 60° above the horizontal, what
must the initial speed be if it were to go through the basket?
41. At a particular instant, a hot air balloon is 100 m inthe air and descending at a constant speed of 2.0 m/s. Atthis exact instant, a girl throws a ball horizontally, relativeto herself, with an initial speed of 20 m/s. When she lands,where will she find the ball? Ignore air resistance.
42. A man on a motorcycle traveling at a uniform speedof 10 m/s throws an empty can straight upward relative tohimself with an initial speed of 3.0 m/s. Find the equationof the trajectory as seen by a police officer on the side ofthe road. Assume the initial position of the can is the pointwhere it is thrown. Ignore air resistance.
43. An athlete can jump a distance of 8.0 m in the broadjump. What is the maximum distance the athlete can jumpon the Moon, where the gravitational acceleration is one-sixth that of Earth?


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44. The maximum horizontal distance a boy can throw aball is 50 m. Assume he can throw with the same initialspeed at all angles. How high does he throw the ball whenhe throws it straight upward?
45. A rock is thrown off a cliff at an angle of 53° with
respect to the horizontal. The cliff is 100 m high. The initialspeed of the rock is 30 m/s. (a) How high above the edgeof the cliff does the rock rise? (b) How far has it movedhorizontally when it is at maximum altitude? (c) How longafter the release does it hit the ground? (d) What is therange of the rock? (e) What are the horizontal and verticalpositions of the rock relative to the edge of the cliff at t =2.0 s, t = 4.0 s, and t = 6.0 s?
46. Trying to escape his pursuers, a secret agent skis off aslope inclined at 30° below the horizontal at 60 km/h. To
survive and land on the snow 100 m below, he must clear agorge 60 m wide. Does he make it? Ignore air resistance.


47. A golfer on a fairway is 70 m away from the green,which sits below the level of the fairway by 20 m. If thegolfer hits the ball at an angle of 40° with an initial speed
of 20 m/s, how close to the green does she come?
48. A projectile is shot at a hill, the base of which is 300m away. The projectile is shot at 60° above the horizontal
with an initial speed of 75 m/s. The hill can beapproximated by a plane sloped at 20° to the horizontal.
Relative to the coordinate system shown in the followingfigure, the equation of this straight line is
y = (tan20°)x − 109. Where on the hill does the
projectile land?


49. An astronaut on Mars kicks a soccer ball at an angle of
45° with an initial velocity of 15 m/s. If the acceleration
of gravity on Mars is 3.7 m/s, (a) what is the range of thesoccer kick on a flat surface? (b) What would be the rangeof the same kick on the Moon, where gravity is one-sixththat of Earth?
50. Mike Powell holds the record for the long jump of 8.95m, established in 1991. If he left the ground at an angle of
15°, what was his initial speed?
51. MIT’s robot cheetah can jump over obstacles 46 cmhigh and has speed of 12.0 km/h. (a) If the robot launchesitself at an angle of 60° at this speed, what is its maximum
height? (b) What would the launch angle have to be to reacha height of 46 cm?
52. Mt. Asama, Japan, is an active volcano. In 2009,an eruption threw solid volcanic rocks that landed 1 kmhorizontally from the crater. If the volcanic rocks werelaunched at an angle of 40° with respect to the horizontal
and landed 900 m below the crater, (a) what would be theirinitial velocity and (b) what is their time of flight?
53. Drew Brees of the New Orleans Saints can throw afootball 23.0 m/s (50 mph). If he angles the throw at 10°
from the horizontal, what distance does it go if it is to becaught at the same elevation as it was thrown?
54. The Lunar Roving Vehicle used in NASA’s late Apollomissions reached an unofficial lunar land speed of 5.0 m/s by astronaut Eugene Cernan. If the rover was moving atthis speed on a flat lunar surface and hit a small bump thatprojected it off the surface at an angle of 20°, how long
would it be “airborne” on the Moon?
55. A soccer goal is 2.44 m high. A player kicks the ballat a distance 10 m from the goal at an angle of 25°. What
is the initial speed of the soccer ball?
56. Olympus Mons on Mars is the largest volcano in thesolar system, at a height of 25 km and with a radius of 312km. If you are standing on the summit, with what initialvelocity would you have to fire a projectile from a cannonhorizontally to clear the volcano and land on the surface


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of Mars? Note that Mars has an acceleration of gravity of
3.7 m/s2.


57. In 1999, Robbie Knievel was the first to jump theGrand Canyon on a motorcycle. At a narrow part of thecanyon (69.0 m wide) and traveling 35.8 m/s off the takeofframp, he reached the other side. What was his launchangle?
58. You throw a baseball at an initial speed of 15.0 m/s atan angle of 30° with respect to the horizontal. What would
the ball’s initial speed have to be at 30° on a planet that
has twice the acceleration of gravity as Earth to achieve thesame range? Consider launch and impact on a horizontalsurface.
59. Aaron Rogers throws a football at 20.0 m/s to his widereceiver, who runs straight down the field at 9.4 m/s for20.0 m. If Aaron throws the football when the wide receiverhas reached 10.0 m, what angle does Aaron have to launchthe ball so the receiver catches it at the 20.0 m mark?


4.4 Uniform Circular Motion
60. A flywheel is rotating at 30 rev/s. What is the totalangle, in radians, through which a point on the flywheelrotates in 40 s?
61. A particle travels in a circle of radius 10 m at aconstant speed of 20 m/s. What is the magnitude of theacceleration?
62. Cam Newton of the Carolina Panthers throws a perfectfootball spiral at 8.0 rev/s. The radius of a pro football is 8.5cm at the middle of the short side. What is the centripetalacceleration of the laces on the football?
63. A fairground ride spins its occupants inside a flyingsaucer-shaped container. If the horizontal circular path theriders follow has an 8.00-m radius, at how manyrevolutions per minute are the riders subjected to acentripetal acceleration equal to that of gravity?
64. A runner taking part in the 200-m dash must runaround the end of a track that has a circular arc with aradius of curvature of 30.0 m. The runner starts the race ata constant speed. If she completes the 200-m dash in 23.2s and runs at constant speed throughout the race, what isher centripetal acceleration as she runs the curved portionof the track?
65. What is the acceleration of Venus toward the Sun,assuming a circular orbit?
66. An experimental jet rocket travels around Earth along


its equator just above its surface. At what speed must the jettravel if the magnitude of its acceleration is g?
67. A fan is rotating at a constant 360.0 rev/min. What isthe magnitude of the acceleration of a point on one of itsblades 10.0 cm from the axis of rotation?
68. A point located on the second hand of a large clock
has a radial acceleration of 0.1cm/s2. How far is the point
from the axis of rotation of the second hand?


4.5 Relative Motion in One and Two
Dimensions
69. The coordinate axes of the reference frame S′ remain
parallel to those of S, as S′ moves away from S at a
constant velocity v→ S′ = (4.0 i^ + 3.0 j^ + 5.0k^ ) m/s.
(a) If at time t = 0 the origins coincide, what is the positionof the origin O′ in the S frame as a function of time?
(b) How is particle position for r→ (t) and r→ ′(t), as
measured in S and S′, respectively, related? (c) What
is the relationship between particle velocities
v→ (t) and v→ ′(t)? (d) How are accelerations
a→ (t) and a→ ′ (t) related?


70. The coordinate axes of the reference frame S′ remain
parallel to those of S, as S′ moves away from S at a
constant velocity v→ S′S = (1.0 i^ + 2.0 j^ + 3.0k^ )tm/s
. (a) If at time t = 0 the origins coincide, what is theposition of origin O′ in the S frame as a function of time?
(b) How is particle position for r→ (t) and r→ ′(t) , as
measured in S and S′, respectively, related? (c) What
is the relationship between particle velocities
v→ (t) and v→ ′(t)? (d) How are accelerations
a→ (t) and a→ ′ (t) related?


71. The velocity of a particle in reference frame A is
(2.0 i


^
+ 3.0 j


^
) m/s. The velocity of reference frame A


with respect to reference frame B is 4.0k^m/s, and the
velocity of reference frame B with respect to C is
2.0 j


^
m/s. What is the velocity of the particle in reference


frame C?
72. Raindrops fall vertically at 4.5 m/s relative to theearth. What does an observer in a car moving at 22.0 m/s ina straight line measure as the velocity of the raindrops?


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73. A seagull can fly at a velocity of 9.00 m/s in still air.(a) If it takes the bird 20.0 min to travel 6.00 km straightinto an oncoming wind, what is the velocity of the wind?(b) If the bird turns around and flies with the wind, howlong will it take the bird to return 6.00 km?
74. A ship sets sail from Rotterdam, heading due north at7.00 m/s relative to the water. The local ocean current is1.50 m/s in a direction 40.0° north of east. What is the
velocity of the ship relative to Earth?
75. A boat can be rowed at 8.0 km/h in still water. (a) Howmuch time is required to row 1.5 km downstream in a rivermoving 3.0 km/h relative to the shore? (b) How much timeis required for the return trip? (c) In what direction must theboat be aimed to row straight across the river? (d) Supposethe river is 0.8 km wide. What is the velocity of the boatwith respect to Earth and how much time is required toget to the opposite shore? (e) Suppose, instead, the boat isaimed straight across the river. How much time is required


to get across and how far downstream is the boat when itreaches the opposite shore?
76. A small plane flies at 200 km/h in still air. If thewind blows directly out of the west at 50 km/h, (a) in whatdirection must the pilot head her plane to move directlynorth across land and (b) how long does it take her to reacha point 300 km directly north of her starting point?
77. A cyclist traveling southeast along a road at 15 km/hfeels a wind blowing from the southwest at 25 km/h. To astationary observer, what are the speed and direction of thewind?
78. A river is moving east at 4 m/s. A boat starts fromthe dock heading 30° north of west at 7 m/s. If the river
is 1800 m wide, (a) what is the velocity of the boat withrespect to Earth and (b) how long does it take the boat tocross the river?


ADDITIONAL PROBLEMS
79. A Formula One race car is traveling at 89.0 m/s alonga straight track enters a turn on the race track with radiusof curvature of 200.0 m. What centripetal acceleration mustthe car have to stay on the track?
80. A particle travels in a circular orbit of radius 10 m.
Its speed is changing at a rate of 15.0 m/s2 at an instant
when its speed is 40.0 m/s. What is the magnitude of theacceleration of the particle?
81. The driver of a car moving at 90.0 km/h presses downon the brake as the car enters a circular curve of radius150.0 m. If the speed of the car is decreasing at a rateof 9.0 km/h each second, what is the magnitude of theacceleration of the car at the instant its speed is 60.0 km/h?
82. A race car entering the curved part of the track atthe Daytona 500 drops its speed from 85.0 m/s to 80.0 m/s in 2.0 s. If the radius of the curved part of the track is316.0 m, calculate the total acceleration of the race car atthe beginning and ending of reduction of speed.
83. An elephant is located on Earth’s surface at a latitude
λ. Calculate the centripetal acceleration of the elephant
resulting from the rotation of Earth around its polar axis.Express your answer in terms of λ, the radius RE of
Earth, and time T for one rotation of Earth. Compare youranswer with g for λ = 40°.


84. A proton in a synchrotron is moving in a circle ofradius 1 km and increasing its speed by
v(t) = c1 + c2 t


2, where c1 = 2.0 × 10
5m/s,


c2 = 10
5m/s3. (a) What is the proton’s total acceleration


at t = 5.0 s? (b) At what time does the expression for thevelocity become unphysical?


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85. A propeller blade at rest starts to rotate from t = 0 s to t= 5.0 s with a tangential acceleration of the tip of the blade
at 3.00 m/s2. The tip of the blade is 1.5 m from the axis of
rotation. At t = 5.0 s, what is the total acceleration of the tipof the blade?
86. A particle is executing circular motion with a constantangular frequency of ω = 4.00 rad/s. If time t = 0
corresponds to the position of the particle being located aty = 0 m and x = 5 m, (a) what is the position of the particleat t = 10 s? (b) What is its velocity at this time? (c) What isits acceleration?
87. A particle’s centripetal acceleration is aC = 4.0 m/s2
at t = 0 s. It is executing uniform circular motion about anaxis at a distance of 5.0 m. What is its velocity at t = 10 s?
88. A rod 3.0 m in length is rotating at 2.0 rev/s aboutan axis at one end. Compare the centripetal accelerations atradii of (a) 1.0 m, (b) 2.0 m, and (c) 3.0 m.


89. A particle located initially at (1.5 j^ + 4.0k^ )m
undergoes a displacement of (2.5 i^ + 3.2 j^ − 1.2k^ ) m.
What is the final position of the particle?
90. The position of a particle is given by
r→ (t) = (50 m/s)t i


^
− (4.9 m/s2)t2 j


^
. (a) What are the


particle’s velocity and acceleration as functions of time? (b)What are the initial conditions to produce the motion?
91. A spaceship is traveling at a constant velocity of
v→ (t) = 250.0 i


^
m/s when its rockets fire, giving it an


acceleration of a→ (t) = (3.0 i^ + 4.0k^ )m/s2. What is its


velocity 5 s after the rockets fire?
92. A crossbow is aimed horizontally at a target 40 maway. The arrow hits 30 cm below the spot at which it wasaimed. What is the initial velocity of the arrow?
93. A long jumper can jump a distance of 8.0 m when hetakes off at an angle of 45° with respect to the horizontal.
Assuming he can jump with the same initial speed at allangles, how much distance does he lose by taking off at
30°?


94. On planet Arcon, the maximum horizontal range ofa projectile launched at 10 m/s is 20 m. What is theacceleration of gravity on this planet?
95. A mountain biker encounters a jump on a race coursethat sends him into the air at 60° to the horizontal. If he
lands at a horizontal distance of 45.0 m and 20 m below hislaunch point, what is his initial speed?
96. Which has the greater centripetal acceleration, a carwith a speed of 15.0 m/s along a circular track of radius100.0 m or a car with a speed of 12.0 m/s along a circulartrack of radius 75.0 m?
97. A geosynchronous satellite orbits Earth at a distanceof 42,250.0 km and has a period of 1 day. What is thecentripetal acceleration of the satellite?
98. Two speedboats are traveling at the same speedrelative to the water in opposite directions in a movingriver. An observer on the riverbank sees the boats movingat 4.0 m/s and 5.0 m/s. (a) What is the speed of the boatsrelative to the river? (b) How fast is the river movingrelative to the shore?


CHALLENGE PROBLEMS
99. World’s Longest Par 3. The tee of the world’s longestpar 3 sits atop South Africa’s Hanglip Mountain at 400.0m above the green and can only be reached by helicopter.The horizontal distance to the green is 359.0 m. Neglectair resistance and answer the following questions. (a) Ifa golfer launches a shot that is 40° with respect to the
horizontal, what initial velocity must she give the ball? (b)What is the time to reach the green?
100. When a field goal kicker kicks a football as hard ashe can at 45° to the horizontal, the ball just clears the 3-m-
high crossbar of the goalposts 45.7 m away. (a) What isthe maximum speed the kicker can impart to the football?


(b) In addition to clearing the crossbar, the football mustbe high enough in the air early during its flight to clear thereach of the onrushing defensive lineman. If the lineman is4.6 m away and has a vertical reach of 2.5 m, can he blockthe 45.7-m field goal attempt? (c) What if the lineman is 1.0m away?


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101. A truck is traveling east at 80 km/h. At anintersection 32 km ahead, a car is traveling north at 50km/h. (a) How long after this moment will the vehicles beclosest to each other? (b) How far apart will they be at thatpoint?


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5 | NEWTON'S LAWS OFMOTION


Figure 5.1 The Golden Gate Bridge, one of the greatest works of modern engineering, was the longest suspension bridge in theworld in the year it opened, 1937. It is still among the 10 longest suspension bridges as of this writing. In designing and buildinga bridge, what physics must we consider? What forces act on the bridge? What forces keep the bridge from falling? How do thetowers, cables, and ground interact to maintain stability?


Chapter Outline
5.1 Forces
5.2 Newton's First Law
5.3 Newton's Second Law
5.4 Mass and Weight
5.5 Newton’s Third Law
5.6 Common Forces
5.7 Drawing Free-Body Diagrams


Introduction
When you drive across a bridge, you expect it to remain stable. You also expect to speed up or slow your car in responseto traffic changes. In both cases, you deal with forces. The forces on the bridge are in equilibrium, so it stays in place. Incontrast, the force produced by your car engine causes a change in motion. Isaac Newton discovered the laws of motion thatdescribe these situations.
Forces affect every moment of your life. Your body is held to Earth by force and held together by the forces of chargedparticles. When you open a door, walk down a street, lift your fork, or touch a baby’s face, you are applying forces. Zoomingin deeper, your body’s atoms are held together by electrical forces, and the core of the atom, called the nucleus, is heldtogether by the strongest force we know—strong nuclear force.


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5.1 | Forces
Learning Objectives


By the end of the section, you will be able to:
• Distinguish between kinematics and dynamics
• Understand the definition of force
• Identify simple free-body diagrams
• Define the SI unit of force, the newton
• Describe force as a vector


The study of motion is called kinematics, but kinematics only describes the way objects move—their velocity and theiracceleration. Dynamics is the study of how forces affect the motion of objects and systems. It considers the causes of motionof objects and systems of interest, where a system is anything being analyzed. The foundation of dynamics are the laws ofmotion stated by Isaac Newton (1642–1727). These laws provide an example of the breadth and simplicity of principlesunder which nature functions. They are also universal laws in that they apply to situations on Earth and in space.
Newton’s laws of motion were just one part of the monumental work that has made him legendary (Figure 5.2). Thedevelopment of Newton’s laws marks the transition from the Renaissance to the modern era. Not until the advent of modernphysics was it discovered that Newton’s laws produce a good description of motion only when the objects are moving at
speeds much less than the speed of light and when those objects are larger than the size of most molecules (about 10−9
m in diameter). These constraints define the realm of Newtonian mechanics. At the beginning of the twentieth century,Albert Einstein (1879–1955) developed the theory of relativity and, along with many other scientists, quantum mechanics.Quantum mechanics does not have the constraints present in Newtonian physics. All of the situations we consider in thischapter, and all those preceding the introduction of relativity in Relativity (http://cnx.org/content/m58555/latest/) ,are in the realm of Newtonian physics.


Figure 5.2 Isaac Newton (1642–1727) published his amazingwork, Philosophiae Naturalis Principia Mathematica, in 1687.It proposed scientific laws that still apply today to describe themotion of objects (the laws of motion). Newton also discoveredthe law of gravity, invented calculus, and made greatcontributions to the theories of light and color.


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Working Definition of Force
Dynamics is the study of the forces that cause objects and systems to move. To understand this, we need a working definitionof force. An intuitive definition of force—that is, a push or a pull—is a good place to start. We know that a push or a pullhas both magnitude and direction (therefore, it is a vector quantity), so we can define force as the push or pull on an objectwith a specific magnitude and direction. Force can be represented by vectors or expressed as a multiple of a standard force.
The push or pull on an object can vary considerably in either magnitude or direction. For example, a cannon exerts a strongforce on a cannonball that is launched into the air. In contrast, Earth exerts only a tiny downward pull on a flea. Our everydayexperiences also give us a good idea of how multiple forces add. If two people push in different directions on a third person,as illustrated in Figure 5.3, we might expect the total force to be in the direction shown. Since force is a vector, it addsjust like other vectors. Forces, like other vectors, are represented by arrows and can be added using the familiar head-to-tailmethod or trigonometric methods. These ideas were developed in Vectors.


Figure 5.3 (a) An overhead view of two ice skaters pushing on a third skater.Forces are vectors and add like other vectors, so the total force on the thirdskater is in the direction shown. (b) A free-body diagram representing the forcesacting on the third skater.


Figure 5.3(b) is our first example of a free-body diagram, which is a sketch showing all external forces acting on anobject or system. The object or system is represented by a single isolated point (or free body), and only those forces actingon it that originate outside of the object or system—that is, external forces—are shown. (These forces are the only onesshown because only external forces acting on the free body affect its motion. We can ignore any internal forces within thebody.) The forces are represented by vectors extending outward from the free body.
Free-body diagrams are useful in analyzing forces acting on an object or system, and are employed extensively in the studyand application of Newton’s laws of motion. You will see them throughout this text and in all your studies of physics. Thefollowing steps briefly explain how a free-body diagram is created; we examine this strategy in more detail in DrawingFree-Body Diagrams.


Problem-Solving Strategy: Drawing Free-Body Diagrams
1. Draw the object under consideration. If you are treating the object as a particle, represent the object as a point.Place this point at the origin of an xy-coordinate system.
2. Include all forces that act on the object, representing these forces as vectors. However, do not include the netforce on the object or the forces that the object exerts on its environment.
3. Resolve all force vectors into x- and y-components.
4. Draw a separate free-body diagram for each object in the problem.


We illustrate this strategy with two examples of free-body diagrams (Figure 5.4). The terms used in this figure areexplained in more detail later in the chapter.


Chapter 5 | Newton's Laws of Motion 209




Figure 5.4 In these free-body diagrams, N→ is the normal force, w→ is the weight of
the object, and f→ is the friction.


The steps given here are sufficient to guide you in this important problem-solving strategy. The final section of this chapterexplains in more detail how to draw free-body diagrams when working with the ideas presented in this chapter.
Development of the Force Concept
A quantitative definition of force can be based on some standard force, just as distance is measured in units relative to astandard length. One possibility is to stretch a spring a certain fixed distance (Figure 5.5) and use the force it exerts to pullitself back to its relaxed shape—called a restoring force—as a standard. The magnitude of all other forces can be consideredas multiples of this standard unit of force. Many other possibilities exist for standard forces. Some alternative definitions offorce will be given later in this chapter.


Figure 5.5 The force exerted by a stretched spring can be used as a standard unit of force. (a)This spring has a length x when undistorted. (b) When stretched a distance Δx , the spring exerts
a restoring force F→ restore, which is reproducible. (c) A spring scale is one device that uses a
spring to measure force. The force F→ restore is exerted on whatever is attached to the hook.
Here, this force has a magnitude of six units of the force standard being employed.


Let’s analyze force more deeply. Suppose a physics student sits at a table, working diligently on his homework (Figure5.6). What external forces act on him? Can we determine the origin of these forces?


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Figure 5.6 (a) The forces acting on the student are due to thechair, the table, the floor, and Earth’s gravitational attraction. (b) Insolving a problem involving the student, we may want to considerthe forces acting along the line running through his torso. A free-body diagram for this situation is shown.


In most situations, forces are grouped into two categories: contact forces and field forces. As you might guess, contactforces are due to direct physical contact between objects. For example, the student in Figure 5.6 experiences the contact
forces C→ , F→ , and T→ , which are exerted by the chair on his posterior, the floor on his feet, and the table on his
forearms, respectively. Field forces, however, act without the necessity of physical contact between objects. They dependon the presence of a “field” in the region of space surrounding the body under consideration. Since the student is in Earth’s
gravitational field, he feels a gravitational force w→ ; in other words, he has weight.
You can think of a field as a property of space that is detectable by the forces it exerts. Scientists think there are only fourfundamental force fields in nature. These are the gravitational, electromagnetic, strong nuclear, and weak fields (we consider
these four forces in nature later in this text). As noted for w→ in Figure 5.6, the gravitational field is responsible for the
weight of a body. The forces of the electromagnetic field include those of static electricity and magnetism; they are alsoresponsible for the attraction among atoms in bulk matter. Both the strong nuclear and the weak force fields are effective
only over distances roughly equal to a length of scale no larger than an atomic nucleus ( 10−15 m ). Their range is so small
that neither field has influence in the macroscopic world of Newtonian mechanics.
Contact forces are fundamentally electromagnetic. While the elbow of the student in Figure 5.6 is in contact with thetabletop, the atomic charges in his skin interact electromagnetically with the charges in the surface of the table. The net
(total) result is the force T→ . Similarly, when adhesive tape sticks to a piece of paper, the atoms of the tape are intermingled
with those of the paper to cause a net electromagnetic force between the two objects. However, in the context of Newtonianmechanics, the electromagnetic origin of contact forces is not an important concern.
Vector Notation for Force
As previously discussed, force is a vector; it has both magnitude and direction. The SI unit of force is called the
newton (abbreviated N), and 1 N is the force needed to accelerate an object with a mass of 1 kg at a rate of 1 m/s2 :
1 N = 1 kg ·m/s2. An easy way to remember the size of a newton is to imagine holding a small apple; it has a weight of
about 1 N.
We can thus describe a two-dimensional force in the form F→ = a i^ + b j^ (the unit vectors i^ and j^ indicate the


Chapter 5 | Newton's Laws of Motion 211




5.1


direction of these forces along the x-axis and the y-axis, respectively) and a three-dimensional force in the form
F


= a i
^


+ b j
^


+ ck
^
. In Figure 5.3, let’s suppose that ice skater 1, on the left side of the figure, pushes horizontally


with a force of 30.0 N to the right; we represent this as F→ 1 = 30.0 i^ N. Similarly, if ice skater 2 pushes with a force of
40.0 N in the positive vertical direction shown, we would write F→ 2 = 40.0 j^ N. The resultant of the two forces causes a
mass to accelerate—in this case, the third ice skater. This resultant is called the net external force F→ net and is found by
taking the vector sum of all external forces acting on an object or system (thus, we can also represent net external force as
∑ F→ ):


(5.1)F→ net = ∑ F→ = F→ 1 + F→ 2 +⋯


This equation can be extended to any number of forces.
In this example, we have F→ net = ∑ F→ = F→ 1 + F→ 2 = 30.0 i^ + 40.0 j^ N . The hypotenuse of the triangle shown
in Figure 5.3 is the resultant force, or net force. It is a vector. To find its magnitude (the size of the vector, without regardto direction), we use the rule given in Vectors, taking the square root of the sum of the squares of the components:


Fnet = (30.0 N)
2 + (40.0 N)2 = 50.0 N.


The direction is given by
θ = tan−1




F2
F1



⎠ = tan


−1 ⎛

40.0
30.0

⎠ = 53.1°,


measured from the positive x-axis, as shown in the free-body diagram in Figure 5.3(b).
Let’s suppose the ice skaters now push the third ice skater with F→ 1 = 3.0 i^ + 8.0 j^ N and F→ 2 = 5.0 i^ + 4.0 j^ N .
What is the resultant of these two forces? We must recognize that force is a vector; therefore, we must add using the rulesfor vector addition:


F


net = F


1 + F


2 =

⎝3.0 i


^
+ 8.0 j


^⎞
⎠+

⎝5.0 i


^
+ 4.0 j


^⎞
⎠ = 8.0 i


^
+ 12 j


^
N


Check Your Understanding Find the magnitude and direction of the net force in the ice skater examplejust given.


View this interactive simulation (https://openstaxcollege.org/l/21addvectors) to learn how to addvectors. Drag vectors onto a graph, change their length and angle, and sum them together. The magnitude, angle,and components of each vector can be displayed in several formats.


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5.2 | Newton's First Law
Learning Objectives


By the end of the section, you will be able to:
• Describe Newton's first law of motion
• Recognize friction as an external force
• Define inertia
• Identify inertial reference frames
• Calculate equilibrium for a system


Experience suggests that an object at rest remains at rest if left alone and that an object in motion tends to slow downand stop unless some effort is made to keep it moving. However, Newton’s first law gives a deeper explanation of thisobservation.
Newton’s First Law of Motion
A body at rest remains at rest or, if in motion, remains in motion at constant velocity unless acted on by a net externalforce.


Note the repeated use of the verb “remains.” We can think of this law as preserving the status quo of motion. Also note theexpression “constant velocity;” this means that the object maintains a path along a straight line, since neither the magnitudenor the direction of the velocity vector changes. We can use Figure 5.7 to consider the two parts of Newton’s first law.


Figure 5.7 (a) A hockey puck is shown at rest; it remains at rest until an outsideforce such as a hockey stick changes its state of rest; (b) a hockey puck is shown inmotion; it continues in motion in a straight line until an outside force causes it tochange its state of motion. Although it is slick, an ice surface provides some frictionthat slows the puck.


Rather than contradicting our experience, Newton’s first law says that there must be a cause for any change in velocity(a change in either magnitude or direction) to occur. This cause is a net external force, which we defined earlier in thechapter. An object sliding across a table or floor slows down due to the net force of friction acting on the object. If frictiondisappears, will the object still slow down?
The idea of cause and effect is crucial in accurately describing what happens in various situations. For example, considerwhat happens to an object sliding along a rough horizontal surface. The object quickly grinds to a halt. If we spray thesurface with talcum powder to make the surface smoother, the object slides farther. If we make the surface even smoother byrubbing lubricating oil on it, the object slides farther yet. Extrapolating to a frictionless surface and ignoring air resistance,


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we can imagine the object sliding in a straight line indefinitely. Friction is thus the cause of slowing (consistent withNewton’s first law). The object would not slow down if friction were eliminated.
Consider an air hockey table (Figure 5.8). When the air is turned off, the puck slides only a short distance before frictionslows it to a stop. However, when the air is turned on, it creates a nearly frictionless surface, and the puck glides longdistances without slowing down. Additionally, if we know enough about the friction, we can accurately predict how quicklythe object slows down.


Figure 5.8 An air hockey table is useful in illustrating Newton’s laws. When the air is off,friction quickly slows the puck; but when the air is on, it minimizes contact between the puckand the hockey table, and the puck glides far down the table.


Newton’s first law is general and can be applied to anything from an object sliding on a table to a satellite in orbit toblood pumped from the heart. Experiments have verified that any change in velocity (speed or direction) must be causedby an external force. The idea of generally applicable or universal laws is important—it is a basic feature of all laws ofphysics. Identifying these laws is like recognizing patterns in nature from which further patterns can be discovered. Thegenius of Galileo, who first developed the idea for the first law of motion, and Newton, who clarified it, was to ask thefundamental question: “What is the cause?” Thinking in terms of cause and effect is fundamentally different from the typicalancient Greek approach, when questions such as “Why does a tiger have stripes?” would have been answered in Aristotelianfashion, such as “That is the nature of the beast.” The ability to think in terms of cause and effect is the ability to make aconnection between an observed behavior and the surrounding world.
Gravitation and Inertia
Regardless of the scale of an object, whether a molecule or a subatomic particle, two properties remain valid and thus ofinterest to physics: gravitation and inertia. Both are connected to mass. Roughly speaking, mass is a measure of the amountof matter in something. Gravitation is the attraction of one mass to another, such as the attraction between yourself andEarth that holds your feet to the floor. The magnitude of this attraction is your weight, and it is a force.
Mass is also related to inertia, the ability of an object to resist changes in its motion—in other words, to resist acceleration.Newton’s first law is often called the law of inertia. As we know from experience, some objects have more inertia thanothers. It is more difficult to change the motion of a large boulder than that of a basketball, for example, because the boulderhas more mass than the basketball. In other words, the inertia of an object is measured by its mass. The relationship betweenmass and weight is explored later in this chapter.
Inertial Reference Frames
Earlier, we stated Newton’s first law as “A body at rest remains at rest or, if in motion, remains in motion at constant velocityunless acted on by a net external force.” It can also be stated as “Every body remains in its state of uniform motion in astraight line unless it is compelled to change that state by forces acting on it.” To Newton, “uniform motion in a straightline” meant constant velocity, which includes the case of zero velocity, or rest. Therefore, the first law says that the velocityof an object remains constant if the net force on it is zero.
Newton’s first law is usually considered to be a statement about reference frames. It provides a method for identifying aspecial type of reference frame: the inertial reference frame. In principle, we can make the net force on a body zero. Ifits velocity relative to a given frame is constant, then that frame is said to be inertial. So by definition, an inertial referenceframe is a reference frame in which Newton’s first law is valid. Newton’s first law applies to objects with constant velocity.From this fact, we can infer the following statement.


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Inertial Reference Frame
A reference frame moving at constant velocity relative to an inertial frame is also inertial. A reference frameaccelerating relative to an inertial frame is not inertial.


Are inertial frames common in nature? It turns out that well within experimental error, a reference frame at rest relativeto the most distant, or “fixed,” stars is inertial. All frames moving uniformly with respect to this fixed-star frame are alsoinertial. For example, a nonrotating reference frame attached to the Sun is, for all practical purposes, inertial, because its
velocity relative to the fixed stars does not vary by more than one part in 1010. Earth accelerates relative to the fixed stars
because it rotates on its axis and revolves around the Sun; hence, a reference frame attached to its surface is not inertial.For most problems, however, such a frame serves as a sufficiently accurate approximation to an inertial frame, because
the acceleration of a point on Earth’s surface relative to the fixed stars is rather small ( < 3.4 × 10−2 m/s2 ). Thus, unless
indicated otherwise, we consider reference frames fixed on Earth to be inertial.
Finally, no particular inertial frame is more special than any other. As far as the laws of nature are concerned, allinertial frames are equivalent. In analyzing a problem, we choose one inertial frame over another simply on the basis ofconvenience.
Newton’s First Law and Equilibrium
Newton’s first law tells us about the equilibrium of a system, which is the state in which the forces on the system are
balanced. Returning to Forces and the ice skaters in Figure 5.3, we know that the forces F→ 1 and F→ 2 combine
to form a resultant force, or the net external force: F→ R = F→ net = F→ 1 + F→ 2. To create equilibrium, we require
a balancing force that will produce a net force of zero. This force must be equal in magnitude but opposite in direction
to F→ R, which means the vector must be − F→ R. Referring to the ice skaters, for which we found F→ R to be
30.0 i


^
+ 40.0 j


^
N , we can determine the balancing force by simply finding − F→ R = −30.0 i^ − 40.0 j^ N. See the free-


body diagram in Figure 5.3(b).
We can give Newton’s first law in vector form:


(5.2)v→ = constant when F→ net = 0→ N.


This equation says that a net force of zero implies that the velocity v→ of the object is constant. (The word “constant” can
indicate zero velocity.)
Newton’s first law is deceptively simple. If a car is at rest, the only forces acting on the car are weight and the contact forceof the pavement pushing up on the car (Figure 5.9). It is easy to understand that a nonzero net force is required to changethe state of motion of the car. However, if the car is in motion with constant velocity, a common misconception is that theengine force propelling the car forward is larger in magnitude than the friction force that opposes forward motion. In fact,the two forces have identical magnitude.


Figure 5.9 A car is shown (a) parked and (b) moving at constant velocity.How do Newton’s laws apply to the parked car? What does the knowledgethat the car is moving at constant velocity tell us about the net horizontal forceon the car?


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5.2


Example 5.1
When Does Newton’s First Law Apply to Your Car?
Newton’s laws can be applied to all physical processes involving force and motion, including something asmundane as driving a car.
(a) Your car is parked outside your house. Does Newton’s first law apply in this situation? Why or why not?
(b) Your car moves at constant velocity down the street. Does Newton’s first law apply in this situation? Why orwhy not?
Strategy
In (a), we are considering the first part of Newton’s first law, dealing with a body at rest; in (b), we look at thesecond part of Newton’s first law for a body in motion.
Solutiona. When your car is parked, all forces on the car must be balanced; the vector sum is 0 N. Thus, the net forceis zero, and Newton’s first law applies. The acceleration of the car is zero, and in this case, the velocity isalso zero.


b. When your car is moving at constant velocity down the street, the net force must also be zero accordingto Newton’s first law. The car’s engine produces a forward force; friction, a force between the road andthe tires of the car that opposes forward motion, has exactly the same magnitude as the engine force,producing the net force of zero. The body continues in its state of constant velocity until the net forcebecomes nonzero. Realize that a net force of zero means that an object is either at rest or moving withconstant velocity, that is, it is not accelerating. What do you suppose happens when the car accelerates?We explore this idea in the next section.
Significance
As this example shows, there are two kinds of equilibrium. In (a), the car is at rest; we say it is in staticequilibrium. In (b), the forces on the car are balanced, but the car is moving; we say that it is in dynamicequilibrium. (We examine this idea in more detail in Static Equilibrium and Elasticity.) Again, it is possiblefor two (or more) forces to act on an object yet for the object to move. In addition, a net force of zero cannotproduce acceleration.


Check Your Understanding A skydiver opens his parachute, and shortly thereafter, he is moving atconstant velocity. (a) What forces are acting on him? (b) Which force is bigger?


Engage this simulation (https://openstaxcollege.org/l/21forcemotion) to predict, qualitatively, how anexternal force will affect the speed and direction of an object’s motion. Explain the effects with the help of a free-body diagram. Use free-body diagrams to draw position, velocity, acceleration, and force graphs, and vice versa.Explain how the graphs relate to one another. Given a scenario or a graph, sketch all four graphs.


5.3 | Newton's Second Law
Learning Objectives


By the end of the section, you will be able to:
• Distinguish between external and internal forces
• Describe Newton's second law of motion
• Explain the dependence of acceleration on net force and mass


Newton’s second law is closely related to his first law. It mathematically gives the cause-and-effect relationship betweenforce and changes in motion. Newton’s second law is quantitative and is used extensively to calculate what happens in


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situations involving a force. Before we can write down Newton’s second law as a simple equation that gives the exactrelationship of force, mass, and acceleration, we need to sharpen some ideas we mentioned earlier.
Force and Acceleration
First, what do we mean by a change in motion? The answer is that a change in motion is equivalent to a change in velocity.A change in velocity means, by definition, that there is acceleration. Newton’s first law says that a net external force causesa change in motion; thus, we see that a net external force causes nonzero acceleration.
We defined external force in Forces as force acting on an object or system that originates outside of the object or system.Let’s consider this concept further. An intuitive notion of external is correct—it is outside the system of interest. Forexample, in Figure 5.10(a), the system of interest is the car plus the person within it. The two forces exerted by the twostudents are external forces. In contrast, an internal force acts between elements of the system. Thus, the force the personin the car exerts to hang on to the steering wheel is an internal force between elements of the system of interest. Onlyexternal forces affect the motion of a system, according to Newton’s first law. (The internal forces cancel each other out,as explained in the next section.) Therefore, we must define the boundaries of the system before we can determine whichforces are external. Sometimes, the system is obvious, whereas at other times, identifying the boundaries of a system ismore subtle. The concept of a system is fundamental to many areas of physics, as is the correct application of Newton’slaws. This concept is revisited many times in the study of physics.


Figure 5.10 Different forces exerted on the same mass produce different accelerations. (a) Two students push a stalled car.All external forces acting on the car are shown. (b) The forces acting on the car are transferred to a coordinate plane (free-bodydiagram) for simpler analysis. (c) The tow truck can produce greater external force on the same mass, and thus greateracceleration.


