College Physics

College Physics




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Table of Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1 Introduction: The Nature of Science and Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5


Physics: An Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Physical Quantities and Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
Accuracy, Precision, and Significant Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24


2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
Displacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
Vectors, Scalars, and Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
Time, Velocity, and Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
Motion Equations for Constant Acceleration in One Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
Problem-Solving Basics for One-Dimensional Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
Falling Objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
Graphical Analysis of One-Dimensional Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69


3 Two-Dimensional Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
Kinematics in Two Dimensions: An Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
Vector Addition and Subtraction: Graphical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
Vector Addition and Subtraction: Analytical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
Projectile Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
Addition of Velocities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113


4 Dynamics: Force and Newton's Laws of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
Development of Force Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
Newton’s First Law of Motion: Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
Newton’s Second Law of Motion: Concept of a System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
Newton’s Third Law of Motion: Symmetry in Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
Normal, Tension, and Other Examples of Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
Problem-Solving Strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
Further Applications of Newton’s Laws of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
Extended Topic: The Four Basic Forces—An Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159


5 Further Applications of Newton's Laws: Friction, Drag, and Elasticity . . . . . . . . . . . . . . . . . . . . . . . 173
Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
Drag Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179
Elasticity: Stress and Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184


6 Uniform Circular Motion and Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
Rotation Angle and Angular Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
Centripetal Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
Centripetal Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
Fictitious Forces and Non-inertial Frames: The Coriolis Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
Newton’s Universal Law of Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
Satellites and Kepler’s Laws: An Argument for Simplicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222


7 Work, Energy, and Energy Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237
Work: The Scientific Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
Kinetic Energy and the Work-Energy Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240
Gravitational Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
Conservative Forces and Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250
Nonconservative Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
Conservation of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258
Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262
Work, Energy, and Power in Humans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266
World Energy Use . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269


8 Linear Momentum and Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283
Linear Momentum and Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284
Impulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286
Conservation of Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
Elastic Collisions in One Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292
Inelastic Collisions in One Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
Collisions of Point Masses in Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298
Introduction to Rocket Propulsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301


9 Statics and Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313
The First Condition for Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314
The Second Condition for Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315
Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320
Applications of Statics, Including Problem-Solving Strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323
Simple Machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326
Forces and Torques in Muscles and Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330




10 Rotational Motion and Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343
Angular Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344
Kinematics of Rotational Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348
Dynamics of Rotational Motion: Rotational Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353
Rotational Kinetic Energy: Work and Energy Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356
Angular Momentum and Its Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
Collisions of Extended Bodies in Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369
Gyroscopic Effects: Vector Aspects of Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373


11 Fluid Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385
What Is a Fluid? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386
Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387
Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389
Variation of Pressure with Depth in a Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391
Pascal’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395
Gauge Pressure, Absolute Pressure, and Pressure Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . 398
Archimedes’ Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401
Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action . . . . . . . . . . . . . . . . . . . . . . . 408
Pressures in the Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417


12 Fluid Dynamics and Its Biological and Medical Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431
Flow Rate and Its Relation to Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432
Bernoulli’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436
The Most General Applications of Bernoulli’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 440
Viscosity and Laminar Flow; Poiseuille’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443
The Onset of Turbulence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450
Motion of an Object in a Viscous Fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452
Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes . . . . . . . . . . . . . . . . . . . . 454


13 Temperature, Kinetic Theory, and the Gas Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467
Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468
Thermal Expansion of Solids and Liquids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475
The Ideal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481
Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature . . . . . . . . . . . . . . . . . . . 487
Phase Changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494
Humidity, Evaporation, and Boiling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498


14 Heat and Heat Transfer Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513
Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514
Temperature Change and Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515
Phase Change and Latent Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521
Heat Transfer Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527
Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 528
Convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 532
Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537


15 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553
The First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554
The First Law of Thermodynamics and Some Simple Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . 559
Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency . . . . . . . . . . . . . . . 566
Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated . . . . . . . . . . . . . . . . . . . . 572
Applications of Thermodynamics: Heat Pumps and Refrigerators . . . . . . . . . . . . . . . . . . . . . . . . . . . 576
Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy . . . . . . . . . . . . 581
Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation . . . . . 588


16 Oscillatory Motion and Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603
Hooke’s Law: Stress and Strain Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604
Period and Frequency in Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608
Simple Harmonic Motion: A Special Periodic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 609
The Simple Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 614
Energy and the Simple Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 616
Uniform Circular Motion and Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 618
Damped Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 621
Forced Oscillations and Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625
Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 627
Superposition and Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 630
Energy in Waves: Intensity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 635


17 Physics of Hearing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 647
Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 648
Speed of Sound, Frequency, and Wavelength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 650
Sound Intensity and Sound Level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653
Doppler Effect and Sonic Booms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 657
Sound Interference and Resonance: Standing Waves in Air Columns . . . . . . . . . . . . . . . . . . . . . . . . . 662
Hearing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 669


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Ultrasound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674
18 Electric Charge and Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 689


Static Electricity and Charge: Conservation of Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 692
Conductors and Insulators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 696
Coulomb’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 700
Electric Field: Concept of a Field Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702
Electric Field Lines: Multiple Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 704
Electric Forces in Biology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 707
Conductors and Electric Fields in Static Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 708
Applications of Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 712


19 Electric Potential and Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 729
Electric Potential Energy: Potential Difference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 730
Electric Potential in a Uniform Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 736
Electrical Potential Due to a Point Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 739
Equipotential Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 741
Capacitors and Dielectrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 743
Capacitors in Series and Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 751
Energy Stored in Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 754


20 Electric Current, Resistance, and Ohm's Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 765
Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 766
Ohm’s Law: Resistance and Simple Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 771
Resistance and Resistivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 773
Electric Power and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 779
Alternating Current versus Direct Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 781
Electric Hazards and the Human Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 785
Nerve Conduction–Electrocardiograms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 790


21 Circuits and DC Instruments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805
Resistors in Series and Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 806
Electromotive Force: Terminal Voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 815
Kirchhoff’s Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 823
DC Voltmeters and Ammeters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 828
Null Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 831
DC Circuits Containing Resistors and Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 834


22 Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 851
Magnets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 853
Ferromagnets and Electromagnets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 855
Magnetic Fields and Magnetic Field Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 858
Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . 859
Force on a Moving Charge in a Magnetic Field: Examples and Applications . . . . . . . . . . . . . . . . . . . . . 861
The Hall Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 866
Magnetic Force on a Current-Carrying Conductor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 868
Torque on a Current Loop: Motors and Meters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 870
Magnetic Fields Produced by Currents: Ampere’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 874
Magnetic Force between Two Parallel Conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 878
More Applications of Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 879


23 Electromagnetic Induction, AC Circuits, and Electrical Technologies . . . . . . . . . . . . . . . . . . . . . . . 897
Induced Emf and Magnetic Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 899
Faraday’s Law of Induction: Lenz’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 901
Motional Emf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 903
Eddy Currents and Magnetic Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 906
Electric Generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 909
Back Emf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 913
Transformers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 914
Electrical Safety: Systems and Devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 918
Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 922
RL Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 926
Reactance, Inductive and Capacitive . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 928
RLC Series AC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 931


24 Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 951
Maxwell’s Equations: Electromagnetic Waves Predicted and Observed . . . . . . . . . . . . . . . . . . . . . . . . 952
Production of Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 954
The Electromagnetic Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 957
Energy in Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 970


25 Geometric Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 981
The Ray Aspect of Light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 982
The Law of Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 983
The Law of Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 985
Total Internal Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 990




Dispersion: The Rainbow and Prisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 995
Image Formation by Lenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1000
Image Formation by Mirrors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1013


26 Vision and Optical Instruments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1029
Physics of the Eye . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1030
Vision Correction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1034
Color and Color Vision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1038
Microscopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1040
Telescopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1046
Aberrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1049


27 Wave Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1057
The Wave Aspect of Light: Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1058
Huygens's Principle: Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1059
Young’s Double Slit Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1062
Multiple Slit Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1066
Single Slit Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1069
Limits of Resolution: The Rayleigh Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1072
Thin Film Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1076
Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1081
*Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light . . . . . . . . . . . . . . . . . . . 1089


28 Special Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1103
Einstein’s Postulates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1104
Simultaneity And Time Dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1106
Length Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1112
Relativistic Addition of Velocities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1116
Relativistic Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1122
Relativistic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1123


29 Introduction to Quantum Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1141
Quantization of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1143
The Photoelectric Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1145
Photon Energies and the Electromagnetic Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1148
Photon Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1155
The Particle-Wave Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1158
The Wave Nature of Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1159
Probability: The Heisenberg Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1163
The Particle-Wave Duality Reviewed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1168


30 Atomic Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1179
Discovery of the Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1180
Discovery of the Parts of the Atom: Electrons and Nuclei . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1182
Bohr’s Theory of the Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1188
X Rays: Atomic Origins and Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1195
Applications of Atomic Excitations and De-Excitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1200
The Wave Nature of Matter Causes Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1208
Patterns in Spectra Reveal More Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1211
Quantum Numbers and Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1213
The Pauli Exclusion Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1218


31 Radioactivity and Nuclear Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1233
Nuclear Radioactivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1234
Radiation Detection and Detectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1239
Substructure of the Nucleus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1241
Nuclear Decay and Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1246
Half-Life and Activity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1253
Binding Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1259
Tunneling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1263


32 Medical Applications of Nuclear Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1277
Medical Imaging and Diagnostics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1278
Biological Effects of Ionizing Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1282
Therapeutic Uses of Ionizing Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1288
Food Irradiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1290
Fusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1291
Fission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1297
Nuclear Weapons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1302


33 Particle Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1317
The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited . . . . . . . . . . . . . . . . . . . . . . 1318
The Four Basic Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1320
Accelerators Create Matter from Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1322
Particles, Patterns, and Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1325
Quarks: Is That All There Is? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1331


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GUTs: The Unification of Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1338
34 Frontiers of Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1351


Cosmology and Particle Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1352
General Relativity and Quantum Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1359
Superstrings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1364
Dark Matter and Closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1365
Complexity and Chaos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1369
High-temperature Superconductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1370
Some Questions We Know to Ask . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1372


A Atomic Masses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1381
B Selected Radioactive Isotopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1387
C Useful Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1391
D Glossary of Key Symbols and Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1395
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1407




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PREFACE


About OpenStax College
OpenStax College is a non-profit organization committed to improving student access to quality learning materials. Our free
textbooks are developed and peer-reviewed by educators to ensure they are readable, accurate, and meet the scope and
sequence requirements of modern college courses. Unlike traditional textbooks, OpenStax College resources live online and are
owned by the community of educators using them. Through our partnerships with companies and foundations committed to
reducing costs for students, OpenStax College is working to improve access to higher education for all. OpenStax College is an
initiative of Rice University and is made possible through the generous support of several philanthropic foundations.


About This Book
Welcome to College Physics, an OpenStax College resource created with several goals in mind: accessibility, affordability,
customization, and student engagement—all while encouraging learners toward high levels of learning. Instructors and students
alike will find that this textbook offers a strong foundation in introductory physics, with algebra as a prerequisite. It is available for
free online and in low-cost print and e-book editions.


To broaden access and encourage community curation, College Physics is “open source” licensed under a Creative Commons
Attribution (CC-BY) license. Everyone is invited to submit examples, emerging research, and other feedback to enhance and
strengthen the material and keep it current and relevant for today’s students. You can make suggestions by contacting us at
info@openstaxcollege.org. You can find the status of the project, as well as alternate versions, corrections, etc., on the StaxDash
at http://openstaxcollege.org (http://openstaxcollege.org) .


To the Student
This book is written for you. It is based on the teaching and research experience of numerous physicists and influenced by a
strong recollection of their own struggles as students. After reading this book, we hope you see that physics is visible
everywhere. Applications range from driving a car to launching a rocket, from a skater whirling on ice to a neutron star spinning
in space, and from taking your temperature to taking a chest X-ray.


To the Instructor
This text is intended for one-year introductory courses requiring algebra and some trigonometry, but no calculus. OpenStax
College provides the essential supplemental resources at http://openstaxcollege.org ; however, we have pared down the number
of supplements to keep costs low. College Physics can be easily customized for your course using Connexions (http://cnx.org/
content/col11406). Simply select the content most relevant to your curriculum and create a textbook that speaks directly to the
needs of your class.


General Approach
College Physics is organized such that topics are introduced conceptually with a steady progression to precise definitions and
analytical applications. The analytical aspect (problem solving) is tied back to the conceptual before moving on to another topic.
Each introductory chapter, for example, opens with an engaging photograph relevant to the subject of the chapter and interesting
applications that are easy for most students to visualize.


Organization, Level, and Content
There is considerable latitude on the part of the instructor regarding the use, organization, level, and content of this book. By
choosing the types of problems assigned, the instructor can determine the level of sophistication required of the student.


Concepts and Calculations
The ability to calculate does not guarantee conceptual understanding. In order to unify conceptual, analytical, and calculation
skills within the learning process, we have integrated Strategies and Discussions throughout the text.


Modern Perspective
The chapters on modern physics are more complete than many other texts on the market, with an entire chapter devoted to
medical applications of nuclear physics and another to particle physics. The final chapter of the text, “Frontiers of Physics,” is
devoted to the most exciting endeavors in physics. It ends with a module titled “Some Questions We Know to Ask.”


Preface 1




2 Preface


Supplements
Accompanying the main text are a Student Solutions Manual and an Instructor Solutions Manual
(http://openstaxcollege.org/textbooks/college-physics) . The Student Solutions Manual provides worked-out solutions to
select end-of-module Problems and Exercises. The Instructor Solutions Manual provides worked-out solutions to all Exercises.


Features of OpenStax College Physics
The following briefly describes the special features of this text.


Modularity
This textbook is organized on Connexions (http://cnx.org) as a collection of modules that can be rearranged and modified to suit
the needs of a particular professor or class. That being said, modules often contain references to content in other modules, as
most topics in physics cannot be discussed in isolation.


Learning Objectives
Every module begins with a set of learning objectives. These objectives are designed to guide the instructor in deciding what
content to include or assign, and to guide the student with respect to what he or she can expect to learn. After completing the
module and end-of-module exercises, students should be able to demonstrate mastery of the learning objectives.


Call-Outs
Key definitions, concepts, and equations are called out with a special design treatment. Call-outs are designed to catch readers’
attention, to make it clear that a specific term, concept, or equation is particularly important, and to provide easy reference for a
student reviewing content.


Key Terms
Key terms are in bold and are followed by a definition in context. Definitions of key terms are also listed in the Glossary, which
appears at the end of the module.


Worked Examples
Worked examples have four distinct parts to promote both analytical and conceptual skills. Worked examples are introduced in
words, always using some application that should be of interest. This is followed by a Strategy section that emphasizes the
concepts involved and how solving the problem relates to those concepts. This is followed by the mathematical Solution and
Discussion.


Many worked examples contain multiple-part problems to help the students learn how to approach normal situations, in which
problems tend to have multiple parts. Finally, worked examples employ the techniques of the problem-solving strategies so that
students can see how those strategies succeed in practice as well as in theory.


Problem-Solving Strategies
Problem-solving strategies are first presented in a special section and subsequently appear at crucial points in the text where
students can benefit most from them. Problem-solving strategies have a logical structure that is reinforced in the worked
examples and supported in certain places by line drawings that illustrate various steps.


Misconception Alerts
Students come to physics with preconceptions from everyday experiences and from previous courses. Some of these
preconceptions are misconceptions, and many are very common among students and the general public. Some are inadvertently
picked up through misunderstandings of lectures and texts. The Misconception Alerts feature is designed to point these out and
correct them explicitly.


Take-Home Investigations
Take Home Investigations provide the opportunity for students to apply or explore what they have learned with a hands-on
activity.


Things Great and Small
In these special topic essays, macroscopic phenomena (such as air pressure) are explained with submicroscopic phenomena
(such as atoms bouncing off walls). These essays support the modern perspective by describing aspects of modern physics
before they are formally treated in later chapters. Connections are also made between apparently disparate phenomena.


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Simulations
Where applicable, students are directed to the interactive PHeT physics simulations developed by the University of Colorado
(http://phet.colorado.edu (http://phet.colorado.edu) ). There they can further explore the physics concepts they have learned
about in the module.


Summary
Module summaries are thorough and functional and present all important definitions and equations. Students are able to find the
definitions of all terms and symbols as well as their physical relationships. The structure of the summary makes plain the
fundamental principles of the module or collection and serves as a useful study guide.


Glossary
At the end of every module or chapter is a glossary containing definitions of all of the key terms in the module or chapter.


End-of-Module Problems
At the end of every chapter is a set of Conceptual Questions and/or skills-based Problems & Exercises. Conceptual Questions
challenge students’ ability to explain what they have learned conceptually, independent of the mathematical details. Problems &
Exercises challenge students to apply both concepts and skills to solve mathematical physics problems. Online, every other
problem includes an answer that students can reveal immediately by clicking on a “Show Solution” button. Fully worked solutions
to select problems are available in the Student Solutions Manual and the Teacher Solutions Manual.


In addition to traditional skills-based problems, there are three special types of end-of-module problems: Integrated Concept
Problems, Unreasonable Results Problems, and Construct Your Own Problems. All of these problems are indicated with a
subtitle preceding the problem.


Integrated Concept Problems
In Integrated Concept Problems, students are asked to apply what they have learned about two or more concepts to arrive at a
solution to a problem. These problems require a higher level of thinking because, before solving a problem, students have to
recognize the combination of strategies required to solve it.


Unreasonable Results
In Unreasonable Results Problems, students are challenged to not only apply concepts and skills to solve a problem, but also to
analyze the answer with respect to how likely or realistic it really is. These problems contain a premise that produces an
unreasonable answer and are designed to further emphasize that properly applied physics must describe nature accurately and
is not simply the process of solving equations.


Construct Your Own Problem
These problems require students to construct the details of a problem, justify their starting assumptions, show specific steps in
the problem’s solution, and finally discuss the meaning of the result. These types of problems relate well to both conceptual and
analytical aspects of physics, emphasizing that physics must describe nature. Often they involve an integration of topics from
more than one chapter. Unlike other problems, solutions are not provided since there is no single correct answer. Instructors
should feel free to direct students regarding the level and scope of their considerations. Whether the problem is solved and
described correctly will depend on initial assumptions.


Appendices
Appendix A: Atomic Masses
Appendix B: Selected Radioactive Isotopes
Appendix C: Useful Information
Appendix D: Glossary of Key Symbols and Notation


Acknowledgements
This text is based on the work completed by Dr. Paul Peter Urone in collaboration with Roger Hinrichs, Kim Dirks, and Manjula
Sharma. We would like to thank the authors as well as the numerous professors (a partial list follows) who have contributed their
time and energy to review and provide feedback on the manuscript. Their input has been critical in maintaining the pedagogical
integrity and accuracy of the text.