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From this example, you can see that different forces exerted on the same mass produce different accelerations. In Figure5.10(a), the two students push a car with a driver in it. Arrows representing all external forces are shown. The system of
interest is the car and its driver. The weight w→ of the system and the support of the ground N→ are also shown for
completeness and are assumed to cancel (because there was no vertical motion and no imbalance of forces in the vertical
direction to create a change in motion). The vector f→ represents the friction acting on the car, and it acts to the left,
opposing the motion of the car. (We discuss friction in more detail in the next chapter.) In Figure 5.10(b), all external
forces acting on the system add together to produce the net force F→ net. The free-body diagram shows all of the forces
acting on the system of interest. The dot represents the center of mass of the system. Each force vector extends from thisdot. Because there are two forces acting to the right, the vectors are shown collinearly. Finally, in Figure 5.10(c), a larger
net external force produces a larger acceleration ( a′→ > a→ ) when the tow truck pulls the car.
It seems reasonable that acceleration would be directly proportional to and in the same direction as the net external forceacting on a system. This assumption has been verified experimentally and is illustrated in Figure 5.10. To obtain an
equation for Newton’s second law, we first write the relationship of acceleration a→ and net external force F→ net as the
proportionality


a→ ∝ F


net


where the symbol ∝ means “proportional to.” (Recall from Forces that the net external force is the vector sum
of all external forces and is sometimes indicated as ∑ F→ . ) This proportionality shows what we have said in
words—acceleration is directly proportional to net external force. Once the system of interest is chosen, identify the externalforces and ignore the internal ones. It is a tremendous simplification to disregard the numerous internal forces actingbetween objects within the system, such as muscular forces within the students’ bodies, let alone the myriad forces betweenthe atoms in the objects. Still, this simplification helps us solve some complex problems.
It also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, thelarger the mass (the inertia), the smaller the acceleration produced by a given force. As illustrated in Figure 5.11, thesame net external force applied to a basketball produces a much smaller acceleration when it is applied to an SUV. Theproportionality is written as


a ∝ 1m,


where m is the mass of the system and a is the magnitude of the acceleration. Experiments have shown that acceleration isexactly inversely proportional to mass, just as it is directly proportional to net external force.


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Figure 5.11 The same force exerted on systems of different masses produces different accelerations. (a) A basketballplayer pushes on a basketball to make a pass. (Ignore the effect of gravity on the ball.) (b) The same player exerts anidentical force on a stalled SUV and produces far less acceleration. (c) The free-body diagrams are identical, permittingdirect comparison of the two situations. A series of patterns for free-body diagrams will emerge as you do more problemsand learn how to draw them in Drawing Free-Body Diagrams.


It has been found that the acceleration of an object depends only on the net external force and the mass of the object.Combining the two proportionalities just given yields Newton’s second law.
Newton’s Second Law of Motion
The acceleration of a system is directly proportional to and in the same direction as the net external force acting on thesystem and is inversely proportion to its mass. In equation form, Newton’s second law is


a→ =
F


net
m ,


where a→ is the acceleration, F→ net is the net force, and m is the mass. This is often written in the more familiar
form


(5.3)F→ net = ∑ F→ = m a→ ,
but the first equation gives more insight into what Newton’s second law means. When only the magnitude of force andacceleration are considered, this equation can be written in the simpler scalar form:


(5.4)Fnet = ma.
The law is a cause-and-effect relationship among three quantities that is not simply based on their definitions. The validityof the second law is based on experimental verification. The free-body diagram, which you will learn to draw in DrawingFree-Body Diagrams, is the basis for writing Newton’s second law.
Example 5.2


What Acceleration Can a Person Produce When Pushing a Lawn Mower?
Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N (about 11 lb.) parallelto the ground (Figure 5.12). The mass of the mower is 24 kg. What is its acceleration?


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5.3


Figure 5.12 (a) The net force on a lawn mower is 51 N to the right. At what rate does the lawn moweraccelerate to the right? (b) The free-body diagram for this problem is shown.


Strategy
This problem involves only motion in the horizontal direction; we are also given the net force, indicated by thesingle vector, but we can suppress the vector nature and concentrate on applying Newton’s second law. Since
Fnet and m are given, the acceleration can be calculated directly from Newton’s second law as Fnet = ma.
Solution
The magnitude of the acceleration a is a = Fnet/m . Entering known values gives


a = 51 N
24 kg


.


Substituting the unit of kilograms times meters per square second for newtons yields
a =


51 kg ·m/s2


24 kg
= 2.1 m/s2.


Significance
The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground.This is a result of the vector relationship expressed in Newton’s second law, that is, the vector representing netforce is the scalar multiple of the acceleration vector. There is no information given in this example about theindividual external forces acting on the system, but we can say something about their relative magnitudes. Forexample, the force exerted by the person pushing the mower must be greater than the friction opposing the motion(since we know the mower moved forward), and the vertical forces must cancel because no acceleration occursin the vertical direction (the mower is moving only horizontally). The acceleration found is small enough to bereasonable for a person pushing a mower. Such an effort would not last too long, because the person’s top speedwould soon be reached.


Check Your Understanding At the time of its launch, the HMS Titanic was the most massive mobile
object ever built, with a mass of 6.0 × 107 kg . If a force of 6 MN (6 × 106 N) was applied to the ship, what
acceleration would it experience?


In the preceding example, we dealt with net force only for simplicity. However, several forces act on the lawn mower. The
weight w→ (discussed in detail inMass and Weight) pulls down on the mower, toward the center of Earth; this produces
a contact force on the ground. The ground must exert an upward force on the lawn mower, known as the normal force N→
, which we define in Common Forces. These forces are balanced and therefore do not produce vertical acceleration. Inthe next example, we show both of these forces. As you continue to solve problems using Newton’s second law, be sure toshow multiple forces.


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Example 5.3
Which Force Is Bigger?
(a) The car shown in Figure 5.13 is moving at a constant speed. Which force is bigger, F→ engine or F→ friction
? Explain.
(b) The same car is now accelerating to the right. Which force is bigger, F→ engine or F→ friction? Explain.


Figure 5.13 A car is shown (a) moving at constant speed and(b) accelerating. How do the forces acting on the car compare ineach case? (a) What does the knowledge that the car is movingat constant velocity tell us about the net horizontal force on thecar compared to the friction force? (b) What does the knowledgethat the car is accelerating tell us about the horizontal force onthe car compared to the friction force?


Strategy
We must consider Newton’s first and second laws to analyze the situation. We need to decide which law applies;this, in turn, will tell us about the relationship between the forces.
Solutiona. The forces are equal. According to Newton’s first law, if the net force is zero, the velocity is constant.


b. In this case, F→ engine must be larger than F→ friction. According to Newton’s second law, a net force is
required to cause acceleration.


Significance
These questions may seem trivial, but they are commonly answered incorrectly. For a car or any other object tomove, it must be accelerated from rest to the desired speed; this requires that the engine force be greater thanthe friction force. Once the car is moving at constant velocity, the net force must be zero; otherwise, the carwill accelerate (gain speed). To solve problems involving Newton’s laws, we must understand whether to apply
Newton’s first law (where ∑ F→ = 0→ ) or Newton’s second law (where ∑ F→ is not zero). This will be
apparent as you see more examples and attempt to solve problems on your own.


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Example 5.4
What Rocket Thrust Accelerates This Sled?
Before manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effectson human subjects at high speeds. They consisted of a platform that was mounted on one or two rails andpropelled by several rockets.
Calculate the magnitude of force exerted by each rocket, called its thrust T, for the four-rocket propulsion system
shown in Figure 5.14. The sled’s initial acceleration is 49 m/s2 , the mass of the system is 2100 kg, and the
force of friction opposing the motion is 650 N.


Figure 5.14 A sled experiences a rocket thrust that accelerates it to the right.Each rocket creates an identical thrust T. The system here is the sled, its rockets,and its rider, so none of the forces between these objects are considered. The
arrow representing friction ( f→ ) is drawn larger than scale.


Strategy
Although forces are acting both vertically and horizontally, we assume the vertical forces cancel because thereis no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem.Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-bodydiagram in Figure 5.14.
Solution
Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for waysto find the thrust of the engines. We have defined the direction of the force and acceleration as acting “to theright,” so we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with


Fnet = ma


where Fnet is the net force along the horizontal direction. We can see from the figure that the engine thrusts add,
whereas friction opposes the thrust. In equation form, the net external force is


Fnet = 4T − f .


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5.4


Substituting this into Newton’s second law gives us
Fnet = ma = 4T − f .


Using a little algebra, we solve for the total thrust 4T:
4T = ma + f .


Substituting known values yields
4T = ma + f = ⎛⎝2100 kg⎞⎠⎛⎝49 m/s


2⎞
⎠+ 650 N.


Therefore, the total thrust is
4T = 1.0 × 105 N,


and the individual thrusts are
T = 1.0 × 10


5 N
4


= 2.5 × 104 N.


Significance
The numbers are quite large, so the result might surprise you. Experiments such as this were performed in theearly 1960s to test the limits of human endurance, and the setup was designed to protect human subjects in jetfighter emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45 g’s. (Recall that g,
acceleration due to gravity, is 9.80 m/s2 . When we say that acceleration is 45 g’s, it is 45 × 9.8 m/s2, which
is approximately 440 m/s2 .) Although living subjects are not used anymore, land speeds of 10,000 km/h have
been obtained with a rocket sled.
In this example, as in the preceding one, the system of interest is obvious. We see in later examples that choosingthe system of interest is crucial—and the choice is not always obvious.
Newton’s second law is more than a definition; it is a relationship among acceleration, force, and mass. It canhelp us make predictions. Each of those physical quantities can be defined independently, so the second law tellsus something basic and universal about nature.


Check Your Understanding A 550-kg sports car collides with a 2200-kg truck, and during the collision,the net force on each vehicle is the force exerted by the other. If the magnitude of the truck’s acceleration is
10 m/s2, what is the magnitude of the sports car’s acceleration?


Component Form of Newton’s Second Law
We have developed Newton’s second law and presented it as a vector equation in Equation 5.3. This vector equation canbe written as three component equations:


(5.5)∑ F→ x = m a→ x, ∑ F→ y = m a→ y, and∑ F→ z = m a→ z.


The second law is a description of how a body responds mechanically to its environment. The influence of the environment
is the net force F→ net, the body’s response is the acceleration a→ , and the strength of the response is inversely
proportional to the mass m. The larger the mass of an object, the smaller its response (its acceleration) to the influence of theenvironment (a given net force). Therefore, a body’s mass is a measure of its inertia, as we explained in Newton’s FirstLaw.


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Example 5.5
Force on a Soccer Ball
A 0.400-kg soccer ball is kicked across the field by a player; it undergoes acceleration given by
a→ = 3.00 i


^
+ 7.00 j


^
m/s2. Find (a) the resultant force acting on the ball and (b) the magnitude and direction


of the resultant force.
Strategy
The vectors in i^ and j^ format, which indicate force direction along the x-axis and the y-axis, respectively, are
involved, so we apply Newton’s second law in vector form.
Solutiona. We apply Newton’s second law:


F


net = m a
→ = ⎛⎝0.400 kg⎞⎠



⎝3.00 i


^
+ 7.00 j


^
m/s2

⎠ = 1.20 i


^
+ 2.80 j


^
N.


b. Magnitude and direction are found using the components of F→ net :
Fnet = (1.20 N)


2 + (2.80 N)2 = 3.05 N and θ = tan−1 ⎛⎝
2.80
1.20

⎠ = 66.8°.


Significance
We must remember that Newton’s second law is a vector equation. In (a), we are multiplying a vector by a scalarto determine the net force in vector form. While the vector form gives a compact representation of the forcevector, it does not tell us how “big” it is, or where it goes, in intuitive terms. In (b), we are determining the actualsize (magnitude) of this force and the direction in which it travels.


Example 5.6
Mass of a Car
Find the mass of a car if a net force of −600.0 j^ N produces an acceleration of −0.2 j^ m/s2 .
Strategy
Vector division is not defined, so m = F→ net/ a→ cannot be performed. However, mass m is a scalar, so we can
use the scalar form of Newton’s second law, m = Fnet/a .
Solution
We use m = Fnet/a and substitute the magnitudes of the two vectors: Fnet = 600.0 N and a = 0.2 m/s2.
Therefore,


m =
Fnet
a =


600.0 N
0.2 m/s2


= 3000 kg.


Significance
Force and acceleration were given in the i^ and j^ format, but the answer, mass m, is a scalar and thus is not
given in i^ and j^ form.


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Example 5.7
Several Forces on a Particle
A particle of mass m = 4.0 kg is acted upon by four forces of magnitudes.
F1 = 10.0 N, F2 = 40.0 N, F3 = 5.0 N, and F4 = 2.0 N , with the directions as shown in the free-body
diagram in Figure 5.15. What is the acceleration of the particle?


Figure 5.15 Four forces in the xy-plane are applied to a 4.0-kgparticle.


Strategy
Because this is a two-dimensional problem, we must use a free-body diagram. First, F→ 1 must be resolved into
x- and y-components. We can then apply the second law in each direction.
Solution
We draw a free-body diagram as shown in Figure 5.15. Now we apply Newton’s second law. We consider allvectors resolved into x- and y-components:


∑ Fx = max ∑ Fy = may
F1x − F3x = max F1y + F4y − F2y = may


F1 cos 30° − F3x = max F1 sin 30° + F4y − F2y = may


(10.0 N)(cos 30°) − 5.0 N = ⎛⎝4.0 kg⎞⎠ax (10.0 N)(sin 30°) + 2.0 N − 40.0 N =

⎝4.0 kg⎞⎠ay


ax = 0.92 m/s
2. ay = −8.3 m/s


2.


Thus, the net acceleration is
a→ =

⎝0.92 i


^
− 8.3 j


^⎞
⎠m/s


2,


which is a vector of magnitude 8.4 m/s2 directed at 276° to the positive x-axis.
Significance
Numerous examples in everyday life can be found that involve three or more forces acting on a single object,


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5.5


such as cables running from the Golden Gate Bridge or a football player being tackled by three defenders. We cansee that the solution of this example is just an extension of what we have already done.
Check Your Understanding A car has forces acting on it, as shown below. The mass of the car is 1000.0kg. The road is slick, so friction can be ignored. (a) What is the net force on the car? (b) What is the accelerationof the car?


Newton’s Second Law and Momentum
Newton actually stated his second law in terms of momentum: “The instantaneous rate at which a body’s momentumchanges is equal to the net force acting on the body.” (“Instantaneous rate” implies that the derivative is involved.) This canbe given by the vector equation


(5.6)
F


net =
d p→


dt
.


This means that Newton’s second law addresses the central question of motion: What causes a change in motion of anobject? Momentum was described by Newton as “quantity of motion,” a way of combining both the velocity of an objectand its mass. We devote Linear Momentum and Collisions to the study of momentum.
For now, it is sufficient to define momentum p→ as the product of the mass of the object m and its velocity v→ :


(5.7)p→ = m v→ .
Since velocity is a vector, so is momentum.
It is easy to visualize momentum. A train moving at 10 m/s has more momentum than one that moves at 2 m/s. In everydaylife, we speak of one sports team as “having momentum” when they score points against the opposing team.
If we substitute Equation 5.7 into Equation 5.6, we obtain


F


net =
d p→


dt
=


d⎛⎝m v
→ ⎞


dt
.


When m is constant, we have
F


net = m
d⎛⎝ v


→ ⎞


dt
= m a→ .


Thus, we see that the momentum form of Newton’s second law reduces to the form given earlier in this section.


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Explore the forces at work (https://openstaxcollege.org/l/21forcesatwork) when pulling a cart(https://openstaxcollege.org/l/21pullacart) or pushing a refrigerator, crate, or person. Create an appliedforce (https://openstaxcollege.org/l/21forcemotion) and see how it makes objects move. Put an objecton a ramp (https://openstaxcollege.org/l/21ramp) and see how it affects its motion.


5.4 | Mass and Weight
Learning Objectives


By the end of the section, you will be able to:
• Explain the difference between mass and weight
• Explain why falling objects on Earth are never truly in free fall
• Describe the concept of weightlessness


Mass and weight are often used interchangeably in everyday conversation. For example, our medical records often showour weight in kilograms but never in the correct units of newtons. In physics, however, there is an important distinction.Weight is the pull of Earth on an object. It depends on the distance from the center of Earth. Unlike weight, mass does notvary with location. The mass of an object is the same on Earth, in orbit, or on the surface of the Moon.
Units of Force
The equation Fnet = ma is used to define net force in terms of mass, length, and time. As explained earlier, the SI unit of
force is the newton. Since Fnet = ma,


1 N = 1 kg ·m/s2.


Although almost the entire world uses the newton for the unit of force, in the United States, the most familiar unit of forceis the pound (lb), where 1 N = 0.225 lb. Thus, a 225-lb person weighs 1000 N.
Weight and Gravitational Force
When an object is dropped, it accelerates toward the center of Earth. Newton’s second law says that a net force on an objectis responsible for its acceleration. If air resistance is negligible, the net force on a falling object is the gravitational force,
commonly called its weight w→ , or its force due to gravity acting on an object of mass m. Weight can be denoted as a
vector because it has a direction; down is, by definition, the direction of gravity, and hence, weight is a downward force. Themagnitude of weight is denoted as w. Galileo was instrumental in showing that, in the absence of air resistance, all objectsfall with the same acceleration g. Using Galileo’s result and Newton’s second law, we can derive an equation for weight.
Consider an object with mass m falling toward Earth. It experiences only the downward force of gravity, which is the weight
w→ . Newton’s second law says that the magnitude of the net external force on an object is F→ net = m a→ . We know that
the acceleration of an object due to gravity is g→ , or a→ = g→ . Substituting these into Newton’s second law gives us
the following equations.


Weight
The gravitational force on a mass is its weight. We can write this in vector form, where w→ is weight and m is mass,
as


(5.8)w→ = m g→ .
In scalar form, we can write


(5.9)w = mg.


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Since g = 9.80 m/s2 on Earth, the weight of a 1.00-kg object on Earth is 9.80 N:
w = mg = (1.00 kg)(9.80 m/s2) = 9.80 N.


When the net external force on an object is its weight, we say that it is in free fall, that is, the only force acting on the objectis gravity. However, when objects on Earth fall downward, they are never truly in free fall because there is always someupward resistance force from the air acting on the object.
Acceleration due to gravity g varies slightly over the surface of Earth, so the weight of an object depends on its location andis not an intrinsic property of the object. Weight varies dramatically if we leave Earth’s surface. On the Moon, for example,
acceleration due to gravity is only 1.67 m/s2 . A 1.0-kg mass thus has a weight of 9.8 N on Earth and only about 1.7 N on
the Moon.
The broadest definition of weight in this sense is that the weight of an object is the gravitational force on it from the nearestlarge body, such as Earth, the Moon, or the Sun. This is the most common and useful definition of weight in physics. Itdiffers dramatically, however, from the definition of weight used by NASA and the popular media in relation to space traveland exploration. When they speak of “weightlessness” and “microgravity,” they are referring to the phenomenon we call
“free fall” in physics. We use the preceding definition of weight, force w→ due to gravity acting on an object of mass m,
and we make careful distinctions between free fall and actual weightlessness.
Be aware that weight and mass are different physical quantities, although they are closely related. Mass is an intrinsicproperty of an object: It is a quantity of matter. The quantity or amount of matter of an object is determined by the numbersof atoms and molecules of various types it contains. Because these numbers do not vary, in Newtonian physics, mass doesnot vary; therefore, its response to an applied force does not vary. In contrast, weight is the gravitational force acting on anobject, so it does vary depending on gravity. For example, a person closer to the center of Earth, at a low elevation such asNew Orleans, weighs slightly more than a person who is located in the higher elevation of Denver, even though they mayhave the same mass.
It is tempting to equate mass to weight, because most of our examples take place on Earth, where the weight of an objectvaries only a little with the location of the object. In addition, it is difficult to count and identify all of the atoms and
molecules in an object, so mass is rarely determined in this manner. If we consider situations in which g→ is a constant on
Earth, we see that weight w→ is directly proportional to mass m, since w→ = m g→ , that is, the more massive an object
is, the more it weighs. Operationally, the masses of objects are determined by comparison with the standard kilogram, aswe discussed in Units and Measurement. But by comparing an object on Earth with one on the Moon, we can easilysee a variation in weight but not in mass. For instance, on Earth, a 5.0-kg object weighs 49 N; on the Moon, where g is
1.67 m/s2 , the object weighs 8.4 N. However, the mass of the object is still 5.0 kg on the Moon.
Example 5.8


Clearing a Field
A farmer is lifting some moderately heavy rocks from a field to plant crops. He lifts a stone that weighs 40.0 lb.
(about 180 N). What force does he apply if the stone accelerates at a rate of 1.5 m/s2?
Strategy
We were given the weight of the stone, which we use in finding the net force on the stone. However, we also needto know its mass to apply Newton’s second law, so we must apply the equation for weight, w = mg , to determine
the mass.
Solution
No forces act in the horizontal direction, so we can concentrate on vertical forces, as shown in the following free-body diagram. We label the acceleration to the side; technically, it is not part of the free-body diagram, but it helpsto remind us that the object accelerates upward (so the net force is upward).


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5.6


w = mg


m = wg =
180 N


9.8 m/s2
= 18 kg


∑ F = ma
F − w = ma


F − 180 N = (18 kg)(1.5 m/s2)


F − 180 N = 27 N
F = 207 N = 210 N to two significant fig es


Significance
To apply Newton’s second law as the primary equation in solving a problem, we sometimes have to rely on otherequations, such as the one for weight or one of the kinematic equations, to complete the solution.


Check Your Understanding For Example 5.8, find the acceleration when the farmer’s applied force is230.0 N.


Can you avoid the boulder field and land safely just before your fuel runs out, as Neil Armstrong did in 1969? Thisversion of the classic video game (https://openstaxcollege.org/l/21lunarlander) accurately simulatesthe real motion of the lunar lander, with the correct mass, thrust, fuel consumption rate, and lunar gravity. The reallunar lander is hard to control.
Use this interactive simulation (https://openstaxcollege.org/l/21gravityorbits) to move the Sun, Earth,Moon, and space station to see the effects on their gravitational forces and orbital paths. Visualize the sizes anddistances between different heavenly bodies, and turn off gravity to see what would happen without it.


5.5 | Newton’s Third Law
Learning Objectives


By the end of the section, you will be able to:
• State Newton’s third law of motion
• Identify the action and reaction forces in different situations
• Apply Newton’s third law to define systems and solve problems of motion


We have thus far considered force as a push or a pull; however, if you think about it, you realize that no push or pull everoccurs by itself. When you push on a wall, the wall pushes back on you. This brings us to Newton’s third law.
Newton’s Third Law of Motion
Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and
opposite in direction to the force that it exerts. Mathematically, if a body A exerts a force F→ on body B, then B


Chapter 5 | Newton's Laws of Motion 229




simultaneously exerts a force − F→ on A, or in vector equation form,
(5.10)F→ AB = − F→ BA.


Newton’s third law represents a certain symmetry in nature: Forces always occur in pairs, and one body cannot exert a forceon another without experiencing a force itself. We sometimes refer to this law loosely as “action-reaction,” where the forceexerted is the action and the force experienced as a consequence is the reaction. Newton’s third law has practical uses inanalyzing the origin of forces and understanding which forces are external to a system.
We can readily see Newton’s third law at work by taking a look at how people move about. Consider a swimmer pushingoff the side of a pool (Figure 5.16). She pushes against the wall of the pool with her feet and accelerates in the directionopposite that of her push. The wall has exerted an equal and opposite force on the swimmer. You might think that two equaland opposite forces would cancel, but they do not because they act on different systems. In this case, there are two systemsthat we could investigate: the swimmer and the wall. If we select the swimmer to be the system of interest, as in the figure,then Fwall on feet is an external force on this system and affects its motion. The swimmer moves in the direction of this
force. In contrast, the force Ffeet on wall acts on the wall, not on our system of interest. Thus, Ffeet on wall does not directly
affect the motion of the system and does not cancel Fwall on feet. The swimmer pushes in the direction opposite that in
which she wishes to move. The reaction to her push is thus in the desired direction. In a free-body diagram, such as the oneshown in Figure 5.16, we never include both forces of an action-reaction pair; in this case, we only use Fwall on feet , not
Ffeet on wall .


Figure 5.16 When the swimmer exerts a force on the wall, she accelerates in the opposite direction; in otherwords, the net external force on her is in the direction opposite of Ffeet on wall. This opposition occurs because,
in accordance with Newton’s third law, the wall exerts a force Fwall on feet on the swimmer that is equal in
magnitude but in the direction opposite to the one she exerts on it. The line around the swimmer indicates thesystem of interest. Thus, the free-body diagram shows only Fwall on feet, w (the gravitational force), and BF,
which is the buoyant force of the water supporting the swimmer’s weight. The vertical forces w and BF cancelbecause there is no vertical acceleration.


Other examples of Newton’s third law are easy to find:
• As a professor paces in front of a whiteboard, he exerts a force backward on the floor. The floor exerts a reactionforce forward on the professor that causes him to accelerate forward.
• A car accelerates forward because the ground pushes forward on the drive wheels, in reaction to the drive wheelspushing backward on the ground. You can see evidence of the wheels pushing backward when tires spin on a gravelroad and throw the rocks backward.
• Rockets move forward by expelling gas backward at high velocity. This means the rocket exerts a large backwardforce on the gas in the rocket combustion chamber; therefore, the gas exerts a large reaction force forward on therocket. This reaction force, which pushes a body forward in response to a backward force, is called thrust. It is acommon misconception that rockets propel themselves by pushing on the ground or on the air behind them. They


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actually work better in a vacuum, where they can more readily expel the exhaust gases.
• Helicopters create lift by pushing air down, thereby experiencing an upward reaction force.
• Birds and airplanes also fly by exerting force on the air in a direction opposite that of whatever force they need. Forexample, the wings of a bird force air downward and backward to get lift and move forward.
• An octopus propels itself in the water by ejecting water through a funnel from its body, similar to a jet ski.
• When a person pulls down on a vertical rope, the rope pulls up on the person (Figure 5.17).


Figure 5.17 When the mountain climber pulls down on the rope, the rope pulls up on themountain climber.


There are two important features of Newton’s third law. First, the forces exerted (the action and reaction) are always equalin magnitude but opposite in direction. Second, these forces are acting on different bodies or systems: A’s force acts on Band B’s force acts on A. In other words, the two forces are distinct forces that do not act on the same body. Thus, they donot cancel each other.
For the situation shown in Figure 5.6, the third law indicates that because the chair is pushing upward on the boy with
force C→ , he is pushing downward on the chair with force − C→ . Similarly, he is pushing downward with forces − F→
and − T→ on the floor and table, respectively. Finally, since Earth pulls downward on the boy with force w→ , he pulls
upward on Earth with force − w→ . If that student were to angrily pound the table in frustration, he would quickly learn the
painful lesson (avoidable by studying Newton’s laws) that the table hits back just as hard.
A person who is walking or running applies Newton’s third law instinctively. For example, the runner in Figure 5.18pushes backward on the ground so that it pushes him forward.


Chapter 5 | Newton's Laws of Motion 231




Figure 5.18 The runner experiences Newton’s third law. (a) A force isexerted by the runner on the ground. (b) The reaction force of the ground onthe runner pushes him forward.


Example 5.9
Forces on a Stationary Object
The package in Figure 5.19 is sitting on a scale. The forces on the package are S→ , which is due to the scale,
and − w→ , which is due to Earth’s gravitational field. The reaction forces that the package exerts are − S→ on
the scale and w→ on Earth. Because the package is not accelerating, application of the second law yields


S


− w→ = m a→ = 0


,


so
S


= w→ .


Thus, the scale reading gives the magnitude of the package’s weight. However, the scale does not measure the
weight of the package; it measures the force − S→ on its surface. If the system is accelerating, S→ and − w→
would not be equal, as explained in Applications of Newton’s Laws.


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Figure 5.19 (a) The forces on a package sitting on a scale, along with their reaction forces. The force w→ is the
weight of the package (the force due to Earth’s gravity) and S→ is the force of the scale on the package. (b) Isolation
of the package-scale system and the package-Earth system makes the action and reaction pairs clear.


Example 5.10
Getting Up to Speed: Choosing the Correct System
A physics professor pushes a cart of demonstration equipment to a lecture hall (Figure 5.20). Her mass is 65.0kg, the cart’s mass is 12.0 kg, and the equipment’s mass is 7.0 kg. Calculate the acceleration produced when theprofessor exerts a backward force of 150 N on the floor. All forces opposing the motion, such as friction on thecart’s wheels and air resistance, total 24.0 N.


Chapter 5 | Newton's Laws of Motion 233




Figure 5.20 A professor pushes the cart with her demonstration equipment. The lengths of the arrows are
proportional to the magnitudes of the forces (except for f→ , because it is too small to drawn to scale). System 1
is appropriate for this example, because it asks for the acceleration of the entire group of objects. Only F→ floo
and f→ are external forces acting on System 1 along the line of motion. All other forces either cancel or act on
the outside world. System 2 is chosen for the next example so that F→ prof is an external force and enters into
Newton’s second law. The free-body diagrams, which serve as the basis for Newton’s second law, vary with thesystem chosen.


Strategy
Since they accelerate as a unit, we define the system to be the professor, cart, and equipment. This is System1 in Figure 5.20. The professor pushes backward with a force Ffoot of 150 N. According to Newton’s third
law, the floor exerts a forward reaction force Ffloo of 150 N on System 1. Because all motion is horizontal,
we can assume there is no net force in the vertical direction. Therefore, the problem is one-dimensional alongthe horizontal direction. As noted, friction f opposes the motion and is thus in the opposite direction of Ffloo .
We do not include the forces Fprof or Fcart because these are internal forces, and we do not include Ffoot
because it acts on the floor, not on the system. There are no other significant forces acting on System 1. If the netexternal force can be found from all this information, we can use Newton’s second law to find the acceleration asrequested. See the free-body diagram in the figure.
Solution
Newton’s second law is given by


a =
Fnet
m .


The net external force on System 1 is deduced from Figure 5.20 and the preceding discussion to be
Fnet = Ffloo − f = 150 N − 24.0 N = 126 N.


The mass of System 1 is
m = (65.0 + 12.0 + 7.0) kg = 84 kg.


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These values of Fnet and m produce an acceleration of
a =


Fnet
m =


126 N
84 kg


= 1.5 m/s2.


Significance
None of the forces between components of System 1, such as between the professor’s hands and the cart,contribute to the net external force because they are internal to System 1. Another way to look at this is thatforces between components of a system cancel because they are equal in magnitude and opposite in direction.For example, the force exerted by the professor on the cart results in an equal and opposite force back on theprofessor. In this case, both forces act on the same system and therefore cancel. Thus, internal forces (betweencomponents of a system) cancel. Choosing System 1 was crucial to solving this problem.


Example 5.11
Force on the Cart: Choosing a New System
Calculate the force the professor exerts on the cart in Figure 5.20, using data from the previous example ifneeded.
Strategy
If we define the system of interest as the cart plus the equipment (System 2 in Figure 5.20), then the net externalforce on System 2 is the force the professor exerts on the cart minus friction. The force she exerts on the cart,
Fprof , is an external force acting on System 2. Fprof was internal to System 1, but it is external to System 2 and
thus enters Newton’s second law for this system.
Solution
Newton’s second law can be used to find Fprof. We start with


a =
Fnet
m .


The magnitude of the net external force on System 2 is
Fnet = Fprof − f .


We solve for Fprof , the desired quantity:
Fprof = Fnet + f .


The value of f is given, so we must calculate net Fnet. That can be done because both the acceleration and the
mass of System 2 are known. Using Newton’s second law, we see that


Fnet = ma,


where the mass of System 2 is 19.0 kg (m = 12.0 kg + 7.0 kg ) and its acceleration was found to be
a = 1.5 m/s2 in the previous example. Thus,


Fnet = ma =

⎝19.0 kg⎞⎠⎛⎝1.5 m/s


2⎞
⎠ = 29 N.


Now we can find the desired force:
Fprof = Fnet + f = 29 N + 24.0 N = 53 N.


Significance
This force is significantly less than the 150-N force the professor exerted backward on the floor. Not all of that150-N force is transmitted to the cart; some of it accelerates the professor. The choice of a system is an important


Chapter 5 | Newton's Laws of Motion 235




5.7


analytical step both in solving problems and in thoroughly understanding the physics of the situation (which arenot necessarily the same things).
Check Your Understanding Two blocks are at rest and in contact on a frictionless surface as shownbelow, with m1 = 2.0 kg, m2 = 6.0 kg, and applied force 24 N. (a) Find the acceleration of the system of


blocks. (b) Suppose that the blocks are later separated. What force will give the second block, with the mass of6.0 kg, the same acceleration as the system of blocks?


View this video (https://openstaxcollege.org/l/21actionreact) to watch examples of action and reaction.


View this video (https://openstaxcollege.org/l/21NewtonsLaws) to watch examples of Newton’s laws andinternal and external forces.


5.6 | Common Forces
Learning Objectives


By the end of the section, you will be able to:
• Define normal and tension forces
• Distinguish between real and fictitious forces
• Apply Newton’s laws of motion to solve problems involving a variety of forces


Forces are given many names, such as push, pull, thrust, and weight. Traditionally, forces have been grouped into severalcategories and given names relating to their source, how they are transmitted, or their effects. Several of these categoriesare discussed in this section, together with some interesting applications. Further examples of forces are discussed later inthis text.
A Catalog of Forces: Normal, Tension, and Other Examples of Forces
A catalog of forces will be useful for reference as we solve various problems involving force and motion. These forcesinclude normal force, tension, friction, and spring force.
Normal force
Weight (also called the force of gravity) is a pervasive force that acts at all times and must be counteracted to keep an objectfrom falling. You must support the weight of a heavy object by pushing up on it when you hold it stationary, as illustratedin Figure 5.21(a). But how do inanimate objects like a table support the weight of a mass placed on them, such as shownin Figure 5.21(b)? When the bag of dog food is placed on the table, the table sags slightly under the load. This would benoticeable if the load were placed on a card table, but even a sturdy oak table deforms when a force is applied to it. Unlessan object is deformed beyond its limit, it will exert a restoring force much like a deformed spring (or a trampoline or divingboard). The greater the deformation, the greater the restoring force. Thus, when the load is placed on the table, the table


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sags until the restoring force becomes as large as the weight of the load. At this point, the net external force on the load iszero. That is the situation when the load is stationary on the table. The table sags quickly and the sag is slight, so we do notnotice it. But it is similar to the sagging of a trampoline when you climb onto it.


Figure 5.21 (a) The person holding the bag of dog food must supply an
upward force F→ hand equal in magnitude and opposite in direction to
the weight of the food w→ so that it doesn’t drop to the ground. (b) The
card table sags when the dog food is placed on it, much like a stifftrampoline. Elastic restoring forces in the table grow as it sags until they
supply a force N→ equal in magnitude and opposite in direction to the
weight of the load.


We must conclude that whatever supports a load, be it animate or not, must supply an upward force equal to the weight ofthe load, as we assumed in a few of the previous examples. If the force supporting the weight of an object, or a load, isperpendicular to the surface of contact between the load and its support, this force is defined as a normal force and here is
given by the symbol N→ . (This is not the newton unit for force, or N.) The word normal means perpendicular to a surface.
This means that the normal force experienced by an object resting on a horizontal surface can be expressed in vector formas follows:


(5.11)N→ = −m g→ .


In scalar form, this becomes


(5.12)N = mg.


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The normal force can be less than the object’s weight if the object is on an incline.
Example 5.12


Weight on an Incline
Consider the skier on the slope in Figure 5.22. Her mass including equipment is 60.0 kg. (a) What is heracceleration if friction is negligible? (b) What is her acceleration if friction is 45.0 N?


Figure 5.22 Since the acceleration is parallel to the slope and acting down the slope, it is most convenient toproject all forces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular to it
(axes shown to the left of the skier). N→ is perpendicular to the slope and f→ is parallel to the slope, but w→
has components along both axes, namely, wy and wx . Here, w→ has a squiggly line to show that it has been
replaced by these components. The force N→ is equal in magnitude to wy , so there is no acceleration
perpendicular to the slope, but f is less than wx , so there is a downslope acceleration (along the axis parallel to the
slope).


Strategy
This is a two-dimensional problem, since not all forces on the skier (the system of interest) are parallel. Theapproach we have used in two-dimensional kinematics also works well here. Choose a convenient coordinatesystem and project the vectors onto its axes, creating two one-dimensional problems to solve. The mostconvenient coordinate system for motion on an incline is one that has one coordinate parallel to the slope and oneperpendicular to the slope. (Motions along mutually perpendicular axes are independent.) We use x and y for theparallel and perpendicular directions, respectively. This choice of axes simplifies this type of problem, becausethere is no motion perpendicular to the slope and the acceleration is downslope. Regarding the forces, friction isdrawn in opposition to motion (friction always opposes forward motion) and is always parallel to the slope, wx
is drawn parallel to the slope and downslope (it causes the motion of the skier down the slope), and wy is drawn
as the component of weight perpendicular to the slope. Then, we can consider the separate problems of forcesparallel to the slope and forces perpendicular to the slope.
Solution
The magnitude of the component of weight parallel to the slope is


wx = w sin 25° = mg sin 25°,


and the magnitude of the component of the weight perpendicular to the slope is


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wy = w cos 25° = mg cos 25°.


a. Neglect friction. Since the acceleration is parallel to the slope, we need only consider forces parallel to theslope. (Forces perpendicular to the slope add to zero, since there is no acceleration in that direction.) The forcesparallel to the slope are the component of the skier’s weight parallel to slope wx and friction f. Using Newton’s
second law, with subscripts to denote quantities parallel to the slope,


ax =
Fnet x
m


where Fnet x = wx − mg sin 25°, assuming no friction for this part. Therefore,


ax =
Fnet x
m =


mg sin 25°
m = g sin 25°



⎝9.80 m/s


2⎞
⎠(0.4226) = 4.14 m/s


2


is the acceleration.
b. Include friction. We have a given value for friction, and we know its direction is parallel to the slope and itopposes motion between surfaces in contact. So the net external force is


Fnet x = wx − f .


Substituting this into Newton’s second law, ax = Fnet x/m, gives
ax =


Fnet x
m =


wx − f
m =


mg sin 25° − f
m .


We substitute known values to obtain
ax =



⎝60.0 kg⎞⎠⎛⎝9.80 m/s


2⎞
⎠(0.4226) − 45.0 N


60.0 kg
.


This gives us
ax = 3.39 m/s


2,


which is the acceleration parallel to the incline when there is 45.0 N of opposing friction.
Significance
Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction thanwhen there is none. It is a general result that if friction on an incline is negligible, then the acceleration down theincline is a = g sin θ , regardless of mass. As discussed previously, all objects fall with the same acceleration in
the absence of air resistance. Similarly, all objects, regardless of mass, slide down a frictionless incline with thesame acceleration (if the angle is the same).


When an object rests on an incline that makes an angle θ with the horizontal, the force of gravity acting on the object
is divided into two components: a force acting perpendicular to the plane, wy , and a force acting parallel to the plane,
wx (Figure 5.23). The normal force N→ is typically equal in magnitude and opposite in direction to the perpendicular
component of the weight wy . The force acting parallel to the plane, wx , causes the object to accelerate down the incline.


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5.8


Figure 5.23 An object rests on an incline that makes an angle θ with the
horizontal.


Be careful when resolving the weight of the object into components. If the incline is at an angle θ to the horizontal, then
the magnitudes of the weight components are


wx = w sin θ = mg sin θ


and
wy = w cos θ = mg cos θ.


We use the second equation to write the normal force experienced by an object resting on an inclined plane:


(5.13)N = mg cos θ.


Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, we draw the rightangle formed by the three weight vectors. The angle θ of the incline is the same as the angle formed between w and wy .
Knowing this property, we can use trigonometry to determine the magnitude of the weight components:


cos θ =
wy
w , wy = w cos θ = mg sin θ


sin θ = wxw , wx = w sin θ = mg sin θ.


Check Your Understanding A force of 1150 N acts parallel to a ramp to push a 250-kg gun safe into amoving van. The ramp is frictionless and inclined at 17°. (a) What is the acceleration of the safe up the ramp?
(b) If we consider friction in this problem, with a friction force of 120 N, what is the acceleration of the safe?


Tension
A tension is a force along the length of a medium; in particular, it is a pulling force that acts along a stretched flexibleconnector, such as a rope or cable. The word “tension” comes from a Latin word meaning “to stretch.” Not coincidentally,the flexible cords that carry muscle forces to other parts of the body are called tendons.
Any flexible connector, such as a string, rope, chain, wire, or cable, can only exert a pull parallel to its length; thus, aforce carried by a flexible connector is a tension with a direction parallel to the connector. Tension is a pull in a connector.Consider the phrase: “You can’t push a rope.” Instead, tension force pulls outward along the two ends of a rope.
Consider a person holding a mass on a rope, as shown in Figure 5.24. If the 5.00-kg mass in the figure is stationary, thenits acceleration is zero and the net force is zero. The only external forces acting on the mass are its weight and the tensionsupplied by the rope. Thus,


Fnet = T − w = 0,


where T and w are the magnitudes of the tension and weight, respectively, and their signs indicate direction, with up beingpositive. As we proved using Newton’s second law, the tension equals the weight of the supported mass:


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(5.14)T = w = mg.


Thus, for a 5.00-kg mass (neglecting the mass of the rope), we see that
T = mg = ⎛⎝5.00 kg⎞⎠⎛⎝9.80 m/s


2⎞
⎠ = 49.0 N.


If we cut the rope and insert a spring, the spring would extend a length corresponding to a force of 49.0 N, providing a directobservation and measure of the tension force in the rope.


Figure 5.24 When a perfectly flexible connector (onerequiring no force to bend it) such as this rope transmits a force
T


, that force must be parallel to the length of the rope, as
shown. By Newton’s third law, the rope pulls with equal forcebut in opposite directions on the hand and the supported mass(neglecting the weight of the rope). The rope is the medium thatcarries the equal and opposite forces between the two objects.The tension anywhere in the rope between the hand and themass is equal. Once you have determined the tension in onelocation, you have determined the tension at all locations alongthe rope.


Flexible connectors are often used to transmit forces around corners, such as in a hospital traction system, a tendon, or abicycle brake cable. If there is no friction, the tension transmission is undiminished; only its direction changes, and it isalways parallel to the flexible connector, as shown in Figure 5.25.


Chapter 5 | Newton's Laws of Motion 241




Figure 5.25 (a) Tendons in the finger carry force T from the muscles to other parts of the finger, usually changingthe force’s direction but not its magnitude (the tendons are relatively friction free). (b) The brake cable on a bicyclecarries the tension T from the brake lever on the handlebars to the brake mechanism. Again, the direction but not themagnitude of T is changed.


Example 5.13
What Is the Tension in a Tightrope?
Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in Figure 5.26.


Figure 5.26 The weight of a tightrope walker causes a wire to sag by 5.0° . The system of interest is the point in
the wire at which the tightrope walker is standing.


Strategy
As you can see in Figure 5.26, the wire is bent under the person’s weight. Thus, the tension on either side of theperson has an upward component that can support his weight. As usual, forces are vectors represented pictoriallyby arrows that have the same direction as the forces and lengths proportional to their magnitudes. The system is
the tightrope walker, and the only external forces acting on him are his weight w→ and the two tensions T→ L
(left tension) and T→ R (right tension). It is reasonable to neglect the weight of the wire. The net external force
is zero, because the system is static. We can use trigonometry to find the tensions. One conclusion is possible atthe outset—we can see from Figure 5.26(b) that the magnitudes of the tensions TL and TR must be equal. We
know this because there is no horizontal acceleration in the rope and the only forces acting to the left and right are


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TL and TR . Thus, the magnitude of those horizontal components of the forces must be equal so that they cancel
each other out.
Whenever we have two-dimensional vector problems in which no two vectors are parallel, the easiest methodof solution is to pick a convenient coordinate system and project the vectors onto its axes. In this case, the bestcoordinate system has one horizontal axis (x) and one vertical axis (y).
Solution
First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to look at a newfree-body diagram showing all horizontal and vertical components of each force acting on the system (Figure5.27).


Figure 5.27 When the vectors are projected onto vertical and horizontal axes, their components along theseaxes must add to zero, since the tightrope walker is stationary. The small angle results in T being muchgreater than w.


Consider the horizontal components of the forces (denoted with a subscript x):
Fnet x = TRx − TLx.


The net external horizontal force Fnet x = 0, since the person is stationary. Thus,
Fnet x = 0 = TRx − TLx
TLx = TRx.


Now observe Figure 5.27. You can use trigonometry to determine the magnitude of TL and TR :


cos 5.0° =
TLx
TL


, TLx = TL cos 5.0°


cos 5.0° =
TRx
TR


, TRx = TR cos 5.0°.


Equating TLx and TRx:
TL cos 5.0° = TR cos 5.0°.


Thus,
TL = TR = T ,


as predicted. Now, considering the vertical components (denoted by a subscript y), we can solve for T. Again,since the person is stationary, Newton’s second law implies that Fnet y = 0 . Thus, as illustrated in the free-body
diagram,


Fnet y = TLy + TRy − w = 0.


We can use trigonometry to determine the relationships among TLy, TRy, and T. As we determined from the
analysis in the horizontal direction, TL = TR = T :


Chapter 5 | Newton's Laws of Motion 243




sin 5.0° =
TLy
TL


, TLy = TL sin 5.0° = T sin 5.0°


sin 5.0° =
TRy
TR


, TRy = TR sin 5.0° = T sin 5.0°.


Now we can substitute the vales for TLy and TRy , into the net force equation in the vertical direction:
Fnet y = TLy + TRy − w = 0


Fnet y = T sin 5.0° + T sin 5.0° − w = 0


2T sin 5.0° − w = 0
2T sin 5.0° = w


and
T = w


2 sin 5.0°
=


mg
2 sin 5.0°


,


so
T =



⎝70.0 kg⎞⎠⎛⎝9.80 m/s


2⎞


2(0.0872)
,


and the tension is
T = 3930 N.


Significance
The vertical tension in the wire acts as a force that supports the weight of the tightrope walker. The tensionis almost six times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the verticalcomponent of its tension is only a fraction of the tension in the wire. The large horizontal components are inopposite directions and cancel, so most of the tension in the wire is not used to support the weight of the tightropewalker.


If we wish to create a large tension, all we have to do is exert a force perpendicular to a taut flexible connector, as illustratedin Figure 5.26. As we saw in Example 5.13, the weight of the tightrope walker acts as a force perpendicular to the rope.We saw that the tension in the rope is related to the weight of the tightrope walker in the following way:
T = w


2 sin θ
.


We can extend this expression to describe the tension T created when a perpendicular force ⎛⎝F⊥ ⎞⎠ is exerted at the middle
of a flexible connector:


T =
F⊥


2 sin θ
.


The angle between the horizontal and the bent connector is represented by θ . In this case, T becomes large as θ approaches
zero. Even the relatively small weight of any flexible connector will cause it to sag, since an infinite tension would resultif it were horizontal (i.e., θ = 0 and sin θ = 0 ). For example, Figure 5.28 shows a situation where we wish to pull a car
out of the mud when no tow truck is available. Each time the car moves forward, the chain is tightened to keep it as straight
as possible. The tension in the chain is given by T = F⊥


2 sin θ
, and since θ is small, T is large. This situation is analogous


to the tightrope walker, except that the tensions shown here are those transmitted to the car and the tree rather than thoseacting at the point where F⊥ is applied.


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5.9
Figure 5.28 We can create a large tension in the chain—and potentially a big mess—by pushing on it perpendicular to itslength, as shown.


Check Your Understanding One end of a 3.0-m rope is tied to a tree; the other end is tied to a car stuckin the mud. The motorist pulls sideways on the midpoint of the rope, displacing it a distance of 0.25 m. If heexerts a force of 200.0 N under these conditions, determine the force exerted on the car.


In Applications of Newton’s Laws, we extend the discussion on tension in a cable to include cases in which the anglesshown are not equal.
Friction
Friction is a resistive force opposing motion or its tendency. Imagine an object at rest on a horizontal surface. The net forceacting on the object must be zero, leading to equality of the weight and the normal force, which act in opposite directions. Ifthe surface is tilted, the normal force balances the component of the weight perpendicular to the surface. If the object doesnot slide downward, the component of the weight parallel to the inclined plane is balanced by friction. Friction is discussedin greater detail in the next chapter.
Spring force
A spring is a special medium with a specific atomic structure that has the ability to restore its shape, if deformed. To restoreits shape, a spring exerts a restoring force that is proportional to and in the opposite direction in which it is stretched orcompressed. This is the statement of a law known as Hooke’s law, which has the mathematical form


F


= −k x→ .


The constant of proportionality k is a measure of the spring’s stiffness. The line of action of this force is parallel to the springaxis, and the sense of the force is in the opposite direction of the displacement vector (Figure 5.29). The displacementmust be measured from the relaxed position; x = 0 when the spring is relaxed.


Figure 5.29 A spring exerts its force proportional to adisplacement, whether it is compressed or stretched. (a) Thespring is in a relaxed position and exerts no force on the block.
(b) The spring is compressed by displacement Δ x→ 1 of the
object and exerts restoring force −kΔ x→ 1. (c) The spring is
stretched by displacement Δ x→ 2 of the object and exerts
restoring force −kΔ x→ 2.


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Real Forces and Inertial Frames
There is another distinction among forces: Some forces are real, whereas others are not. Real forces have some physicalorigin, such as a gravitational pull. In contrast, fictitious forces arise simply because an observer is in an accelerating ornoninertial frame of reference, such as one that rotates (like a merry-go-round) or undergoes linear acceleration (like a carslowing down). For example, if a satellite is heading due north above Earth’s Northern Hemisphere, then to an observer onEarth, it will appear to experience a force to the west that has no physical origin. Instead, Earth is rotating toward the eastand moves east under the satellite. In Earth’s frame, this looks like a westward force on the satellite, or it can be interpretedas a violation of Newton’s first law (the law of inertia). We can identify a fictitious force by asking the question, “What isthe reaction force?” If we cannot name the reaction force, then the force we are considering is fictitious. In the example ofthe satellite, the reaction force would have to be an eastward force on Earth. Recall that an inertial frame of reference is onein which all forces are real and, equivalently, one in which Newton’s laws have the simple forms given in this chapter.
Earth’s rotation is slow enough that Earth is nearly an inertial frame. You ordinarily must perform precise experiments toobserve fictitious forces and the slight departures from Newton’s laws, such as the effect just described. On a large scale,such as for the rotation of weather systems and ocean currents, the effects can be easily observed (Figure 5.30).


Figure 5.30 Hurricane Fran is shown heading toward thesoutheastern coast of the United States in September 1996.Notice the characteristic “eye” shape of the hurricane. This is aresult of the Coriolis effect, which is the deflection of objects (inthis case, air) when considered in a rotating frame of reference,like the spin of Earth.


The crucial factor in determining whether a frame of reference is inertial is whether it accelerates or rotates relative to aknown inertial frame. Unless stated otherwise, all phenomena discussed in this text are in inertial frames.
The forces discussed in this section are real forces, but they are not the only real forces. Lift and thrust, for example, aremore specialized real forces. In the long list of forces, are some more basic than others? Are some different manifestationsof the same underlying force? The answer to both questions is yes, as you will see in the treatment of modern physics laterin the text.


Explore forces and motion in this interactive simulation (https://openstaxcollege.org/l/21ramp) as youpush household objects up and down a ramp. Lower and raise the ramp to see how the angle of inclination affectsthe parallel forces. Graphs show forces, energy, and work.
Stretch and compress springs in this activity (https://openstaxcollege.org/l/21hookeslaw) to explore therelationships among force, spring constant, and displacement. Investigate what happens when two springs areconnected in series and in parallel.


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5.7 | Drawing Free-Body Diagrams
Learning Objectives


By the end of the section, you will be able to:
• Explain the rules for drawing a free-body diagram
• Construct free-body diagrams for different situations


The first step in describing and analyzing most phenomena in physics involves the careful drawing of a free-body diagram.Free-body diagrams have been used in examples throughout this chapter. Remember that a free-body diagram must onlyinclude the external forces acting on the body of interest. Once we have drawn an accurate free-body diagram, we can applyNewton’s first law if the body is in equilibrium (balanced forces; that is, Fnet = 0 ) or Newton’s second law if the body is
accelerating (unbalanced force; that is, Fnet ≠ 0 ).
In Forces, we gave a brief problem-solving strategy to help you understand free-body diagrams. Here, we add some detailsto the strategy that will help you in constructing these diagrams.


Problem-Solving Strategy: Constructing Free-Body Diagrams
Observe the following rules when constructing a free-body diagram:


1. Draw the object under consideration; it does not have to be artistic. At first, you may want to draw a circlearound the object of interest to be sure you focus on labeling the forces acting on the object. If you are treatingthe object as a particle (no size or shape and no rotation), represent the object as a point. We often place thispoint at the origin of an xy-coordinate system.
2. Include all forces that act on the object, representing these forces as vectors. Consider the types of forcesdescribed in Common Forces—normal force, friction, tension, and spring force—as well as weight andapplied force. Do not include the net force on the object. With the exception of gravity, all of the forces wehave discussed require direct contact with the object. However, forces that the object exerts on its environmentmust not be included. We never include both forces of an action-reaction pair.
3. Convert the free-body diagram into a more detailed diagram showing the x- and y-components of a givenforce (this is often helpful when solving a problem using Newton’s first or second law). In this case, place asquiggly line through the original vector to show that it is no longer in play—it has been replaced by its x- andy-components.
4. If there are two or more objects, or bodies, in the problem, draw a separate free-body diagram for each object.


Note: If there is acceleration, we do not directly include it in the free-body diagram; however, it may help to indicateacceleration outside the free-body diagram. You can label it in a different color to indicate that it is separate from thefree-body diagram.


Let’s apply the problem-solving strategy in drawing a free-body diagram for a sled. In Figure 5.31(a), a sled is pulled byforce P at an angle of 30° . In part (b), we show a free-body diagram for this situation, as described by steps 1 and 2 of the
problem-solving strategy. In part (c), we show all forces in terms of their x- and y-components, in keeping with step 3.


Chapter 5 | Newton's Laws of Motion 247




Figure 5.31 (a) A moving sled is shown as (b) a free-body diagram and (c) a free-bodydiagram with force components.


Example 5.14
Two Blocks on an Inclined Plane
Construct the free-body diagram for object A and object B in Figure 5.32.
Strategy
We follow the four steps listed in the problem-solving strategy.
Solution
We start by creating a diagram for the first object of interest. In Figure 5.32(a), object A is isolated (circled) andrepresented by a dot.


Figure 5.32 (a) The free-body diagram for isolated object A. (b) The free-body diagram for isolated object B.Comparing the two drawings, we see that friction acts in the opposite direction in the two figures. Because object Aexperiences a force that tends to pull it to the right, friction must act to the left. Because object B experiences acomponent of its weight that pulls it to the left, down the incline, the friction force must oppose it and act up the ramp.Friction always acts opposite the intended direction of motion.


We now include any force that acts on the body. Here, no applied force is present. The weight of the object acts asa force pointing vertically downward, and the presence of the cord indicates a force of tension pointing away fromthe object. Object A has one interface and hence experiences a normal force, directed away from the interface.The source of this force is object B, and this normal force is labeled accordingly. Since object B has a tendency


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to slide down, object A has a tendency to slide up with respect to the interface, so the friction fBA is directed
downward parallel to the inclined plane.
As noted in step 4 of the problem-solving strategy, we then construct the free-body diagram in Figure 5.32(b)using the same approach. Object B experiences two normal forces and two friction forces due to the presence oftwo contact surfaces. The interface with the inclined plane exerts external forces of NB and fB , and the interface
with object B exerts the normal force NAB and friction fAB ; NAB is directed away from object B, and fAB
is opposing the tendency of the relative motion of object B with respect to object A.
Significance
The object under consideration in each part of this problem was circled in gray. When you are first learning howto draw free-body diagrams, you will find it helpful to circle the object before deciding what forces are acting onthat particular object. This focuses your attention, preventing you from considering forces that are not acting onthe body.


Example 5.15
Two Blocks in Contact
A force is applied to two blocks in contact, as shown.
Strategy
Draw a free-body diagram for each block. Be sure to consider Newton’s third law at the interface where the twoblocks touch.


Solution


Significance
A


21 is the action force of block 2 on block 1. A→ 12 is the reaction force of block 1 on block 2. We use these
free-body diagrams in Applications of Newton’s Laws.


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Example 5.16
Block on the Table (Coupled Blocks)
A block rests on the table, as shown. A light rope is attached to it and runs over a pulley. The other end of the ropeis attached to a second block. The two blocks are said to be coupled. Block m2 exerts a force due to its weight,
which causes the system (two blocks and a string) to accelerate.
Strategy
We assume that the string has no mass so that we do not have to consider it as a separate object. Draw a free-bodydiagram for each block.


Solution


Significance
Each block accelerates (notice the labels shown for a→ 1 and a→ 2 ); however, assuming the string remains
taut, they accelerate at the same rate. Thus, we have a→ 1 = a→ 2 . If we were to continue solving the problem,
we could simply call the acceleration a→ . Also, we use two free-body diagrams because we are usually finding
tension T, which may require us to use a system of two equations in this type of problem. The tension is the sameon both m1 andm2 .


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5.10 Check Your Understanding (a) Draw the free-body diagram for the situation shown. (b) Redraw itshowing components; use x-axes parallel to the two ramps.


View this simulation (https://openstaxcollege.org/l/21forcemotion) to predict, qualitatively, how anexternal force will affect the speed and direction of an object’s motion. Explain the effects with the help of a free-body diagram. Use free-body diagrams to draw position, velocity, acceleration, and force graphs, and vice versa.Explain how the graphs relate to one another. Given a scenario or a graph, sketch all four graphs.


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dynamics
external force
force
free fall
free-body diagram
Hooke’s law
inertia
inertial reference frame
law of inertia
net external force
newton
Newton’s first law of motion
Newton’s second law of motion
Newton’s third law of motion
normal force
tension
thrust
weight


CHAPTER 5 REVIEW
KEY TERMS


study of how forces affect the motion of objects and systems
force acting on an object or system that originates outside of the object or system


push or pull on an object with a specific magnitude and direction; can be represented by vectors or expressed as amultiple of a standard force
situation in which the only force acting on an object is gravity


sketch showing all external forces acting on an object or system; the system is represented by asingle isolated point, and the forces are represented by vectors extending outward from that point
in a spring, a restoring force proportional to and in the opposite direction of the imposed displacement


ability of an object to resist changes in its motion
reference frame moving at constant velocity relative to an inertial frame is also inertial; areference frame accelerating relative to an inertial frame is not inertial


see Newton’s first law of motion
vector sum of all external forces acting on an object or system; causes a mass to accelerate


SI unit of force; 1 N is the force needed to accelerate an object with a mass of 1 kg at a rate of 1 m/s2
body at rest remains at rest or, if in motion, remains in motion at constant velocity unlessacted on by a net external force; also known as the law of inertia


acceleration of a system is directly proportional to and in the same direction as thenet external force acting on the system and is inversely proportional to its mass
whenever one body exerts a force on a second body, the first body experiences a forcethat is equal in magnitude and opposite in direction to the force that it exerts


force supporting the weight of an object, or a load, that is perpendicular to the surface of contact betweenthe load and its support; the surface applies this force to an object to support the weight of the object
pulling force that acts along a stretched flexible connector, such as a rope or cable


reaction force that pushes a body forward in response to a backward force
force w→ due to gravity acting on an object of mass m


KEY EQUATIONS
Net external force F→ net = ∑ F→ = F→ 1 + F→ 2 +⋯
Newton’s first law v→ = constant when F→ net = 0→ N
Newton’s second law, vector form F→ net = ∑ F→ = m a→
Newton’s second law, scalar form Fnet = ma
Newton’s second law, component form ∑ F→ x = m a→ x ,∑ F→ y = m a→ y, and∑ F→ z = m a→ z.
Newton’s second law, momentum form F→ net = d p→dt
Definition of weight, vector form w→ = m g→


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Definition of weight, scalar form w = mg
Newton’s third law F→ AB = − F→ BA
Normal force on an object resting on ahorizontal surface, vector form N→ = −m g→
Normal force on an object resting on ahorizontal surface, scalar form N = mg
Normal force on an object resting on aninclined plane, scalar form N = mgcos θ
Tension in a cable supporting an objectof mass m at rest, scalar form T = w = mg


SUMMARY
5.1 Forces


• Dynamics is the study of how forces affect the motion of objects, whereas kinematics simply describes the wayobjects move.
• Force is a push or pull that can be defined in terms of various standards, and it is a vector that has both magnitudeand direction.
• External forces are any outside forces that act on a body. A free-body diagram is a drawing of all external forcesacting on a body.
• The SI unit of force is the newton (N).


5.2 Newton's First Law
• According to Newton’s first law, there must be a cause for any change in velocity (a change in either magnitude ordirection) to occur. This law is also known as the law of inertia.
• Friction is an external force that causes an object to slow down.
• Inertia is the tendency of an object to remain at rest or remain in motion. Inertia is related to an object’s mass.
• If an object’s velocity relative to a given frame is constant, then the frame is inertial. This means that for an inertialreference frame, Newton’s first law is valid.
• Equilibrium is achieved when the forces on a system are balanced.
• A net force of zero means that an object is either at rest or moving with constant velocity; that is, it is notaccelerating.


5.3 Newton's Second Law
• An external force acts on a system from outside the system, as opposed to internal forces, which act betweencomponents within the system.
• Newton’s second law of motion says that the net external force on an object with a certain mass is directlyproportional to and in the same direction as the acceleration of the object.
• Newton’s second law can also describe net force as the instantaneous rate of change of momentum. Thus, a netexternal force causes nonzero acceleration.


5.4 Mass and Weight
• Mass is the quantity of matter in a substance.
• The weight of an object is the net force on a falling object, or its gravitational force. The object experiencesacceleration due to gravity.
• Some upward resistance force from the air acts on all falling objects on Earth, so they can never truly be in free fall.


Chapter 5 | Newton's Laws of Motion 253




• Careful distinctions must be made between free fall and weightlessness using the definition of weight as force dueto gravity acting on an object of a certain mass.
5.5 Newton’s Third Law


• Newton’s third law of motion represents a basic symmetry in nature, with an experienced force equal in magnitudeand opposite in direction to an exerted force.
• Two equal and opposite forces do not cancel because they act on different systems.
• Action-reaction pairs include a swimmer pushing off a wall, helicopters creating lift by pushing air down, and anoctopus propelling itself forward by ejecting water from its body. Rockets, airplanes, and cars are pushed forwardby a thrust reaction force.
• Choosing a system is an important analytical step in understanding the physics of a problem and solving it.


5.6 Common Forces
• When an object rests on a surface, the surface applies a force to the object that supports the weight of the object.This supporting force acts perpendicular to and away from the surface. It is called a normal force.
• When an object rests on a nonaccelerating horizontal surface, the magnitude of the normal force is equal to theweight of the object.
• When an object rests on an inclined plane that makes an angle θ with the horizontal surface, the weight of the
object can be resolved into components that act perpendicular and parallel to the surface of the plane.


• The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension. When arope supports the weight of an object at rest, the tension in the rope is equal to the weight of the object. If the objectis accelerating, tension is greater than weight, and if it is decelerating, tension is less than weight.
• The force of friction is a force experienced by a moving object (or an object that has a tendency to move) parallel tothe interface opposing the motion (or its tendency).
• The force developed in a spring obeys Hooke’s law, according to which its magnitude is proportional to thedisplacement and has a sense in the opposite direction of the displacement.
• Real forces have a physical origin, whereas fictitious forces occur because the observer is in an accelerating ornoninertial frame of reference.


5.7 Drawing Free-Body Diagrams
• To draw a free-body diagram, we draw the object of interest, draw all forces acting on that object, and resolve allforce vectors into x- and y-components. We must draw a separate free-body diagram for each object in the problem.
• A free-body diagram is a useful means of describing and analyzing all the forces that act on a body to determineequilibrium according to Newton’s first law or acceleration according to Newton’s second law.


CONCEPTUAL QUESTIONS
5.1 Forces
1. What properties do forces have that allow us to classifythem as vectors?


5.2 Newton's First Law
2. Taking a frame attached to Earth as inertial, which ofthe following objects cannot have inertial frames attachedto them, and which are inertial reference frames?
(a) A car moving at constant velocity
(b) A car that is accelerating


(c) An elevator in free fall
(d) A space capsule orbiting Earth
(e) An elevator descending uniformly
3. A woman was transporting an open box of cupcakes toa school party. The car in front of her stopped suddenly;she applied her brakes immediately. She was wearing herseat belt and suffered no physical harm (just a great deal ofembarrassment), but the cupcakes flew into the dashboardand became “smushcakes.” Explain what happened.


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5.3 Newton's Second Law
4. Why can we neglect forces such as those holding a bodytogether when we apply Newton’s second law?
5. A rock is thrown straight up. At the top of the trajectory,the velocity is momentarily zero. Does this imply that theforce acting on the object is zero? Explain your answer.


5.4 Mass and Weight
6. What is the relationship between weight and mass?Which is an intrinsic, unchanging property of a body?
7. How much does a 70-kg astronaut weight in space, farfrom any celestial body? What is her mass at this location?
8. Which of the following statements is accurate?
(a) Mass and weight are the same thing expressed indifferent units.
(b) If an object has no weight, it must have no mass.
(c) If the weight of an object varies, so must the mass.
(d) Mass and inertia are different concepts.
(e) Weight is always proportional to mass.
9. When you stand on Earth, your feet push against it witha force equal to your weight. Why doesn’t Earth accelerateaway from you?
10. How would you give the value of g→ in vector
form?


5.5 Newton’s Third Law
11. Identify the action and reaction forces in the following


situations: (a) Earth attracts the Moon, (b) a boy kicksa football, (c) a rocket accelerates upward, (d) a caraccelerates forward, (e) a high jumper leaps, and (f) a bulletis shot from a gun.
12. Suppose that you are holding a cup of coffee in yourhand. Identify all forces on the cup and the reaction to eachforce.
13. (a) Why does an ordinary rifle recoil (kick backward)when fired? (b) The barrel of a recoilless rifle is open atboth ends. Describe how Newton’s third law applies whenone is fired. (c) Can you safely stand close behind one whenit is fired?


5.6 Common Forces
14. A table is placed on a rug. Then a book is placed onthe table. What does the floor exert a normal force on?
15. A particle is moving to the right. (a) Can the force onit to be acting to the left? If yes, what would happen? (b)Can that force be acting downward? If yes, why?


5.7 Drawing Free-Body Diagrams
16. In completing the solution for a problem involvingforces, what do we do after constructing the free-bodydiagram? That is, what do we apply?
17. If a book is located on a table, how many forces shouldbe shown in a free-body diagram of the book? Describethem.
18. If the book in the previous question is in free fall, howmany forces should be shown in a free-body diagram of thebook? Describe them.


PROBLEMS
5.1 Forces
19. Two ropes are attached to a tree, and forces of
F


1 = 2.0 i
^


+ 4.0 j
^


N and F→ 2 = 3.0 i^ + 6.0 j^ N
are applied. The forces are coplanar (in the same plane). (a)What is the resultant (net force) of these two force vectors?(b) Find the magnitude and direction of this net force.
20. A telephone pole has three cables pulling as shown
from above, with F→ 1 = ⎛⎝300.0 i^ + 500.0 j^⎞⎠ ,


F


2 = −200.0 i
^ , and F→ 3 = −800.0 j^ . (a) Find the


net force on the telephone pole in component form. (b) Findthe magnitude and direction of this net force.


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21. Two teenagers are pulling on ropes attached to a tree.The angle between the ropes is 30.0° . David pulls with a
force of 400.0 N and Stephanie pulls with a force of 300.0N. (a) Find the component form of the net force. (b) Findthe magnitude of the resultant (net) force on the tree and theangle it makes with David’s rope.


5.2 Newton's First Law
22. Two forces of F→ 1 = 75.02



⎝ i
^


− j
^⎞
⎠N and


F


2 =
150.0


2

⎝ i
^


− j
^⎞
⎠N act on an object. Find the third


force F→ 3 that is needed to balance the first two forces.
23. While sliding a couch across a floor, Andrea and
Jennifer exert forces F→ A and F→ J on the couch.
Andrea’s force is due north with a magnitude of 130.0 Nand Jennifer’s force is 32° east of north with a magnitude
of 180.0 N. (a) Find the net force in component form. (b)Find the magnitude and direction of the net force. (c) IfAndrea and Jennifer’s housemates, David and Stephanie,disagree with the move and want to prevent its relocation,
with what combined force F→ DS should they push so that
the couch does not move?


5.3 Newton's Second Law
24. Andrea, a 63.0-kg sprinter, starts a race with an
acceleration of 4.200 m/s2 . What is the net external force
on her?
25. If the sprinter from the previous problem acceleratesat that rate for 20.00 m and then maintains that velocity forthe remainder of a 100.00-m dash, what will her time be forthe race?
26. A cleaner pushes a 4.50-kg laundry cart in such a waythat the net external force on it is 60.0 N. Calculate the


magnitude of his cart’s acceleration.
27. Astronauts in orbit are apparently weightless. Thismeans that a clever method of measuring the mass ofastronauts is needed to monitor their mass gains or losses,and adjust their diet. One way to do this is to exert aknown force on an astronaut and measure the accelerationproduced. Suppose a net external force of 50.0 N is exerted,and an astronaut’s acceleration is measured to be
0.893 m/s2 . (a) Calculate her mass. (b) By exerting a force
on the astronaut, the vehicle in which she orbits experiencesan equal and opposite force. Use this knowledge to find anequation for the acceleration of the system (astronaut andspaceship) that would be measured by a nearby observer.(c) Discuss how this would affect the measurement of theastronaut’s acceleration. Propose a method by which recoilof the vehicle is avoided.
28. In Figure 5.12, the net external force on the 24-kgmower is given as 51 N. If the force of friction opposingthe motion is 24 N, what force F (in newtons) is the personexerting on the mower? Suppose the mower is moving at1.5 m/s when the force F is removed. How far will themower go before stopping?
29. The rocket sled shown below decelerates at a rate
of 196 m/s2 . What force is necessary to produce this
deceleration? Assume that the rockets are off. The mass of
the system is 2.10 × 103 kg.


30. If the rocket sled shown in the previous problem startswith only one rocket burning, what is the magnitude ofthis acceleration? Assume that the mass of the system is
2.10 × 103 kg, the thrust T is 2.40 × 104 N, and the
force of friction opposing the motion is 650.0 N. (b) Why isthe acceleration not one-fourth of what it is with all rocketsburning?
31. What is the deceleration of the rocket sled if it comesto rest in 1.10 s from a speed of 1000.0 km/h? (Suchdeceleration caused one test subject to black out and havetemporary blindness.)
32. Suppose two children push horizontally, but in exactlyopposite directions, on a third child in a wagon. The firstchild exerts a force of 75.0 N, the second exerts a forceof 90.0 N, friction is 12.0 N, and the mass of the thirdchild plus wagon is 23.0 kg. (a) What is the system of


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interest if the acceleration of the child in the wagon is to becalculated? (See the free-body diagram.) (b) Calculate theacceleration. (c) What would the acceleration be if frictionwere 15.0 N?


33. A powerful motorcycle can produce an acceleration
of 3.50 m/s2 while traveling at 90.0 km/h. At that speed,
the forces resisting motion, including friction and airresistance, total 400.0 N. (Air resistance is analogous to airfriction. It always opposes the motion of an object.) What isthe magnitude of the force that motorcycle exerts backwardon the ground to produce its acceleration if the mass of themotorcycle with rider is 245 kg?
34. A car with a mass of 1000.0 kg accelerates from 0 to90.0 km/h in 10.0 s. (a) What is its acceleration? (b) Whatis the net force on the car?
35. The driver in the previous problem applies the brakeswhen the car is moving at 90.0 km/h, and the car comes torest after traveling 40.0 m. What is the net force on the carduring its deceleration?
36. An 80.0-kg passenger in an SUV traveling at
1.00 × 103 km/h is wearing a seat belt. The driver slams
on the brakes and the SUV stops in 45.0 m. Find the forceof the seat belt on the passenger.
37. A particle of mass 2.0 kg is acted on by a single force
F


1 = 18 i
^


N. (a) What is the particle’s acceleration?
(b) If the particle starts at rest, how far does it travel in thefirst 5.0 s?
38. Suppose that the particle of the previous problem
also experiences forces F→ 2 = −15 i^ N and
F


3 = 6.0 j
^


N. What is its acceleration in this case?
39. Find the acceleration of the body of mass 5.0 kg shownbelow.


40. In the following figure, the horizontal surface onwhich this block slides is frictionless. If the two forcesacting on it each have magnitude F = 30.0 N and
M = 10.0 kg , what is the magnitude of the resulting
acceleration of the block?


5.4 Mass and Weight
41. The weight of an astronaut plus his space suit onthe Moon is only 250 N. (a) How much does the suitedastronaut weigh on Earth? (b) What is the mass on theMoon? On Earth?
42. Suppose the mass of a fully loaded module in which
astronauts take off from the Moon is 1.00 × 104 kg. The
thrust of its engines is 3.00 × 104 N. (a) Calculate the
module’s magnitude of acceleration in a vertical takeofffrom the Moon. (b) Could it lift off from Earth? If not, whynot? If it could, calculate the magnitude of its acceleration.
43. A rocket sled accelerates at a rate of 49.0 m/s2 .
Its passenger has a mass of 75.0 kg. (a) Calculate thehorizontal component of the force the seat exerts againsthis body. Compare this with his weight using a ratio. (b)Calculate the direction and magnitude of the total force theseat exerts against his body.
44. Repeat the previous problem for a situation in which
the rocket sled decelerates at a rate of 201 m/s2 . In this
problem, the forces are exerted by the seat and the seat belt.
45. A body of mass 2.00 kg is pushed straight upward bya 25.0 N vertical force. What is its acceleration?
46. A car weighing 12,500 N starts from rest andaccelerates to 83.0 km/h in 5.00 s. The friction force is 1350N. Find the applied force produced by the engine.
47. A body with a mass of 10.0 kg is assumed to be in


Chapter 5 | Newton's Laws of Motion 257




Earth’s gravitational field with g = 9.80 m/s2 . What is its
acceleration?
48. A fireman has mass m; he hears the fire alarm andslides down the pole with acceleration a (which is less thang in magnitude). (a) Write an equation giving the verticalforce he must apply to the pole. (b) If his mass is 90.0 kg
and he accelerates at 5.00 m/s2, what is the magnitude of
his applied force?
49. A baseball catcher is performing a stunt for atelevision commercial. He will catch a baseball (mass 145g) dropped from a height of 60.0 m above his glove. Hisglove stops the ball in 0.0100 s. What is the force exertedby his glove on the ball?
50. When the Moon is directly overhead at sunset, theforce by Earth on the Moon, FEM , is essentially at 90°
to the force by the Sun on the Moon, FSM , as shown
below. Given that FEM = 1.98 × 1020 N and
FSM = 4.36 × 10


20 N, all other forces on the Moon are
negligible, and the mass of the Moon is 7.35 × 1022 kg,
determine the magnitude of the Moon’s acceleration.


5.5 Newton’s Third Law
51. (a) What net external force is exerted on a 1100.0-kgartillery shell fired from a battleship if the shell is
accelerated at 2.40 × 104 m/s2? (b) What is the
magnitude of the force exerted on the ship by the artilleryshell, and why?
52. A brave but inadequate rugby player is being pushedbackward by an opposing player who is exerting a forceof 800.0 N on him. The mass of the losing player plusequipment is 90.0 kg, and he is accelerating backward at
1.20 m/s2 . (a) What is the force of friction between the
losing player’s feet and the grass? (b) What force does thewinning player exert on the ground to move forward if hismass plus equipment is 110.0 kg?
53. A history book is lying on top of a physics bookon a desk, as shown below; a free-body diagram is alsoshown. The history and physics books weigh 14 N and 18N, respectively. Identify each force on each book with a


double subscript notation (for instance, the contact forceof the history book pressing against physics book can be
described as F→ HP ), and determine the value of each of
these forces, explaining the process used.


54. A truck collides with a car, and during the collision,the net force on each vehicle is essentially the force exertedby the other. Suppose the mass of the car is 550 kg, themass of the truck is 2200 kg, and the magnitude of the
truck’s acceleration is 10 m/s2 . Find the magnitude of the
car’s acceleration.


5.6 Common Forces
55. A leg is suspended in a traction system, as shownbelow. (a) Which part of the figure is used to calculate theforce exerted on the foot? (b) What is the tension in the
rope? Here T→ is the tension, w→ leg is the weight of the
leg, and w→ is the weight of the load that provides the
tension.


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56. Suppose the shinbone in the preceding image was afemur in a traction setup for a broken bone, with pulleysand rope available. How might we be able to increase theforce along the femur using the same weight?
57. Two teams of nine members each engage in tug-of-war. Each of the first team’s members has an average massof 68 kg and exerts an average force of 1350 N horizontally.Each of the second team’s members has an average mass of73 kg and exerts an average force of 1365 N horizontally.(a) What is magnitude of the acceleration of the two teams,and which team wins? (b) What is the tension in the sectionof rope between the teams?
58. What force does a trampoline have to apply toJennifer, a 45.0-kg gymnast, to accelerate her straight up
at 7.50 m/s2 ? The answer is independent of the velocity
of the gymnast—she can be moving up or down or can beinstantly stationary.
59. (a) Calculate the tension in a vertical strand of spider
web if a spider of mass 2.00 × 10−5 kg hangs motionless
on it. (b) Calculate the tension in a horizontal strand ofspider web if the same spider sits motionless in the middleof it much like the tightrope walker in Figure 5.26. Thestrand sags at an angle of 12° below the horizontal.
Compare this with the tension in the vertical strand (findtheir ratio).
60. Suppose Kevin, a 60.0-kg gymnast, climbs a rope. (a)What is the tension in the rope if he climbs at a constantspeed? (b) What is the tension in the rope if he accelerates
upward at a rate of 1.50 m/s2 ?
61. Show that, as explained in the text, a force F⊥
exerted on a flexible medium at its center and perpendicularto its length (such as on the tightrope wire in Figure 5.26)


gives rise to a tension of magnitude T = F⊥ /2 sin(θ) .
62. Consider Figure 5.28. The driver attempts to get thecar out of the mud by exerting a perpendicular force of610.0 N, and the distance she pushes in the middle of therope is 1.00 m while she stands 6.00 m away from the caron the left and 6.00 m away from the tree on the right.What is the tension T in the rope, and how do you find theanswer?
63. A bird has a mass of 26 g and perches in the middleof a stretched telephone line. (a) Show that the tension in
the line can be calculated using the equation T = mg


2 sin θ
.