Senior Contributing Authors
Dr. Paul Peter Urone
Dr. Roger Hinrichs, State University of New York, College at Oswego


Preface 3




Contributing Authors
Dr. Kim Dirks, University of Auckland, New Zealand
Dr. Manjula Sharma, University of Sydney, Australia


Expert Reviewers
Erik Christensen, P.E, South Florida Community College
Dr. Eric Kincanon, Gonzaga University
Dr. Douglas Ingram, Texas Christian University
Lee H. LaRue, Paris Junior College
Dr. Marc Sher, College of William and Mary
Dr. Ulrich Zurcher, Cleveland State University
Dr. Matthew Adams, Crafton Hills College, San Bernardino Community College District
Dr. Chuck Pearson, Virginia Intermont College


Our Partners


WebAssign
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WebAssign has recently begun to support the Open Education Resource community by creating a high quality online homework
solution for selected open-source textbooks, available at an affordable price to students. These question collections include
randomized values and variables, immediate feedback, links to the open-source textbook, and a variety of text-specific resources
and tools; as well as the same level of rigorous coding and accuracy-checking as any commercially available online homework
solution supporting traditionally available textbooks.


Sapling Learning
Sapling Learning provides the most effective interactive homework and instruction that improve student learning outcomes for the
problem-solving disciplines. They offer an enjoyable teaching and effective learning experience that is distinctive in three
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Expert TA
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Expert TA is used by universities, community colleges, and high schools.


4 Preface


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1 INTRODUCTION: THE NATURE OF
SCIENCE AND PHYSICS


Figure 1.1 Galaxies are as immense as atoms are small. Yet the same laws of physics describe both, and all the rest of nature—an indication of the
underlying unity in the universe. The laws of physics are surprisingly few in number, implying an underlying simplicity to nature’s apparent complexity.
(credit: NASA, JPL-Caltech, P. Barmby, Harvard-Smithsonian Center for Astrophysics)


Chapter Outline
1.1. Physics: An Introduction


• Explain the difference between a principle and a law.
• Explain the difference between a model and a theory.


1.2. Physical Quantities and Units
• Perform unit conversions both in the SI and English units.
• Explain the most common prefixes in the SI units and be able to write them in scientific notation.


1.3. Accuracy, Precision, and Significant Figures
• Determine the appropriate number of significant figures in both addition and subtraction, as well as multiplication and


division calculations.
• Calculate the percent uncertainty of a measurement.


1.4. Approximation
• Make reasonable approximations based on given data.


Introduction to Science and the Realm of Physics, Physical Quantities, and Units
What is your first reaction when you hear the word “physics”? Did you imagine working through difficult equations or memorizing
formulas that seem to have no real use in life outside the physics classroom? Many people come to the subject of physics with a
bit of fear. But as you begin your exploration of this broad-ranging subject, you may soon come to realize that physics plays a
much larger role in your life than you first thought, no matter your life goals or career choice.


For example, take a look at the image above. This image is of the Andromeda Galaxy, which contains billions of individual stars,
huge clouds of gas, and dust. Two smaller galaxies are also visible as bright blue spots in the background. At a staggering 2.5
million light years from the Earth, this galaxy is the nearest one to our own galaxy (which is called the Milky Way). The stars and
planets that make up Andromeda might seem to be the furthest thing from most people’s regular, everyday lives. But Andromeda
is a great starting point to think about the forces that hold together the universe. The forces that cause Andromeda to act as it
does are the same forces we contend with here on Earth, whether we are planning to send a rocket into space or simply raise
the walls for a new home. The same gravity that causes the stars of Andromeda to rotate and revolve also causes water to flow
over hydroelectric dams here on Earth. Tonight, take a moment to look up at the stars. The forces out there are the same as the
ones here on Earth. Through a study of physics, you may gain a greater understanding of the interconnectedness of everything
we can see and know in this universe.


Think now about all of the technological devices that you use on a regular basis. Computers, smart phones, GPS systems, MP3
players, and satellite radio might come to mind. Next, think about the most exciting modern technologies that you have heard
about in the news, such as trains that levitate above tracks, “invisibility cloaks” that bend light around them, and microscopic
robots that fight cancer cells in our bodies. All of these groundbreaking advancements, commonplace or unbelievable, rely on the
principles of physics. Aside from playing a significant role in technology, professionals such as engineers, pilots, physicians,
physical therapists, electricians, and computer programmers apply physics concepts in their daily work. For example, a pilot must
understand how wind forces affect a flight path and a physical therapist must understand how the muscles in the body
experience forces as they move and bend. As you will learn in this text, physics principles are propelling new, exciting
technologies, and these principles are applied in a wide range of careers.


Chapter 1 | Introduction: The Nature of Science and Physics 5




In this text, you will begin to explore the history of the formal study of physics, beginning with natural philosophy and the ancient
Greeks, and leading up through a review of Sir Isaac Newton and the laws of physics that bear his name. You will also be
introduced to the standards scientists use when they study physical quantities and the interrelated system of measurements
most of the scientific community uses to communicate in a single mathematical language. Finally, you will study the limits of our
ability to be accurate and precise, and the reasons scientists go to painstaking lengths to be as clear as possible regarding their
own limitations.


1.1 Physics: An Introduction


6 Chapter 1 | Introduction: The Nature of Science and Physics


Figure 1.2 The flight formations of migratory birds such as Canada geese are governed by the laws of physics. (credit: David Merrett)


The physical universe is enormously complex in its detail. Every day, each of us observes a great variety of objects and
phenomena. Over the centuries, the curiosity of the human race has led us collectively to explore and catalog a tremendous
wealth of information. From the flight of birds to the colors of flowers, from lightning to gravity, from quarks to clusters of galaxies,
from the flow of time to the mystery of the creation of the universe, we have asked questions and assembled huge arrays of
facts. In the face of all these details, we have discovered that a surprisingly small and unified set of physical laws can explain
what we observe. As humans, we make generalizations and seek order. We have found that nature is remarkably cooperative—it
exhibits the underlying order and simplicity we so value.


It is the underlying order of nature that makes science in general, and physics in particular, so enjoyable to study. For example,
what do a bag of chips and a car battery have in common? Both contain energy that can be converted to other forms. The law of
conservation of energy (which says that energy can change form but is never lost) ties together such topics as food calories,
batteries, heat, light, and watch springs. Understanding this law makes it easier to learn about the various forms energy takes
and how they relate to one another. Apparently unrelated topics are connected through broadly applicable physical laws,
permitting an understanding beyond just the memorization of lists of facts.


The unifying aspect of physical laws and the basic simplicity of nature form the underlying themes of this text. In learning to apply
these laws, you will, of course, study the most important topics in physics. More importantly, you will gain analytical abilities that
will enable you to apply these laws far beyond the scope of what can be included in a single book. These analytical skills will help
you to excel academically, and they will also help you to think critically in any professional career you choose to pursue. This
module discusses the realm of physics (to define what physics is), some applications of physics (to illustrate its relevance to
other disciplines), and more precisely what constitutes a physical law (to illuminate the importance of experimentation to theory).


Science and the Realm of Physics
Science consists of the theories and laws that are the general truths of nature as well as the body of knowledge they encompass.
Scientists are continually trying to expand this body of knowledge and to perfect the expression of the laws that describe it.
Physics is concerned with describing the interactions of energy, matter, space, and time, and it is especially interested in what
fundamental mechanisms underlie every phenomenon. The concern for describing the basic phenomena in nature essentially
defines the realm of physics.


Physics aims to describe the function of everything around us, from the movement of tiny charged particles to the motion of
people, cars, and spaceships. In fact, almost everything around you can be described quite accurately by the laws of physics.
Consider a smart phone (Figure 1.3). 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 the smart phone. Next, consider a
GPS system. Physics describes the relationship between the speed of an object, the distance over which it travels, and the time
it takes to travel that distance. When you use a GPS device in a vehicle, it utilizes these physics equations to determine the
travel time from one location to another.


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Figure 1.3 The Apple “iPhone” is a common smart phone with a GPS function. Physics describes the way that electricity flows through the circuits of
this device. Engineers use their knowledge of physics to construct an iPhone with features that consumers will enjoy. One specific feature of an iPhone
is the GPS function. GPS uses physics equations to determine the driving time between two locations on a map. (credit: @gletham GIS, Social, Mobile
Tech Images)


Applications of Physics
You need not be a scientist to use physics. On the contrary, knowledge of physics is useful in everyday situations as well as in
nonscientific professions. It can help you understand how microwave ovens work, why metals should not be put into them, and
why they might affect pacemakers. (See Figure 1.4 and Figure 1.5.) Physics allows you to understand the hazards of radiation
and rationally evaluate these hazards more easily. Physics also explains the reason why a black car radiator helps remove heat
in a car engine, and it explains why a white roof helps keep the inside of a house cool. Similarly, the operation of a car’s ignition
system as well as the transmission of electrical signals through our body’s nervous system are much easier to understand when
you think about them in terms of basic physics.


Physics is the foundation of many important disciplines and contributes directly to others. Chemistry, for example—since it deals
with the interactions of atoms and molecules—is rooted in atomic and molecular physics. Most branches of engineering are
applied physics. In architecture, physics is at the heart of structural stability, and is involved in the acoustics, heating, lighting,
and cooling of buildings. Parts of geology rely heavily on physics, such as radioactive dating of rocks, earthquake analysis, and
heat transfer in the 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 of cell walls
and cell membranes (Figure 1.6 and Figure 1.7). On the macroscopic level, it can explain the heat, work, and power associated
with the human body. Physics is involved in medical diagnostics, such as x-rays, magnetic resonance imaging (MRI), and
ultrasonic blood flow measurements. Medical therapy sometimes directly involves physics; for example, cancer radiotherapy
uses ionizing radiation. Physics can also explain sensory phenomena, such as how musical instruments make sound, how the
eye detects color, and how lasers can transmit information.