Determine the tension when (b) θ = 5° and (c) θ = 0.5° .
Assume that each half of the line is straight.


64. One end of a 30-m rope is tied to a tree; the otherend is tied to a car stuck in the mud. The motorist pullssideways on the midpoint of the rope, displacing it adistance of 2 m. If he exerts a force of 80 N under theseconditions, determine the force exerted on the car.
65. Consider the baby being weighed in the followingfigure. (a) What is the mass of the infant and basket if ascale reading of 55 N is observed? (b) What is tension T1
in the cord attaching the baby to the scale? (c) What istension T2 in the cord attaching the scale to the ceiling, if
the scale has a mass of 0.500 kg? (d) Sketch the situation,indicating the system of interest used to solve each part.The masses of the cords are negligible.


Chapter 5 | Newton's Laws of Motion 259




66. What force must be applied to a 100.0-kg crate on africtionless plane inclined at 30° to cause an acceleration
of 2.0 m/s2 up the plane?


67. A 2.0-kg block is on a perfectly smooth ramp thatmakes an angle of 30° with the horizontal. (a) What is
the block’s acceleration down the ramp and the force of theramp on the block? (b) What force applied upward alongand parallel to the ramp would allow the block to movewith constant velocity?


5.7 Drawing Free-Body Diagrams
68. A ball of mass m hangs at rest, suspended by a string.(a) Sketch all forces. (b) Draw the free-body diagram forthe ball.


69. A car moves along a horizontal road. Draw a free-body diagram; be sure to include the friction of the road thatopposes the forward motion of the car.
70. A runner pushes against the track, as shown. (a)Provide a free-body diagram showing all the forces on therunner. (Hint: Place all forces at the center of his body, andinclude his weight.) (b) Give a revised diagram showing thexy-component form.


71. The traffic light hangs from the cables as shown. Drawa free-body diagram on a coordinate plane for this situation.


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ADDITIONAL PROBLEMS
72. Two small forces, F→ 1 = −2.40 i^ − 6.10t j^ N and
F


2 = 8.50 i
^


− 9.70 j
^ N, are exerted on a rogue


asteroid by a pair of space tractors. (a) Find the net force.(b) What are the magnitude and direction of the net force?(c) If the mass of the asteroid is 125 kg, what accelerationdoes it experience (in vector form)? (d) What are themagnitude and direction of the acceleration?
73. Two forces of 25 and 45 N act on an object. Theirdirections differ by 70° . The resulting acceleration has
magnitude of 10.0 m/s2. What is the mass of the body?
74. A force of 1600 N acts parallel to a ramp to push a300-kg piano into a moving van. The ramp is inclined at
20° . (a) What is the acceleration of the piano up the ramp?
(b) What is the velocity of the piano when it reaches the topif the ramp is 4.0 m long and the piano starts from rest?
75. Draw a free-body diagram of a diver who has enteredthe water, moved downward, and is acted on by an upwardforce due to the water which balances the weight (that is,the diver is suspended).
76. For a swimmer who has just jumped off a divingboard, assume air resistance is negligible. The swimmer hasa mass of 80.0 kg and jumps off a board 10.0 m abovethe water. Three seconds after entering the water, herdownward motion is stopped. What average upward forcedid the water exert on her?
77. (a) Find an equation to determine the magnitude of thenet force required to stop a car of mass m, given that theinitial speed of the car is v0 and the stopping distance is x.
(b) Find the magnitude of the net force if the mass of the caris 1050 kg, the initial speed is 40.0 km/h, and the stoppingdistance is 25.0 m.
78. A sailboat has a mass of 1.50 × 103 kg and is acted
on by a force of 2.00 × 103 N toward the east, while the
wind acts behind the sails with a force of 3.00 × 103 N
in a direction 45° north of east. Find the magnitude and
direction of the resulting acceleration.
79. Find the acceleration of the body of mass 10.0 kgshown below.


80. A body of mass 2.0 kg is moving along the x-axiswith a speed of 3.0 m/s at the instant represented below. (a)What is the acceleration of the body? (b) What is the body’svelocity 10.0 s later? (c) What is its displacement after 10.0s?


81. Force F→ B has twice the magnitude of force F→ A.
Find the direction in which the particle accelerates in thisfigure.


82. Shown below is a body of mass 1.0 kg under the
influence of the forces F→ A , F→ B , and m g→ . If the
body accelerates to the left at 0.20 m/s2 , what are F→ A
and F→ B ?


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83. A force acts on a car of mass m so that the speed v
of the car increases with position x as v = kx2 , where k
is constant and all quantities are in SI units. Find the forceacting on the car as a function of position.
84. A 7.0-N force parallel to an incline is applied to a1.0-kg crate. The ramp is tilted at 20° and is frictionless.
(a) What is the acceleration of the crate? (b) If all otherconditions are the same but the ramp has a friction force of1.9 N, what is the acceleration?
85. Two boxes, A and B, are at rest. Box A is on levelground, while box B rests on an inclined plane tilted atangle θ with the horizontal. (a) Write expressions for the
normal force acting on each block. (b) Compare the twoforces; that is, tell which one is larger or whether they areequal in magnitude. (c) If the angle of incline is 10° , which
force is greater?
86. A mass of 250.0 g is suspended from a spring hangingvertically. The spring stretches 6.00 cm. How much will thespring stretch if the suspended mass is 530.0 g?
87. As shown below, two identical springs, each with thespring constant 20 N/m, support a 15.0-N weight. (a) Whatis the tension in spring A? (b) What is the amount of stretchof spring A from the rest position?


88. Shown below is a 30.0-kg block resting on africtionless ramp inclined at 60° to the horizontal. The
block is held by a spring that is stretched 5.0 cm. What isthe force constant of the spring?


89. In building a house, carpenters use nails from a largebox. The box is suspended from a spring twice during theday to measure the usage of nails. At the beginning of theday, the spring stretches 50 cm. At the end of the day, thespring stretches 30 cm. What fraction or percentage of thenails have been used?
90. A force is applied to a block to move it up a 30°
incline. The incline is frictionless. If F = 65.0 N and
M = 5.00 kg , what is the magnitude of the acceleration of
the block?


91. Two forces are applied to a 5.0-kg object, and it
accelerates at a rate of 2.0 m/s2 in the positive y-direction.
If one of the forces acts in the positive x-direction withmagnitude 12.0 N, find the magnitude of the other force.
92. The block on the right shown below has more massthan the block on the left (m2 > m1 ). Draw free-body
diagrams for each block.


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CHALLENGE PROBLEMS
93. If two tugboats pull on a disabled vessel, as shownhere in an overhead view, the disabled vessel will be pulledalong the direction indicated by the result of the exertedforces. (a) Draw a free-body diagram for the vessel.Assume no friction or drag forces affect the vessel. (b) Didyou include all forces in the overhead view in your free-body diagram? Why or why not?


94. A 10.0-kg object is initially moving east at 15.0 m/s. Then a force acts on it for 2.00 s, after which it movesnorthwest, also at 15.0 m/s. What are the magnitude anddirection of the average force that acted on the object overthe 2.00-s interval?
95. On June 25, 1983, shot-putter Udo Beyer of EastGermany threw the 7.26-kg shot 22.22 m, which at thattime was a world record. (a) If the shot was released at aheight of 2.20 m with a projection angle of 45.0° , what
was its initial velocity? (b) If while in Beyer’s hand the shotwas accelerated uniformly over a distance of 1.20 m, whatwas the net force on it?


96. A body of mass m moves in a horizontal directionsuch that at time t its position is given by
x(t) = at4 + bt3 + ct, where a, b, and c are constants. (a)
What is the acceleration of the body? (b) What is the time-dependent force acting on the body?
97. A body of mass m has initial velocity v0 in the
positive x-direction. It is acted on by a constant force F fortime t until the velocity becomes zero; the force continuesto act on the body until its velocity becomes −v0 in the
same amount of time. Write an expression for the totaldistance the body travels in terms of the variablesindicated.
98. The velocities of a 3.0-kg object at t = 6.0 s and
t = 8.0 s are (3.0 i^ − 6.0 j^ + 4.0k^ ) m/s and
(−2.0 i


^
+ 4.0k


^
) m/s , respectively. If the object is moving


at constant acceleration, what is the force acting on it?
99. A 120-kg astronaut is riding in a rocket sled that issliding along an inclined plane. The sled has a horizontal
component of acceleration of 5.0 m/s2 and a downward
component of 3.8 m/s2 . Calculate the magnitude of the
force on the rider by the sled. (Hint: Remember thatgravitational acceleration must be considered.)
100. Two forces are acting on a 5.0-kg object that moves
with acceleration 2.0 m/s2 in the positive y-direction. If
one of the forces acts in the positive x-direction and hasmagnitude of 12 N, what is the magnitude of the other


Chapter 5 | Newton's Laws of Motion 263




force?
101. Suppose that you are viewing a soccer game froma helicopter above the playing field. Two soccer playerssimultaneously kick a stationary soccer ball on the flatfield; the soccer ball has mass 0.420 kg. The first playerkicks with force 162 N at 9.0° north of west. At the same
instant, the second player kicks with force 215 N at 15°
east of south. Find the acceleration of the ball in i^ and j^
form.
102. A 10.0-kg mass hangs from a spring that has thespring constant 535 N/m. Find the position of the end of the
spring away from its rest position. (Use g = 9.80 m/s2 .)
103. A 0.0502-kg pair of fuzzy dice is attached to therearview mirror of a car by a short string. The caraccelerates at constant rate, and the dice hang at an angle of
3.20° from the vertical because of the car’s acceleration.
What is the magnitude of the acceleration of the car?
104. At a circus, a donkey pulls on a sled carrying a small
clown with a force given by 2.48 i^ + 4.33 j^ N . A horse
pulls on the same sled, aiding the hapless donkey, with a
force of 6.56 i^ + 5.33 j^ N . The mass of the sled is 575
kg. Using i^ and j^ form for the answer to each problem,
find (a) the net force on the sled when the two animalsact together, (b) the acceleration of the sled, and (c) thevelocity after 6.50 s.
105. Hanging from the ceiling over a baby bed, well outof baby’s reach, is a string with plastic shapes, as shownhere. The string is taut (there is no slack), as shown by thestraight segments. Each plastic shape has the same massm, and they are equally spaced by a distance d, as shown.The angles labeled θ describe the angle formed by the
end of the string and the ceiling at each end. The centerlength of sting is horizontal. The remaining two segmentseach form an angle with the horizontal, labeled ϕ . Let T1
be the tension in the leftmost section of the string, T2 be
the tension in the section adjacent to it, and T3 be the
tension in the horizontal segment. (a) Find an equation forthe tension in each section of the string in terms of thevariables m, g, and θ . (b) Find the angle ϕ in terms of


the angle θ . (c) If θ = 5.10° , what is the value of ϕ ? (d)
Find the distance x between the endpoints in terms of d and
θ .


106. A bullet shot from a rifle has mass of 10.0 g andtravels to the right at 350 m/s. It strikes a target, a largebag of sand, penetrating it a distance of 34.0 cm. Find themagnitude and direction of the retarding force that slowsand stops the bullet.
107. An object is acted on by three simultaneous forces:
F


1 =

⎝−3.00 i


^
+ 2.00 j


^⎞
⎠N ,


F


2 =

⎝6.00 i


^
− 4.00 j


^⎞
⎠N , and


F


3 =

⎝2.00 i


^
+ 5.00 j


^⎞
⎠N . The object experiences


acceleration of 4.23 m/s2 . (a) Find the acceleration vector
in terms of m. (b) Find the mass of the object. (c) If theobject begins from rest, find its speed after 5.00 s. (d) Findthe components of the velocity of the object after 5.00s.
108. In a particle accelerator, a proton has mass
1.67 × 10−27 kg and an initial speed of 2.00 × 105 m/s.
It moves in a straight line, and its speed increases to
9.00 × 105 m/s in a distance of 10.0 cm. Assume that the
acceleration is constant. Find the magnitude of the forceexerted on the proton.
109. A drone is being directed across a frictionless ice-covered lake. The mass of the drone is 1.50 kg, and its
velocity is 3.00 i^m/s . After 10.0 s, the velocity is
9.00 i


^
+ 4.00 j


^
m/s . If a constant force in the horizontal


direction is causing this change in motion, find (a) thecomponents of the force and (b) the magnitude of the force.


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6 | APPLICATIONS OFNEWTON'S LAWS


Figure 6.1 Stock cars racing in the Grand National Divisional race at Iowa Speedway in May, 2015. Cars often reach speeds of200 mph (320 km/h).


Chapter Outline
6.1 Solving Problems with Newton’s Laws
6.2 Friction
6.3 Centripetal Force
6.4 Drag Force and Terminal Speed


Introduction
Car racing has grown in popularity in recent years. As each car moves in a curved path around the turn, its wheels also spinrapidly. The wheels complete many revolutions while the car makes only part of one (a circular arc). How can we describethe velocities, accelerations, and forces involved? What force keeps a racecar from spinning out, hitting the wall borderingthe track? What provides this force? Why is the track banked? We answer all of these questions in this chapter as we expandour consideration of Newton’s laws of motion.


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6.1 | Solving Problems with Newton’s Laws
Learning Objectives


By the end of the section, you will be able to:
• Apply problem-solving techniques to solve for quantities in more complex systems of forces
• Use concepts from kinematics to solve problems using Newton’s laws of motion
• Solve more complex equilibrium problems
• Solve more complex acceleration problems
• Apply calculus to more advanced dynamics problems


Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzingand setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion; in this chapter, wecontinue to discuss these strategies and apply a step-by-step process.
Problem-Solving Strategies
We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies thatare useful in applying Newton’s laws of motion. Once you identify the physical principles involved in the problem anddetermine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques alsoreinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in theworked examples, so the following techniques should reinforce skills you have already begun to develop.


Problem-Solving Strategy: Applying Newton’s Laws of Motion
1. Identify the physical principles involved by listing the givens and the quantities to be calculated.
2. Sketch the situation, using arrows to represent all forces.
3. Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
4. Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from thechapter on motion along a straight line.
5. Check the solution to see whether it is reasonable.


Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once wehave determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important todraw a careful sketch of the situation. Such a sketch is shown in Figure 6.2(a). Then, as in Figure 6.2(b), we can representall forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the lengthand direction of each correspond to the represented force.


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Figure 6.2 (a) A grand piano is being lifted to a second-story apartment. (b) Arrows are used to represent all forces: T→ is
the tension in the rope above the piano, F→ T is the force that the piano exerts on the rope, and w→ is the weight of the piano.
All other forces, such as the nudge of a breeze, are assumed to be negligible. (c) Suppose we are given the piano’s mass and
asked to find the tension in the rope. We then define the system of interest as shown and draw a free-body diagram. Now F→ T
is no longer shown, because it is not a force acting on the system of interest; rather, F→ T acts on the outside world. (d)
Showing only the arrows, the head-to-tail method of addition is used. It is apparent that if the piano is stationary, T→ = − w→ .


As with most problems, we next need to identify what needs to be determined and what is known or can be inferred fromthe problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system ofinterest, since Newton’s second law involves only external forces. We can then determine which forces are external andwhich are internal, a necessary step to employ Newton’s second law. (See Figure 6.2(c).) Newton’s third law may be usedto identify whether forces are exerted between components of a system (internal) or between the system and somethingoutside (external). As illustrated in Newton’s Laws of Motion, the system of interest depends on the question we needto answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-bodydiagrams in previous worked examples. Figure 6.2(c) shows a free-body diagram for the system of interest. Note that nointernal forces are shown in a free-body diagram.
Once a free-body diagram is drawn, we apply Newton’s second law. This is done in Figure 6.2(d) for a particular situation.In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put theminto equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—thatis, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must bebroken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosenfor convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an inclineis involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almostalways convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’ssecond law in components along the different directions. Then, you have the following equations:


∑ Fx = max, ∑ Fy = may.


(If, for example, the system is accelerating horizontally, then you can then set ay = 0. ) We need this information to
determine unknown forces acting on a system.
As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example,it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. Inpractice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judgewhether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end


Chapter 6 | Applications of Newton's Laws 267




up with units of millimeters per second, then we have made a mistake.
There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section.These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first atproblems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration,which involves Newton’s second law.
Particle Equilibrium
Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objectsat rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that theseconditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same objectwould appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledgeattained in Newton’s Laws of Motion, regarding the different types of forces and the use of free-body diagrams, to solveadditional problems in particle equilibrium.
Example 6.1


Different Tensions at Different Angles
Consider the traffic light (mass of 15.0 kg) suspended from two wires as shown in Figure 6.3. Find the tensionin each wire, neglecting the masses of the wires.


Figure 6.3 A traffic light is suspended from two wires. (b) Some of the forces involved. (c) Only forces acting onthe system are shown here. The free-body diagram for the traffic light is also shown. (d) The forces projected ontovertical (y) and horizontal (x) axes. The horizontal components of the tensions must cancel, and the sum of the verticalcomponents of the tensions must equal the weight of the traffic light. (e) The free-body diagram shows the vertical andhorizontal forces acting on the traffic light.


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Strategy
The system of interest is the traffic light, and its free-body diagram is shown in Figure 6.3(c). The three forcesinvolved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinatesystem has one axis vertical and one horizontal, and the vector projections on it are shown in Figure 6.3(d).There are two unknowns in this problem ( T1 and T2 ), so two equations are needed to find them. These two
equations come from applying Newton’s second law along the vertical and horizontal axes, noting that the netexternal force is zero along each axis because acceleration is zero.
Solution
First consider the horizontal or x-axis:


Fnet x = T2x − T1x = 0.


Thus, as you might expect,
T1x = T2x.


This gives us the following relationship:
T1 cos 30° = T2 cos 45°.


Thus,
T2 = 1.225T1.


Note that T1 and T2 are not equal in this case because the angles on either side are not equal. It is reasonable
that T2 ends up being greater than T1 because it is exerted more vertically than T1.
Now consider the force components along the vertical or y-axis:


Fnet y = T1y + T2y − w = 0.


This implies
T1y + T2y = w.


Substituting the expressions for the vertical components gives
T1 sin 30° + T2 sin 45° = w.


There are two unknowns in this equation, but substituting the expression for T2 in terms of T1 reduces this to
one equation with one unknown:


T1(0.500) + (1.225T1)(0.707) = w = mg,


which yields
1.366T1 = (15.0 kg)(9.80 m/s


2).


Solving this last equation gives the magnitude of T1 to be
T1 = 108 N.


Finally, we find the magnitude of T2 by using the relationship between them, T2 = 1.225T1 , found above. Thus
we obtain


T2 = 132 N.


Significance
Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angleson either side are the same (as they were in the earlier example of a tightrope walker in Newton’s Laws of


Chapter 6 | Applications of Newton's Laws 269




Motion.
Particle Acceleration
We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems,which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice howthey are applied to the following examples.
Example 6.2


Drag Force on a Barge
Two tugboats push on a barge at different angles (Figure 6.4). The first tugboat exerts a force of 2.7 × 105 N
in the x-direction, and the second tugboat exerts a force of 3.6 × 105 N in the y-direction. The mass of the barge
is 5.0 × 106 kg and its acceleration is observed to be 7.5 × 10−2 m/s2 in the direction shown. What is the drag
force of the water on the barge resisting the motion? (Note: Drag force is a frictional force exerted by fluids, suchas air or water. The drag force opposes the motion of the object. Since the barge is flat bottomed, we can assumethat the drag force is in the direction opposite of motion of the barge.)


Figure 6.4 (a) A view from above of two tugboats pushing on a barge. (b) The free-body diagram for the shipcontains only forces acting in the plane of the water. It omits the two vertical forces—the weight of the barge and the
buoyant force of the water supporting it cancel and are not shown. Note that F→ app is the total applied force of the
tugboats.


Strategy
The directions and magnitudes of acceleration and the applied forces are given in Figure 6.4(a). We define the
total force of the tugboats on the barge as F→ app so that


F


app = F


1 + F


2.


The drag of the water F→ D is in the direction opposite to the direction of motion of the boat; this force thus
works against F→ app, as shown in the free-body diagram in Figure 6.4(b). The system of interest here is the
barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular,
the x- and y-axes are in the same direction as F→ 1 and F→ 2. The problem quickly becomes a one-dimensional


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problem along the direction of F→ app , since friction is in the direction opposite to F→ app. Our strategy is to
find the magnitude and direction of the net applied force F→ app and then apply Newton’s second law to solve
for the drag force F→ D.
Solution
Since Fx and Fy are perpendicular, we can find the magnitude and direction of F→ app directly. First, the
resultant magnitude is given by the Pythagorean theorem:


Fapp = F1
2 + F2


2 = (2.7 × 105N)2 + (3.6 × 105 N)2 = 4.5 × 105 N.


The angle is given by
θ = tan−1




F2
F1



⎠ = tan


−1⎛

3.6 × 105 N
2.7 × 105 N



⎠ = 53.1°.


From Newton’s first law, we know this is the same direction as the acceleration. We also know that F→ D is in
the opposite direction of F→ app, since it acts to slow down the acceleration. Therefore, the net external force
is in the same direction as F→ app, but its magnitude is slightly less than F→ app. The problem is now one-
dimensional. From the free-body diagram, we can see that


Fnet = Fapp − FD.


However, Newton’s second law states that
Fnet = ma.


Thus,
Fapp − FD = ma.


This can be solved for the magnitude of the drag force of the water FD in terms of known quantities:
FD = Fapp − ma.


Substituting known values gives
FD =



⎝4.5 × 10


5 N⎞⎠−

⎝5.0 × 10


6 kg⎞⎠

⎝7.5 × 10


−2 m/s2⎞⎠ = 7.5 × 10
4 N.


The direction of F→ D has already been determined to be in the direction opposite to F→ app, or at an angle of
53° south of west.
Significance
The numbers used in this example are reasonable for a moderately large barge. It is certainly difficult to obtainlarger accelerations with tugboats, and small speeds are desirable to avoid running the barge into the docks. Dragis relatively small for a well-designed hull at low speeds, consistent with the answer to this example, where FD
is less than 1/600th of the weight of the ship.


In Newton’s Laws of Motion, we discussed the normal force, which is a contact force that acts normal to the surface sothat an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of anormal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight ofan object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride?Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward ata constant speed? Take a guess before reading the next example.


Chapter 6 | Applications of Newton's Laws 271




Example 6.3
What Does the Bathroom Scale Read in an Elevator?
Figure 6.5 shows a 75.0-kg man (weight of about 165 lb.) standing on a bathroom scale in an elevator. Calculate
the scale reading: (a) if the elevator accelerates upward at a rate of 1.20 m/s2, and (b) if the elevator moves
upward at a constant speed of 1 m/s.


Figure 6.5 (a) The various forces acting when a person stands on a bathroom scale in an elevator. Thearrows are approximately correct for when the elevator is accelerating upward—broken arrows represent
forces too large to be drawn to scale. T→ is the tension in the supporting cable, w→ is the weight of the
person, w→ s is the weight of the scale, w→ e is the weight of the elevator, F→ s is the force of the scale
on the person, F→ p is the force of the person on the scale, F→ t is the force of the scale on the floor of
the elevator, and N→ is the force of the floor upward on the scale. (b) The free-body diagram shows only
the external forces acting on the designated system of interest—the person—and is the diagram we use forthe solution of the problem.


Strategy
If the scale at rest is accurate, its reading equals F→ p , the magnitude of the force the person exerts downward
on it. Figure 6.5(a) shows the numerous forces acting on the elevator, scale, and person. It makes this one-dimensional problem look much more formidable than if the person is chosen to be the system of interest and afree-body diagram is drawn, as in Figure 6.5(b). Analysis of the free-body diagram using Newton’s laws canproduce answers to both Figure 6.5(a) and (b) of this example, as well as some other questions that might arise.


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The only forces acting on the person are his weight w→ and the upward force of the scale F→ s. According to
Newton’s third law, F→ p and F→ s are equal in magnitude and opposite in direction, so that we need to find
Fs in order to find what the scale reads. We can do this, as usual, by applying Newton’s second law,


F


net = m a
→ .


From the free-body diagram, we see that F→ net = F→ s − w→ , so we have
Fs − w = ma.


Solving for Fs gives us an equation with only one unknown:
Fs = ma + w,


or, because w = mg, simply
Fs = ma + mg.


No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerationsin addition to those in this situation. (Note:We are considering the case when the elevator is accelerating upward.If the elevator is accelerating downward, Newton’s second law becomes Fs − w = −ma. )
Solution


a. We have a = 1.20 m/s2, so that
Fs = (75.0 kg)(9.80 m/s


2) + (75.0 kg)(1.20 m/s2)


yielding
Fs = 825 N.b. Now, what happens when the elevator reaches a constant upward velocity? Will the scale still readmore than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because


a = Δv
Δt


and Δv = 0. Thus,
Fs = ma + mg = 0 + mg


or
Fs = (75.0 kg)(9.80 m/s


2),


which gives
Fs = 735 N.


Significance
The scale reading in Figure 6.5(a) is about 185 lb. What would the scale have read if he were stationary? Sincehis acceleration would be zero, the force of the scale would be equal to his weight:


Fnet = ma = 0 = Fs − w


Fs = w = mg


Fs = (75.0 kg)(9.80 m/s
2) = 735 N.


Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale ispushing up on the person with a force greater than his weight, as it must in order to accelerate him upward.Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feelin rapidly accelerating versus slowly accelerating elevators. In Figure 6.5(b), the scale reading is 735 N, which


Chapter 6 | Applications of Newton's Laws 273




6.1


equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, movingdown, or stationary.
Check Your Understanding Now calculate the scale reading when the elevator accelerates downward at


a rate of 1.20 m/s2.


The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevatoraccelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downwardvelocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall andaccelerating downward at g, then the scale reading is zero and the person appears to be weightless.
Example 6.4


Two Attached Blocks
Figure 6.6 shows a block of mass m1 on a frictionless, horizontal surface. It is pulled by a light string that
passes over a frictionless and massless pulley. The other end of the string is connected to a block of mass m2.
Find the acceleration of the blocks and the tension in the string in terms of m1, m2, and g.


Figure 6.6 (a) Block 1 is connected by a light string to block 2. (b) The free-body diagrams ofthe blocks.


Strategy
We draw a free-body diagram for each mass separately, as shown in Figure 6.6. Then we analyze each one tofind the required unknowns. The forces on block 1 are the gravitational force, the contact force of the surface, andthe tension in the string. Block 2 is subjected to the gravitational force and the string tension. Newton’s secondlaw applies to each, so we write two vector equations:
For block 1: T→ + w→ 1 + N→ = m1 a→ 1
For block 2: T→ + w→ 2 = m2 a→ 2.


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6.2


Notice that T→ is the same for both blocks. Since the string and the pulley have negligible mass, and since there
is no friction in the pulley, the tension is the same throughout the string. We can now write component equationsfor each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinatesystem for both objects
Solution
The component equations follow from the vector equations above. We see that block 1 has the vertical forcesbalanced, so we ignore them and write an equation relating the x-components. There are no horizontal forces onblock 2, so only the y-equation is written. We obtain these results:


Block 1 Block 2


∑ Fx = max ∑ Fy = may
Tx = m1a1x Ty − m2g = m2a2y.


When block 1 moves to the right, block 2 travels an equal distance downward; thus, a1x = −a2y. Writing the
common acceleration of the blocks as a = a1x = −a2y, we now have


T = m1a


and
T − m2g = −m2a.


From these two equations, we can express a and T in terms of the masses m1 andm2, and g :
a =


m2
m1 + m2


g


and
T =


m1m2
m1 + m2


g.


Significance
Notice that the tension in the string is less than the weight of the block hanging from the end of it. A commonerror in problems like this is to set T = m2g. You can see from the free-body diagram of block 2 that cannot be
correct if the block is accelerating.


Check Your Understanding Calculate the acceleration of the system, and the tension in the string, whenthe masses are m1 = 5.00 kg and m2 = 3.00 kg.


Example 6.5
Atwood Machine
A classic problem in physics, similar to the one we just solved, is that of the Atwood machine, which consists ofa rope running over a pulley, with two objects of different mass attached. It is particularly useful in understandingthe connection between force and motion. In Figure 6.7, m1 = 2.00 kg and m2 = 4.00 kg. Consider the
pulley to be frictionless. (a) If m2 is released, what will its acceleration be? (b) What is the tension in the string?


Chapter 6 | Applications of Newton's Laws 275




Figure 6.7 An Atwood machine and free-body diagrams for each of the twoblocks.


Strategy
We draw a free-body diagram for each mass separately, as shown in the figure. Then we analyze each diagram tofind the required unknowns. This may involve the solution of simultaneous equations. It is also important to notethe similarity with the previous example. As block 2 accelerates with acceleration a2 in the downward direction,
block 1 accelerates upward with acceleration a1 . Thus, a = a1 = −a2.
Solutiona. We have


Form1, ∑ Fy = T − m1g = m1a. Form2, ∑ Fy = T − m2g = −m2a.


(The negative sign in front of m2a indicates that m2 accelerates downward; both blocks accelerate at
the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and theresult is


(m2 − m1)g = (m1 + m2)a.


Solving for a:
a =


m2 − m1
m1 + m2


g =
4 kg − 2 kg
4 kg + 2 kg


(9.8 m/s2) = 3.27 m/s2.


b. Observing the first block, we see that
T − m1g = m1a


T = m1(g + a) = (2 kg)(9.8 m/s
2 + 3.27 m/s2) = 26.1 N.


Significance
The result for the acceleration given in the solution can be interpreted as the ratio of the unbalanced force onthe system, (m2 − m1)g , to the total mass of the system, m1 + m2 . We can also use the Atwood machine to
measure local gravitational field strength.


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6.3 Check Your Understanding Determine a general formula in terms of m1, m2 and g for calculating the
tension in the string for the Atwood machine shown above.


Newton’s Laws of Motion and Kinematics
Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set ofphysical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previouslyin this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics, and hence therelevance of earlier chapters.
When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givensand the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chaptersthat deal with a particular topic and solve the problem using strategies outlined in the text. The following worked exampleillustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, isapplied to an integrated concept problem.
Example 6.6


What Force Must a Soccer Player Exert to Reach Top Speed?
A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What is heraverage acceleration? (b) What average force does the ground exert forward on the runner so that she achievesthis acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.
Strategy
To find the answers to this problem, we use the problem-solving strategy given earlier in this chapter. Thesolutions to each part of the example illustrate how to apply specific problem-solving steps. In this case, we donot need to use all of the steps. We simply identify the physical principles, and thus the knowns and unknowns;apply Newton’s second law; and check to see whether the answer is reasonable.
Solutiona. We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is


Δv = 8.00 m/s . We are given the elapsed time, so Δt = 2.50 s. The unknown is acceleration, which
can be found from its definition:


a = Δv
Δt


.


Substituting the known values yields
a = 8.00 m/s


2.50 s
= 3.20 m/s2.


b. Here we are asked to find the average force the ground exerts on the runner to produce this acceleration.(Remember that we are dealing with the force or forces acting on the object of interest.) This isthe reaction force to that exerted by the player backward against the ground, by Newton’s third law.Neglecting air resistance, this would be equal in magnitude to the net external force on the player, sincethis force causes her acceleration. Since we now know the player’s acceleration and are given her mass,we can use Newton’s second law to find the force exerted. That is,
Fnet = ma.


Substituting the known values of m and a gives
Fnet = (70.0 kg)(3.20 m/s


2) = 224 N.


This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50pounds, a reasonable average force.
Significance
This example illustrates how to apply problem-solving strategies to situations that include topics from different


Chapter 6 | Applications of Newton's Laws 277




6.4


6.5


chapters. The first step is to identify the physical principles, the knowns, and the unknowns involved in theproblem. The second step is to solve for the unknown, in this case using Newton’s second law. Finally, wecheck our answer to ensure it is reasonable. These techniques for integrated concept problems will be useful inapplications of physics outside of a physics course, such as in your profession, in other science disciplines, and ineveryday life.
Check Your Understanding The soccer player stops after completing the play described above, but nownotices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal theball, which is 2.00 m away from her, how long will it take her to get to the ball?


Example 6.7
What Force Acts on a Model Helicopter?
A 1.50-kg model helicopter has a velocity of 5.00 j^ m/s at t = 0. It is accelerated at a constant rate for two
seconds (2.00 s) after which it has a velocity of ⎛⎝6.00 i^ + 12.00 j^⎞⎠m/s. What is the magnitude of the resultant
force acting on the helicopter during this time interval?
Strategy
We can easily set up a coordinate system in which the x-axis ( i^ direction) is horizontal, and the y-axis ( j^
direction) is vertical. We know that Δt = 2.00s and (6.00 i^ + 12.00 j^ m/s) − (5.00 j^ m/s). From this, we can
calculate the acceleration by the definition; we can then apply Newton’s second law.
Solution
We have


a = Δv
Δt


=
(6.00 i


^
+ 12.00 j


^
m/s) − (5.00 j


^
m/s)


2.00 s
= 3.00 i


^
+ 3.50 j


^
m/s2


∑ F→ = m a→ = (1.50 kg)(3.00 i^ + 3.50 j^ m/s2) = 4.50 i^ + 5.25 j^ N.


The magnitude of the force is now easily found:
F = (4.50 N)2 + (5.25 N)2 = 6.91 N.


Significance
The original problem was stated in terms of i^ − j^ vector components, so we used vector methods. Compare
this example with the previous example.


Check Your Understanding Find the direction of the resultant for the 1.50-kg model helicopter.


Example 6.8
Baggage Tractor
Figure 6.8(a) shows a baggage tractor pulling luggage carts from an airplane. The tractor has mass 650.0 kg,while cart A has mass 250.0 kg and cart B has mass 150.0 kg. The driving force acting for a brief period of time


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accelerates the system from rest and acts for 3.00 s. (a) If this driving force is given by F = (820.0t) N, find the
speed after 3.00 seconds. (b) What is the horizontal force acting on the connecting cable between the tractor andcart A at this instant?


Figure 6.8 (a) A free-body diagram is shown, which indicates all the external forces on the systemconsisting of the tractor and baggage carts for carrying airline luggage. (b) A free-body diagram of the tractoronly is shown isolated in order to calculate the tension in the cable to the carts.


Strategy
A free-body diagram shows the driving force of the tractor, which gives the system its acceleration. We only needto consider motion in the horizontal direction. The vertical forces balance each other and it is not necessary toconsider them. For part b, we make use of a free-body diagram of the tractor alone to determine the force between
it and cart A. This exposes the coupling force T→ , which is our objective.
Solution


a. ∑ Fx = msystem ax and ∑ Fx = 820.0t, so
820.0t = (650.0 + 250.0 + 150.0)a


a = 0.7809t.


Since acceleration is a function of time, we can determine the velocity of the tractor by using a = dv
dt


with the initial condition that v0 = 0 at t = 0. We integrate from t = 0 to t = 3:
dv = adt, ∫


0


3
dv = ∫


0


3.00
adt = ∫


0


3.00
0.7809tdt, v = 0.3905t2⎤⎦0


3.00
= 3.51 m/s.


b. Refer to the free-body diagram in Figure 6.8(b).
∑ Fx = mtractorax


820.0t − T = mtractor(0.7805)t


(820.0)(3.00) − T = (650.0)(0.7805)(3.00)
T = 938 N.


Significance
Since the force varies with time, we must use calculus to solve this problem. Notice how the total mass of thesystem was important in solving Figure 6.8(a), whereas only the mass of the truck (since it supplied the force)was of use in Figure 6.8(b).


Recall that v = ds
dt


and a = dv
dt
. If acceleration is a function of time, we can use the calculus forms developed in Motion


Along a Straight Line, as shown in this example. However, sometimes acceleration is a function of displacement. In


Chapter 6 | Applications of Newton's Laws 279




this case, we can derive an important result from these calculus relations. Solving for dt in each, we have dt = dsv and
dt = dva . Now, equating these expressions, we have dsv = dva . We can rearrange this to obtain a ds = v dv.
Example 6.9


Motion of a Projectile Fired Vertically
A 10.0-kg mortar shell is fired vertically upward from the ground, with an initial velocity of 50.0 m/s (see Figure
6.9). Determine the maximum height it will travel if atmospheric resistance is measured as FD = (0.0100v2) N,
where v is the speed at any instant.


Figure 6.9 (a) The mortar fires a shell straight up; weconsider the friction force provided by the air. (b) A free-bodydiagram is shown which indicates all the forces on the mortarshell.


Strategy
The known force on the mortar shell can be related to its acceleration using the equations of motion. Kinematicscan then be used to relate the mortar shell’s acceleration to its position.
Solution
Initially, y0 = 0 and v0 = 50.0 m/s. At the maximum height y = h, v = 0. The free-body diagram shows FD
to act downward, because it slows the upward motion of the mortar shell. Thus, we can write


∑ Fy = may
−FD − w = may


−0.0100v2 − 98.0 = 10.0a


a = −0.00100v2 − 9.80.


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6.6


The acceleration depends on v and is therefore variable. Since a = f (v), we can relate a to v using the
rearrangement described above,


a ds = v dv.


We replace ds with dy because we are dealing with the vertical direction,
ady = vdv,   (−0.00100v2 − 9.80)dy = vdv.


We now separate the variables (v’s and dv’s on one side; dy on the other):


0


h
dy = ⌠
⌡50.0


0
vdv


(−0.00100v2 − 9.80)



0


h
dy = −⌠


⌡50.0


0
vdv


(0.00100v2 + 9.80)
= (−5 × 103)ln(0.00100v2 + 9.80)|50.0


0


.


Thus, h = 114 m.
Significance
Notice the need to apply calculus since the force is not constant, which also means that acceleration is notconstant. To make matters worse, the force depends on v (not t), and so we must use the trick explained prior tothe example. The answer for the height indicates a lower elevation if there were air resistance. We will deal withthe effects of air resistance and other drag forces in greater detail in Drag Force and Terminal Speed.


Check Your Understanding If atmospheric resistance is neglected, find the maximum height for themortar shell. Is calculus required for this solution?


Explore the forces at work in this simulation (https://openstaxcollege.org/l/21forcesatwork) when youtry to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting onthe cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of allthe forces (including gravitational and normal forces).


6.2 | Friction
Learning Objectives


By the end of the section, you will be able to:
• Describe the general characteristics of friction
• List the various types of friction
• Calculate the magnitude of static and kinetic friction, and use these in problems involvingNewton’s laws of motion


When a body is in motion, it has resistance because the body interacts with its surroundings. This resistance is a force offriction. Friction opposes relative motion between systems in contact but also allows us to move, a concept that becomesobvious if you try to walk on ice. Friction is a common yet complex force, and its behavior still not completely understood.Still, it is possible to understand the circumstances in which it behaves.
Static and Kinetic Friction
The basic definition of friction is relatively simple to state.


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Friction
Friction is a force that opposes relative motion between systems in contact.


There are several forms of friction. One of the simpler characteristics of sliding friction is that it is parallel to the contactsurfaces between systems and is always in a direction that opposes motion or attempted motion of the systems relativeto each other. If two systems are in contact and moving relative to one another, then the friction between them is calledkinetic friction. For example, friction slows a hockey puck sliding on ice. When objects are stationary, static friction can actbetween them; the static friction is usually greater than the kinetic friction between two objects.
Static and Kinetic Friction
If two systems are in contact and stationary relative to one another, then the friction between them is called staticfriction. If two systems are in contact and moving relative to one another, then the friction between them is calledkinetic friction.


Imagine, for example, trying to slide a heavy crate across a concrete floor—you might push very hard on the crate and notmove it at all. This means that the static friction responds to what you do—it increases to be equal to and in the oppositedirection of your push. If you finally push hard enough, the crate seems to slip suddenly and starts to move. Now staticfriction gives way to kinetic friction. Once in motion, it is easier to keep it in motion than it was to get it started, indicatingthat the kinetic frictional force is less than the static frictional force. If you add mass to the crate, say by placing a box ontop of it, you need to push even harder to get it started and also to keep it moving. Furthermore, if you oiled the concreteyou would find it easier to get the crate started and keep it going (as you might expect).
Figure 6.10 is a crude pictorial representation of how friction occurs at the interface between two objects. Close-upinspection of these surfaces shows them to be rough. Thus, when you push to get an object moving (in this case, a crate),you must raise the object until it can skip along with just the tips of the surface hitting, breaking off the points, or both. Aconsiderable force can be resisted by friction with no apparent motion. The harder the surfaces are pushed together (such asif another box is placed on the crate), the more force is needed to move them. Part of the friction is due to adhesive forcesbetween the surface molecules of the two objects, which explains the dependence of friction on the nature of the substances.For example, rubber-soled shoes slip less than those with leather soles. Adhesion varies with substances in contact andis a complicated aspect of surface physics. Once an object is moving, there are fewer points of contact (fewer moleculesadhering), so less force is required to keep the object moving. At small but nonzero speeds, friction is nearly independentof speed.


Figure 6.10 Frictional forces, such as f→ , always oppose motion or
attempted motion between objects in contact. Friction arises in part becauseof the roughness of the surfaces in contact, as seen in the expanded view.For the object to move, it must rise to where the peaks of the top surface canskip along the bottom surface. Thus, a force is required just to set the objectin motion. Some of the peaks will be broken off, also requiring a force tomaintain motion. Much of the friction is actually due to attractive forcesbetween molecules making up the two objects, so that even perfectlysmooth surfaces are not friction-free. (In fact, perfectly smooth, cleansurfaces of similar materials would adhere, forming a bond called a “coldweld.”)


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The magnitude of the frictional force has two forms: one for static situations (static friction), the other for situationsinvolving motion (kinetic friction). What follows is an approximate empirical (experimentally determined) model only.These equations for static and kinetic friction are not vector equations.
Magnitude of Static Friction
The magnitude of static friction fs is


(6.1)fs ≤ µsN,
where µs is the coefficient of static friction and N is the magnitude of the normal force.


The symbol ≤ means less than or equal to, implying that static friction can have a maximum value of µsN. Static
friction is a responsive force that increases to be equal and opposite to whatever force is exerted, up to its maximum limit.Once the applied force exceeds
fs (max), the object moves. Thus,


fs(max) = µsN.


Magnitude of Kinetic Friction
The magnitude of kinetic friction fk is given by


(6.2)fk = µkN,
where µk is the coefficient of kinetic friction.


A system in which fk = µkN is described as a system in which friction behaves simply. The transition from static friction
to kinetic friction is illustrated in Figure 6.11.


Figure 6.11 (a) The force of friction f→ between the block and the rough surface opposes the direction of the applied force
F


. The magnitude of the static friction balances that of the applied force. This is shown in the left side of the graph in (c). (b)
At some point, the magnitude of the applied force is greater than the force of kinetic friction, and the block moves to the right.This is shown in the right side of the graph. (c) The graph of the frictional force versus the applied force; note that
fs(max) > fk. This means that µs > µk.


Chapter 6 | Applications of Newton's Laws 283




As you can see in Table 6.1, the coefficients of kinetic friction are less than their static counterparts. The approximatevalues of µ are stated to only one or two digits to indicate the approximate description of friction given by the preceding
two equations.


System Static Friction µs Kinetic Friction µk
Rubber on dry concrete 1.0 0.7
Rubber on wet concrete 0.5-0.7 0.3-0.5
Wood on wood 0.5 0.3
Waxed wood on wet snow 0.14 0.1
Metal on wood 0.5 0.3
Steel on steel (dry) 0.6 0.3
Steel on steel (oiled) 0.05 0.03
Teflon on steel 0.04 0.04
Bone lubricated by synovial fluid 0.016 0.015
Shoes on wood 0.9 0.7
Shoes on ice 0.1 0.05
Ice on ice 0.1 0.03
Steel on ice 0.4 0.02


Table 6.1 Approximate Coefficients of Static and Kinetic Friction
Equation 6.1 and Equation 6.2 include the dependence of friction on materials and the normal force. The direction offriction is always opposite that of motion, parallel to the surface between objects, and perpendicular to the normal force.For example, if the crate you try to push (with a force parallel to the floor) has a mass of 100 kg, then the normal force isequal to its weight,


w = mg = ⎛⎝100 kg⎞⎠⎛⎝9.80 m/s
2⎞
⎠ = 980 N,


perpendicular to the floor. If the coefficient of static friction is 0.45, you would have to exert a force parallel to the floorgreater than
fs(max) = µsN = (0.45)(980 N) = 440 N


to move the crate. Once there is motion, friction is less and the coefficient of kinetic friction might be 0.30, so that a forceof only
fk = µkN = (0.30)(980 N) = 290 N


keeps it moving at a constant speed. If the floor is lubricated, both coefficients are considerably less than they would bewithout lubrication. Coefficient of friction is a unitless quantity with a magnitude usually between 0 and 1.0. The actualvalue depends on the two surfaces that are in contact.
Many people have experienced the slipperiness of walking on ice. However, many parts of the body, especially the joints,have much smaller coefficients of friction—often three or four times less than ice. A joint is formed by the ends of twobones, which are connected by thick tissues. The knee joint is formed by the lower leg bone (the tibia) and the thighbone(the femur). The hip is a ball (at the end of the femur) and socket (part of the pelvis) joint. The ends of the bones in thejoint are covered by cartilage, which provides a smooth, almost-glassy surface. The joints also produce a fluid (synovialfluid) that reduces friction and wear. A damaged or arthritic joint can be replaced by an artificial joint (Figure 6.12). Thesereplacements can be made of metals (stainless steel or titanium) or plastic (polyethylene), also with very small coefficientsof friction.


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Figure 6.12 Artificial knee replacement is a procedure that has been performed for more than 20 years. These post-operativeX-rays show a right knee joint replacement. (credit: Mike Baird)


Natural lubricants include saliva produced in our mouths to aid in the swallowing process, and the slippery mucus foundbetween organs in the body, allowing them to move freely past each other during heartbeats, during breathing, and when aperson moves. Hospitals and doctor’s clinics commonly use artificial lubricants, such as gels, to reduce friction.
The equations given for static and kinetic friction are empirical laws that describe the behavior of the forces of friction.While these formulas are very useful for practical purposes, they do not have the status of mathematical statements thatrepresent general principles (e.g., Newton’s second law). In fact, there are cases for which these equations are not even goodapproximations. For instance, neither formula is accurate for lubricated surfaces or for two surfaces siding across each otherat high speeds. Unless specified, we will not be concerned with these exceptions.
Example 6.10


Static and Kinetic Friction
A 20.0-kg crate is at rest on a floor as shown in Figure 6.13. The coefficient of static friction between the crate
and floor is 0.700 and the coefficient of kinetic friction is 0.600. A horizontal force P→ is applied to the crate.
Find the force of friction if (a) P→ = 20.0 N, (b) P→ = 30.0 N, (c) P→ = 120.0 N, and (d) P→ = 180.0 N.


Figure 6.13 (a) A crate on a horizontal surface is pushed with a force P→ . (b) The forces on the crate. Here,
f
→ may represent either the static or the kinetic frictional force.


Chapter 6 | Applications of Newton's Laws 285




6.7


Strategy
The free-body diagram of the crate is shown in Figure 6.13(b). We apply Newton’s second law in the horizontaland vertical directions, including the friction force in opposition to the direction of motion of the box.
Solution
Newton’s second law gives


∑ Fx = max ∑ Fy = may
P − f = max N − w = 0.


Here we are using the symbol f to represent the frictional force since we have not yet determined whether thecrate is subject to station friction or kinetic friction. We do this whenever we are unsure what type of friction isacting. Now the weight of the crate is
w = (20.0 kg)(9.80 m/s2) = 196 N,


which is also equal to N. The maximum force of static friction is therefore (0.700)(196 N) = 137 N. As long
as P→ is less than 137 N, the force of static friction keeps the crate stationary and fs = P→ . Thus, (a)
fs = 20.0 N, (b) fs = 30.0 N, and (c) fs = 120.0 N.
(d) If P→ = 180.0 N, the applied force is greater than the maximum force of static friction (137 N), so the crate
can no longer remain at rest. Once the crate is in motion, kinetic friction acts. Then


fk = µkN = (0.600)(196 N) = 118 N,


and the acceleration is
ax =


P


− fk
m =


180.0 N − 118 N
20.0 kg


= 3.10 m/s2 .


Significance
This example illustrates how we consider friction in a dynamics problem. Notice that static friction has a valuethat matches the applied force, until we reach the maximum value of static friction. Also, no motion can occuruntil the applied force equals the force of static friction, but the force of kinetic friction will then become smaller.


Check Your Understanding A block of mass 1.0 kg rests on a horizontal surface. The frictionalcoefficients for the block and surface are µs = 0.50 and µk = 0.40. (a) What is the minimum horizontal force
required to move the block? (b) What is the block’s acceleration when this force is applied?


Friction and the Inclined Plane
One situation where friction plays an obvious role is that of an object on a slope. It might be a crate being pushed up a rampto a loading dock or a skateboarder coasting down a mountain, but the basic physics is the same. We usually generalize thesloping surface and call it an inclined plane but then pretend that the surface is flat. Let’s look at an example of analyzingmotion on an inclined plane with friction.
Example 6.11


Downhill Skier
A skier with a mass of 62 kg is sliding down a snowy slope at a constant velocity. Find the coefficient of kineticfriction for the skier if friction is known to be 45.0 N.


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Strategy
The magnitude of kinetic friction is given as 45.0 N. Kinetic friction is related to the normal force N by
fk = µkN ; thus, we can find the coefficient of kinetic friction if we can find the normal force on the skier. The
normal force is always perpendicular to the surface, and since there is no motion perpendicular to the surface,the normal force should equal the component of the skier’s weight perpendicular to the slope. (See Figure 6.14,which repeats a figure from the chapter on Newton’s laws of motion.)


Figure 6.14 The motion of the skier and friction are parallel to the slope, so it is most convenient to project allforces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular (axes shown to
left of skier). The normal force N→ is perpendicular to the slope, and friction f→ is parallel to the slope, but the
skier’s weight w→ has components along both axes, namely w→ y and w→ x. The normal force N→ is equal in
magnitude to w→ y, so there is no motion perpendicular to the slope. However, f→ is less than w→ x in
magnitude, so there is acceleration down the slope (along the x-axis).


We have
N = wy = w cos 25° = mg cos 25°.


Substituting this into our expression for kinetic friction, we obtain
fk = µkmg cos 25°,


which can now be solved for the coefficient of kinetic friction µk.
Solution
Solving for µk gives


µk =
fk
N


=
fk


w cos 25°
=


fk
mg cos 25°


.


Substituting known values on the right-hand side of the equation,
µk =


45.0 N
(62 kg)(9.80 m/s2)(0.906)


= 0.082.


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Significance
This result is a little smaller than the coefficient listed in Table 6.1 for waxed wood on snow, but it is stillreasonable since values of the coefficients of friction can vary greatly. In situations like this, where an object ofmass m slides down a slope that makes an angle θ with the horizontal, friction is given by fk = µkmg cos θ.
All objects slide down a slope with constant acceleration under these circumstances.


We have discussed that when an object rests on a horizontal surface, the normal force supporting it is equal in magnitudeto its weight. Furthermore, simple friction is always proportional to the normal force. When an object is not on a horizontalsurface, as with the inclined plane, we must find the force acting on the object that is directed perpendicular to the surface;it is a component of the weight.
We now derive a useful relationship for calculating coefficient of friction on an inclined plane. Notice that the result appliesonly for situations in which the object slides at constant speed down the ramp.
An object slides down an inclined plane at a constant velocity if the net force on the object is zero. We can use this factto measure the coefficient of kinetic friction between two objects. As shown in Example 6.10, the kinetic friction on aslope is fk = µkmg cos θ . The component of the weight down the slope is equal to mg sin θ (see the free-body diagram in
Figure 6.14). These forces act in opposite directions, so when they have equal magnitude, the acceleration is zero. Writingthese out,


µkmg cos θ = mg sin θ.


Solving for µk, we find that
µk =


mg sin θ
mg cos θ


= tan θ.


Put a coin on a book and tilt it until the coin slides at a constant velocity down the book. You might need to tap the booklightly to get the coin to move. Measure the angle of tilt relative to the horizontal and find µk. Note that the coin does not
start to slide at all until an angle greater than θ is attained, since the coefficient of static friction is larger than the coefficient
of kinetic friction. Think about how this may affect the value for µk and its uncertainty.
Atomic-Scale Explanations of Friction
The simpler aspects of friction dealt with so far are its macroscopic (large-scale) characteristics. Great strides have beenmade in the atomic-scale explanation of friction during the past several decades. Researchers are finding that the atomicnature of friction seems to have several fundamental characteristics. These characteristics not only explain some of thesimpler aspects of friction—they also hold the potential for the development of nearly friction-free environments that couldsave hundreds of billions of dollars in energy which is currently being converted (unnecessarily) into heat.
Figure 6.15 illustrates one macroscopic characteristic of friction that is explained by microscopic (small-scale) research.We have noted that friction is proportional to the normal force, but not to the amount of area in contact, a somewhatcounterintuitive notion. When two rough surfaces are in contact, the actual contact area is a tiny fraction of the total areabecause only high spots touch. When a greater normal force is exerted, the actual contact area increases, and we find thatthe friction is proportional to this area.


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Figure 6.15 Two rough surfaces in contact have a much smaller area of actual contactthan their total area. When the normal force is larger as a result of a larger applied force,the area of actual contact increases, as does friction.


However, the atomic-scale view promises to explain far more than the simpler features of friction. The mechanism forhow heat is generated is now being determined. In other words, why do surfaces get warmer when rubbed? Essentially,atoms are linked with one another to form lattices. When surfaces rub, the surface atoms adhere and cause atomic latticesto vibrate—essentially creating sound waves that penetrate the material. The sound waves diminish with distance, andtheir energy is converted into heat. Chemical reactions that are related to frictional wear can also occur between atomsand molecules on the surfaces. Figure 6.16 shows how the tip of a probe drawn across another material is deformed byatomic-scale friction. The force needed to drag the tip can be measured and is found to be related to shear stress, which is
discussed in Static Equilibrium and Elasticity. The variation in shear stress is remarkable (more than a factor of 1012
) and difficult to predict theoretically, but shear stress is yielding a fundamental understanding of a large-scale phenomenonknown since ancient times—friction.


Figure 6.16 The tip of a probe is deformed sideways by frictionalforce as the probe is dragged across a surface. Measurements of howthe force varies for different materials are yielding fundamentalinsights into the atomic nature of friction.
Describe a model for friction (https://openstaxcollege.org/l/21friction) on a molecular level. Describematter in terms of molecular motion. The description should include diagrams to support the description; howthe temperature affects the image; what are the differences and similarities between solid, liquid, and gas particlemotion; and how the size and speed of gas molecules relate to everyday objects.


Chapter 6 | Applications of Newton's Laws 289




Example 6.12
Sliding Blocks
The two blocks of Figure 6.17 are attached to each other by a massless string that is wrapped around a
frictionless pulley. When the bottom 4.00-kg block is pulled to the left by the constant force P→ , the top 2.00-kg
block slides across it to the right. Find the magnitude of the force necessary to move the blocks at constant speed.Assume that the coefficient of kinetic friction between all surfaces is 0.400.


Figure 6.17 (a) Each block moves at constant velocity. (b) Free-body diagrams for the blocks.


Strategy
We analyze the motions of the two blocks separately. The top block is subjected to a contact force exerted bythe bottom block. The components of this force are the normal force N1 and the frictional force −0.400N1.
Other forces on the top block are the tension Ti in the string and the weight of the top block itself, 19.6 N. The
bottom block is subjected to contact forces due to the top block and due to the floor. The first contact force hascomponents −N1 and 0.400N1, which are simply reaction forces to the contact forces that the bottom block
exerts on the top block. The components of the contact force of the floor are N2 and 0.400N2. Other forces on
this block are −P, the tension Ti, and the weight –39.2 N.
Solution
Since the top block is moving horizontally to the right at constant velocity, its acceleration is zero in both thehorizontal and the vertical directions. From Newton’s second law,


∑ Fx = m1ax ∑ Fy = m1ay
T − 0.400N1 = 0 N1 − 19.6 N = 0.


Solving for the two unknowns, we obtain N1 = 19.6 N and T = 0.40N1 = 7.84 N. The bottom block is also
not accelerating, so the application of Newton’s second law to this block gives


∑ Fx = m2ax ∑ Fy = m2ay
T − P + 0.400 N1 + 0.400 N2 = 0 N2 − 39.2 N − N1 = 0.


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The values of N1 and T were found with the first set of equations. When these values are substituted into the
second set of equations, we can determine N2 and P. They are


N2 = 58.8 N and P = 39.2 N.


Significance
Understanding what direction in which to draw the friction force is often troublesome. Notice that each frictionforce labeled in Figure 6.17 acts in the direction opposite the motion of its corresponding block.


Example 6.13
A Crate on an Accelerating Truck
A 50.0-kg crate rests on the bed of a truck as shown in Figure 6.18. The coefficients of friction between thesurfaces are µk = 0.300 and µs = 0.400. Find the frictional force on the crate when the truck is accelerating
forward relative to the ground at (a) 2.00 m/s2, and (b) 5.00 m/s2.


Figure 6.18 (a) A crate rests on the bed of the truck that is accelerating forward. (b)The free-body diagram of the crate.


Strategy
The forces on the crate are its weight and the normal and frictional forces due to contact with the truck bed.We start by assuming that the crate is not slipping. In this case, the static frictional force fs acts on the crate.
Furthermore, the accelerations of the crate and the truck are equal.
Solutiona. Application of Newton’s second law to the crate, using the reference frame attached to the ground, yields


∑ Fx = max ∑ Fy = may
fs = (50.0 kg)(2.00 m/s


2) N − 4.90 × 102 N = (50.0 kg)(0)


= 1.00 × 102 N N = 4.90 × 102 N.


We can now check the validity of our no-slip assumption. The maximum value of the force of staticfriction is
µsN = (0.400)(4.90 × 10


2 N) = 196 N,


whereas the actual force of static friction that acts when the truck accelerates forward at 2.00 m/s2 is
only 1.00 × 102 N. Thus, the assumption of no slipping is valid.


b. If the crate is to move with the truck when it accelerates at 5.0 m/s2, the force of static friction must be


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fs = max = (50.0 kg)(5.00 m/s
2) = 250 N.


Since this exceeds the maximum of 196 N, the crate must slip. The frictional force is therefore kinetic andis
fk = µkN = (0.300)(4.90 × 10


2 N) = 147 N.


The horizontal acceleration of the crate relative to the ground is now found from
∑ Fx = max
147 N = (50.0 kg)ax,


so ax = 2.94 m/s
2.


Significance
Relative to the ground, the truck is accelerating forward at 5.0 m/s2 and the crate is accelerating forward
at 2.94 m/s2 . Hence the crate is sliding backward relative to the bed of the truck with an acceleration
2.94 m/s2 − 5.00 m/s2 = −2.06 m/s2 .


Example 6.14
Snowboarding
Earlier, we analyzed the situation of a downhill skier moving at constant velocity to determine the coefficientof kinetic friction. Now let’s do a similar analysis to determine acceleration. The snowboarder of Figure 6.19
glides down a slope that is inclined at θ = 130 to the horizontal. The coefficient of kinetic friction between the
board and the snow is µk = 0.20. What is the acceleration of the snowboarder?


Figure 6.19 (a) A snowboarder glides down a slope inclined at 13° to the horizontal. (b) The free-body diagram ofthe snowboarder.


Strategy
The forces acting on the snowboarder are her weight and the contact force of the slope, which has a componentnormal to the incline and a component along the incline (force of kinetic friction). Because she moves along the


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6.8


slope, the most convenient reference frame for analyzing her motion is one with the x-axis along and the y-axisperpendicular to the incline. In this frame, both the normal and the frictional forces lie along coordinate axes,the components of the weight are mg sin θ along the slope andmg cos θ at right angles into the slope , and the
only acceleration is along the x-axis ⎛⎝ay = 0⎞⎠.
Solution
We can now apply Newton’s second law to the snowboarder:


∑ Fx = max ∑ Fy = may
mg sin θ − µkN = max N − mg cos θ = m(0).


From the second equation, N = mg cos θ. Upon substituting this into the first equation, we find
ax = g(sin θ − µk cos θ)


= g(sin 13° − 0.20 cos 13°) = 0.29 m/s2 .


Significance
Notice from this equation that if θ is small enough or µk is large enough, ax is negative, that is, the
snowboarder slows down.


Check Your Understanding The snowboarder is now moving down a hill with incline 10.0° . What is
the skier’s acceleration?


6.3 | Centripetal Force
Learning Objectives


By the end of the section, you will be able to:
• Explain the equation for centripetal acceleration
• Apply Newton’s second law to develop the equation for centripetal force
• Use circular motion concepts in solving problems involving Newton’s laws of motion


In Motion in Two and Three Dimensions, we examined the basic concepts of circular motion. An object undergoingcircular motion, like one of the race cars shown at the beginning of this chapter, must be accelerating because it is changingthe direction of its velocity. We proved that this centrally directed acceleration, called centripetal acceleration, is given bythe formula
ac =


v2
r


where v is the velocity of the object, directed along a tangent line to the curve at any instant. If we know the angular velocity
ω , then we can use


ac = rω
2.


Angular velocity gives the rate at which the object is turning through the curve, in units of rad/s. This acceleration acts alongthe radius of the curved path and is thus also referred to as a radial acceleration.
An acceleration must be produced by a force. Any force or combination of forces can cause a centripetal or radialacceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon,friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinningcentrifuge. Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal forceis toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law


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of motion, net force is mass times acceleration: Fnet = ma. For uniform circular motion, the acceleration is the centripetal
acceleration:. a = ac. Thus, the magnitude of centripetal force Fc is


Fc = mac.


By substituting the expressions for centripetal acceleration ac (ac = v2r ; ac = rω2), we get two expressions for the
centripetal force Fc in terms of mass, velocity, angular velocity, and radius of curvature:


(6.3)
Fc = m


v2
r ; Fc = mrω


2.


You may use whichever expression for centripetal force is more convenient. Centripetal force F→ c is always perpendicular
to the path and points to the center of curvature, because a→ c is perpendicular to the velocity and points to the center of
curvature. Note that if you solve the first expression for r, you get


r = mv
2


Fc
.


This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tightcurve, as in Figure 6.20.


Figure 6.20 The frictional force supplies the centripetal force and isnumerically equal to it. Centripetal force is perpendicular to velocity and causesuniform circular motion. The larger the Fc, the smaller the radius of
curvature r and the sharper the curve. The second curve has the same v, but alarger Fc produces a smaller r′.


Example 6.15
What Coefficient of Friction Do Cars Need on a Flat Curve?
(a) Calculate the centripetal force exerted on a 900.0-kg car that negotiates a 500.0-m radius curve at 25.00 m/s.(b) Assuming an unbanked curve, find the minimum static coefficient of friction between the tires and the road,static friction being the reason that keeps the car from slipping (Figure 6.21).


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Figure 6.21 This car on level ground is moving away andturning to the left. The centripetal force causing the car to turn ina circular path is due to friction between the tires and the road. Aminimum coefficient of friction is needed, or the car will movein a larger-radius curve and leave the roadway.


Strategy
a. We know that Fc = mv2r . Thus,


Fc =
mv2
r =


(900.0 kg)(25.00 m/s)2


(500.0 m)
= 1125 N.


b. Figure 6.21 shows the forces acting on the car on an unbanked (level ground) curve. Friction is to theleft, keeping the car from slipping, and because it is the only horizontal force acting on the car, the frictionis the centripetal force in this case. We know that the maximum static friction (at which the tires roll butdo not slip) is µsN, where µs is the static coefficient of friction and N is the normal force. The normal
force equals the car’s weight on level ground, so N = mg. Thus the centripetal force in this situation is


Fc ≡ f = µsN = µsmg.


Now we have a relationship between centripetal force and the coefficient of friction. Using the equation
Fc = m


v2
r ,


we obtain
mv


2


r = µsmg.


We solve this for µs, noting that mass cancels, and obtain
µs =


v2
rg.


Substituting the knowns,
µs =


(25.00 m/s)2


(500.0 m)(9.80 m/s2)
= 0.13.


(Because coefficients of friction are approximate, the answer is given to only two digits.)


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6.9


Significance
The coefficient of friction found in Figure 6.21(b) is much smaller than is typically found between tires androads. The car still negotiates the curve if the coefficient is greater than 0.13, because static friction is a responsiveforce, able to assume a value less than but no more than µsN. A higher coefficient would also allow the car to
negotiate the curve at a higher speed, but if the coefficient of friction is less, the safe speed would be less than25 m/s. Note that mass cancels, implying that, in this example, it does not matter how heavily loaded the car isto negotiate the turn. Mass cancels because friction is assumed proportional to the normal force, which in turn isproportional to mass. If the surface of the road were banked, the normal force would be less, as discussed next.


Check Your UnderstandingA car moving at 96.8 km/h travels around a circular curve of radius 182.9 mon a flat country road. What must be the minimum coefficient of static friction to keep the car from slipping?


Banked Curves
Let us now consider banked curves, where the slope of the road helps you negotiate the curve (Figure 6.22). The greaterthe angle θ , the faster you can take the curve. Race tracks for bikes as well as cars, for example, often have steeply banked
curves. In an “ideally banked curve,” the angle θ is such that you can negotiate the curve at a certain speed without the aid
of friction between the tires and the road. We will derive an expression for θ for an ideally banked curve and consider an
example related to it.


Figure 6.22 The car on this banked curve is moving away andturning to the left.


For ideal banking, the net external force equals the horizontal centripetal force in the absence of friction. The componentsof the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car,respectively. In cases in which forces are not parallel, it is most convenient to consider components along perpendicularaxes—in this case, the vertical and horizontal directions.
Figure 6.22 shows a free-body diagram for a car on a frictionless banked curve. If the angle θ is ideal for the speed and
radius, then the net external force equals the necessary centripetal force. The only two external forces acting on the car are
its weight w→ and the normal force of the road N→ . (A frictionless surface can only exert a force perpendicular to the
surface—that is, a normal force.) These two forces must add to give a net external force that is horizontal toward the center
of curvature and has magnitude mv2 /r. Because this is the crucial force and it is horizontal, we use a coordinate system
with vertical and horizontal axes. Only the normal force has a horizontal component, so this must equal the centripetal force,that is,


N sin θ = mv
2


r .


Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the verticalcomponents of the two external forces must be equal in magnitude and opposite in direction. From Figure 6.22, we seethat the vertical component of the normal force is N cos θ, and the only other vertical force is the car’s weight. These


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must be equal in magnitude; thus,
N cos θ = mg.


Now we can combine these two equations to eliminate N and get an expression for θ , as desired. Solving the second
equation for N = mg/(cosθ) and substituting this into the first yields


mg sin θ
cos θ


= mv
2


r


mg tan θ = mv
2


r


tan θ = v
2


rg.


Taking the inverse tangent gives


(6.4)
θ = tan−1




v2
rg

⎠.


This expression can be understood by considering how θ depends on v and r. A large θ is obtained for a large v and a
small r. That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you totake the curve at greater or lower speed than if the curve were frictionless. Note that θ does not depend on the mass of the
vehicle.
Example 6.16


What Is the Ideal Speed to Take a Steeply Banked Tight Curve?
Curves on some test tracks and race courses, such as Daytona International Speedway in Florida, are very steeplybanked. This banking, with the aid of tire friction and very stable car configurations, allows the curves to be takenat very high speed. To illustrate, calculate the speed at which a 100.0-m radius curve banked at 31.0° should be
driven if the road were frictionless.
Strategy
We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known;thus, we need only rearrange it so that speed appears on the left-hand side and then substitute known quantities.
Solution
Starting with


tan θ = v
2


rg,


we get
v = rg tan θ.


Noting that tan 31.0° = 0.609, we obtain
v = (100.0 m)(9.80 m/s2)(0.609) = 24.4 m/s.


Significance
This is just about 165 km/h, consistent with a very steeply banked and rather sharp curve. Tire friction enables avehicle to take the curve at significantly higher speeds.


Airplanes also make turns by banking. The lift force, due to the force of the air on the wing, acts at right angles to the wing.When the airplane banks, the pilot is obtaining greater lift than necessary for level flight. The vertical component of lift


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balances the airplane’s weight, and the horizontal component accelerates the plane. The banking angle shown in Figure6.23 is given by θ . We analyze the forces in the same way we treat the case of the car rounding a banked curve.


Figure 6.23 In a banked turn, the horizontal component of lift is unbalanced andaccelerates the plane. The normal component of lift balances the plane’s weight. Thebanking angle is given by θ . Compare the vector diagram with that shown in Figure
6.22.


Join the ladybug (https://openstaxcollege.org/l/21ladybug) in an exploration of rotational motion. Rotatethe merry-go-round to change its angle or choose a constant angular velocity or angular acceleration. Explore howcircular motion relates to the bug’s xy-position, velocity, and acceleration using vectors or graphs.
A circular motion requires a force, the so-called centripetal force, which is directed to the axis of rotation. Thissimplified model of a carousel (https://openstaxcollege.org/l/21carousel) demonstrates this force.


Inertial Forces and Noninertial (Accelerated) Frames: The CoriolisForce
What do taking off in a jet airplane, turning a corner in a car, riding a merry-go-round, and the circular motion of atropical cyclone have in common? Each exhibits inertial forces—forces that merely seem to arise from motion, because theobserver’s frame of reference is accelerating or rotating. When taking off in a jet, most people would agree it feels as if youare being pushed back into the seat as the airplane accelerates down the runway. Yet a physicist would say that you tend toremain stationary while the seat pushes forward on you. An even more common experience occurs when you make a tightcurve in your car—say, to the right (Figure 6.24). You feel as if you are thrown (that is, forced) toward the left relative tothe car. Again, a physicist would say that you are going in a straight line (recall Newton’s first law) but the car moves to theright, not that you are experiencing a force from the left.


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Figure 6.24 (a) The car driver feels herself forced to the leftrelative to the car when she makes a right turn. This is an inertialforce arising from the use of the car as a frame of reference. (b) InEarth’s frame of reference, the driver moves in a straight line,obeying Newton’s first law, and the car moves to the right. There isno force to the left on the driver relative to Earth. Instead, there is aforce to the right on the car to make it turn.


We can reconcile these points of view by examining the frames of reference used. Let us concentrate on people in a car.Passengers instinctively use the car as a frame of reference, whereas a physicist might use Earth. The physicist might makethis choice because Earth is nearly an inertial frame of reference, in which all forces have an identifiable physical origin.In such a frame of reference, Newton’s laws of motion take the form given in Newton’s Laws of Motion. The car is anoninertial frame of reference because it is accelerated to the side. The force to the left sensed by car passengers is aninertial force having no physical origin (it is due purely to the inertia of the passenger, not to some physical cause such astension, friction, or gravitation). The car, as well as the driver, is actually accelerating to the right. This inertial force is saidto be an inertial force because it does not have a physical origin, such as gravity.
A physicist will choose whatever reference frame is most convenient for the situation being analyzed. There is no problemto a physicist in including inertial forces and Newton’s second law, as usual, if that is more convenient, for example, on amerry-go-round or on a rotating planet. Noninertial (accelerated) frames of reference are used when it is useful to do so.Different frames of reference must be considered in discussing the motion of an astronaut in a spacecraft traveling at speedsnear the speed of light, as you will appreciate in the study of the special theory of relativity.
Let us now take a mental ride on a merry-go-round—specifically, a rapidly rotating playground merry-go-round (Figure6.25). You take the merry-go-round to be your frame of reference because you rotate together. When rotating in thatnoninertial frame of reference, you feel an inertial force that tends to throw you off; this is often referred to as a centrifugalforce (not to be confused with centripetal force). Centrifugal force is a commonly used term, but it does not actually exist.You must hang on tightly to counteract your inertia (which people often refer to as centrifugal force). In Earth’s frame ofreference, there is no force trying to throw you off; we emphasize that centrifugal force is a fiction. You must hang on tomake yourself go in a circle because otherwise you would go in a straight line, right off the merry-go-round, in keeping withNewton’s first law. But the force you exert acts toward the center of the circle.


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Figure 6.25 (a) A rider on a merry-go-round feels as if he is being thrown off. This inertial force is sometimes mistakenlycalled the centrifugal force in an effort to explain the rider’s motion in the rotating frame of reference. (b) In an inertial frameof reference and according to Newton’s laws, it is his inertia that carries him off (the unshaded rider has Fnet = 0 and heads
in a straight line). A force, Fcentripetal , is needed to cause a circular path.


This inertial effect, carrying you away from the center of rotation if there is no centripetal force to cause circular motion, isput to good use in centrifuges (Figure 6.26). A centrifuge spins a sample very rapidly, as mentioned earlier in this chapter.Viewed from the rotating frame of reference, the inertial force throws particles outward, hastening their sedimentation. Thegreater the angular velocity, the greater the centrifugal force. But what really happens is that the inertia of the particlescarries them along a line tangent to the circle while the test tube is forced in a circular path by a centripetal force.


Figure 6.26 Centrifuges use inertia to perform their task.Particles in the fluid sediment settle out because their inertiacarries them away from the center of rotation. The large angularvelocity of the centrifuge quickens the sedimentation.Ultimately, the particles come into contact with the test tubewalls, which then supply the centripetal force needed to makethem move in a circle of constant radius.


Let us now consider what happens if something moves in a rotating frame of reference. For example, what if you slide a balldirectly away from the center of the merry-go-round, as shown in Figure 6.27? The ball follows a straight path relative to


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Earth (assuming negligible friction) and a path curved to the right on the merry-go-round’s surface. A person standing nextto the merry-go-round sees the ball moving straight and the merry-go-round rotating underneath it. In the merry-go-round’sframe of reference, we explain the apparent curve to the right by using an inertial force, called the Coriolis force, whichcauses the ball to curve to the right. The Coriolis force can be used by anyone in that frame of reference to explain whyobjects follow curved paths and allows us to apply Newton’s laws in noninertial frames of reference.


Figure 6.27 Looking down on the counterclockwise rotation of a merry-go-round, we see that a ball slid straight toward theedge follows a path curved to the right. The person slides the ball toward point B, starting at point A. Both points rotate to theshaded positions (A’ and B’) shown in the time that the ball follows the curved path in the rotating frame and a straight path inEarth’s frame.


Up until now, we have considered Earth to be an inertial frame of reference with little or no worry about effects due to itsrotation. Yet such effects do exist—in the rotation of weather systems, for example. Most consequences of Earth’s rotationcan be qualitatively understood by analogy with the merry-go-round. Viewed from above the North Pole, Earth rotatescounterclockwise, as does the merry-go-round in Figure 6.27. As on the merry-go-round, any motion in Earth’s NorthernHemisphere experiences a Coriolis force to the right. Just the opposite occurs in the Southern Hemisphere; there, the forceis to the left. Because Earth’s angular velocity is small, the Coriolis force is usually negligible, but for large-scale motions,such as wind patterns, it has substantial effects.
The Coriolis force causes hurricanes in the Northern Hemisphere to rotate in the counterclockwise direction, whereastropical cyclones in the Southern Hemisphere rotate in the clockwise direction. (The terms hurricane, typhoon, and tropicalstorm are regionally specific names for cyclones, which are storm systems characterized by low pressure centers, strongwinds, and heavy rains.) Figure 6.28 helps show how these rotations take place. Air flows toward any region of lowpressure, and tropical cyclones contain particularly low pressures. Thus winds flow toward the center of a tropical cycloneor a low-pressure weather system at the surface. In the Northern Hemisphere, these inward winds are deflected to the right,as shown in the figure, producing a counterclockwise circulation at the surface for low-pressure zones of any type. Lowpressure at the surface is associated with rising air, which also produces cooling and cloud formation, making low-pressurepatterns quite visible from space. Conversely, wind circulation around high-pressure zones is clockwise in the SouthernHemisphere but is less visible because high pressure is associated with sinking air, producing clear skies.


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Figure 6.28 (a) The counterclockwise rotation of this Northern Hemisphere hurricane is a major consequence of the Coriolisforce. (b) Without the Coriolis force, air would flow straight into a low-pressure zone, such as that found in tropical cyclones.(c) The Coriolis force deflects the winds to the right, producing a counterclockwise rotation. (d) Wind flowing away from ahigh-pressure zone is also deflected to the right, producing a clockwise rotation. (e) The opposite direction of rotation isproduced by the Coriolis force in the Southern Hemisphere, leading to tropical cyclones. (credit a and credit e: modifications ofwork by NASA)


The rotation of tropical cyclones and the path of a ball on a merry-go-round can just as well be explained by inertia andthe rotation of the system underneath. When noninertial frames are used, inertial forces, such as the Coriolis force, must beinvented to explain the curved path. There is no identifiable physical source for these inertial forces. In an inertial frame,inertia explains the path, and no force is found to be without an identifiable source. Either view allows us to describe nature,but a view in an inertial frame is the simplest in the sense that all forces have origins and explanations.
6.4 | Drag Force and Terminal Speed


Learning Objectives
By the end of the section, you will be able to:
• Express the drag force mathematically
• Describe applications of the drag force
• Define terminal velocity
• Determine an object’s terminal velocity given its mass


Another interesting force in everyday life is the force of drag on an object when it is moving in a fluid (either a gas or aliquid). You feel the drag force when you move your hand through water. You might also feel it if you move your handduring a strong wind. The faster you move your hand, the harder it is to move. You feel a smaller drag force when youtilt your hand so only the side goes through the air—you have decreased the area of your hand that faces the direction ofmotion.
Drag Forces
Like friction, the drag force always opposes the motion of an object. Unlike simple friction, the drag force is proportionalto some function of the velocity of the object in that fluid. This functionality is complicated and depends upon the shape ofthe object, its size, its velocity, and the fluid it is in. For most large objects such as cyclists, cars, and baseballs not moving


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too slowly, the magnitude of the drag force FD is proportional to the square of the speed of the object. We can write this
relationship mathematically as FD ∝ v2. When taking into account other factors, this relationship becomes


(6.5)FD = 12CρAv2,


where C is the drag coefficient, A is the area of the object facing the fluid, and ρ is the density of the fluid. (Recall that
density is mass per unit volume.) This equation can also be written in a more generalized fashion as FD = bv2, where b
is a constant equivalent to 0.5CρA. We have set the exponent n for these equations as 2 because when an object is moving
at high velocity through air, the magnitude of the drag force is proportional to the square of the speed. As we shall see inFluid Mechanics, for small particles moving at low speeds in a fluid, the exponent n is equal to 1.