It is not necessary to formally study all applications of physics. What is most useful is knowledge of the basic laws of physics and
a skill in the analytical methods for applying them. The study of physics also can improve your problem-solving skills.
Furthermore, physics has retained the most basic aspects of science, so it is used by all of the sciences, and the study of
physics makes other sciences easier to understand.


Figure 1.4 The laws of physics help us understand how common appliances work. For example, the laws of physics can help explain how microwave
ovens heat up food, and they also help us understand why it is dangerous to place metal objects in a microwave oven. (credit: MoneyBlogNewz)


Chapter 1 | Introduction: The Nature of Science and Physics 7




Figure 1.5 These two applications of physics have more in common than meets the eye. Microwave ovens use electromagnetic waves to heat food.
Magnetic resonance imaging (MRI) also uses electromagnetic waves to yield an image of the brain, from which the exact location of tumors can be
determined. (credit: Rashmi Chawla, Daniel Smith, and Paul E. Marik)


Figure 1.6 Physics, chemistry, and biology help describe the properties of cell walls in plant cells, such as the onion cells seen here. (credit: Umberto
Salvagnin)


8 Chapter 1 | Introduction: The Nature of Science and Physics


Figure 1.7 An artist’s rendition of the the structure of a cell membrane. Membranes form the boundaries of animal cells and are complex in structure
and function. Many of the most fundamental properties of life, such as the firing of nerve cells, are related to membranes. The disciplines of biology,
chemistry, and physics all help us understand the membranes of animal cells. (credit: Mariana Ruiz)


Models, Theories, and Laws; The Role of Experimentation
The laws of nature are concise descriptions of the universe around us; they are human statements of the underlying laws or rules
that all natural processes follow. Such laws are intrinsic to the universe; humans did not create them and so cannot change
them. We can only discover and 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. (See Figure 1.8 and Figure 1.9.) The
cornerstone of discovering natural laws is observation; science must describe the universe as it is, not as we may imagine it to
be.


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/content/col11406/1.9




Figure 1.8 Isaac Newton (1642–1727) was very reluctant to publish his revolutionary work and had to be convinced to do so. In his later years, he
stepped down from his academic post and became exchequer of the Royal Mint. He took this post seriously, inventing reeding (or creating ridges) on
the edge of coins to prevent unscrupulous people from trimming the silver off of them before using them as currency. (credit: Arthur Shuster and Arthur
E. Shipley: Britain’s Heritage of Science. London, 1917.)


Figure 1.9 Marie Curie (1867–1934) sacrificed monetary assets to help finance her early research and damaged her physical well-being with radiation
exposure. She is the only person to win Nobel prizes in both physics and chemistry. One of her daughters also won a Nobel Prize. (credit: Wikimedia
Commons)


We all are curious to some extent. We look around, make generalizations, and try to understand what we see—for example, we
look up and wonder whether one type of cloud signals an oncoming storm. As we become serious about exploring nature, we
become more organized and formal in collecting and analyzing data. We attempt greater precision, perform controlled
experiments (if we can), and write down ideas about how the data may be organized and unified. We then formulate models,
theories, and laws based on the data we have collected and analyzed to generalize and communicate the results of these
experiments.


A model is a representation of something that is often too difficult (or impossible) to display directly. While a model is justified
with experimental proof, it is only accurate under limited situations. An example is the planetary model of the atom in which
electrons are pictured as orbiting the nucleus, analogous to the way planets orbit the Sun. (See Figure 1.10.) We cannot observe
electron orbits directly, but the mental image helps explain the observations we can make, such as the emission of light from hot
gases (atomic spectra). Physicists use models for a variety of purposes. For example, models can help physicists analyze a
scenario and perform a calculation, or they can be used to represent a situation in the form of a computer simulation. A theory is
an explanation for patterns in nature that is supported by scientific evidence and verified multiple times 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 objects directly with our own senses. The
kinetic theory of gases, on the other hand, is a model in which a gas is viewed as being composed of atoms and molecules.
Atoms and molecules are too small to be observed directly with our senses—thus, we picture them mentally to understand what
our instruments tell us about the behavior of gases.


A law uses concise language to describe a generalized pattern in nature that is supported by scientific evidence and repeated
experiments. Often, a law can be expressed in the form of a single mathematical equation. Laws and theories are similar in that
they are both scientific statements that result from a tested hypothesis and are supported by scientific evidence. However, the
designation law is reserved for a concise and very general statement that describes phenomena in nature, such as the law that


Chapter 1 | Introduction: The Nature of Science and Physics 9




energy is conserved during any process, or Newton’s second law of motion, which relates force, mass, and acceleration by the
simple equation F = ma . A theory, in contrast, is a less concise statement of observed phenomena. For example, the Theory of
Evolution and the Theory of Relativity cannot be expressed concisely enough to be considered a law. 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. And, whereas a law is a postulate that forms the foundation of the
scientific method, a theory is the end result of that process.


Less broadly applicable statements are usually called principles (such as Pascal’s principle, which is applicable only in fluids),
but the distinction between laws and principles often is not carefully made.


Figure 1.10What is a model? This planetary model of the atom shows electrons orbiting the nucleus. It is a drawing that we use to form a mental
image of the atom that we cannot see directly with our eyes because it is too small.


Models, Theories, and Laws


Models, theories, and laws are used to help scientists analyze the data they have already collected. However, often after a
model, theory, or law has been developed, it points scientists toward new discoveries they would not otherwise have made.


The models, theories, and laws we devise sometimes imply the existence of objects or phenomena as yet unobserved. These
predictions are remarkable triumphs and tributes to the power of science. It is the underlying order in the universe that enables
scientists to make such spectacular predictions. However, if experiment does not verify our predictions, then the theory or law is
wrong, no matter how elegant or convenient it is. Laws can never be known with absolute certainty because it is impossible to
perform every imaginable experiment in order to confirm a law in every possible scenario. Physicists operate under the
assumption that all scientific laws and theories are valid until a counterexample is observed. If a good-quality, verifiable
experiment contradicts a well-established law, then the law must be modified or overthrown completely.


The study of science in general and physics in particular is an adventure much like the exploration of uncharted ocean.
Discoveries are made; models, theories, and laws are formulated; and the beauty of the physical universe is made more sublime
for the insights gained.


The Scientific Method


As scientists inquire and gather information about the world, they follow a process called the scientific method. This
process typically begins with an observation and question that the scientist will research. Next, the scientist typically
performs some research about the topic and then devises a hypothesis. Then, the scientist will test the hypothesis by
performing an experiment. Finally, the scientist analyzes the results of the experiment and draws a conclusion. Note that the
scientific method can be applied to many situations that are not limited to science, and this method can be modified to suit
the situation.


Consider an example. Let us say that you try to turn on your car, but it will not start. You undoubtedly wonder: Why will the
car not start? You can follow a scientific method to answer this question. First off, you may perform some research to
determine a variety of reasons why the car will not start. Next, you will state a hypothesis. For example, you may believe that
the car is not starting because it has no engine oil. To test this, you open the hood of the car and examine the oil level. You
observe that the oil is at an acceptable level, and you thus conclude that the oil level is not contributing to your car issue. To
troubleshoot the issue further, you may devise a new hypothesis to test and then repeat the process again.


10 Chapter 1 | Introduction: The Nature of Science and Physics


The Evolution of Natural Philosophy into Modern Physics
Physics was not always a separate and distinct discipline. It remains connected to other sciences to this day. The word physics
comes from Greek, meaning nature. The study of nature came to be called “natural philosophy.” From ancient times through the
Renaissance, natural philosophy encompassed many fields, including astronomy, biology, chemistry, physics, mathematics, and
medicine. Over the last few centuries, the growth of knowledge has resulted in ever-increasing specialization and branching of
natural philosophy into separate fields, with physics retaining the most basic facets. (See Figure 1.11, Figure 1.12, and Figure
1.13.) Physics as it developed from the Renaissance to the end of the 19th century is called classical physics. It was
transformed into modern physics by revolutionary discoveries made starting at the beginning of the 20th century.


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Figure 1.11 Over the centuries, natural philosophy has evolved into more specialized disciplines, as illustrated by the contributions of some of the
greatest minds in history. The Greek philosopher Aristotle (384–322 B.C.) wrote on a broad range of topics including physics, animals, the soul,
politics, and poetry. (credit: Jastrow (2006)/Ludovisi Collection)


Figure 1.12 Galileo Galilei (1564–1642) laid the foundation of modern experimentation and made contributions in mathematics, physics, and
astronomy. (credit: Domenico Tintoretto)


Figure 1.13 Niels Bohr (1885–1962) made fundamental contributions to the development of quantum mechanics, one part of modern physics. (credit:
United States Library of Congress Prints and Photographs Division)


Classical physics is not an exact description of the universe, but it is an excellent approximation under the following conditions:
Matter must be moving at speeds less than about 1% of the speed of light, the objects dealt with must be large enough to be
seen with a microscope, and only weak gravitational fields, such as the field generated by the Earth, can be involved. Because
humans live under such circumstances, classical physics seems intuitively reasonable, while many aspects of modern physics
seem bizarre. This is why models are so useful in modern physics—they let us conceptualize phenomena we do not ordinarily
experience. We can relate to models in human terms and visualize what happens when objects move at high speeds or imagine
what objects too small to observe with our senses might be like. For example, we can understand an atom’s properties because
we can picture it in our minds, although we have never seen an atom with our eyes. New tools, of course, allow us to better
picture phenomena we cannot see. In fact, new instrumentation has allowed us in recent years to actually “picture” the atom.