Drag Force
Drag force FD is proportional to the square of the speed of the object. Mathematically,


FD =
1
2
C ρ Av2,


where C is the drag coefficient, A is the area of the object facing the fluid, and ρ is the density of the fluid.


Athletes as well as car designers seek to reduce the drag force to lower their race times (Figure 6.29). Aerodynamicshaping of an automobile can reduce the drag force and thus increase a car’s gas mileage.


Figure 6.29 From racing cars to bobsled racers, aerodynamicshaping is crucial to achieving top speeds. Bobsleds aredesigned for speed and are shaped like a bullet with tapered fins.(credit: “U.S. Army”/Wikimedia Commons)


The value of the drag coefficient C is determined empirically, usually with the use of a wind tunnel (Figure 6.30).


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Figure 6.30 NASA researchers test a model plane in a windtunnel. (credit: NASA/Ames)


The drag coefficient can depend upon velocity, but we assume that it is a constant here. Table 6.2 lists some typical dragcoefficients for a variety of objects. Notice that the drag coefficient is a dimensionless quantity. At highway speeds, over
50% of the power of a car is used to overcome air drag. The most fuel-efficient cruising speed is about 70–80 km/h (about
45–50 mi/h). For this reason, during the 1970s oil crisis in the United States, maximum speeds on highways were set atabout 90 km/h (55 mi/h).


Object C
Airfoil 0.05
Toyota Camry 0.28
Ford Focus 0.32
Honda Civic 0.36
Ferrari Testarossa 0.37
Dodge Ram Pickup 0.43
Sphere 0.45
Hummer H2 SUV 0.64
Skydiver (feet first) 0.70
Bicycle 0.90
Skydiver (horizontal) 1.0
Circular flat plate 1.12


Table 6.2 Typical Values of DragCoefficient C


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Substantial research is under way in the sporting world to minimize drag. The dimples on golf balls are being redesigned,as are the clothes that athletes wear. Bicycle racers and some swimmers and runners wear full bodysuits. Australian CathyFreeman wore a full body suit in the 2000 Sydney Olympics and won a gold medal in the 400-m race. Many swimmersin the 2008 Beijing Olympics wore (Speedo) body suits; it might have made a difference in breaking many world records(Figure 6.31). Most elite swimmers (and cyclists) shave their body hair. Such innovations can have the effect of slicingaway milliseconds in a race, sometimes making the difference between a gold and a silver medal. One consequence is thatcareful and precise guidelines must be continuously developed to maintain the integrity of the sport.


Figure 6.31 Body suits, such as this LZR Racer Suit, havebeen credited with aiding in many world records after theirrelease in 2008. Smoother “skin” and more compression forceson a swimmer’s body provide at least 10% less drag. (credit:
NASA/Kathy Barnstorff)


Terminal Velocity
Some interesting situations connected to Newton’s second law occur when considering the effects of drag forces upon amoving object. For instance, consider a skydiver falling through air under the influence of gravity. The two forces actingon him are the force of gravity and the drag force (ignoring the small buoyant force). The downward force of gravityremains constant regardless of the velocity at which the person is moving. However, as the person’s velocity increases, themagnitude of the drag force increases until the magnitude of the drag force is equal to the gravitational force, thus producinga net force of zero. A zero net force means that there is no acceleration, as shown by Newton’s second law. At this point,the person’s velocity remains constant and we say that the person has reached his terminal velocity (vT). Since FD is
proportional to the speed squared, a heavier skydiver must go faster for FD to equal his weight. Let’s see how this works
out more quantitatively.
At the terminal velocity,


Fnet = mg − FD = ma = 0.


Thus,
mg = FD.


Using the equation for drag force, we have
mg = 1


2
CρAvT


2 .


Solving for the velocity, we obtain
vT =


2mg
ρCA


.


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6.10


Assume the density of air is ρ = 1.21 kg/m3. A 75-kg skydiver descending head first has a cross-sectional area of
approximately A = 0.18 m2 and a drag coefficient of approximately C = 0.70 . We find that


vT =
2(75 kg)(9.80 m/s2)


(1.21 kg/m3)(0.70)(0.18 m2)
= 98 m/s = 350 km/h.


This means a skydiver with a mass of 75 kg achieves a terminal velocity of about 350 km/h while traveling in a pike (headfirst) position, minimizing the area and his drag. In a spread-eagle position, that terminal velocity may decrease to about200 km/h as the area increases. This terminal velocity becomes much smaller after the parachute opens.
Example 6.17


Terminal Velocity of a Skydiver
Find the terminal velocity of an 85-kg skydiver falling in a spread-eagle position.
Strategy
At terminal velocity, Fnet = 0. Thus, the drag force on the skydiver must equal the force of gravity (the person’s
weight). Using the equation of drag force, we find mg = 1


2
ρCAv2.


Solution
The terminal velocity vT can be written as


vT =
2mg
ρCA


=
2(85 kg)(9.80 m/s2)


(1.21 kg/m3)(1.0)(0.70 m2)
= 44 m/s.


Significance
This result is consistent with the value for vT mentioned earlier. The 75-kg skydiver going feet first had a
terminal velocity of vT = 98 m/s. He weighed less but had a smaller frontal area and so a smaller drag due to
the air.


Check Your Understanding Find the terminal velocity of a 50-kg skydiver falling in spread-eaglefashion.


The size of the object that is falling through air presents another interesting application of air drag. If you fall from a 5-m-high branch of a tree, you will likely get hurt—possibly fracturing a bone. However, a small squirrel does this all the time,without getting hurt. You do not reach a terminal velocity in such a short distance, but the squirrel does.
The following interesting quote on animal size and terminal velocity is from a 1928 essay by a British biologist, J. B. S.Haldane, titled “On Being the Right Size.”
“To the mouse and any smaller animal, [gravity] presents practically no dangers. You can drop a mouse down a thousand-yard mine shaft; and, on arriving at the bottom, it gets a slight shock and walks away, provided that the ground is fairly soft.A rat is killed, a man is broken, and a horse splashes. For the resistance presented to movement by the air is proportionalto the surface of the moving object. Divide an animal’s length, breadth, and height each by ten; its weight is reduced to athousandth, but its surface only to a hundredth. So the resistance to falling in the case of the small animal is relatively tentimes greater than the driving force.”
The above quadratic dependence of air drag upon velocity does not hold if the object is very small, is going very slow, or isin a denser medium than air. Then we find that the drag force is proportional just to the velocity. This relationship is givenby Stokes’ law.


Stokes’ Law
For a spherical object falling in a medium, the drag force is


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(6.6)Fs = 6πrηv,
where r is the radius of the object, η is the viscosity of the fluid, and v is the object’s velocity.


Good examples of Stokes’ law are provided by microorganisms, pollen, and dust particles. Because each of these objectsis so small, we find that many of these objects travel unaided only at a constant (terminal) velocity. Terminal velocities forbacteria (size about 1 µm) can be about 2 µm/s. To move at a greater speed, many bacteria swim using flagella (organelles
shaped like little tails) that are powered by little motors embedded in the cell.
Sediment in a lake can move at a greater terminal velocity (about 5 µm/s), so it can take days for it to reach the bottom of
the lake after being deposited on the surface.
If we compare animals living on land with those in water, you can see how drag has influenced evolution. Fish, dolphins,and even massive whales are streamlined in shape to reduce drag forces. Birds are streamlined and migratory species thatfly large distances often have particular features such as long necks. Flocks of birds fly in the shape of a spearhead as theflock forms a streamlined pattern (Figure 6.32). In humans, one important example of streamlining is the shape of sperm,which need to be efficient in their use of energy.


Figure 6.32 Geese fly in a V formation during their longmigratory travels. This shape reduces drag and energyconsumption for individual birds, and also allows them a betterway to communicate. (credit: “Julo”/Wikimedia Commons)
In lecture demonstrations, we do measurements of the drag force (https://openstax.org/l/21dragforce)on different objects. The objects are placed in a uniform airstream created by a fan. Calculate the Reynolds numberand the drag coefficient.


The Calculus of Velocity-Dependent Frictional Forces
When a body slides across a surface, the frictional force on it is approximately constant and given by µkN. Unfortunately,
the frictional force on a body moving through a liquid or a gas does not behave so simply. This drag force is generally acomplicated function of the body’s velocity. However, for a body moving in a straight line at moderate speeds through aliquid such as water, the frictional force can often be approximated by


fR = −bv,


where b is a constant whose value depends on the dimensions and shape of the body and the properties of the liquid, andv is the velocity of the body. Two situations for which the frictional force can be represented this equation are a motorboatmoving through water and a small object falling slowly through a liquid.
Let’s consider the object falling through a liquid. The free-body diagram of this object with the positive direction downwardis shown in Figure 6.33. Newton’s second law in the vertical direction gives the differential equation


Chapter 6 | Applications of Newton's Laws 307




mg − bv = mdv
dt


,


where we have written the acceleration as dv/dt. As v increases, the frictional force –bv increases until it matches mg.
At this point, there is no acceleration and the velocity remains constant at the terminal velocity vT. From the previous
equation,


mg − bvT = 0,


so
vT =


mg
b


.


Figure 6.33 Free-body diagram of an object falling through aresistive medium.


We can find the object’s velocity by integrating the differential equation for v. First, we rearrange terms in this equation toobtain
dv


g − (b/m)v
= dt.


Assuming that v = 0 at t = 0, integration of this equation yields

⌡0


v
dv′


g − (b/m)v′
= ∫


0


t
dt′,


or
−m
b
ln⎛⎝g −


b
mv′

⎠|0
v
= t′|0


t ,


where v ' and t ' are dummy variables of integration. With the limits given, we find
−m
b
[ln⎛⎝g −


b
mv

⎠− lng] = t.


Since lnA − lnB = ln(A/B), and ln(A/B) = x implies ex = A/B, we obtain
g − (bv/m)


g = e
−bt/m,


and
v =


mg
b


(1 − e−bt/m).


Notice that as t → ∞, v → mg/b = vT, which is the terminal velocity.


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The position at any time may be found by integrating the equation for v. With v = dy/dt,
dy =


mg
b


(1 − e−bt/m)dt.


Assuming y = 0 when t = 0,


0


y
dy′ =


mg
b ∫0


t
(1 − e−bt '/m)dt′,


which integrates to
y =


mg
b
t +


m2g


b2
(e−bt/m − 1).


Example 6.18
Effect of the Resistive Force on a Motorboat
A motorboat is moving across a lake at a speed v0 when its motor suddenly freezes up and stops. The boat then
slows down under the frictional force fR = −bv. (a) What are the velocity and position of the boat as functions
of time? (b) If the boat slows down from 4.0 to 1.0 m/s in 10 s, how far does it travel before stopping?
Solutiona. With the motor stopped, the only horizontal force on the boat is fR = −bv, so from Newton’s second


law,
mdv
dt


= −bv,


which we can write as
dv
v = −


b
mdt.


Integrating this equation between the time zero when the velocity is v0 and the time t when the velocity
is v , we have



⌡0


v
dv′
v′


= − bm∫
0


t
dt′.


Thus,
ln vv0


= − bmt,


which, since lnA = x implies ex = A, we can write this as
v = v0 e


−bt/m.


Now from the definition of velocity,
dx
dt


= v0 e
−bt/m,


so we have
dx = v0 e


−bt/mdt.


With the initial position zero, we have


Chapter 6 | Applications of Newton's Laws 309




6.11



⌡0


x


dx ' = v0∫
0


t
e−bt '/mdt ',


and
x = −


mv0
b


e−bt '/m|0
t
=


mv0
b


(1 − e−bt/m).


As time increases, e−bt/m → 0, and the position of the boat approaches a limiting value
xmax =


mv0
b


.


Although this tells us that the boat takes an infinite amount of time to reach xmax, the boat effectively
stops after a reasonable time. For example, at t = 10m/b, we have


v = v0 e
−10 ≃ 4.5 × 10−5 v0,


whereas we also have
x = xmax(1 − e


−10) ≃ 0.99995xmax.


Therefore, the boat’s velocity and position have essentially reached their final values.
b. With v0 = 4.0 m/s and v = 1.0 m/s, we have 1.0 m/s = (4.0 m/s)e−(b/m)(10 s), so


ln 0.25 = −ln 4.0 = − bm(10 s),


and
b
m =


1
10


ln 4.0 s-1 = 0.14 s-1 .


Now the boat’s limiting position is
xmax =


mv0
b


= 4.0 m/s
0.14 s−1


= 29 m.


Significance
In the both of the previous examples, we found “limiting” values. The terminal velocity is the same as the limitingvelocity, which is the velocity of the falling object after a (relatively) long time has passed. Similarly, the limitingdistance of the boat is the distance the boat will travel after a long amount of time has passed. Due to the propertiesof exponential decay, the time involved to reach either of these values is actually not too long (certainly not aninfinite amount of time!) but they are quickly found by taking the limit to infinity.


Check Your Understanding Suppose the resistive force of the air on a skydiver can be approximated
by f = −bv2 . If the terminal velocity of a 100-kg skydiver is 60 m/s, what is the value of b?


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banked curve
centripetal force
Coriolis force
drag force
friction
ideal banking


inertial force
kinetic friction
noninertial frame of reference
static friction
terminal velocity


CHAPTER 6 REVIEW
KEY TERMS


curve in a road that is sloping in a manner that helps a vehicle negotiate the curve
any net force causing uniform circular motion


inertial force causing the apparent deflection of moving objects when viewed in a rotating frame ofreference
force that always opposes the motion of an object in a fluid; unlike simple friction, the drag force isproportional to some function of the velocity of the object in that fluid


force that opposes relative motion or attempts at motion between systems in contact
sloping of a curve in a road, where the angle of the slope allows the vehicle to negotiate the curve at acertain speed without the aid of friction between the tires and the road; the net external force on the vehicle equals thehorizontal centripetal force in the absence of friction
force that has no physical origin
force that opposes the motion of two systems that are in contact and moving relative to each other


accelerated frame of reference
force that opposes the motion of two systems that are in contact and are not moving relative to each other


constant velocity achieved by a falling object, which occurs when the weight of the object is balancedby the upward drag force


KEY EQUATIONS
Magnitude of static friction fs ≤ µsN
Magnitude of kinetic friction fk = µkN
Centripetal force


Fc = m
v2
r or Fc = mrω


2


Ideal angle of a banked curve
tan θ = v


2


rg


Drag force FD = 12CρAv2
Stokes’ law Fs = 6πrηv


SUMMARY
6.1 Solving Problems with Newton’s Laws


• Newton’s laws of motion can be applied in numerous situations to solve motion problems.
• Some problems contain multiple force vectors acting in different directions on an object. Be sure to draw diagrams,resolve all force vectors into horizontal and vertical components, and draw a free-body diagram. Always analyzethe direction in which an object accelerates so that you can determine whether Fnet = ma or Fnet = 0.
• The normal force on an object is not always equal in magnitude to the weight of the object. If an object isaccelerating vertically, the normal force is less than or greater than the weight of the object. Also, if the object is onan inclined plane, the normal force is always less than the full weight of the object.
• Some problems contain several physical quantities, such as forces, acceleration, velocity, or position. You can applyconcepts from kinematics and dynamics to solve these problems.


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6.2 Friction
• Friction is a contact force that opposes the motion or attempted motion between two systems. Simple friction isproportional to the normal force N supporting the two systems.
• The magnitude of static friction force between two materials stationary relative to each other is determined usingthe coefficient of static friction, which depends on both materials.
• The kinetic friction force between two materials moving relative to each other is determined using the coefficient ofkinetic friction, which also depends on both materials and is always less than the coefficient of static friction.


6.3 Centripetal Force
• Centripetal force F→ c is a “center-seeking” force that always points toward the center of rotation. It is
perpendicular to linear velocity and has the magnitude


Fc = mac.• Rotating and accelerated frames of reference are noninertial. Inertial forces, such as the Coriolis force, are neededto explain motion in such frames.
6.4 Drag Force and Terminal Speed


• Drag forces acting on an object moving in a fluid oppose the motion. For larger objects (such as a baseball) movingat a velocity in air, the drag force is determined using the drag coefficient (typical values are given in Table 6.2),the area of the object facing the fluid, and the fluid density.
• For small objects (such as a bacterium) moving in a denser medium (such as water), the drag force is given byStokes’ law.


CONCEPTUAL QUESTIONS
6.1 Solving Problems with Newton’s Laws
1. To simulate the apparent weightlessness of space orbit,astronauts are trained in the hold of a cargo aircraft thatis accelerating downward at g. Why do they appear tobe weightless, as measured by standing on a bathroomscale, in this accelerated frame of reference? Is there anydifference between their apparent weightlessness in orbitand in the aircraft?


6.2 Friction
2. The glue on a piece of tape can exert forces. Can theseforces be a type of simple friction? Explain, consideringespecially that tape can stick to vertical walls and even toceilings.
3. When you learn to drive, you discover that you need tolet up slightly on the brake pedal as you come to a stop orthe car will stop with a jerk. Explain this in terms of therelationship between static and kinetic friction.
4. When you push a piece of chalk across a chalkboard, itsometimes screeches because it rapidly alternates betweenslipping and sticking to the board. Describe this processin more detail, in particular, explaining how it is relatedto the fact that kinetic friction is less than static friction.


(The same slip-grab process occurs when tires screech onpavement.)
5. A physics major is cooking breakfast when she noticesthat the frictional force between her steel spatula and Teflonfrying pan is only 0.200 N. Knowing the coefficient ofkinetic friction between the two materials, she quicklycalculates the normal force. What is it?


6.3 Centripetal Force
6. If you wish to reduce the stress (which is related tocentripetal force) on high-speed tires, would you use large-or small-diameter tires? Explain.
7. Define centripetal force. Can any type of force (forexample, tension, gravitational force, friction, and so on)be a centripetal force? Can any combination of forces be acentripetal force?
8. If centripetal force is directed toward the center, why doyou feel that you are ‘thrown’ away from the center as a cargoes around a curve? Explain.
9. Race car drivers routinely cut corners, as shown below(Path 2). Explain how this allows the curve to be taken atthe greatest speed.


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10. Many amusement parks have rides that make verticalloops like the one shown below. For safety, the cars areattached to the rails in such a way that they cannot fall off.If the car goes over the top at just the right speed, gravityalone will supply the centripetal force. What other forceacts and what is its direction if:
(a) The car goes over the top at faster than this speed?
(b) The car goes over the top at slower than this speed?


11. What causes water to be removed from clothes in aspin-dryer?
12. As a skater forms a circle, what force is responsible formaking his turn? Use a free-body diagram in your answer.
13. Suppose a child is riding on a merry-go-round at a


distance about halfway between its center and edge. Shehas a lunch box resting on wax paper, so that there is verylittle friction between it and the merry-go-round. Whichpath shown below will the lunch box take when she lets go?The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved tothe right? Explain your answer.


14. Do you feel yourself thrown to either side when younegotiate a curve that is ideally banked for your car’sspeed? What is the direction of the force exerted on you bythe car seat?
15. Suppose a mass is moving in a circular path on africtionless table as shown below. In Earth’s frame ofreference, there is no centrifugal force pulling the massaway from the center of rotation, yet there is a forcestretching the string attaching the mass to the nail. Usingconcepts related to centripetal force and Newton’s thirdlaw, explain what force stretches the string, identifying itsphysical origin.


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16. When a toilet is flushed or a sink is drained, the water(and other material) begins to rotate about the drain on theway down. Assuming no initial rotation and a flow initiallydirectly straight toward the drain, explain what causes therotation and which direction it has in the NorthernHemisphere. (Note that this is a small effect and in mosttoilets the rotation is caused by directional water jets.)Would the direction of rotation reverse if water were forcedup the drain?
17. A car rounds a curve and encounters a patch of icewith a very low coefficient of kinetic fiction. The car slidesoff the road. Describe the path of the car as it leaves theroad.
18. In one amusement park ride, riders enter a largevertical barrel and stand against the wall on its horizontalfloor. The barrel is spun up and the floor drops away. Ridersfeel as if they are pinned to the wall by a force somethinglike the gravitational force. This is an inertial force sensedand used by the riders to explain events in the rotatingframe of reference of the barrel. Explain in an inertial frameof reference (Earth is nearly one) what pins the riders to thewall, and identify all forces acting on them.
19. Two friends are having a conversation. Anna says asatellite in orbit is in free fall because the satellite keepsfalling toward Earth. Tom says a satellite in orbit is not


in free fall because the acceleration due to gravity is not
9.80 m/s2 . Who do you agree with and why?
20. A nonrotating frame of reference placed at the centerof the Sun is very nearly an inertial one. Why is it notexactly an inertial frame?


6.4 Drag Force and Terminal Speed
21. Athletes such as swimmers and bicyclists wear bodysuits in competition. Formulate a list of pros and cons ofsuch suits.
22. Two expressions were used for the drag forceexperienced by a moving object in a liquid. One dependedupon the speed, while the other was proportional to thesquare of the speed. In which types of motion would eachof these expressions be more applicable than the other one?
23. As cars travel, oil and gasoline leaks onto the roadsurface. If a light rain falls, what does this do to the controlof the car? Does a heavy rain make any difference?
24. Why can a squirrel jump from a tree branch to theground and run away undamaged, while a human couldbreak a bone in such a fall?


PROBLEMS
6.1 Solving Problems with Newton’s Laws
25. A 30.0-kg girl in a swing is pushed to one side and
held at rest by a horizontal force F→ so that the swing
ropes are 30.0° with respect to the vertical. (a) Calculate
the tension in each of the two ropes supporting the swingunder these conditions. (b) Calculate the magnitude of
F


.


26. Find the tension in each of the three cables supportingthe traffic light if it weighs 2.00 × 102 N.


27. Three forces act on an object, considered to be a


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particle, which moves with constant velocity
v = (3 i


^
− 2 j


^
) m/s. Two of the forces are


F


1 = (3 i
^


+ 5 j
^


− 6k
^
) N and


F


2 = (4 i
^


− 7 j
^


+ 2k
^
) N. Find the third force.


28. A flea jumps by exerting a force of 1.20 × 10−5 N
straight down on the ground. A breeze blowing on the flea
parallel to the ground exerts a force of 0.500 × 10−6 N
on the flea while the flea is still in contact with the ground.Find the direction and magnitude of the acceleration of the
flea if its mass is 6.00 × 10−7 kg . Do not neglect the
gravitational force.
29. Two muscles in the back of the leg pull upward on theAchilles tendon, as shown below. (These muscles are calledthe medial and lateral heads of the gastrocnemius muscle.)Find the magnitude and direction of the total force on theAchilles tendon. What type of movement could be causedby this force?


30. After a mishap, a 76.0-kg circus performer clings to atrapeze, which is being pulled to the side by another circusartist, as shown here. Calculate the tension in the two ropesif the person is momentarily motionless. Include a free-body diagram in your solution.


31. A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to join another dolphin in play. What averageforce was exerted to slow the first dolphin if it was movinghorizontally? (The gravitational force is balanced by thebuoyant force of the water.)
32. When starting a foot race, a 70.0-kg sprinter exerts anaverage force of 650 N backward on the ground for 0.800s. (a) What is his final speed? (b) How far does he travel?
33. A large rocket has a mass of 2.00 × 106 kg at
takeoff, and its engines produce a thrust of 3.50 × 107 N.
(a) Find its initial acceleration if it takes off vertically. (b)How long does it take to reach a velocity of 120 km/hstraight up, assuming constant mass and thrust?
34. A basketball player jumps straight up for a ball. Todo this, he lowers his body 0.300 m and then acceleratesthrough this distance by forcefully straightening his legs.This player leaves the floor with a vertical velocitysufficient to carry him 0.900 m above the floor. (a)Calculate his velocity when he leaves the floor. (b)Calculate his acceleration while he is straightening his legs.He goes from zero to the velocity found in (a) in a distanceof 0.300 m. (c) Calculate the force he exerts on the floor todo this, given that his mass is 110.0 kg.
35. A 2.50-kg fireworks shell is fired straight up froma mortar and reaches a height of 110.0 m. (a) Neglectingair resistance (a poor assumption, but we will make it forthis example), calculate the shell’s velocity when it leavesthe mortar. (b) The mortar itself is a tube 0.450 m long.Calculate the average acceleration of the shell in the tube as


Chapter 6 | Applications of Newton's Laws 315




it goes from zero to the velocity found in (a). (c) What isthe average force on the shell in the mortar? Express youranswer in newtons and as a ratio to the weight of the shell.
36. A 0.500-kg potato is fired at an angle of 80.0° above
the horizontal from a PVC pipe used as a “potato gun” andreaches a height of 110.0 m. (a) Neglecting air resistance,calculate the potato’s velocity when it leaves the gun. (b)The gun itself is a tube 0.450 m long. Calculate the averageacceleration of the potato in the tube as it goes from zero tothe velocity found in (a). (c) What is the average force onthe potato in the gun? Express your answer in newtons andas a ratio to the weight of the potato.
37. An elevator filled with passengers has a mass of
1.70 × 103 kg . (a) The elevator accelerates upward from
rest at a rate of 1.20 m/s2 for 1.50 s. Calculate the tension
in the cable supporting the elevator. (b) The elevatorcontinues upward at constant velocity for 8.50 s. What isthe tension in the cable during this time? (c) The elevator
decelerates at a rate of 0.600 m/s2 for 3.00 s. What is the
tension in the cable during deceleration? (d) How high hasthe elevator moved above its original starting point, andwhat is its final velocity?
38. A 20.0-g ball hangs from the roof of a freight car bya string. When the freight car begins to move, the stringmakes an angle of 35.0° with the vertical. (a) What is the
acceleration of the freight car? (b) What is the tension in thestring?
39. A student’s backpack, full of textbooks, is hung from aspring scale attached to the ceiling of an elevator. When the
elevator is accelerating downward at 3.8 m/s2 , the scale
reads 60 N. (a) What is the mass of the backpack? (b) Whatdoes the scale read if the elevator moves upward while
slowing down at a rate 3.8 m/s2 ? (c) What does the scale
read if the elevator moves upward at constant velocity? (d)If the elevator had no brakes and the cable supporting itwere to break loose so that the elevator could fall freely,what would the spring scale read?
40. A service elevator takes a load of garbage, mass 10.0kg, from a floor of a skyscraper under construction, downto ground level, accelerating downward at a rate of
1.2 m/s2 . Find the magnitude of the force the garbage
exerts on the floor of the service elevator?
41. A roller coaster car starts from rest at the top of atrack 30.0 m long and inclined at 20.0° to the horizontal.
Assume that friction can be ignored. (a) What is theacceleration of the car? (b) How much time elapses beforeit reaches the bottom of the track?


42. The device shown below is the Atwood’s machineconsidered in Example 6.5. Assuming that the masses ofthe string and the frictionless pulley are negligible, (a) findan equation for the acceleration of the two blocks; (b) findan equation for the tension in the string; and (c) find boththe acceleration and tension when block 1 has mass 2.00 kgand block 2 has mass 4.00 kg.


43. Two blocks are connected by a massless rope as shownbelow. The mass of the block on the table is 4.0 kg andthe hanging mass is 1.0 kg. The table and the pulley arefrictionless. (a) Find the acceleration of the system. (b) Findthe tension in the rope. (c) Find the speed with which thehanging mass hits the floor if it starts from rest and isinitially located 1.0 m from the floor.


44. Shown below are two carts connected by a cord thatpasses over a small frictionless pulley. Each cart rolls freelywith negligible friction. Calculate the acceleration of thecarts and the tension in the cord.


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45. A 2.00 kg block (mass 1) and a 4.00 kg block (mass2) are connected by a light string as shown; the inclinationof the ramp is 40.0° . Friction is negligible. What is (a) the
acceleration of each block and (b) the tension in the string?


6.2 Friction
46. (a) When rebuilding his car’s engine, a physics major
must exert 3.00 × 102 N of force to insert a dry steel
piston into a steel cylinder. What is the normal forcebetween the piston and cylinder? (b) What force would hehave to exert if the steel parts were oiled?
47. (a) What is the maximum frictional force in the kneejoint of a person who supports 66.0 kg of her mass on thatknee? (b) During strenuous exercise, it is possible to exertforces to the joints that are easily 10 times greater thanthe weight being supported. What is the maximum forceof friction under such conditions? The frictional forces injoints are relatively small in all circumstances except whenthe joints deteriorate, such as from injury or arthritis.Increased frictional forces can cause further damage andpain.
48. Suppose you have a 120-kg wooden crate resting on awood floor, with coefficient of static friction 0.500 betweenthese wood surfaces. (a) What maximum force can youexert horizontally on the crate without moving it? (b) Ifyou continue to exert this force once the crate starts to slip,what will its acceleration then be? The coefficient of slidingfriction is known to be 0.300 for this situation.


49. (a) If half of the weight of a small 1.00 × 103 -kg
utility truck is supported by its two drive wheels, what isthe maximum acceleration it can achieve on dry concrete?(b) Will a metal cabinet lying on the wooden bed of thetruck slip if it accelerates at this rate? (c) Solve bothproblems assuming the truck has four-wheel drive.
50. A team of eight dogs pulls a sled with waxed woodrunners on wet snow (mush!). The dogs have averagemasses of 19.0 kg, and the loaded sled with its rider has amass of 210 kg. (a) Calculate the acceleration of the dogsstarting from rest if each dog exerts an average force of185 N backward on the snow. (b) Calculate the force in thecoupling between the dogs and the sled.
51. Consider the 65.0-kg ice skater being pushed by twoothers shown below. (a) Find the direction and magnitudeof Ftot, the total force exerted on her by the others, given
that the magnitudes F1 and F2 are 26.4 N and 18.6
N, respectively. (b) What is her initial acceleration if sheis initially stationary and wearing steel-bladed skates thatpoint in the direction of Ftot? (c) What is her acceleration
assuming she is already moving in the direction of Ftot?
(Remember that friction always acts in the directionopposite that of motion or attempted motion betweensurfaces in contact.)


52. Show that the acceleration of any object down a


Chapter 6 | Applications of Newton's Laws 317




frictionless incline that makes an angle θ with the
horizontal is a = g sin θ . (Note that this acceleration is
independent of mass.)


53. Show that the acceleration of any object down anincline where friction behaves simply (that is, where
fk = µkN) is a = g(sin θ − µk cos θ). Note that the
acceleration is independent of mass and reduces to theexpression found in the previous problem when frictionbecomes negligibly small (µk = 0).


54. Calculate the deceleration of a snow boarder goingup a 5.00° slope, assuming the coefficient of friction for
waxed wood on wet snow. The result of the precedingproblem may be useful, but be careful to consider the factthat the snow boarder is going uphill.
55. A machine at a post office sends packages out a chuteand down a ramp to be loaded into delivery vehicles. (a)Calculate the acceleration of a box heading down a 10.0°
slope, assuming the coefficient of friction for a parcel onwaxed wood is 0.100. (b) Find the angle of the slope downwhich this box could move at a constant velocity. You canneglect air resistance in both parts.
56. If an object is to rest on an incline without slipping,then friction must equal the component of the weight ofthe object parallel to the incline. This requires greater andgreater friction for steeper slopes. Show that the maximumangle of an incline above the horizontal for which an object
will not slide down is θ = tan−1 µs. You may use the
result of the previous problem. Assume that a = 0 and that
static friction has reached its maximum value.


57. Calculate the maximum acceleration of a car that isheading down a 6.00° slope (one that makes an angle
of 6.00° with the horizontal) under the following road
conditions. You may assume that the weight of the car isevenly distributed on all four tires and that the coefficient ofstatic friction is involved—that is, the tires are not allowedto slip during the deceleration. (Ignore rolling.) Calculatefor a car: (a) On dry concrete. (b) On wet concrete. (c) Onice, assuming that µs = 0.100 , the same as for shoes on
ice.
58. Calculate the maximum acceleration of a car that isheading up a 4.00° slope (one that makes an angle of
4.00° with the horizontal) under the following road
conditions. Assume that only half the weight of the car issupported by the two drive wheels and that the coefficientof static friction is involved—that is, the tires are notallowed to slip during the acceleration. (Ignore rolling.) (a)On dry concrete. (b) On wet concrete. (c) On ice, assumingthat µs = 0.100 , the same as for shoes on ice.
59. Repeat the preceding problem for a car with four-wheel drive.
60. A freight train consists of two 8.00 × 105 -kg
engines and 45 cars with average masses of
5.50 × 105 kg. (a) What force must each engine exert
backward on the track to accelerate the train at a rate of
5.00 × 10−2m/s2 if the force of friction is 7.50 × 105N
, assuming the engines exert identical forces? This is nota large frictional force for such a massive system. Rollingfriction for trains is small, and consequently, trains arevery energy-efficient transportation systems. (b) What isthe force in the coupling between the 37th and 38th cars(this is the force each exerts on the other), assuming all carshave the same mass and that friction is evenly distributedamong all of the cars and engines?
61. Consider the 52.0-kg mountain climber shown below.(a) Find the tension in the rope and the force that themountain climber must exert with her feet on the verticalrock face to remain stationary. Assume that the force isexerted parallel to her legs. Also, assume negligible forceexerted by her arms. (b) What is the minimum coefficientof friction between her shoes and the cliff?


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62. A contestant in a winter sporting event pushes a45.0-kg block of ice across a frozen lake as shown below.(a) Calculate the minimum force F he must exert to get theblock moving. (b) What is its acceleration once it starts tomove, if that force is maintained?


63. The contestant now pulls the block of ice with a ropeover his shoulder at the same angle above the horizontal asshown below. Calculate the minimum force F he must exertto get the block moving. (b) What is its acceleration once itstarts to move, if that force is maintained?


64. At a post office, a parcel that is a 20.0-kg box slidesdown a ramp inclined at 30.0° with the horizontal. The
coefficient of kinetic friction between the box and plane is0.0300. (a) Find the acceleration of the box. (b) Find thevelocity of the box as it reaches the end of the plane, if thelength of the plane is 2 m and the box starts at rest.


6.3 Centripetal Force
65. (a) A 22.0-kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetalforce is exerted if he is 1.25 m from its center? (b) Whatcentripetal force is exerted if the merry-go-round rotates at3.00 rev/min and he is 8.00 m from its center? (c) Compareeach force with his weight.
66. Calculate the centripetal force on the end of a 100-m(radius) wind turbine blade that is rotating at 0.5 rev/s.Assume the mass is 4 kg.
67. What is the ideal banking angle for a gentle turn of1.20-km radius on a highway with a 105 km/h speed limit(about 65 mi/h), assuming everyone travels at the limit?
68. What is the ideal speed to take a 100.0-m-radius curvebanked at a 20.0° angle?
69. (a) What is the radius of a bobsled turn banked at
75.0° and taken at 30.0 m/s, assuming it is ideally banked?
(b) Calculate the centripetal acceleration. (c) Does thisacceleration seem large to you?
70. Part of riding a bicycle involves leaning at the correctangle when making a turn, as seen below. To be stable, theforce exerted by the ground must be on a line going throughthe center of gravity. The force on the bicycle wheel canbe resolved into two perpendicular components—frictionparallel to the road (this must supply the centripetal force)and the vertical normal force (which must equal thesystem’s weight). (a) Show that θ (as defined as shown) is
related to the speed v and radius of curvature r of the turnin the same way as for an ideally banked roadway—that is,
θ = tan−1(v2 /rg). (b) Calculate θ for a 12.0-m/s turn of
radius 30.0 m (as in a race).


71. If a car takes a banked curve at less than the idealspeed, friction is needed to keep it from sliding toward theinside of the curve (a problem on icy mountain roads). (a)


Chapter 6 | Applications of Newton's Laws 319




Calculate the ideal speed to take a 100.0 m radius curvebanked at 15.0° . (b) What is the minimum coefficient of
friction needed for a frightened driver to take the samecurve at 20.0 km/h?
72. Modern roller coasters have vertical loops like theone shown here. The radius of curvature is smaller at thetop than on the sides so that the downward centripetalacceleration at the top will be greater than the accelerationdue to gravity, keeping the passengers pressed firmly intotheir seats. (a) What is the speed of the roller coaster at thetop of the loop if the radius of curvature there is 15.0 m andthe downward acceleration of the car is 1.50 g? (b) Howhigh above the top of the loop must the roller coaster startfrom rest, assuming negligible friction? (c) If it actuallystarts 5.00 m higher than your answer to (b), how much
energy did it lose to friction? Its mass is 1.50 × 103 kg .


73. A child of mass 40.0 kg is in a roller coaster car thattravels in a loop of radius 7.00 m. At point A the speed ofthe car is 10.0 m/s, and at point B, the speed is 10.5 m/s.Assume the child is not holding on and does not wear a seatbelt. (a) What is the force of the car seat on the child atpoint A? (b) What is the force of the car seat on the child atpoint B? (c) What minimum speed is required to keep thechild in his seat at point A?


74. In the simple Bohr model of the ground state of thehydrogen atom, the electron travels in a circular orbitaround a fixed proton. The radius of the orbit is
5.28 × 10−11 m, and the speed of the electron is
2.18 × 106 m/s. The mass of an electron is
9.11 × 10−31 kg . What is the force on the electron?
75. Railroad tracks follow a circular curve of radius 500.0m and are banked at an angle of 5.0° . For trains of what
speed are these tracks designed?
76. The CERN particle accelerator is circular with acircumference of 7.0 km. (a) What is the acceleration of
the protons (m = 1.67 × 10−27 kg) that move around the
accelerator at 5% of the speed of light? (The speed of
light is v = 3.00 × 108 m/s. ) (b) What is the force on the
protons?
77. A car rounds an unbanked curve of radius 65 m. Ifthe coefficient of static friction between the road and car is0.70, what is the maximum speed at which the car traversethe curve without slipping?
78. A banked highway is designed for traffic moving at90.0 km/h. The radius of the curve is 310 m. What is theangle of banking of the highway?


6.4 Drag Force and Terminal Speed
79. The terminal velocity of a person falling in air dependsupon the weight and the area of the person facing thefluid. Find the terminal velocity (in meters per second andkilometers per hour) of an 80.0-kg skydiver falling in a pike
(headfirst) position with a surface area of 0.140 m2 .


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80. A 60.0-kg and a 90.0-kg skydiver jump from an
airplane at an altitude of 6.00 × 103m , both falling in the
pike position. Make some assumption on their frontal areasand calculate their terminal velocities. How long will it takefor each skydiver to reach the ground (assuming the timeto reach terminal velocity is small)? Assume all values areaccurate to three significant digits.
81. A 560-g squirrel with a surface area of 930 cm2
falls from a 5.0-m tree to the ground. Estimate its terminalvelocity. (Use a drag coefficient for a horizontal skydiver.)What will be the velocity of a 56-kg person hitting theground, assuming no drag contribution in such a shortdistance?
82. To maintain a constant speed, the force provided bya car’s engine must equal the drag force plus the force offriction of the road (the rolling resistance). (a) What are thedrag forces at 70 km/h and 100 km/h for a Toyota Camry?
(Drag area is 0.70 m2 ) (b) What is the drag force at 70 km/
h and 100 km/h for a Hummer H2? (Drag area is 2.44 m2)
Assume all values are accurate to three significant digits.
83. By what factor does the drag force on a car increase asit goes from 65 to 110 km/h?
84. Calculate the velocity a spherical rain drop wouldachieve falling from 5.00 km (a) in the absence of air drag(b) with air drag. Take the size across of the drop to be 4
mm, the density to be 1.00 × 103 kg/m3 , and the surface
area to be πr2 .
85. Using Stokes’ law, verify that the units for viscosityare kilograms per meter per second.
86. Find the terminal velocity of a spherical bacterium(diameter 2.00 µm ) falling in water. You will first need to
note that the drag force is equal to the weight at terminalvelocity. Take the density of the bacterium to be
1.10 × 103 kg/m3 .
87. Stokes’ law describes sedimentation of particles inliquids and can be used to measure viscosity. Particles inliquids achieve terminal velocity quickly. One can measurethe time it takes for a particle to fall a certain distance andthen use Stokes’ law to calculate the viscosity of the liquid.
Suppose a steel ball bearing (density 7.8 × 103 kg/m3 ,
diameter 3.0 mm) is dropped in a container of motor oil.It takes 12 s to fall a distance of 0.60 m. Calculate theviscosity of the oil.
88. Suppose that the resistive force of the air on a skydiver


can be approximated by f = −bv2. If the terminal
velocity of a 50.0-kg skydiver is 60.0 m/s, what is the valueof b?
89. A small diamond of mass 10.0 g drops from aswimmer’s earring and falls through the water, reaching aterminal velocity of 2.0 m/s. (a) Assuming the frictionalforce on the diamond obeys f = −bv, what is b? (b) How
far does the diamond fall before it reaches 90 percent of itsterminal speed?
90. (a) What is the final velocity of a car originallytraveling at 50.0 km/h that decelerates at a rate of
0.400 m/s2 for 50.0 s? Assume a coefficient of friction of
1.0. (b) What is unreasonable about the result? (c) Whichpremise is unreasonable, or which premises areinconsistent?
91. A 75.0-kg woman stands on a bathroom scale in anelevator that accelerates from rest to 30.0 m/s in 2.00 s. (a)Calculate the scale reading in newtons and compare it withher weight. (The scale exerts an upward force on her equalto its reading.) (b) What is unreasonable about the result?(c) Which premise is unreasonable, or which premises areinconsistent?
92. (a) Calculate the minimum coefficient of frictionneeded for a car to negotiate an unbanked 50.0 m radiuscurve at 30.0 m/s. (b) What is unreasonable about theresult? (c) Which premises are unreasonable orinconsistent?
93. As shown below, if M = 5.50 kg, what is the tension
in string 1?


94. As shown below, if F = 60.0 N and M = 4.00 kg,
what is the magnitude of the acceleration of the suspendedobject? All surfaces are frictionless.


Chapter 6 | Applications of Newton's Laws 321




95. As shown below, if M = 6.0 kg, what is the tension
in the connecting string? The pulley and all surfaces arefrictionless.


96. A small space probe is released from a spaceship.The space probe has mass 20.0 kg and contains 90.0 kgof fuel. It starts from rest in deep space, from the originof a coordinate system based on the spaceship, and burnsfuel at the rate of 3.00 kg/s. The engine provides a constantthrust of 120.0 N. (a) Write an expression for the mass ofthe space probe as a function of time, between 0 and 30seconds, assuming that the engine ignites fuel beginning at
t = 0. (b) What is the velocity after 15.0 s? (c) What is the
position of the space probe after 15.0 s, with initial positionat the origin? (d) Write an expression for the position as afunction of time, for t > 30.0 s.
97. A half-full recycling bin has mass 3.0 kg and is pushedup a 40.0° incline with constant speed under the action
of a 26-N force acting up and parallel to the incline. Theincline has friction. What magnitude force must act up andparallel to the incline for the bin to move down the inclineat constant velocity?
98. A child has mass 6.0 kg and slides down a 35° incline
with constant speed under the action of a 34-N force actingup and parallel to the incline. What is the coefficient ofkinetic friction between the child and the surface of theincline?


ADDITIONAL PROBLEMS
99. The two barges shown here are coupled by a cableof negligible mass. The mass of the front barge is
2.00 × 103 kg and the mass of the rear barge is
3.00 × 103 kg. A tugboat pulls the front barge with a
horizontal force of magnitude 20.0 × 103 N, and the
frictional forces of the water on the front and rear barges
are 8.00 × 103 N and 10.0 × 103 N, respectively. Find
the horizontal acceleration of the barges and the tension inthe connecting cable.


100. If the order of the barges of the preceding exercise is
reversed so that the tugboat pulls the 3.00 × 103 -kg barge
with a force of 20.0 × 103 N, what are the acceleration
of the barges and the tension in the coupling cable?
101. An object with mass m moves along the x-axis. Its


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position at any time is given by x(t) = pt3 + qt2 where p
and q are constants. Find the net force on this object for anytime t.
102. A helicopter with mass 2.35 × 104 kg has a
position given by
r→ (t) = (0.020 t3) i


^
+ (2.2t) j


^
− (0.060 t2)k


^
. Find the


net force on the helicopter at t = 3.0 s.
103. Located at the origin, an electric car of mass mis at rest and in equilibrium. A time dependent force of
F


(t) is applied at time t = 0 , and its components are
Fx(t) = p + nt and Fy(t) = qt where p, q, and n are
constants. Find the position r→ (t) and velocity v→ (t) as
functions of time t.
104. A particle of mass m is located at the origin. Itis at rest and in equilibrium. A time-dependent force of
F


(t) is applied at time t = 0 , and its components are
Fx(t) = pt and Fy(t) = n + qt where p, q, and n are
constants. Find the position r→ (t) and velocity v→ (t) as
functions of time t.


105. A 2.0-kg object has a velocity of 4.0 i^ m/s at
t = 0. A constant resultant force of (2.0 i^ + 4.0 j^ ) N
then acts on the object for 3.0 s. What is the magnitude ofthe object’s velocity at the end of the 3.0-s interval?
106. A 1.5-kg mass has an acceleration of
(4.0 i


^
− 3.0 j


^
) m/s2. Only two forces act on the mass.


If one of the forces is (2.0 i^ − 1.4 j^ ) N, what is the
magnitude of the other force?
107. A box is dropped onto a conveyor belt moving at 3.4m/s. If the coefficient of friction between the box and thebelt is 0.27, how long will it take before the box moveswithout slipping?
108. Shown below is a 10.0-kg block being pushed by
a horizontal force F→ of magnitude 200.0 N. The
coefficient of kinetic friction between the two surfaces is0.50. Find the acceleration of the block.


109. As shown below, the mass of block 1 is
m1 = 4.0 kg, while the mass of block 2 is m2 = 8.0 kg.
The coefficient of friction between m1 and the inclined
surface is µk = 0.40. What is the acceleration of the
system?


110. A student is attempting to move a 30-kg mini-fridgeinto her dorm room. During a moment of inattention, themini-fridge slides down a 35 degree incline at constantspeed when she applies a force of 25 N acting up andparallel to the incline. What is the coefficient of kineticfriction between the fridge and the surface of the incline?
111. A crate of mass 100.0 kg rests on a rough surfaceinclined at an angle of 37.0° with the horizontal. A
massless rope to which a force can be applied parallel to thesurface is attached to the crate and leads to the top of theincline. In its present state, the crate is just ready to slip andstart to move down the plane. The coefficient of friction is
80% of that for the static case. (a) What is the coefficient
of static friction? (b) What is the maximum force that canbe applied upward along the plane on the rope and notmove the block? (c) With a slightly greater applied force,the block will slide up the plane. Once it begins to move,what is its acceleration and what reduced force is necessaryto keep it moving upward at constant speed? (d) If the blockis given a slight nudge to get it started down the plane, whatwill be its acceleration in that direction? (e) Once the blockbegins to slide downward, what upward force on the rope isrequired to keep the block from accelerating downward?
112. A car is moving at high speed along a highwaywhen the driver makes an emergency braking. The wheelsbecome locked (stop rolling), and the resulting skid marks


Chapter 6 | Applications of Newton's Laws 323




are 32.0 meters long. If the coefficient of kinetic frictionbetween tires and road is 0.550, and the acceleration wasconstant during braking, how fast was the car going whenthe wheels became locked?
113. A crate having mass 50.0 kg falls horizontally off theback of the flatbed truck, which is traveling at 100 km/h.Find the value of the coefficient of kinetic friction betweenthe road and crate if the crate slides 50 m on the road incoming to rest. The initial speed of the crate is the same asthe truck, 100 km/h.


114. A 15-kg sled is pulled across a horizontal, snow-covered surface by a force applied to a rope at 30 degreeswith the horizontal. The coefficient of kinetic frictionbetween the sled and the snow is 0.20. (a) If the force is33 N, what is the horizontal acceleration of the sled? (b)What must the force be in order to pull the sled at constantvelocity?
115. A 30.0-g ball at the end of a string is swung ina vertical circle with a radius of 25.0 cm. The rotationalvelocity is 200.0 cm/s. Find the tension in the string: (a)at the top of the circle, (b) at the bottom of the circle, and(c) at a distance of 12.5 cm from the center of the circle
(r = 12.5 cm).


116. A particle of mass 0.50 kg starts moves througha circular path in the xy-plane with a position given by
r→ (t) = (4.0 cos 3t) i


^
+ (4.0 sin 3t) j


^ where r is in
meters and t is in seconds. (a) Find the velocity andacceleration vectors as functions of time. (b) Show thatthe acceleration vector always points toward the center ofthe circle (and thus represents centripetal acceleration). (c)Find the centripetal force vector as a function of time.
117. A stunt cyclist rides on the interior of a cylinder 12 min radius. The coefficient of static friction between the tiresand the wall is 0.68. Find the value of the minimum speedfor the cyclist to perform the stunt.
118. When a body of mass 0.25 kg is attached to a verticalmassless spring, it is extended 5.0 cm from its unstretchedlength of 4.0 cm. The body and spring are placed on ahorizontal frictionless surface and rotated about the heldend of the spring at 2.0 rev/s. How far is the springstretched?


119. Railroad tracks follow a circular curve of radius500.0 m and are banked at an angle of 5.00° . For trains of
what speed are these tracks designed?
120. A plumb bob hangs from the roof of a railroad car.The car rounds a circular track of radius 300.0 m at a speedof 90.0 km/h. At what angle relative to the vertical does theplumb bob hang?
121. An airplane flies at 120.0 m/s and banks at a 30°
angle. If its mass is 2.50 × 103 kg, (a) what is the
magnitude of the lift force? (b) what is the radius of theturn?
122. The position of a particle is given by
r→ (t) = A



⎝cosωt i


^
+ sinωt j


^⎞
⎠, where ω is a


constant. (a) Show that the particle moves in a circle of
radius A. (b) Calculate d r→ /dt and then show that the
speed of the particle is a constant Aω. (c) Determine
d2 r→ /dt2 and show that a is given by ac = rω2. (d)
Calculate the centripetal force on the particle. [Hint: For (b)and (c), you will need to use (d/dt)(cosωt) = −ω sinωt
and (d/dt)(sinωt) = ω cosωt.
123. Two blocks connected by a string are pulled across ahorizontal surface by a force applied to one of the blocks,as shown below. The coefficient of kinetic friction betweenthe blocks and the surface is 0.25. If each block has an
acceleration of 2.0 m/s2 to the right, what is the
magnitude F of the applied force?


124. As shown below, the coefficient of kinetic frictionbetween the surface and the larger block is 0.20, and thecoefficient of kinetic friction between the surface and thesmaller block is 0.30. If F = 10 N and M = 1.0 kg , what
is the tension in the connecting string?


125. In the figure, the coefficient of kinetic friction


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between the surface and the blocks is µk. If M = 1.0 kg,
find an expression for the magnitude of the acceleration ofeither block (in terms of F, µk, and g).


126. Two blocks are stacked as shown below, and reston a frictionless surface. There is friction between the twoblocks (coefficient of friction µ ). An external force is
applied to the top block at an angle θ with the horizontal.
What is the maximum force F that can be applied for thetwo blocks to move together?


127. A box rests on the (horizontal) back of a truck. Thecoefficient of static friction between the box and the surfaceon which it rests is 0.24. What maximum distance can thetruck travel (starting from rest and moving horizontallywith constant acceleration) in 3.0 s without having the boxslide?
128. A double-incline plane is shown below. Thecoefficient of friction on the left surface is 0.30, and on theright surface 0.16. Calculate the acceleration of the system.


CHALLENGE PROBLEMS
129. In a later chapter, you will find that the weight of a
particle varies with altitude such that w = mgr0 2


r2
where


r0 is the radius of Earth and r is the distance from Earth’s
center. If the particle is fired vertically with velocity v0
from Earth’s surface, determine its velocity as a functionof position r. (Hint: use a dr = v dv, the rearrangement
mentioned in the text.)
130. A large centrifuge, like the one shown below, isused to expose aspiring astronauts to accelerations similarto those experienced in rocket launches and atmosphericreentries. (a) At what angular velocity is the centripetalacceleration 10g if the rider is 15.0 m from the center ofrotation? (b) The rider’s cage hangs on a pivot at the end


of the arm, allowing it to swing outward during rotationas shown in the bottom accompanying figure. At whatangle θ below the horizontal will the cage hang when the
centripetal acceleration is 10g? (Hint: The arm suppliescentripetal force and supports the weight of the cage. Drawa free-body diagram of the forces to see what the angle θ
should be.)


Chapter 6 | Applications of Newton's Laws 325




131. A car of mass 1000.0 kg is traveling along a levelroad at 100.0 km/h when its brakes are applied. Calculatethe stopping distance if the coefficient of kinetic friction ofthe tires is 0.500. Neglect air resistance. (Hint: since thedistance traveled is of interest rather than the time, x is the


desired independent variable and not t. Use the Chain Rule
to change the variable: dv


dt
= dv


dx
dx
dt


= vdv
dx


.)


132. An airplane flying at 200.0 m/s makes a turn thattakes 4.0 min. What bank angle is required? What is thepercentage increase in the perceived weight of thepassengers?
133. A skydiver is at an altitude of 1520 m. After 10.0seconds of free fall, he opens his parachute and finds thatthe air resistance, FD , is given by the formula
FD = −bv, where b is a constant and v is the velocity.
If b = 0.750, and the mass of the skydiver is 82.0 kg,
first set up differential equations for the velocity and theposition, and then find: (a) the speed of the skydiver whenthe parachute opens, (b) the distance fallen before theparachute opens, (c) the terminal velocity after theparachute opens (find the limiting velocity), and (d) thetime the skydiver is in the air after the parachute opens.
134. In a television commercial, a small, spherical beadof mass 4.00 g is released from rest at t = 0 in a bottle of
liquid shampoo. The terminal speed is observed to be 2.00cm/s. Find (a) the value of the constant b in the equation
v =


mg
b


(1 − e−bt/m), and (b) the value of the resistive
force when the bead reaches terminal speed.
135. A boater and motor boat are at rest on a lake.Together, they have mass 200.0 kg. If the thrust of themotor is a constant force of 40.0 N in the direction ofmotion, and if the resistive force of the water is numericallyequivalent to 2 times the speed v of the boat, set up andsolve the differential equation to find: (a) the velocity of theboat at time t; (b) the limiting velocity (the velocity after along time has passed).


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7 | WORK AND KINETICENERGY


Figure 7.1 A sprinter exerts her maximum power to do as much work on herself as possible in the short time that her foot is incontact with the ground. This adds to her kinetic energy, preventing her from slowing down during the race. Pushing back hardon the track generates a reaction force that propels the sprinter forward to win at the finish. (credit: modification of work byMarie-Lan Nguyen)


Chapter Outline
7.1 Work
7.2 Kinetic Energy
7.3 Work-Energy Theorem
7.4 Power


Introduction
In this chapter, we discuss some basic physical concepts involved in every physical motion in the universe, going beyond theconcepts of force and change in motion, which we discussed in Motion in Two and Three Dimensions and Newton’sLaws of Motion. These concepts are work, kinetic energy, and power. We explain how these quantities are related to oneanother, which will lead us to a fundamental relationship called the work-energy theorem. In the next chapter, we generalizethis idea to the broader principle of conservation of energy.
The application of Newton’s laws usually requires solving differential equations that relate the forces acting on an objectto the accelerations they produce. Often, an analytic solution is intractable or impossible, requiring lengthy numericalsolutions or simulations to get approximate results. In such situations, more general relations, like the work-energy theorem(or the conservation of energy), can still provide useful answers to many questions and require a more modest amountof mathematical calculation. In particular, you will see how the work-energy theorem is useful in relating the speedsof a particle, at different points along its trajectory, to the forces acting on it, even when the trajectory is otherwisetoo complicated to deal with. Thus, some aspects of motion can be addressed with fewer equations and without vectordecompositions.


Chapter 7 | Work and Kinetic Energy 327




7.1 | Work
Learning Objectives


By the end of this section, you will be able to:
• Represent the work done by any force
• Evaluate the work done for various forces


In physics, work represents a type of energy. Work is done when a force acts on something that undergoes a displacementfrom one position to another. Forces can vary as a function of position, and displacements can be along various paths
between two points. We first define the increment of work dW done by a force F→ acting through an infinitesimal
displacement d r→ as the dot product of these two vectors:


(7.1)dW = F→ · d r→ = | F→ ||d r→ |cos θ.


Then, we can add up the contributions for infinitesimal displacements, along a path between two positions, to get the totalwork.
Work Done by a Force
The work done by a force is the integral of the force with respect to displacement along the path of the displacement:


(7.2)WAB = ∫
path AB


F


· d r→ .


The vectors involved in the definition of the work done by a force acting on a particle are illustrated in Figure 7.2.


Figure 7.2 Vectors used to define work. The force acting on aparticle and its infinitesimal displacement are shown at onepoint along the path between A and B. The infinitesimal work isthe dot product of these two vectors; the total work is theintegral of the dot product along the path.


We choose to express the dot product in terms of the magnitudes of the vectors and the cosine of the angle betweenthem, because the meaning of the dot product for work can be put into words more directly in terms of magnitudes andangles. We could equally well have expressed the dot product in terms of the various components introduced in Vectors.In two dimensions, these were the x- and y-components in Cartesian coordinates, or the r- and φ -components in polar
coordinates; in three dimensions, it was just x-, y-, and z-components. Which choice is more convenient depends on thesituation. In words, you can express Equation 7.1 for the work done by a force acting over a displacement as a productof one component acting parallel to the other component. From the properties of vectors, it doesn’t matter if you take thecomponent of the force parallel to the displacement or the component of the displacement parallel to the force—you get thesame result either way.
Recall that the magnitude of a force times the cosine of the angle the force makes with a given direction is the component


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of the force in the given direction. The components of a vector can be positive, negative, or zero, depending on whetherthe angle between the vector and the component-direction is between 0° and 90° or 90° and 180° , or is equal to 90° .
As a result, the work done by a force can be positive, negative, or zero, depending on whether the force is generally in thedirection of the displacement, generally opposite to the displacement, or perpendicular to the displacement. The maximumwork is done by a given force when it is along the direction of the displacement ( cos θ = ± 1 ), and zero work is done
when the force is perpendicular to the displacement ( cos θ = 0 ).
The units of work are units of force multiplied by units of length, which in the SI system is newtons times meters, N ·m.
This combination is called a joule, for historical reasons that we will mention later, and is abbreviated as J. In the Englishsystem, still used in the United States, the unit of force is the pound (lb) and the unit of distance is the foot (ft), so the unitof work is the foot-pound (ft · lb).
Work Done by Constant Forces and Contact Forces
The simplest work to evaluate is that done by a force that is constant in magnitude and direction. In this case, we can factorout the force; the remaining integral is just the total displacement, which only depends on the end points A and B, but noton the path between them:


WAB = F


·∫
A


B
d r→ = F



· ⎛⎝ r


B − r


A

⎠ = | F→ || r→ B − r→ A|cos θ (constant force).


We can also see this by writing out Equation 7.2 in Cartesian coordinates and using the fact that the components of theforce are constant:
WAB = ∫


path AB


F


· d r→ = ∫
path AB



⎝Fx dx + Fydy + Fzdz



⎠ = Fx∫


A


B
dx + Fy∫


A


B
dy + Fz∫


A


B
dz


= Fx (xB − xA) + Fy (yB − yA) + Fz (zB − zA) = F


· ( r→ B − r


A).


Figure 7.3(a) shows a person exerting a constant force F→ along the handle of a lawn mower, which makes an angle θ
with the horizontal. The horizontal displacement of the lawn mower, over which the force acts, is d→ . The work done on
the lawn mower isW = F→ · d→ = Fd cos θ , which the figure also illustrates as the horizontal component of the force
times the magnitude of the displacement.


Chapter 7 | Work and Kinetic Energy 329




Figure 7.3 Work done by a constant force. (a) A person pushes a lawnmower with a constant force. The component of the force parallel to thedisplacement is the work done, as shown in the equation in the figure. (b) Aperson holds a briefcase. No work is done because the displacement is zero. (c)The person in (b) walks horizontally while holding the briefcase. No work isdone because cos θ is zero.


Figure 7.3(b) shows a person holding a briefcase. The person must exert an upward force, equal in magnitude to the weightof the briefcase, but this force does no work, because the displacement over which it acts is zero. So why do you eventuallyfeel tired just holding the briefcase, if you’re not doing any work on it? The answer is that muscle fibers in your arm arecontracting and doing work inside your arm, even though the force your muscles exert externally on the briefcase doesn’t doany work on it. (Part of the force you exert could also be tension in the bones and ligaments of your arm, but other musclesin your body would be doing work to maintain the position of your arm.)
In Figure 7.3(c), where the person in (b) is walking horizontally with constant speed, the work done by the person on the
briefcase is still zero, but now because the angle between the force exerted and the displacement is 90° ( F→ perpendicular
to d→ ) and cos 90° = 0 .
Example 7.1


Calculating the Work You Do to Push a Lawn Mower
How much work is done on the lawn mower by the person in Figure 7.3(a) if he exerts a constant force of 75.0


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N at an angle 35° below the horizontal and pushes the mower 25.0 m on level ground?
Strategy
We can solve this problem by substituting the given values into the definition of work done on an object by aconstant force, stated in the equation W = Fd cos θ . The force, angle, and displacement are given, so that only
the work W is unknown.
Solution
The equation for the work is


W = Fd cos θ.


Substituting the known values gives
W = (75.0 N)(25.0 m)cos(35.0°) = 1.54 × 103 J.


Significance
Even though one and a half kilojoules may seem like a lot of work, we will see in Potential Energy andConservation of Energy that it’s only about as much work as you could do by burning one sixth of a gram offat.


When you mow the grass, other forces act on the lawn mower besides the force you exert—namely, the contact force of theground and the gravitational force of Earth. Let’s consider the work done by these forces in general. For an object moving on
a surface, the displacement d r→ is tangent to the surface. The part of the contact force on the object that is perpendicular
to the surface is the normal force N→ . Since the cosine of the angle between the normal and the tangent to a surface is
zero, we have


dWN = N


· d r→ = 0


.


The normal force never does work under these circumstances. (Note that if the displacement d r→ did have a relative
component perpendicular to the surface, the object would either leave the surface or break through it, and there would nolonger be any normal contact force. However, if the object is more than a particle, and has an internal structure, the normalcontact force can do work on it, for example, by displacing it or deforming its shape. This will be mentioned in the nextchapter.)
The part of the contact force on the object that is parallel to the surface is friction, f→ . For this object sliding along the
surface, kinetic friction f→ k is opposite to d r→ , relative to the surface, so the work done by kinetic friction is negative.
If the magnitude of f→ k is constant (as it would be if all the other forces on the object were constant), then the work done
by friction is


(7.3)
Wfr = ∫


A


B
f


k · d r
→ = − fk∫


A


B


|dr| = − fk |lAB|,


where |lAB| is the path length on the surface. (Note that, especially if the work done by a force is negative, people may refer
to the work done against this force, where dWagainst = −dWby . The work done against a force may also be viewed as
the work required to overcome this force, as in “How much work is required to overcome…?”) The force of static friction,however, can do positive or negative work. When you walk, the force of static friction exerted by the ground on your backfoot accelerates you for part of each step. If you’re slowing down, the force of the ground on your front foot deceleratesyou. If you’re driving your car at the speed limit on a straight, level stretch of highway, the negative work done by kineticfriction of air resistance is balanced by the positive work done by the static friction of the road on the drive wheels. Youcan pull the rug out from under an object in such a way that it slides backward relative to the rug, but forward relative tothe floor. In this case, kinetic friction exerted by the rug on the object could be in the same direction as the displacement


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7.1


of the object, relative to the floor, and do positive work. The bottom line is that you need to analyze each particular case todetermine the work done by the forces, whether positive, negative or zero.
Example 7.2


Moving a Couch
You decide to move your couch to a new position on your horizontal living room floor. The normal force on thecouch is 1 kN and the coefficient of friction is 0.6. (a) You first push the couch 3 m parallel to a wall and then 1 mperpendicular to the wall (A to B in Figure 7.4). How much work is done by the frictional force? (b) You don’tlike the new position, so you move the couch straight back to its original position (B to A in Figure 7.4). Whatwas the total work done against friction moving the couch away from its original position and back again?


Figure 7.4 Top view of paths for moving a couch.


Strategy
The magnitude of the force of kinetic friction on the couch is constant, equal to the coefficient of friction times thenormal force, fK = µK N . Therefore, the work done by it is Wfr = − fK d , where d is the path length traversed.
The segments of the paths are the sides of a right triangle, so the path lengths are easily calculated. In part (b),you can use the fact that the work done against a force is the negative of the work done by the force.
Solutiona. The work done by friction is


W = − (0.6)(1 kN)(3 m + 1m) = − 2.4 kJ.


b. The length of the path along the hypotenuse is 10 m , so the total work done against friction is
W = (0.6)(1 kN)(3 m + 1m + 10 m) = 4.3 kJ.


Significance
The total path over which the work of friction was evaluated began and ended at the same point (it was a closedpath), so that the total displacement of the couch was zero. However, the total work was not zero. The reason isthat forces like friction are classified as nonconservative forces, or dissipative forces, as we discuss in the nextchapter.


Check Your Understanding Can kinetic friction ever be a constant force for all paths?


The other force on the lawn mower mentioned above was Earth’s gravitational force, or the weight of the mower. Nearthe surface of Earth, the gravitational force on an object of mass m has a constant magnitude, mg, and constant direction,vertically down. Therefore, the work done by gravity on an object is the dot product of its weight and its displacement. Inmany cases, it is convenient to express the dot product for gravitational work in terms of the x-, y-, and z-components of thevectors. A typical coordinate system has the x-axis horizontal and the y-axis vertically up. Then the gravitational force is
−mg j


^
, so the work done by gravity, over any path from A to B, is


(7.4)
Wgrav, AB = −mg j


^
· ( r→ B − r



A) = −mg(yB − yA).


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7.2


The work done by a constant force of gravity on an object depends only on the object’s weight and the difference in heightthrough which the object is displaced. Gravity does negative work on an object that moves upward ( yB > yA ), or, in other
words, you must do positive work against gravity to lift an object upward. Alternately, gravity does positive work on anobject that moves downward ( yB < yA ), or you do negative work against gravity to “lift” an object downward, controlling
its descent so it doesn’t drop to the ground. (“Lift” is used as opposed to “drop”.)
Example 7.3


Shelving a Book
You lift an oversized library book, weighing 20 N, 1 m vertically down from a shelf, and carry it 3 m horizontallyto a table (Figure 7.5). How much work does gravity do on the book? (b) When you’re finished, you move thebook in a straight line back to its original place on the shelf. What was the total work done against gravity, movingthe book away from its original position on the shelf and back again?


Figure 7.5 Side view of the paths for moving a book to andfrom a shelf.


Strategy
We have just seen that the work done by a constant force of gravity depends only on the weight of the objectmoved and the difference in height for the path taken, WAB = −mg(yB − yA) . We can evaluate the difference in
height to answer (a) and (b).
Solutiona. Since the book starts on the shelf and is lifted down yB − yA = −1m , we have


W = −(20 N)( − 1 m) = 20 J.


b. There is zero difference in height for any path that begins and ends at the same place on the shelf, so
W = 0.


Significance
Gravity does positive work (20 J) when the book moves down from the shelf. The gravitational force betweentwo objects is an attractive force, which does positive work when the objects get closer together. Gravity doeszero work (0 J) when the book moves horizontally from the shelf to the table and negative work (−20 J) when thebook moves from the table back to the shelf. The total work done by gravity is zero [20 J + 0 J + (−20 J) = 0].
Unlike friction or other dissipative forces, described in Example 7.2, the total work done against gravity, overany closed path, is zero. Positive work is done against gravity on the upward parts of a closed path, but an equalamount of negative work is done against gravity on the downward parts. In other words, work done againstgravity, lifting an object up, is “given back” when the object comes back down. Forces like gravity (those that dozero work over any closed path) are classified as conservative forces and play an important role in physics.


Check Your Understanding Can Earth’s gravity ever be a constant force for all paths?


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Work Done by Forces that Vary
In general, forces may vary in magnitude and direction at points in space, and paths between two points may be curved. Theinfinitesimal work done by a variable force can be expressed in terms of the components of the force and the displacementalong the path,


dW = Fx dx + Fydy + Fzdz.


Here, the components of the force are functions of position along the path, and the displacements depend on the equationsof the path. (Although we chose to illustrate dW in Cartesian coordinates, other coordinates are better suited to somesituations.) Equation 7.2 defines the total work as a line integral, or the limit of a sum of infinitesimal amounts of work.The physical concept of work is straightforward: you calculate the work for tiny displacements and add them up. Sometimesthe mathematics can seem complicated, but the following example demonstrates how cleanly they can operate.
Example 7.4


Work Done by a Variable Force over a Curved Path
An object moves along a parabolic path y = (0.5 m−1)x2 from the origin A = (0, 0) to the point
B = (2 m, 2 m) under the action of a force F→ = (5 N/m)y i^ + (10 N/m)x j^ (Figure 7.6). Calculate the
work done.


Figure 7.6 The parabolic path of a particle acted on by agiven force.


Strategy
The components of the force are given functions of x and y. We can use the equation of the path to express y anddy in terms of x and dx; namely,


y = (0.5 m−1)x2 and dy = 2(0.5 m−1)xdx.


Then, the integral for the work is just a definite integral of a function of x.
Solution
The infinitesimal element of work is


dW = Fx dx + Fydy = (5 N/m)ydx + (10 N/m)xdy


= (5 N/m)(0.5 m−1)x2dx + (10 N/m)2(0.5 m−1)x2dx = (12.5 N/m2)x2dx.


The integral of x2 is x3 /3, so
W = ∫


0


2 m
(12.5 N/m2)x2dx = (12.5 N/m2)x


3


3 |0
2 m


= (12.5 N/m2)⎛⎝
8
3

⎠ = 33.3 J.


Significance
This integral was not hard to do. You can follow the same steps, as in this example, to calculate line integralsrepresenting work for more complicated forces and paths. In this example, everything was given in terms of x-and y-components, which are easiest to use in evaluating the work in this case. In other situations, magnitudes


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7.3
and angles might be easier.


Check Your Understanding Find the work done by the same force in Example 7.4 over a cubic path,
y = (0.25 m−2)x3 , between the same points A = (0, 0) and B = (2 m, 2 m).


You saw in Example 7.4 that to evaluate a line integral, you could reduce it to an integral over a single variable orparameter. Usually, there are several ways to do this, which may be more or less convenient, depending on the particularcase. In Example 7.4, we reduced the line integral to an integral over x, but we could equally well have chosen to reduceeverything to a function of y. We didn’t do that because the functions in y involve the square root and fractional exponents,which may be less familiar, but for illustrative purposes, we do this now. Solving for x and dx, in terms of y, along theparabolic path, we get
x = y/(0.5 m−1) = (2 m)y and dx = (2 m) × 1


2
dy/ y = dy/ (2 m−1)y.


The components of the force, in terms of y, are
Fx = (5 N/m)y and Fy = (10 N/m)x = (10 N/m) (2 m)y,


so the infinitesimal work element becomes
dW = Fx dx + Fydy =


(5 N/m)y dy


(2 m−1)y
+ (10 N/m) (2 m)y dy


= (5 N ·m−1/2)


1
2
+ 2 2

⎠ y dy = (17.7 N ·m


−1/2)y1/2dy.


The integral of y1/2 is 2
3
y3/2 , so the work done from A to B is


W = ∫
0


2 m
(17.7 N ·m−1/2)y1/2dy = (17.7 N ·m−1/2)2


3
(2 m)3/2 = 33.3 J.


As expected, this is exactly the same result as before.
One very important and widely applicable variable force is the force exerted by a perfectly elastic spring, which satisfies
Hooke’s law F→ = −kΔ x→ , where k is the spring constant, and Δ x→ = x→ − x→ eq is the displacement from the
spring’s unstretched (equilibrium) position (Newton’s Laws of Motion). Note that the unstretched position is only thesame as the equilibrium position if no other forces are acting (or, if they are, they cancel one another). Forces betweenmolecules, or in any system undergoing small displacements from a stable equilibrium, behave approximately like a springforce.
To calculate the work done by a spring force, we can choose the x-axis along the length of the spring, in the direction ofincreasing length, as in Figure 7.7, with the origin at the equilibrium position xeq = 0. (Then positive x corresponds
to a stretch and negative x to a compression.) With this choice of coordinates, the spring force has only an x-component,
Fx = −kx , and the work done when x changes from xA to xB is


(7.5)
Wspring, AB = ∫


A


B
Fx dx = − k∫


A


B
xdx = −kx


2


2 |A
B


= − 1
2
k⎛⎝xB


2 − xA
2 ⎞
⎠.


Chapter 7 | Work and Kinetic Energy 335




Figure 7.7 (a) The spring exerts no force at its equilibriumposition. The spring exerts a force in the opposite direction to(b) an extension or stretch, and (c) a compression.


Notice that WAB depends only on the starting and ending points, A and B, and is independent of the actual path between
them, as long as it starts at A and ends at B. That is, the actual path could involve going back and forth before ending.
Another interesting thing to notice about Equation 7.5 is that, for this one-dimensional case, you can readily see thecorrespondence between the work done by a force and the area under the curve of the force versus its displacement. Recallthat, in general, a one-dimensional integral is the limit of the sum of infinitesimals, f (x)dx , representing the area of strips,
as shown in Figure 7.8. In Equation 7.5, since F = −kx is a straight line with slope −k , when plotted versus x, the
“area” under the line is just an algebraic combination of triangular “areas,” where “areas” above the x-axis are positive andthose below are negative, as shown in Figure 7.9. The magnitude of one of these “areas” is just one-half the triangle’sbase, along the x-axis, times the triangle’s height, along the force axis. (There are quotation marks around “area” becausethis base-height product has the units of work, rather than square meters.)


Figure 7.8 A curve of f(x) versus x showing the area of aninfinitesimal strip, f(x)dx, and the sum of such areas, which isthe integral of f(x) from x1 to x2 .


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7.4


Figure 7.9 Curve of the spring force f (x) = −kx versus x,
showing areas under the line, between xA and xB , for both
positive and negative values of xA . When xA is negative, the
total area under the curve for the integral in Equation 7.5 is thesum of positive and negative triangular areas. When xA is
positive, the total area under the curve is the difference betweentwo negative triangles.


Example 7.5
Work Done by a Spring Force
A perfectly elastic spring requires 0.54 J of work to stretch 6 cm from its equilibrium position, as in Figure7.7(b). (a) What is its spring constant k? (b) How much work is required to stretch it an additional 6 cm?
Strategy
Work “required” means work done against the spring force, which is the negative of the work in Equation 7.5,that is


W = 1
2
k(xB


2 − xA
2 ).


For part (a), xA = 0 and xB = 6cm ; for part (b), xB = 6cm and xB = 12cm . In part (a), the work is given
and you can solve for the spring constant; in part (b), you can use the value of k, from part (a), to solve for thework.
Solution


a. W = 0.54 J = 1
2
k[(6 cm)2 − 0] , so k = 3 N/cm.


b. W = 1
2
(3 N/cm)[(12 cm)2 − (6 cm)2] = 1.62 J.


Significance
Since the work done by a spring force is independent of the path, you only needed to calculate the difference in
the quantity ½kx2 at the end points. Notice that the work required to stretch the spring from 0 to 12 cm is four
times that required to stretch it from 0 to 6 cm, because that work depends on the square of the amount of stretch
from equilibrium, ½kx2 . In this circumstance, the work to stretch the spring from 0 to 12 cm is also equal to
the work for a composite path from 0 to 6 cm followed by an additional stretch from 6 cm to 12 cm. Therefore,
4W(0 cm to 6 cm) = W(0 cm to 6 cm) +W(6 cm to 12 cm) , or W(6 cm to 12 cm) = 3W(0 cm to 6 cm) , as
we found above.


Check Your Understanding The spring in Example 7.5 is compressed 6 cm from its equilibriumlength. (a) Does the spring force do positive or negative work and (b) what is the magnitude?


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7.2 | Kinetic Energy
Learning Objectives


By the end of this section, you will be able to:
• Calculate the kinetic energy of a particle given its mass and its velocity or momentum
• Evaluate the kinetic energy of a body, relative to different frames of reference


It’s plausible to suppose that the greater the velocity of a body, the greater effect it could have on other bodies. This does notdepend on the direction of the velocity, only its magnitude. At the end of the seventeenth century, a quantity was introducedinto mechanics to explain collisions between two perfectly elastic bodies, in which one body makes a head-on collision withan identical body at rest. The first body stops, and the second body moves off with the initial velocity of the first body. (Ifyou have ever played billiards or croquet, or seen a model of Newton’s Cradle, you have observed this type of collision.)The idea behind this quantity was related to the forces acting on a body and was referred to as “the energy of motion.” Lateron, during the eighteenth century, the name kinetic energy was given to energy of motion.
With this history in mind, we can now state the classical definition of kinetic energy. Note that when we say “classical,”we mean non-relativistic, that is, at speeds much less that the speed of light. At speeds comparable to the speed of light,the special theory of relativity requires a different expression for the kinetic energy of a particle, as discussed in Relativity(http://cnx.org/content/m58555/latest/) .
Since objects (or systems) of interest vary in complexity, we first define the kinetic energy of a particle with mass m.


Kinetic Energy
The kinetic energy of a particle is one-half the product of the particle’s mass m and the square of its speed v:


(7.6)K = 1
2
mv2.


We then extend this definition to any system of particles by adding up the kinetic energies of all the constituent particles:
(7.7)K = ∑ 1


2
mv2.