Chapter 1 | Introduction: The Nature of Science and Physics 11




Limits on the Laws of Classical Physics


For the laws of classical physics to apply, the following criteria must be met: Matter must be moving at speeds less than
about 1% of the speed of light, the objects dealt with must be large enough to be seen with a microscope, and only weak
gravitational fields (such as the field generated by the Earth) can be involved.


Figure 1.14 Using a scanning tunneling microscope (STM), scientists can see the individual atoms that compose this sheet of gold. (credit:
Erwinrossen)


Some of the most spectacular advances in science have been made in modern physics. Many of the laws of classical physics
have been modified or rejected, and revolutionary changes in technology, society, and our view of the universe have resulted.
Like science fiction, modern physics is filled with fascinating objects beyond our normal experiences, but it has the advantage
over science fiction of being very real. Why, then, is the majority of this text devoted to topics of classical physics? There are two
main reasons: Classical physics gives an extremely accurate description of the universe under a wide range of everyday
circumstances, and knowledge of classical physics is necessary to understand modern physics.


Modern physics itself consists of the two revolutionary theories, relativity and quantum mechanics. These theories deal with the
very fast and the very small, respectively. Relativity must be used whenever an object is traveling at greater than about 1% of
the speed of light or experiences a strong gravitational field such as that near the Sun. Quantum mechanics must be used for
objects smaller than can be seen with a microscope. The combination of these two theories is relativistic quantum mechanics,
and it describes the behavior of small objects traveling at high speeds or experiencing a strong gravitational field. Relativistic
quantum mechanics is the best universally applicable theory we have. Because of its mathematical complexity, it is used only
when necessary, and the other theories are used whenever they will produce sufficiently accurate results. We will find, however,
that we can do a great deal of modern physics with the algebra and trigonometry used in this text.


Check Your Understanding


A friend tells you he has learned about a new law of nature. What can you know about the information even before your
friend describes the law? How would the information be different if your friend told you he had learned about a scientific
theory rather than a law?


Solution
Without knowing the details of the law, you can still infer that the information your friend has learned conforms to the
requirements of all laws of nature: it will be a concise description of the universe around us; a statement of the underlying
rules that all natural processes follow. If the information had been a theory, you would be able to infer that the information will
be a large-scale, broadly applicable generalization.


PhET Explorations: Equation Grapher


Learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the
individual terms (e.g. y = bx ) to see how they add to generate the polynomial curve.


Figure 1.15 Equation Grapher (http://


/content/m42092/1.4/equation-grapher_en.jar)


12 Chapter 1 | Introduction: The Nature of Science and Physics


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1.2 Physical Quantities and Units


Figure 1.16 The distance from Earth to the Moon may seem immense, but it is just a tiny fraction of the distances from Earth to other celestial bodies.
(credit: NASA)


The range of objects and phenomena studied in physics is immense. From the incredibly short lifetime of a nucleus to the age of
the Earth, from the tiny sizes of sub-nuclear particles to the vast distance to the edges of the known universe, from the force
exerted by a jumping flea to the force between Earth and the Sun, there are enough factors of 10 to challenge the imagination of
even the most experienced scientist. Giving numerical values for physical quantities and equations for physical principles allows
us to understand nature much more deeply than does qualitative description alone. To comprehend these vast ranges, we must
also have accepted units in which to express them. And we shall find that (even in the potentially mundane discussion of meters,
kilograms, and seconds) a profound simplicity of nature appears—all physical quantities can be expressed as combinations of
only four fundamental physical quantities: length, mass, time, and electric current.


We define a physical quantity either by specifying how it is measured or by stating how it is calculated from other
measurements. For example, we define distance and time by specifying methods for measuring them, whereas we define
average speed by stating that it is calculated as distance traveled divided by time of travel.


Measurements of physical quantities are expressed in terms of units, which are standardized values. For example, the length of
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 a meaningful
way. (See Figure 1.17.)


Figure 1.17 Distances given in unknown units are maddeningly useless.


There are two major systems of units used in the world: SI units (also known as the metric system) and English units (also
known as the customary or imperial system). English units were historically used in nations once ruled by the British Empire
and are still widely used in the United States. Virtually every other country in the world now uses SI units as the standard; the
metric system is also the standard system agreed upon by scientists and mathematicians. The acronym “SI” is derived from the
French Système International.


SI Units: Fundamental and Derived Units
Table 1.1 gives the fundamental SI units that are used throughout this textbook. This text uses non-SI units in a few applications
where they are in very common use, such as the measurement of blood pressure in millimeters of mercury (mm Hg). Whenever
non-SI units are discussed, they will be tied to SI units through conversions.


Table 1.1 Fundamental SI Units


Length Mass Time Electric Current


meter (m) kilogram (kg) second (s) ampere (A)


Chapter 1 | Introduction: The Nature of Science and Physics 13




It is an intriguing fact that some physical quantities are more fundamental than others and that the most fundamental physical
quantities can be defined only in terms of the procedure used to measure them. The units in which they are measured are thus
called fundamental units. In this textbook, the fundamental physical quantities are taken to be length, mass, time, and electric
current. (Note that electric current will not be introduced until much later in this text.) All other physical quantities, such as force
and electric charge, can be expressed as algebraic combinations of length, mass, time, and current (for example, speed is length
divided by time); these units are called derived units.


Units of Time, Length, and Mass: The Second, Meter, and Kilogram


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 mean solar
day. More recently, a new standard was adopted to gain greater accuracy and to define the second in terms of a non-varying, or
constant, physical phenomenon (because the solar day is getting longer due to very gradual slowing of the Earth’s rotation).
Cesium atoms can be made to vibrate in a very steady way, and these vibrations can be readily observed and counted. In 1967
the second was redefined as the time required for 9,192,631,770 of these vibrations. (See Figure 1.18.) Accuracy in the
fundamental units is essential, because all measurements are ultimately expressed in terms of fundamental units and can be no
more accurate than are the fundamental units themselves.


Figure 1.18 An atomic clock such as this one uses the vibrations of cesium atoms to keep time to a precision of better than a microsecond per year.
The fundamental unit of time, the second, is based on such clocks. This image is looking down from the top of an atomic fountain nearly 30 feet tall!
(credit: Steve Jurvetson/Flickr)


The Meter


The SI unit for length is the meter (abbreviated m); its definition has also changed over time to become more accurate and
precise. The meter was first defined in 1791 as 1/10,000,000 of the distance from the equator to the North Pole. This
measurement was improved in 1889 by redefining the meter to be the distance between two engraved lines on a platinum-iridium
bar now kept near Paris. By 1960, it had become possible to define the meter even more accurately in terms of the wavelength
of light, so it was again redefined as 1,650,763.73 wavelengths of orange light emitted by krypton atoms. In 1983, the meter was
given its present definition (partly for greater accuracy) as the distance light travels in a vacuum in 1/299,792,458 of a second.
(See Figure 1.19.) This change defines the speed of light to be exactly 299,792,458 meters per second. The length of the meter
will change if the speed of light is someday measured with greater accuracy.


The Kilogram


The SI unit for mass is the kilogram (abbreviated kg); it is defined to be the mass of a platinum-iridium cylinder kept with the old
meter standard at the International Bureau of Weights and Measures near Paris. Exact replicas of the standard kilogram are also
kept at the United States’ National Institute of Standards and Technology, or NIST, located in Gaithersburg, Maryland outside of
Washington D.C., and at other locations around the world. The determination of all other masses can be ultimately traced to a
comparison with the standard mass.


14 Chapter 1 | Introduction: The Nature of Science and Physics


Figure 1.19 The meter is defined to be the distance light travels in 1/299,792,458 of a second in a vacuum. Distance traveled is speed multiplied by
time.


Electric current and its accompanying unit, the ampere, will be introduced in Introduction to Electric Current, Resistance, and
Ohm's Law when electricity and magnetism are covered. The initial modules in this textbook are concerned with mechanics,


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fluids, heat, and waves. In these subjects all pertinent physical quantities can be expressed in terms of the fundamental units of
length, mass, and time.


Metric Prefixes
SI units are part of the metric system. The metric system is convenient for scientific and engineering calculations because the
units are categorized by factors of 10. Table 1.2 gives metric prefixes and symbols used to denote various factors of 10.


Metric systems have the advantage that conversions of units involve only powers of 10. There are 100 centimeters in a meter,
1000 meters in a kilometer, and so on. In nonmetric systems, such as the system of U.S. customary units, the relationships are
not as simple—there are 12 inches in a foot, 5280 feet in a mile, and so on. Another advantage of the metric system is that the
same unit can be used over extremely large ranges of values simply by using an appropriate metric prefix. For example,
distances in meters are suitable in construction, while distances in kilometers are appropriate for air travel, and the tiny measure
of nanometers are convenient in optical design. With the metric system there is no need to invent new units for particular
applications.


The term order of magnitude refers to the scale of a value expressed in the metric system. Each power of 10 in the metric
system represents a different order of magnitude. For example, 101 , 102 , 103 , and so forth are all different orders of
magnitude. All quantities that can be expressed as a product of a specific power of 10 are said to be of the same order of
magnitude. For example, the number 800 can be written as 8×102 , and the number 450 can be written as 4.5×102. Thus,
the numbers 800 and 450 are of the same order of magnitude: 102. Order of magnitude can be thought of as a ballpark
estimate for the scale of a value. The diameter of an atom is on the order of 10−9 m, while the diameter of the Sun is on the


order of 109 m.


The Quest for Microscopic Standards for Basic Units


The fundamental units described in this chapter are those that produce the greatest accuracy and precision in
measurement. There is a sense among physicists that, because there is an underlying microscopic substructure to matter, it
would be most satisfying to base our standards of measurement on microscopic objects and fundamental physical
phenomena such as the speed of light. A microscopic standard has been accomplished for the standard of time, which is
based on the oscillations of the cesium atom.