Note that just as we can express Newton’s second law in terms of either the rate of change of momentum or mass timesthe rate of change of velocity, so the kinetic energy of a particle can be expressed in terms of its mass and momentum
( p→ = m v→ ), instead of its mass and velocity. Since v = p/m , we see that


K = 1
2
m⎛⎝


p
m



2
=


p2


2m


also expresses the kinetic energy of a single particle. Sometimes, this expression is more convenient to use than Equation7.6.
The units of kinetic energy are mass times the square of speed, or kg ·m2 /s2 . But the units of force are mass times
acceleration, kg ·m/s2 , so the units of kinetic energy are also the units of force times distance, which are the units of work,
or joules. You will see in the next section that work and kinetic energy have the same units, because they are different formsof the same, more general, physical property.
Example 7.6


Kinetic Energy of an Object
(a) What is the kinetic energy of an 80-kg athlete, running at 10 m/s? (b) The Chicxulub crater in Yucatan, one ofthe largest existing impact craters on Earth, is thought to have been created by an asteroid, traveling at
22 km/s and releasing 4.2 × 1023 J of kinetic energy upon impact. What was its mass? (c) In nuclear reactors,


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7.5


thermal neutrons, traveling at about 2.2 km/s, play an important role. What is the kinetic energy of such a particle?
Strategy
To answer these questions, you can use the definition of kinetic energy in Equation 7.6. You also have to lookup the mass of a neutron.
Solution
Don’t forget to convert km into m to do these calculations, although, to save space, we omitted showing theseconversions.


a. K = 1
2
(80 kg)(10 m/s)2 = 4.0 kJ.


b. m = 2K/v2 = 2(4.2 × 1023 J)/(22 km/s)2 = 1.7 × 1015 kg.
c. K = 1


2
(1.68 × 10−27 kg)(2.2 km/s)2 = 4.1 × 10−21 J.


Significance
In this example, we used the way mass and speed are related to kinetic energy, and we encountered a verywide range of values for the kinetic energies. Different units are commonly used for such very large andvery small values. The energy of the impactor in part (b) can be compared to the explosive yield of TNT
and nuclear explosions, 1 megaton = 4.18 × 1015 J. The Chicxulub asteroid’s kinetic energy was about a
hundred million megatons. At the other extreme, the energy of subatomic particle is expressed in electron-volts,
1 eV = 1.6 × 10−19 J. The thermal neutron in part (c) has a kinetic energy of about one fortieth of an electron-
volt.


Check Your Understanding (a) A car and a truck are each moving with the same kinetic energy. Assumethat the truck has more mass than the car. Which has the greater speed? (b) A car and a truck are each movingwith the same speed. Which has the greater kinetic energy?
Because velocity is a relative quantity, you can see that the value of kinetic energy must depend on your frame of reference.You can generally choose a frame of reference that is suited to the purpose of your analysis and that simplifies yourcalculations. One such frame of reference is the one in which the observations of the system are made (likely an externalframe). Another choice is a frame that is attached to, or moves with, the system (likely an internal frame). The equations forrelative motion, discussed in Motion in Two and Three Dimensions, provide a link to calculating the kinetic energy ofan object with respect to different frames of reference.
Example 7.7


Kinetic Energy Relative to Different Frames
A 75.0-kg person walks down the central aisle of a subway car at a speed of 1.50 m/s relative to the car, whereasthe train is moving at 15.0 m/s relative to the tracks. (a) What is the person’s kinetic energy relative to the car?(b) What is the person’s kinetic energy relative to the tracks? (c) What is the person’s kinetic energy relative to aframe moving with the person?
Strategy
Since speeds are given, we can use 1


2
mv2 to calculate the person’s kinetic energy. However, in part (a), the


person’s speed is relative to the subway car (as given); in part (b), it is relative to the tracks; and in part (c), it iszero. If we denote the car frame by C, the track frame by T, and the person by P, the relative velocities in part (b)
are related by v→ PT = v→ PC + v→ CT. We can assume that the central aisle and the tracks lie along the same
line, but the direction the person is walking relative to the car isn’t specified, so we will give an answer for eachpossibility, vPT = vCT ± vPC , as shown in Figure 7.10.


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7.6


Figure 7.10 The possible motions of a person walking in a train are (a) toward the front of the car and (b) towardthe back of the car.


Solution
a. K = 1


2
(75.0 kg)(1.50 m/s)2 = 84.4 J.


b. vPT = (15.0 ± 1.50) m/s. Therefore, the two possible values for kinetic energy relative to the car are
K = 1


2
(75.0 kg)(13.5 m/s)2 = 6.83 kJ


and
K = 1


2
(75.0 kg)(16.5 m/s)2 = 10.2 kJ.


c. In a frame where vP = 0, K = 0 as well.
Significance
You can see that the kinetic energy of an object can have very different values, depending on the frame ofreference. However, the kinetic energy of an object can never be negative, since it is the product of the mass andthe square of the speed, both of which are always positive or zero.


Check Your Understanding You are rowing a boat parallel to the banks of a river. Your kinetic energyrelative to the banks is less than your kinetic energy relative to the water. Are you rowing with or against thecurrent?
The kinetic energy of a particle is a single quantity, but the kinetic energy of a system of particles can sometimes be dividedinto various types, depending on the system and its motion. For example, if all the particles in a system have the samevelocity, the system is undergoing translational motion and has translational kinetic energy. If an object is rotating, it couldhave rotational kinetic energy, or if it’s vibrating, it could have vibrational kinetic energy. The kinetic energy of a system,relative to an internal frame of reference, may be called internal kinetic energy. The kinetic energy associated with randommolecular motion may be called thermal energy. These names will be used in later chapters of the book, when appropriate.Regardless of the name, every kind of kinetic energy is the same physical quantity, representing energy associated withmotion.
Example 7.8


Special Names for Kinetic Energy
(a) A player lobs a mid-court pass with a 624-g basketball, which covers 15 m in 2 s. What is the basketball’shorizontal translational kinetic energy while in flight? (b) An average molecule of air, in the basketball in part
(a), has a mass of 29 u, and an average speed of 500 m/s, relative to the basketball. There are about 3 × 1023
molecules inside it, moving in random directions, when the ball is properly inflated. What is the averagetranslational kinetic energy of the random motion of all the molecules inside, relative to the basketball? (c) Howfast would the basketball have to travel relative to the court, as in part (a), so as to have a kinetic energy equal tothe amount in part (b)?


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Strategy
In part (a), first find the horizontal speed of the basketball and then use the definition of kinetic energy in terms of
mass and speed, K = 1


2
mv2 . Then in part (b), convert unified units to kilograms and then use K = 1


2
mv2 to get


the average translational kinetic energy of one molecule, relative to the basketball. Then multiply by the numberof molecules to get the total result. Finally, in part (c), we can substitute the amount of kinetic energy in part (b),
and the mass of the basketball in part (a), into the definition K = 1


2
mv2 , and solve for v.


Solutiona. The horizontal speed is (15 m)/(2 s), so the horizontal kinetic energy of the basketball is
1
2
(0.624 kg)(7.5 m/s)2 = 17.6 J.


b. The average translational kinetic energy of a molecule is
1
2
(29 u)(1.66 × 10−27 kg/u)(500 m/s)2 = 6.02 × 10−21 J,


and the total kinetic energy of all the molecules is
(3 × 1023)(6.02 × 10−21 J) = 1.80 kJ.


c. v = 2(1.8 kJ)/(0.624 kg) = 76.0 m/s.
Significance
In part (a), this kind of kinetic energy can be called the horizontal kinetic energy of an object (the basketball),relative to its surroundings (the court). If the basketball were spinning, all parts of it would have not just theaverage speed, but it would also have rotational kinetic energy. Part (b) reminds us that this kind of kinetic energycan be called internal or thermal kinetic energy. Notice that this energy is about a hundred times the energy in part(a). How to make use of thermal energy will be the subject of the chapters on thermodynamics. In part (c), sincethe energy in part (b) is about 100 times that in part (a), the speed should be about 10 times as big, which it is (76compared to 7.5 m/s).


7.3 | Work-Energy Theorem
Learning Objectives


By the end of this section, you will be able to:
• Apply the work-energy theorem to find information about the motion of a particle, given theforces acting on it
• Use the work-energy theorem to find information about the forces acting on a particle, giveninformation about its motion


We have discussed how to find the work done on a particle by the forces that act on it, but how is that work manifested inthe motion of the particle? According to Newton’s second law of motion, the sum of all the forces acting on a particle, orthe net force, determines the rate of change in the momentum of the particle, or its motion. Therefore, we should considerthe work done by all the forces acting on a particle, or the net work, to see what effect it has on the particle’s motion.
Let’s start by looking at the net work done on a particle as it moves over an infinitesimal displacement, which is
the dot product of the net force and the displacement: dWnet = F→ net · d r→ . Newton’s second law tells us that
F


net = m(d v
→ /dt), , so dWnet = m(d v→ /dt) · d r→ . For the mathematical functions describing the motion of a


physical particle, we can rearrange the differentials dt, etc., as algebraic quantities in this expression, that is,
dWnet = m




d v→
dt

⎠ · d r


→ = md v→ ·


d r→


dt

⎠ = m v


→ · d v→ ,


Chapter 7 | Work and Kinetic Energy 341




where we substituted the velocity for the time derivative of the displacement and used the commutative property of thedot product [Equation 2.30]. Since derivatives and integrals of scalars are probably more familiar to you at this point,we express the dot product in terms of Cartesian coordinates before we integrate between any two points A and B on theparticle’s trajectory. This gives us the net work done on the particle:
(7.8)


Wnet, AB = ∫
A


B
(mvx dvx + mvydvy + mvzdvz)


= 1
2
m|vx2 + vy2 + vz2|A


B
= |12mv2|A


B
= KB − KA.


In the middle step, we used the fact that the square of the velocity is the sum of the squares of its Cartesian components,and in the last step, we used the definition of the particle’s kinetic energy. This important result is called the work-energytheorem (Figure 7.11).
Work-Energy Theorem
The net work done on a particle equals the change in the particle’s kinetic energy:


(7.9)Wnet = KB − KA.


Figure 7.11 Horse pulls are common events at state fairs. The work done bythe horses pulling on the load results in a change in kinetic energy of the load,ultimately going faster. (credit: “Jassen”/ Flickr)


According to this theorem, when an object slows down, its final kinetic energy is less than its initial kinetic energy, thechange in its kinetic energy is negative, and so is the net work done on it. If an object speeds up, the net work done on it ispositive. When calculating the net work, you must include all the forces that act on an object. If you leave out any forcesthat act on an object, or if you include any forces that don’t act on it, you will get a wrong result.
The importance of the work-energy theorem, and the further generalizations to which it leads, is that it makes some typesof calculations much simpler to accomplish than they would be by trying to solve Newton’s second law. For example, inNewton’s Laws of Motion, we found the speed of an object sliding down a frictionless plane by solving Newton’s secondlaw for the acceleration and using kinematic equations for constant acceleration, obtaining


vf
2 = vi


2 + 2g(sf − si)sin θ,


where s is the displacement down the plane.
We can also get this result from the work-energy theorem. Since only two forces are acting on the object—gravity and thenormal force—and the normal force doesn’t do any work, the net work is just the work done by gravity. This only dependson the object’s weight and the difference in height, so


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Wnet = Wgrav = −mg(yf − yi),


where y is positive up. The work-energy theorem says that this equals the change in kinetic energy:
−mg(yf − yi) =


1
2
m(vf


2 − vi
2).


Using a right triangle, we can see that (yf − yi) = (sf − si)sin θ, so the result for the final speed is the same.
What is gained by using the work-energy theorem? The answer is that for a frictionless plane surface, not much. However,Newton’s second law is easy to solve only for this particular case, whereas the work-energy theorem gives the final speedfor any shaped frictionless surface. For an arbitrary curved surface, the normal force is not constant, and Newton’s secondlaw may be difficult or impossible to solve analytically. Constant or not, for motion along a surface, the normal force neverdoes any work, because it’s perpendicular to the displacement. A calculation using the work-energy theorem avoids thisdifficulty and applies to more general situations.


Problem-Solving Strategy: Work-Energy Theorem
1. Draw a free-body diagram for each force on the object.
2. Determine whether or not each force does work over the displacement in the diagram. Be sure to keep anypositive or negative signs in the work done.
3. Add up the total amount of work done by each force.
4. Set this total work equal to the change in kinetic energy and solve for any unknown parameter.
5. Check your answers. If the object is traveling at a constant speed or zero acceleration, the total work doneshould be zero and match the change in kinetic energy. If the total work is positive, the object must have spedup or increased kinetic energy. If the total work is negative, the object must have slowed down or decreasedkinetic energy.


Example 7.9
Loop-the-Loop
The frictionless track for a toy car includes a loop-the-loop of radius R. How high, measured from the bottom ofthe loop, must the car be placed to start from rest on the approaching section of track and go all the way aroundthe loop?


Figure 7.12 A frictionless track for a toy car has a loop-the-loop in it. How high must the car start so that it can go aroundthe loop without falling off?


Strategy
The free-body diagram at the final position of the object is drawn in Figure 7.12. The gravitational work is theonly work done over the displacement that is not zero. Since the weight points in the same direction as the netvertical displacement, the total work done by the gravitational force is positive. From the work-energy theorem,


Chapter 7 | Work and Kinetic Energy 343




7.7


the starting height determines the speed of the car at the top of the loop,
mg(y2 − y1) =


1
2
mv2


2,


where the notation is shown in the accompanying figure. At the top of the loop, the normal force and gravity areboth down and the acceleration is centripetal, so
atop =


F
m =


N + mg
m =


v2
2


R
.


The condition for maintaining contact with the track is that there must be some normal force, however slight; that
is, N > 0 . Substituting for v22 and N, we can find the condition for y1 .
Solution
Implement the steps in the strategy to arrive at the desired result:


N =
−mg + mv2


2


R
=


−mg + 2mg(y1 − 2R)
R


> 0 or y1 > 5R2 .


Significance
On the surface of the loop, the normal component of gravity and the normal contact force must provide thecentripetal acceleration of the car going around the loop. The tangential component of gravity slows down orspeeds up the car. A child would find out how high to start the car by trial and error, but now that you know thework-energy theorem, you can predict the minimum height (as well as other more useful results) from physicalprinciples. By using the work-energy theorem, you did not have to solve a differential equation to determine theheight.


Check Your Understanding Suppose the radius of the loop-the-loop in Example 7.9 is 15 cm and thetoy car starts from rest at a height of 45 cm above the bottom. What is its speed at the top of the loop?


Visit Carleton College’s site to see a video (https://openstaxcollege.org/l/21carcollvidrol) of a loopingrollercoaster.


In situations where the motion of an object is known, but the values of one or more of the forces acting on it are not known,you may be able to use the work-energy theorem to get some information about the forces. Work depends on the force andthe distance over which it acts, so the information is provided via their product.
Example 7.10


Determining a Stopping Force
A bullet from a 0.22LR-caliber cartridge has a mass of 40 grains (2.60 g) and a muzzle velocity of 1100 ft./s (335m/s). It can penetrate eight 1-inch pine boards, each with thickness 0.75 inches. What is the average stoppingforce exerted by the wood, as shown in Figure 7.13?


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Figure 7.13 The boards exert a force to stop the bullet. As a result, the boards do work and thebullet loses kinetic energy.


Strategy
We can assume that under the general conditions stated, the bullet loses all its kinetic energy penetrating theboards, so the work-energy theorem says its initial kinetic energy is equal to the average stopping force timesthe distance penetrated. The change in the bullet’s kinetic energy and the net work done stopping it are bothnegative, so when you write out the work-energy theorem, with the net work equal to the average force times thestopping distance, that’s what you get. The total thickness of eight 1-inch pine boards that the bullet penetrates is
8 × 3


4
in. = 6 in. = 15.2 cm.


Solution
Applying the work-energy theorem, we get


Wnet = −FaveΔsstop = −Kinitial,


so
Fave =


1
2
mv2


Δsstop
=


1
2
(2.6 × 10−3kg)(335 m/s)2


0.152 m
= 960 N.


Significance
We could have used Newton’s second law and kinematics in this example, but the work-energy theorem alsosupplies an answer to less simple situations. The penetration of a bullet, fired vertically upward into a blockof wood, is discussed in one section of Asif Shakur’s recent article [“Bullet-Block Science Video Puzzle.” ThePhysics Teacher (January 2015) 53(1): 15-16]. If the bullet is fired dead center into the block, it loses all its kineticenergy and penetrates slightly farther than if fired off-center. The reason is that if the bullet hits off-center, it hasa little kinetic energy after it stops penetrating, because the block rotates. The work-energy theorem implies that asmaller change in kinetic energy results in a smaller penetration. You will understand more of the physics in thisinteresting article after you finish reading Angular Momentum.
Learn more about work and energy in this PhET simulation (https://openstaxcollege.org/l/21PhETSimRamp) called “the ramp.” Try changing the force pushing the box and the frictional force along theincline. The work and energy plots can be examined to note the total work done and change in kinetic energy ofthe box.


7.4 | Power
Learning Objectives


By the end of this section, you will be able to:
• Relate the work done during a time interval to the power delivered
• Find the power expended by a force acting on a moving body


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The concept of work involves force and displacement; the work-energy theorem relates the net work done on a body to thedifference in its kinetic energy, calculated between two points on its trajectory. None of these quantities or relations involvestime explicitly, yet we know that the time available to accomplish a particular amount of work is frequently just as importantto us as the amount itself. In the chapter-opening figure, several sprinters may have achieved the same velocity at the finish,and therefore did the same amount of work, but the winner of the race did it in the least amount of time.
We express the relation between work done and the time interval involved in doing it, by introducing the concept of power.Since work can vary as a function of time, we first define average power as the work done during a time interval, dividedby the interval,


(7.10)Pave = ΔWΔt .
Then, we can define the instantaneous power (frequently referred to as just plain power).


Power
Power is defined as the rate of doing work, or the limit of the average power for time intervals approaching zero,


(7.11)P = dW
dt


.


If the power is constant over a time interval, the average power for that interval equals the instantaneous power, and thework done by the agent supplying the power is W = PΔt . If the power during an interval varies with time, then the work
done is the time integral of the power,


W = ∫ Pdt.


The work-energy theorem relates how work can be transformed into kinetic energy. Since there are other forms of energyas well, as we discuss in the next chapter, we can also define power as the rate of transfer of energy. Work and energyare measured in units of joules, so power is measured in units of joules per second, which has been given the SI namewatts, abbreviation W: 1 J/s = 1W . Another common unit for expressing the power capability of everyday devices is
horsepower: 1 hp = 746W .
Example 7.11


Pull-Up Power
An 80-kg army trainee does 10 pull-ups in 10 s (Figure 7.14). How much average power do the trainee’s musclessupply moving his body? (Hint:Make reasonable estimates for any quantities needed.)


Figure 7.14 What is the power expended in doing ten pull-upsin ten seconds?


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7.8


Strategy
The work done against gravity, going up or down a distance Δy , is mgΔy. (If you lift and lower yourself at
constant speed, the force you exert cancels gravity over the whole pull-up cycle.) Thus, the work done by thetrainee’s muscles (moving, but not accelerating, his body) for a complete repetition (up and down) is 2mgΔy.
Let’s assume that Δy = 2ft ≈ 60 cm. Also, assume that the arms comprise 10% of the body mass and are not
included in the moving mass. With these assumptions, we can calculate the work done for 10 pull-ups and divideby 10 s to get the average power.
Solution
The result we get, applying our assumptions, is


Pave =
10 × 2(0.9 × 80 kg)(9.8 m/s2)(0.6 m)


10 s
= 850W.


Significance
This is typical for power expenditure in strenuous exercise; in everyday units, it’s somewhat more than onehorsepower (1 hp = 746W).


Check Your Understanding Estimate the power expended by a weightlifter raising a 150-kg barbell 2 min 3 s.


The power involved in moving a body can also be expressed in terms of the forces acting on it. If a force F→ acts on a
body that is displaced d r→ in a time dt, the power expended by the force is


(7.12)
P = dW


dt
= F



· d r→


dt
= F



·


d r→


dt

⎠ = F



· v→ ,


where v→ is the velocity of the body. The fact that the limits implied by the derivatives exist, for the motion of a real body,
justifies the rearrangement of the infinitesimals.
Example 7.12


Automotive Power Driving Uphill
How much power must an automobile engine expend to move a 1200-kg car up a 15% grade at 90 km/h (Figure7.15)? Assume that 25% of this power is dissipated overcoming air resistance and friction.


Figure 7.15 We want to calculate the power needed to move a car up a hill atconstant speed.


Chapter 7 | Work and Kinetic Energy 347




Strategy
At constant velocity, there is no change in kinetic energy, so the net work done to move the car is zero. Thereforethe power supplied by the engine to move the car equals the power expended against gravity and air resistance.
By assumption, 75% of the power is supplied against gravity, which equals m g→ · v→ = mgv sin θ, where θ
is the angle of the incline. A 15% grade means tan θ = 0.15. This reasoning allows us to solve for the power
required.
Solution
Carrying out the suggested steps, we find


0.75 P = mgv sin(tan−1 0.15),


or
P = (1200 × 9.8 N)(90 m/3.6 s)sin(8.53°)


0.75
= 58 kW,


or about 78 hp. (You should supply the steps used to convert units.)
Significance
This is a reasonable amount of power for the engine of a small to mid-size car to supply (1 hp = 0.746 kW).
Note that this is only the power expended to move the car. Much of the engine’s power goes elsewhere, forexample, into waste heat. That’s why cars need radiators. Any remaining power could be used for acceleration, orto operate the car’s accessories.


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average power
kinetic energy
net work
power
work
work done by a force
work-energy theorem


CHAPTER 7 REVIEW
KEY TERMS


work done in a time interval divided by the time interval
energy of motion, one-half an object’s mass times the square of its speed


work done by all the forces acting on an object
(or instantaneous power) rate of doing work
done when a force acts on something that undergoes a displacement from one position to another


integral, from the initial position to the final position, of the dot product of the force and theinfinitesimal displacement along the path over which the force acts
net work done on a particle is equal to the change in its kinetic energy


KEY EQUATIONS
Work done by a force over an infinitesimal displacement dW = F→ · d r→ = | F→ ||d r→ |cos θ


Work done by a force acting along a path from A to B WAB = ∫
pathAB


F


· d r→


Work done by a constant force of kinetic friction Wfr = − fk |lAB|
Work done going from A to B by Earth’s gravity, near its surface Wgrav,AB = −mg(yB − yA)
Work done going from A to B by one-dimensional spring force Wspring,AB = −⎛⎝12k⎞⎠⎛⎝xB2 − xA2 ⎞⎠
Kinetic energy of a non-relativistic particle K = 1


2
mv2 =


p2


2m


Work-energy theorem Wnet = KB − KA
Power as rate of doing work P = dWdt
Power as the dot product of force and velocity P = F→ · v→


SUMMARY
7.1 Work


• The infinitesimal increment of work done by a force, acting over an infinitesimal displacement, is the dot productof the force and the displacement.
• The work done by a force, acting over a finite path, is the integral of the infinitesimal increments of work donealong the path.
• The work done against a force is the negative of the work done by the force.
• The work done by a normal or frictional contact force must be determined in each particular case.
• The work done by the force of gravity, on an object near the surface of Earth, depends only on the weight of theobject and the difference in height through which it moved.
• The work done by a spring force, acting from an initial position to a final position, depends only on the springconstant and the squares of those positions.


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7.2 Kinetic Energy
• The kinetic energy of a particle is the product of one-half its mass and the square of its speed, for non-relativisticspeeds.
• The kinetic energy of a system is the sum of the kinetic energies of all the particles in the system.
• Kinetic energy is relative to a frame of reference, is always positive, and is sometimes given special names fordifferent types of motion.


7.3 Work-Energy Theorem
• Because the net force on a particle is equal to its mass times the derivative of its velocity, the integral for the network done on the particle is equal to the change in the particle’s kinetic energy. This is the work-energy theorem.
• You can use the work-energy theorem to find certain properties of a system, without having to solve the differentialequation for Newton’s second law.


7.4 Power
• Power is the rate of doing work; that is, the derivative of work with respect to time.
• Alternatively, the work done, during a time interval, is the integral of the power supplied over the time interval.
• The power delivered by a force, acting on a moving particle, is the dot product of the force and the particle’svelocity.


CONCEPTUAL QUESTIONS
7.1 Work
1. Give an example of something we think of as work ineveryday circumstances that is not work in the scientificsense. Is energy transferred or changed in form in yourexample? If so, explain how this is accomplished withoutdoing work.
2. Give an example of a situation in which there is a forceand a displacement, but the force does no work. Explainwhy it does no work.
3. Describe a situation in which a force is exerted for along time but does no work. Explain.
4. A body moves in a circle at constant speed. Does thecentripetal force that accelerates the body do any work?Explain.
5. Suppose you throw a ball upward and catch it whenit returns at the same height. How much work does thegravitational force do on the ball over its entire trip?
6. Why is it more difficult to do sit-ups while on a slantboard than on a horizontal surface? (See below.)


7. As a young man, Tarzan climbed up a vine to reach histree house. As he got older, he decided to build and use astaircase instead. Since the work of the gravitational forcemg is path independent, what did the King of the Apes gainin using stairs?


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7.2 Kinetic Energy
8. A particle of m has a velocity of vx i^ + vy j^ + vz k^ .
Is its kinetic energy given by
m(vx


2 i
^


+ vy
2 j
^


+ vz
2 k
^
)/2? If not, what is the correct


expression?
9. One particle has mass m and a second particle has mass2m. The second particle is moving with speed v and the firstwith speed 2v. How do their kinetic energies compare?
10. A person drops a pebble of mass m1 from a height
h, and it hits the floor with kinetic energy K. The persondrops another pebble of mass m2 from a height of 2h, and
it hits the floor with the same kinetic energy K. How do themasses of the pebbles compare?


7.3 Work-Energy Theorem
11. The person shown below does work on the lawnmower. Under what conditions would the mower gainenergy from the person pushing the mower? Under whatconditions would it lose energy?


12. Work done on a system puts energy into it. Work doneby a system removes energy from it. Give an example foreach statement.


13. Two marbles of masses m and 2m are dropped froma height h. Compare their kinetic energies when they reachthe ground.
14. Compare the work required to accelerate a car of mass2000 kg from 30.0 to 40.0 km/h with that required for anacceleration from 50.0 to 60.0 km/h.
15. Suppose you are jogging at constant velocity. Are youdoing any work on the environment and vice versa?
16. Two forces act to double the speed of a particle,initially moving with kinetic energy of 1 J. One of theforces does 4 J of work. How much work does the otherforce do?


7.4 Power
17. Most electrical appliances are rated in watts. Does thisrating depend on how long the appliance is on? (When off,it is a zero-watt device.) Explain in terms of the definitionof power.
18. Explain, in terms of the definition of power, whyenergy consumption is sometimes listed in kilowatt-hoursrather than joules. What is the relationship between thesetwo energy units?
19. A spark of static electricity, such as that you mightreceive from a doorknob on a cold dry day, may carry a fewhundred watts of power. Explain why you are not injuredby such a spark.
20. Does the work done in lifting an object depend on howfast it is lifted? Does the power expended depend on howfast it is lifted?
21. Can the power expended by a force be negative?
22. How can a 50-W light bulb use more energy than a1000-W oven?


PROBLEMS
7.1 Work
23. How much work does a supermarket checkoutattendant do on a can of soup he pushes 0.600 mhorizontally with a force of 5.00 N?
24. A 75.0-kg person climbs stairs, gaining 2.50 m inheight. Find the work done to accomplish this task.
25. (a) Calculate the work done on a 1500-kg elevator car


by its cable to lift it 40.0 m at constant speed, assumingfriction averages 100 N. (b) What is the work done on thelift by the gravitational force in this process? (c) What is thetotal work done on the lift?
26. Suppose a car travels 108 km at a speed of 30.0 m/s,and uses 2.0 gal of gasoline. Only 30% of the gasoline goesinto useful work by the force that keeps the car movingat constant speed despite friction. (The energy content ofgasoline is about 140 MJ/gal.) (a) What is the magnitude ofthe force exerted to keep the car moving at constant speed?


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(b) If the required force is directly proportional to speed,how many gallons will be used to drive 108 km at a speedof 28.0 m/s?
27. Calculate the work done by an 85.0-kg man whopushes a crate 4.00 m up along a ramp that makes an angleof 20.0° with the horizontal (see below). He exerts a force
of 500 N on the crate parallel to the ramp and moves at aconstant speed. Be certain to include the work he does onthe crate and on his body to get up the ramp.


28. How much work is done by the boy pulling his sister30.0 m in a wagon as shown below? Assume no frictionacts on the wagon.


29. A shopper pushes a grocery cart 20.0 m at constantspeed on level ground, against a 35.0 N frictional force. Hepushes in a direction 25.0° below the horizontal. (a) What
is the work done on the cart by friction? (b) What is thework done on the cart by the gravitational force? (c) Whatis the work done on the cart by the shopper? (d) Find theforce the shopper exerts, using energy considerations. (e)What is the total work done on the cart?
30. Suppose the ski patrol lowers a rescue sled and victim,having a total mass of 90.0 kg, down a 60.0° slope at
constant speed, as shown below. The coefficient of frictionbetween the sled and the snow is 0.100. (a) How muchwork is done by friction as the sled moves 30.0 m alongthe hill? (b) How much work is done by the rope on thesled in this distance? (c) What is the work done by thegravitational force on the sled? (d) What is the total workdone?


31. A constant 20-N force pushes a small ball in thedirection of the force over a distance of 5.0 m. What is thework done by the force?
32. A toy cart is pulled a distance of 6.0 m in a straight lineacross the floor. The force pulling the cart has a magnitudeof 20 N and is directed at 37° above the horizontal. What
is the work done by this force?
33. A 5.0-kg box rests on a horizontal surface. Thecoefficient of kinetic friction between the box and surfaceis µK = 0.50. A horizontal force pulls the box at constant
velocity for 10 cm. Find the work done by (a) the appliedhorizontal force, (b) the frictional force, and (c) the netforce.
34. A sled plus passenger with total mass 50 kg is pulled20 m across the snow (µk = 0.20) at constant velocity by
a force directed 25° above the horizontal. Calculate (a) the
work of the applied force, (b) the work of friction, and (c)the total work.
35. Suppose that the sled plus passenger of the precedingproblem is pushed 20 m across the snow at constantvelocity by a force directed 30° below the horizontal.
Calculate (a) the work of the applied force, (b) the work offriction, and (c) the total work.
36. How much work does the force F(x) = (−2.0/x) N
do on a particle as it moves from x = 2.0 m to
x = 5.0 m?


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37. Howmuch work is done against the gravitational forceon a 5.0-kg briefcase when it is carried from the groundfloor to the roof of the Empire State Building, a verticalclimb of 380 m?
38. It takes 500 J of work to compress a spring 10 cm.What is the force constant of the spring?
39. A bungee cord is essentially a very long rubber bandthat can stretch up to four times its unstretched length.However, its spring constant varies over its stretch [seeMenz, P.G. “The Physics of Bungee Jumping.” The PhysicsTeacher (November 1993) 31: 483-487]. Take the length ofthe cord to be along the x-direction and define the stretchx as the length of the cord l minus its un-stretched length
l0; that is, x = l − l0 (see below). Suppose a particular
bungee cord has a spring constant, for 0 ≤ x ≤ 4.88 m , of
k1 = 204 N/m and for 4.88 m ≤ x , of k2 = 111 N/m.
(Recall that the spring constant is the slope of the forceF(x) versus its stretch x.) (a) What is the tension in thecord when the stretch is 16.7 m (the maximum desired for agiven jump)? (b) How much work must be done against theelastic force of the bungee cord to stretch it 16.7 m?


Figure 7.16 (credit: Graeme Churchard)
40. A bungee cord exerts a nonlinear elastic force of
magnitude F(x) = k1 x + k2 x3, where x is the distance
the cord is stretched, k1 = 204 N/m and
k2 = −0.233 N/m


3. How much work must be done on the
cord to stretch it 16.7 m?
41. Engineers desire to model the magnitude of the elasticforce of a bungee cord using the equation


F(x) = a



⎢x + 9m


9m
− ⎛⎝


9 m
x + 9m





2⎤



⎥ ,


where x is the stretch of the cord along its length and a isa constant. If it takes 22.0 kJ of work to stretch the cord by16.7 m, determine the value of the constant a.
42. A particle moving in the xy-plane is subject to a force
F


(x, y) = (50 N ·m2)
(x i
^


+ y j
^
)


(x2 + y2)3/2
,


where x and y are in meters. Calculate the work done on theparticle by this force, as it moves in a straight line from thepoint (3 m, 4 m) to the point (8 m, 6 m).
43. A particle moves along a curved path
y(x) = (10 m)





⎨1 + cos[(0.1 m−1)x]





⎬, from x = 0 to


x = 10πm, subject to a tangential force of variable
magnitude F(x) = (10 N)sin[(0.1 m−1)x]. How much
work does the force do? (Hint: Consult a table of integralsor use a numerical integration program.)


7.2 Kinetic Energy
44. Compare the kinetic energy of a 20,000-kg truckmoving at 110 km/h with that of an 80.0-kg astronaut inorbit moving at 27,500 km/h.
45. (a) How fast must a 3000-kg elephant move to havethe same kinetic energy as a 65.0-kg sprinter running at10.0 m/s? (b) Discuss how the larger energies needed forthe movement of larger animals would relate to metabolicrates.
46. Estimate the kinetic energy of a 90,000-ton aircraftcarrier moving at a speed of at 30 knots. You will need tolook up the definition of a nautical mile to use in convertingthe unit for speed, where 1 knot equals 1 nautical mile perhour.
47. Calculate the kinetic energies of (a) a 2000.0-kgautomobile moving at 100.0 km/h; (b) an 80.-kg runner
sprinting at 10. m/s; and (c) a 9.1 × 10−31 -kg electron
moving at 2.0 × 107 m/s.
48. A 5.0-kg body has three times the kinetic energy ofan 8.0-kg body. Calculate the ratio of the speeds of thesebodies.
49. An 8.0-g bullet has a speed of 800 m/s. (a) What is itskinetic energy? (b) What is its kinetic energy if the speed ishalved?


Chapter 7 | Work and Kinetic Energy 353




7.3 Work-Energy Theorem
50. (a) Calculate the force needed to bring a 950-kg carto rest from a speed of 90.0 km/h in a distance of 120 m(a fairly typical distance for a non-panic stop). (b) Supposeinstead the car hits a concrete abutment at full speed and isbrought to a stop in 2.00 m. Calculate the force exerted onthe car and compare it with the force found in part (a).
51. A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object withoutdamage to the body of the car. The bumper cushions theshock by absorbing the force over a distance. Calculate themagnitude of the average force on a bumper that collapses0.200 m while bringing a 900-kg car to rest from an initialspeed of 1.1 m/s.
52. Boxing gloves are padded to lessen the force of ablow. (a) Calculate the force exerted by a boxing glove onan opponent’s face, if the glove and face compress 7.50cm during a blow in which the 7.00-kg arm and gloveare brought to rest from an initial speed of 10.0 m/s. (b)Calculate the force exerted by an identical blow in the goryold days when no gloves were used, and the knuckles andface would compress only 2.00 cm. Assume the change inmass by removing the glove is negligible. (c) Discuss themagnitude of the force with glove on. Does it seem highenough to cause damage even though it is lower than theforce with no glove?
53. Using energy considerations, calculate the averageforce a 60.0-kg sprinter exerts backward on the track toaccelerate from 2.00 to 8.00 m/s in a distance of 25.0 m,if he encounters a headwind that exerts an average force of30.0 N against him.
54. A 5.0-kg box has an acceleration of 2.0 m/s2 when
it is pulled by a horizontal force across a surface with
µK = 0.50. Find the work done over a distance of 10 cm
by (a) the horizontal force, (b) the frictional force, and (c)the net force. (d) What is the change in kinetic energy of thebox?
55. A constant 10-N horizontal force is applied to a 20-kgcart at rest on a level floor. If friction is negligible, what isthe speed of the cart when it has been pushed 8.0 m?
56. In the preceding problem, the 10-N force is applied atan angle of 45° below the horizontal. What is the speed of
the cart when it has been pushed 8.0 m?
57. Compare the work required to stop a 100-kg cratesliding at 1.0 m/s and an 8.0-g bullet traveling at 500 m/s.
58. A wagon with its passenger sits at the top of a hill.The wagon is given a slight push and rolls 100 m down a


10° incline to the bottom of the hill. What is the wagon’s
speed when it reaches the end of the incline. Assume thatthe retarding force of friction is negligible.
59. An 8.0-g bullet with a speed of 800 m/s is shot into awooden block and penetrates 20 cm before stopping. Whatis the average force of the wood on the bullet? Assume theblock does not move.
60. A 2.0-kg block starts with a speed of 10 m/s at thebottom of a plane inclined at 37° to the horizontal. The
coefficient of sliding friction between the block and plane is
µk = 0.30. (a) Use the work-energy principle to determine
how far the block slides along the plane beforemomentarily coming to rest. (b) After stopping, the blockslides back down the plane. What is its speed when itreaches the bottom? (Hint: For the round trip, only the forceof friction does work on the block.)
61. When a 3.0-kg block is pushed against a massless
spring of force constant constant 4.5 × 103 N/m, the
spring is compressed 8.0 cm. The block is released, and itslides 2.0 m (from the point at which it is released) acrossa horizontal surface before friction stops it. What is thecoefficient of kinetic friction between the block and thesurface?
62. A small block of mass 200 g starts at rest at A, slidesto B where its speed is vB = 8.0 m/s, then slides along the
horizontal surface a distance 10 m before coming to restat C. (See below.) (a) What is the work of friction alongthe curved surface? (b) What is the coefficient of kineticfriction along the horizontal surface?


63. A small object is placed at the top of an incline that isessentially frictionless. The object slides down the inclineonto a rough horizontal surface, where it stops in 5.0 safter traveling 60 m. (a) What is the speed of the objectat the bottom of the incline and its acceleration along thehorizontal surface? (b) What is the height of the incline?
64. When released, a 100-g block slides down the pathshown below, reaching the bottom with a speed of 4.0 m/s.How much work does the force of friction do?


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65. A 0.22LR-caliber bullet like that mentioned inExample 7.10 is fired into a door made of a singlethickness of 1-inch pine boards. How fast would the bulletbe traveling after it penetrated through the door?
66. A sled starts from rest at the top of a snow-coveredincline that makes a 22° angle with the horizontal. After
sliding 75 m down the slope, its speed is 14 m/s. Use thework-energy theorem to calculate the coefficient of kineticfriction between the runners of the sled and the snowysurface.


7.4 Power
67. A person in good physical condition can put out 100W of useful power for several hours at a stretch, perhapsby pedaling a mechanism that drives an electric generator.Neglecting any problems of generator efficiency andpractical considerations such as resting time: (a) How manypeople would it take to run a 4.00-kW electric clothesdryer? (b) How many people would it take to replace a largeelectric power plant that generates 800 MW?
68. What is the cost of operating a 3.00-W electric clockfor a year if the cost of electricity is $0.0900 per kW · h ?
69. A large household air conditioner may consume 15.0kW of power. What is the cost of operating this airconditioner 3.00 h per day for 30.0 d if the cost ofelectricity is $0.110 per kW · h ?
70. (a) What is the average power consumption in watts ofan appliance that uses 5.00 kW · h of energy per day? (b)
How many joules of energy does this appliance consume ina year?
71. (a) What is the average useful power output of a
person who does 6.00 × 106 J of useful work in 8.00 h?
(b) Working at this rate, how long will it take this personto lift 2000 kg of bricks 1.50 m to a platform? (Work doneto lift his body can be omitted because it is not considereduseful output he