The standard for length was once based on the wavelength of light (a small-scale length) emitted by a certain type of atom,
but it has been supplanted by the more precise measurement of the speed of light. If it becomes possible to measure the
mass of atoms or a particular arrangement of atoms such as a silicon sphere to greater precision than the kilogram
standard, it may become possible to base mass measurements on the small scale. There are also possibilities that electrical
phenomena on the small scale may someday allow us to base a unit of charge on the charge of electrons and protons, but
at present current and charge are related to large-scale currents and forces between wires.


Chapter 1 | Introduction: The Nature of Science and Physics 15




Table 1.2 Metric Prefixes for Powers of 10 and their Symbols


Prefix Symbol Value[1] Example (some are approximate)


exa E 1018 exameter Em 1018 m distance light travels in a century


peta P 1015 petasecond Ps 1015 s 30 million years


tera T 1012 terawatt TW 1012 W powerful laser output


giga G 109 gigahertz GHz 109 Hz a microwave frequency


mega M 106 megacurie MCi 106 Ci high radioactivity


kilo k 103 kilometer km 103 m about 6/10 mile


hecto h 102 hectoliter hL 102 L 26 gallons


deka da 101 dekagram dag 101 g teaspoon of butter


— — 100 (=1)


deci d 10−1 deciliter dL 10−1 L less than half a soda


centi c 10−2 centimeter cm 10−2 m fingertip thickness


milli m 10−3 millimeter mm 10−3 m flea at its shoulders


micro µ 10−6 micrometer µm 10−6 m detail in microscope


nano n 10−9 nanogram ng 10−9 g small speck of dust


pico p 10−12 picofarad pF 10−12 F small capacitor in radio


femto f 10−15 femtometer fm 10−15 m size of a proton


atto a 10−18 attosecond as 10−18 s time light crosses an atom


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 of known
lengths, masses, and times in Table 1.3. Examination of this table will give you some feeling for the range of possible topics and
numerical values. (See Figure 1.20 and Figure 1.21.)


Figure 1.20 Tiny phytoplankton swims among crystals of ice in the Antarctic Sea. They range from a few micrometers to as much as 2 millimeters in
length. (credit: Prof. Gordon T. Taylor, Stony Brook University; NOAA Corps Collections)


1. See Appendix A for a discussion of powers of 10.


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Figure 1.21 Galaxies collide 2.4 billion light years away from Earth. The tremendous range of observable phenomena in nature challenges the
imagination. (credit: NASA/CXC/UVic./A. Mahdavi et al. Optical/lensing: CFHT/UVic./H. Hoekstra et al.)


Unit Conversion and Dimensional Analysis
It is often necessary to convert from one type of unit to another. For example, if you are reading a European cookbook, some
quantities may be expressed in units of liters and you need to convert them to cups. Or, perhaps you are reading walking
directions from one location to another and you are interested in how many miles you will be walking. In this case, you will need
to convert units of feet to miles.


Let us consider a simple example of how to convert units. Let us say that we want to convert 80 meters (m) to kilometers (km).


The first thing to do is to list the units that you have and the units that you want to convert to. In this case, we have units in
meters and we want to convert to kilometers.


Next, we need to determine a conversion factor relating meters to kilometers. A conversion factor is a ratio expressing how
many of one unit are equal to another unit. For example, there are 12 inches in 1 foot, 100 centimeters in 1 meter, 60 seconds in
1 minute, and so on. In this case, we know that there are 1,000 meters in 1 kilometer.


Now we can set up our unit conversion. We will write the units that we have and then multiply them by the conversion factor so
that the units cancel out, as shown:


(1.1)80m× 1 km1000m = 0.080 km.


Note that the unwanted m unit cancels, leaving only the desired km unit. You can use this method to convert between any types
of unit.


Click Appendix C for a more complete list of conversion factors.


Chapter 1 | Introduction: The Nature of Science and Physics 17




Table 1.3 Approximate Values of Length, Mass, and Time


Lengths in meters Masses in kilograms (moreprecise values in parentheses)
Times in seconds (more precise
values in parentheses)


10−18 Present experimental limit tosmallest observable detail 10
−30


Mass of an electron

⎝9.11×10−31 kg





10−23 Time for light to cross aproton


10−15 Diameter of a proton 10−27
Mass of a hydrogen atom

⎝1.67×10−27 kg





10−22 Mean life of an extremelyunstable nucleus


10−14 Diameter of a uranium nucleus 10−15 Mass of a bacterium 10−15 Time for one oscillation ofvisible light


10−10 Diameter of a hydrogen atom 10−5 Mass of a mosquito 10−13 Time for one vibration of anatom in a solid


10−8 Thickness of membranes in cells ofliving organisms 10
−2 Mass of a hummingbird 10−8 Time for one oscillation of anFM radio wave


10−6 Wavelength of visible light 1 Mass of a liter of water (abouta quart) 10
−3 Duration of a nerve impulse


10−3 Size of a grain of sand 102 Mass of a person 1 Time for one heartbeat


1 Height of a 4-year-old child 103 Mass of a car 105 One day

⎝8.64×104 s





102 Length of a football field 108 Mass of a large ship 107 One year (y)

⎝3.16×107 s





104 Greatest ocean depth 1012 Mass of a large iceberg 109 About half the lifeexpectancy of a human


107 Diameter of the Earth 1015 Mass of the nucleus of a comet 1011 Recorded history


1011 Distance from the Earth to the Sun 1023
Mass of the Moon

⎝7.35×1022 kg





1017 Age of the Earth


1016 Distance traveled by light in 1 year(a light year) 10
25


Mass of the Earth

⎝5.97×1024 kg





1018 Age of the universe


1021 Diameter of the Milky Way galaxy 1030
Mass of the Sun

⎝1.99×1030 kg





1022 Distance from the Earth to thenearest large galaxy (Andromeda) 10
42 Mass of the Milky Way galaxy


(current upper limit)


1026 Distance from the Earth to theedges of the known universe 10
53 Mass of the known universe


(current upper limit)


Example 1.1 Unit Conversions: A Short Drive Home


Suppose that you drive the 10.0 km from your university to home in 20.0 min. Calculate your average speed (a) in kilometers
per hour (km/h) and (b) 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 desired units by
picking the correct conversion factor and multiplying by it. The correct conversion factor is the one that cancels the
unwanted unit and leaves the desired unit in its place.


Solution for (a)


(1) Calculate average speed. Average speed is distance traveled divided by time of travel. (Take this definition as a given for
now—average speed and other motion concepts will be covered in a later module.) In equation form,


(1.2)average speed =distancetime .


(2) Substitute the given values for distance and time.


18 Chapter 1 | Introduction: The Nature of Science and Physics


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(1.3)average speed = 10.0 km20.0 min = 0.500
km
min.


(3) Convert km/min to km/h: multiply by the conversion factor that will cancel minutes and leave hours. That conversion
factor is 60 min/hr . Thus,


(1.4)average speed =0.500 km min×
60 min
1 h = 30.0


km
h .


Discussion for (a)


To check your answer, consider the following:


(1) Be sure that you have properly cancelled the units in the unit conversion. If you have written the unit conversion factor
upside down, the units will not cancel properly in the equation. If you accidentally get the ratio upside down, then the units
will not cancel; rather, they will give you the wrong units as follows:


(1.5) km
min×


1 hr
60 min =


1
60
km ⋅ hr
min2


,


which are obviously not the desired units of km/h.


(2) Check that the units of the final answer are the desired units. The problem asked us to solve for average speed in units
of km/h and we have indeed obtained these units.


(3) Check the significant figures. Because each of the values given in the problem has three significant figures, the answer
should also have three significant figures. The answer 30.0 km/hr does indeed have three significant figures, so this is
appropriate. Note that the significant figures in the conversion factor are not relevant because an hour is defined to be 60
minutes, so the precision of the conversion factor is perfect.


(4) Next, check whether the answer is reasonable. Let us consider some information from the problem—if you travel 10 km
in a third of an hour (20 min), you would travel three times that far in an hour. The answer does seem reasonable.


Solution for (b)


There are several ways to convert the average speed into meters per second.


(1) Start with the answer to (a) and convert km/h to m/s. Two conversion factors are needed—one to convert hours to
seconds, and another to convert kilometers to meters.


(2) Multiplying by these yields


(1.6)Average speed = 30.0kmh ×
1 h


3,600 s×
1,000 m
1 km ,


(1.7)Average speed = 8.33ms .


Discussion for (b)


If we had started with 0.500 km/min, we would have needed different conversion factors, but the answer would have been
the same: 8.33 m/s.


You may have noted that the answers in the worked example just covered were given to three digits. Why? When do you
need to be concerned about the number of digits in something you calculate? Why not write down all the digits your
calculator produces? The module Accuracy, Precision, and Significant Figures will help you answer these questions.


Nonstandard Units


While there are numerous types of units that we are all familiar with, there are others that are much more obscure. For
example, a firkin is a unit of volume that was once used to measure beer. One firkin equals about 34 liters. To learn more
about nonstandard units, use a dictionary or encyclopedia to research different “weights and measures.” Take note of any
unusual units, such as a barleycorn, that are not listed in the text. Think about how the unit is defined and state its
relationship to SI units.


Check Your Understanding


Some hummingbirds beat their wings more than 50 times per second. A scientist is measuring the time it takes for a
hummingbird to beat its wings once. Which fundamental unit should the scientist use to describe the measurement? Which
factor of 10 is the scientist likely to use to describe the motion precisely? Identify the metric prefix that corresponds to this
factor of 10.


Solution


Chapter 1 | Introduction: The Nature of Science and Physics 19




The scientist will measure the time between each movement using the fundamental unit of seconds. Because the wings beat


so fast, the scientist will probably need to measure in milliseconds, or 10−3 seconds. (50 beats per second corresponds to
20 milliseconds per beat.)


Check Your Understanding


One cubic centimeter is equal to one milliliter. What does this tell you about the different units in the SI metric system?


Solution
The fundamental unit of length (meter) is probably used to create the derived unit of volume (liter). The measure of a milliliter
is dependent on the measure of a centimeter.


1.3 Accuracy, Precision, and Significant Figures


Figure 1.22 A double-pan mechanical balance is used to compare different masses. Usually an object with unknown mass is placed in one pan and
objects of known mass are placed in the other pan. When the bar that connects the two pans is horizontal, then the masses in both pans are equal.
The “known masses” are typically metal cylinders of standard mass such as 1 gram, 10 grams, and 100 grams. (credit: Serge Melki)


20 Chapter 1 | Introduction: The Nature of Science and Physics


Figure 1.23 Many mechanical balances, such as double-pan balances, have been replaced by digital scales, which can typically measure the mass of
an object more precisely. Whereas a mechanical balance may only read the mass of an object to the nearest tenth of a gram, many digital scales can
measure the mass of an object up to the nearest thousandth of a gram. (credit: 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 the
correct value for that measurement. For example, let us say that you are measuring the length of standard computer paper. The
packaging in which you purchased the paper states that it is 11.0 inches long. You measure the length of the paper three times
and obtain the following measurements: 11.1 in., 11.2 in., and 10.9 in. These measurements are quite accurate because they are
very close to the correct value of 11.0 inches. In contrast, if you had obtained a measurement of 12 inches, your measurement
would not be very accurate.


The precision of a measurement system is refers to how close the agreement is between repeated measurements (which are
repeated under the same conditions). Consider the example of the paper measurements. The precision of the measurements
refers to the spread of the measured values. One way to analyze the precision of the measurements would be to determine the


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range, or difference, between the lowest and the highest measured values. In that case, the lowest value was 10.9 in. and the
highest value was 11.2 in. Thus, the measured values deviated from each other by at most 0.3 in. These measurements were
relatively precise because they did not vary too much in value. However, if the measured values had been 10.9, 11.1, and 11.9,
then the measurements would not be very precise because there would be significant variation from one measurement to
another.


The measurements in the paper example are both accurate and precise, but in some cases, measurements are accurate but not
precise, or they are precise but not accurate. Let us consider an example of a GPS system that is attempting to locate the
position of a restaurant in a city. Think of the restaurant location as existing at the center of a bull’s-eye target, and think of each
GPS attempt to locate the restaurant as a black dot. In Figure 1.24, you can see that the GPS measurements are spread out far
apart 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 1.25, the GPS measurements are concentrated
quite closely to one another, but they are far away from the target location. This indicates a high precision, low accuracy
measuring system.


Figure 1.24 A GPS system attempts to locate a restaurant at the center of the bull’s-eye. The black dots represent each attempt to pinpoint the location
of the restaurant. The dots are spread out quite far apart from one another, indicating low precision, but they are each rather close to the actual location
of the restaurant, indicating high accuracy. (credit: Dark Evil)


Figure 1.25 In this figure, the dots are concentrated rather closely to one another, indicating high precision, but they are rather far away from the actual
location of the restaurant, indicating low accuracy. (credit: Dark Evil)


Accuracy, Precision, and Uncertainty
The degree of accuracy and precision of a measuring system are related to the uncertainty in the measurements. Uncertainty is
a quantitative measure of how much your measured values deviate from a standard or expected value. If your measurements
are not very accurate or precise, then the uncertainty of your values will be very high. In more general terms, uncertainty can be
thought of as a disclaimer for your measured values. For example, if someone asked you to provide the mileage on your car, you
might say that it is 45,000 miles, plus or minus 500 miles. The plus or minus amount is the uncertainty in your value. That is, you
are indicating that the actual mileage of your car might be as low as 44,500 miles or as high as 45,500 miles, or anywhere in
between. All measurements contain some amount of uncertainty. In our example of measuring the length of the paper, we might
say that the length of the paper is 11 in., plus or minus 0.2 in. The uncertainty in a measurement, A , is often denoted as δA
(“delta A ”), so the measurement result would be recorded as A ± δA . In our paper example, the length of the paper could be
expressed as 11 in. ± 0.2.


The factors contributing to uncertainty in a measurement include:


1. Limitations of the measuring device,


2. The skill of the person making the measurement,


3. Irregularities in the object being measured,


4. Any other factors that affect the outcome (highly dependent on the situation).


Chapter 1 | Introduction: The Nature of Science and Physics 21




In our example, such factors contributing to the uncertainty could be the following: the smallest division on the ruler is 0.1 in., the
person using the ruler has bad eyesight, or one side of the paper is slightly longer than the other. At any rate, the uncertainty in a
measurement must be based on a careful consideration of all the factors that might contribute and their possible effects.


Making Connections: Real-World Connections – Fevers or Chills?


Uncertainty is a critical piece of information, both in physics and in many other real-world applications. Imagine you are
caring for a sick child. You suspect the child has a fever, so you check his or her temperature with a thermometer. What if
the uncertainty of the thermometer were 3.0ºC ? If the child’s temperature reading was 37.0ºC (which is normal body
temperature), the “true” temperature could be anywhere from a hypothermic 34.0ºC to a dangerously high 40.0ºC . A
thermometer with an uncertainty of 3.0ºC would be useless.


Percent Uncertainty


One method of expressing uncertainty is as a percent of the measured value. If a measurement A is expressed with
uncertainty, δA , the percent uncertainty (%unc) is defined to be


(1.8)% unc =δA
A
×100%.


Example 1.2 Calculating Percent Uncertainty: A Bag of Apples


A grocery store sells 5-lb bags of apples. You purchase four bags over the course of a month and weigh the apples each
time. You 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


You determine that the weight of the 5-lb bag has an uncertainty of ±0.4 lb . What is the percent uncertainty of the bag’s
weight?


Strategy


First, observe that the expected value of the bag’s weight, A , is 5 lb. The uncertainty in this value, δA , is 0.4 lb. We can
use the following equation to determine the percent uncertainty of the weight:


(1.9)% unc =δA
A
×100%.


Solution


Plug the known values into the equation:


(1.10)% unc =0.4 lb5 lb ×100% = 8%.


Discussion


We can conclude that the weight of the apple bag is 5 lb ± 8% . Consider how this percent uncertainty would change if the
bag of apples were half as heavy, but the uncertainty in the weight remained the same. Hint for future calculations: when
calculating percent uncertainty, always remember that you must multiply the fraction by 100%. If you do not do this, you will
have a decimal quantity, not a percent value.


22 Chapter 1 | Introduction: The Nature of Science and Physics


Uncertainties in Calculations


There is an uncertainty in anything calculated from measured quantities. For example, the area of a floor calculated from
measurements of its length and width has an uncertainty because the length and width have uncertainties. How big is the
uncertainty in something you calculate by multiplication or division? If the measurements going into the calculation have small
uncertainties (a few percent or less), then the method of adding percents can be used for multiplication or division. This
method says that the percent uncertainty in a quantity calculated by multiplication or division is the sum of the percent
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 , which we round to 0.4 m2 since the area of the floor is given to a tenth of a square
meter.)


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Check Your Understanding


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


Solution
No, the uncertainty in the stopwatch is too great to effectively differentiate between the sprint times.


Precision of Measuring Tools and Significant Figures
An important factor in the accuracy and precision of measurements involves the precision of the measuring tool. In general, a
precise measuring tool is one that can measure values in very small increments. For example, a standard ruler can measure
length to the nearest millimeter, while a caliper can measure length to the nearest 0.01 millimeter. The caliper is a more precise
measuring tool because it can measure extremely small differences in length. The more precise the measuring tool, the more
precise and accurate the measurements can be.


When we express measured values, we can only list as many digits as we initially measured with our measuring tool. For
example, if you use a standard ruler to measure the length of a stick, you may measure it to be 36.7 cm . You could not express
this value as 36.71 cm because your measuring tool was not precise enough to measure a hundredth of a centimeter. It should
be noted that the last digit in a measured value has been estimated in some way by the person performing the measurement.
For example, the person measuring the length of a stick with a ruler notices that the stick length seems to be somewhere in
between 36.6 cm and 36.7 cm , and he or she must estimate the value of the last digit. Using the method of significant
figures, the rule is that the last digit written down in a measurement is the first digit with some uncertainty. In order to determine
the number of significant digits in a value, start with the first measured value at the left and count the number of digits through the
last digit written on the right. For example, the measured value 36.7 cm has three digits, or significant figures. Significant
figures indicate the precision of a measuring tool that was used to measure a value.


Zeros


Special consideration is given to zeros when counting significant figures. The zeros in 0.053 are not significant, because they are
only placekeepers that locate the decimal point. There are two significant figures in 0.053. The zeros in 10.053 are not
placekeepers but are significant—this number has five significant figures. The zeros in 1300 may or may not be significant
depending on the style of writing numbers. They could mean the number is known to the last digit, or they could be
placekeepers. So 1300 could have two, three, or four significant figures. (To avoid this ambiguity, write 1300 in scientific
notation.) Zeros are significant except when they serve only as placekeepers.


Check Your Understanding


Determine the number of significant figures in the following measurements:


a. 0.0009


b. 15,450.0


c. 6×103


d. 87.990


e. 30.42


Solution


(a) 1; the zeros in this number are placekeepers that indicate the decimal point


(b) 6; here, the zeros indicate that a measurement was made to the 0.1 decimal point, so the zeros are significant


(c) 1; the value 103 signifies the decimal place, not the number of measured values
(d) 5; the final zero indicates that a measurement was made to the 0.001 decimal point, so it is significant


(e) 4; any zeros located in between significant figures in a number are also significant


Significant Figures in Calculations


When combining measurements with different degrees of accuracy and precision, the number of significant digits in the final
answer can be no greater than the number of significant digits in the least precise measured value. There are two different rules,
one for multiplication and division and the other for addition and subtraction, as discussed below.


Chapter 1 | Introduction: The Nature of Science and Physics 23




1. For multiplication and division: The result should have the same number of significant figures as the quantity having the
least significant figures entering into the calculation. For example, the area of a circle can be calculated from its radius using


A = πr2 . Let us see how many significant figures the area has if the radius has only two—say, r = 1.2 m . Then,
(1.11)A = πr2 = (3.1415927...)×(1.2 m)2 = 4.5238934 m2


is what you would get using a calculator that has an eight-digit output. But because the radius has only two significant figures, it
limits the calculated quantity to two significant figures or


(1.12)A=4.5 m2,


even though 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 that you buy 7.56-kg of potatoes in a grocery store as measured with a scale with precision 0.01 kg. Then you drop off
6.052-kg of potatoes at your laboratory as measured by a scale with precision 0.001 kg. Finally, you go home and add 13.7 kg of
potatoes as measured by a bathroom scale with precision 0.1 kg. How many kilograms of potatoes do you now have, and how
many significant figures are appropriate in the answer? The mass is found by simple addition and subtraction:


(1.13)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 significant figures
are used in all worked examples. You will note that 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 also taken that
the number of significant figures is reasonable for the situation posed. In some topics, particularly in optics, more accurate
numbers are needed and more than three significant figures will be used. Finally, if a number is exact, such as the two in the
formula for the circumference of a circle, c = 2πr , it does not affect the number of significant figures in a calculation.


Check Your Understanding


Perform the following calculations and express your answer using the correct number of significant digits.


(a) A woman has two bags weighing 13.5 pounds and one bag with a weight of 10.2 pounds. What is the total weight of the
bags?


(b) The force F on an object is equal to its mass m multiplied by its acceleration a . If a wagon with mass 55 kg


accelerates at a rate of 0.0255 m/s2 , what is the force on the wagon? (The unit of force is called the newton, and it is
expressed with the symbol N.)


Solution


(a) 37.2 pounds; Because the number of bags is an exact value, it is not considered in the significant figures.


(b) 1.4 N; Because the value 55 kg has only two significant figures, the final value must also contain two significant figures.


PhET Explorations: Estimation


Explore size estimation in one, two, and three dimensions! Multiple levels of difficulty allow for progressive skill improvement.


Figure 1.26 Estimation (http://


/content/m42120/1.7/estimation_en.jar)


24 Chapter 1 | Introduction: The Nature of Science and Physics


1.4 Approximation
On many occasions, physicists, other scientists, and engineers need to make approximations or “guesstimates” for a particular
quantity. What is the distance to a certain destination? What is the approximate density of a given item? About how large a


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current will there be in a circuit? Many approximate numbers are based on formulae in which the input quantities are known only
to a limited accuracy. As you develop problem-solving skills (that can be applied to a variety of fields through a study of physics),
you will also develop skills at approximating. You will develop these skills through thinking more quantitatively, and by being
willing to take risks. As with any endeavor, experience helps, as well as familiarity with units. These approximations allow us to
rule out certain scenarios or unrealistic numbers. Approximations also allow us to challenge others and guide us in our
approaches to our scientific world. Let us do two examples to illustrate this concept.


Example 1.3 Approximate the Height of a Building


Can you approximate the height of one of the buildings on your campus, or in your neighborhood? Let us make an
approximation based upon the height of a person. In this example, we will calculate the height of a 39-story building.


Strategy


Think about the average height of an adult male. We can approximate the height of the building by scaling up from the
height of a person.


Solution


Based on information in the example, we know there are 39 stories in the building. If we use the fact that the height of one
story is approximately equal to about the length of two adult humans (each human is about 2-m tall), then we can estimate
the total height of the building to be


(1.14)2 m
1 person×


2 person
1 story ×39 stories = 156 m.


Discussion


You can use known quantities to determine an approximate measurement of unknown quantities. If your hand measures 10
cm across, how many hand lengths equal the width of your desk? What other measurements can you approximate besides
length?


Example 1.4 Approximating Vast Numbers: a Trillion Dollars


Figure 1.27 A bank stack contains one-hundred $100 bills, and is worth $10,000. How many bank stacks make up a trillion dollars? (credit:
Andrew Magill)


The U.S. federal deficit in the 2008 fiscal year was a little greater than $10 trillion. Most of us do not have any concept of
how much even one trillion actually is. Suppose that you were given a trillion dollars in $100 bills. If you made 100-bill stacks
and used them to evenly cover a football field (between the end zones), make an approximation of how high the money pile
would become. (We will use feet/inches rather than meters here because football fields are measured in yards.) One of your
friends says 3 in., while another says 10 ft. What do you think?


Strategy


When you imagine the situation, you probably envision thousands of small stacks of 100 wrapped $100 bills, such as you
might see in movies or at a bank. Since this is an easy-to-approximate quantity, let us start there. We can find the volume of
a stack of 100 bills, find out how many stacks make up one trillion dollars, and then set this volume equal to the area of the
football field multiplied by the unknown height.


Solution


(1) Calculate the volume of a stack of 100 bills. The dimensions of a single bill are approximately 3 in. by 6 in. A stack of 100
of these is about 0.5 in. thick. So the total volume of a stack of 100 bills is:


Chapter 1 | Introduction: The Nature of Science and Physics 25




accuracy:


approximation:


classical physics:


conversion factor:


derived units:


(1.15)volume of stack = length×width×height,
volume of stack = 6 in.×3 in.×0.5 in.,
volume of stack = 9 in.3 .


(2) Calculate the number of stacks. Note that a trillion dollars is equal to $1×1012, and a stack of one-hundred $100


bills is equal to $10,000, or $1×104 . The number of stacks you will have is:


(1.16)$1×1012(a trillion dollars)/ $1×104 per stack = 1×108 stacks.


(3) Calculate the area of a football field in square inches. The area of a football field is 100 yd×50 yd, which gives


5,000 yd2. Because we are working in inches, we need to convert square yards to square inches:


(1.17)Area = 5,000 yd2× 3 ft1 yd×
3 ft
1 yd×


12 in.
1 ft ×


12 in.
1 ft = 6,480,000 in.


2 ,


Area ≈ 6×106 in.2 .


This conversion gives us 6×106 in.2 for the area of the field. (Note that we are using only one significant figure in these
calculations.)


(4) Calculate the total volume of the bills. The volume of all the $100 -bill stacks is


9 in.3 / stack×108 stacks = 9×108 in.3.
(5) Calculate the height. To determine the height of the bills, use the equation:


(1.18)volume of bills = area of field×height of money:


Height of money = volume of billsarea of field ,


Height of money = 9×10
8in.3


6×106in.2
= 1.33×102 in.,


Height of money ≈ 1×102 in. = 100 in.


The height of the money will be about 100 in. high. Converting this value to feet gives


(1.19)100 in.× 1 ft12 in. = 8.33 ft ≈ 8 ft.


Discussion


The final approximate value is much higher than the early estimate of 3 in., but the other early estimate of 10 ft (120 in.) was
roughly correct. How did the approximation measure up to your first guess? What can this exercise tell you in terms of rough
“guesstimates” versus carefully calculated approximations?


Check Your Understanding


Using mental math and your understanding of fundamental units, approximate the area of a regulation basketball court.
Describe the process you used to arrive at your final approximation.


Solution
An average male is about two meters tall. It would take approximately 15 men laid out end to end to cover the length, and


about 7 to cover the width. That gives an approximate area of 420 m2 .


Glossary
the degree to which a measured value agrees with correct value for that measurement


an estimated value based on prior experience and reasoning


physics that was developed from the Renaissance to the end of the 19th century


a ratio expressing how many of one unit are equal to another unit


units that can be calculated using algebraic combinations of the fundamental units


26 Chapter 1 | Introduction: The Nature of Science and Physics


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English units:


fundamental units:


kilogram:


law:


meter:


method of adding percents:


metric system:


model:


modern physics:


order of magnitude:


percent uncertainty:


physical quantity :


physics:


precision:


quantum mechanics:


relativity:


scientific method:


second:


SI units :


significant figures:


theory:


uncertainty:


units :


system of measurement used in the United States; includes units of measurement such as feet, gallons, and
pounds


units that can only be expressed relative to the procedure used to measure them


the SI unit for mass, abbreviated (kg)


a description, using concise language or a mathematical formula, a generalized pattern in nature that is supported by
scientific evidence and repeated experiments


the SI unit for length, abbreviated (m)


the percent uncertainty in a quantity calculated by multiplication or division is the sum of the
percent uncertainties in the items used to make the calculation


a system in which values can be calculated in factors of 10


representation of something that is often too difficult (or impossible) to display directly


the study of relativity, quantum mechanics, or both


refers to 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


a characteristic or property of an object that can be measured or calculated from other measurements


the science concerned with describing the interactions of energy, matter, space, and time; it is especially interested
in what fundamental mechanisms underlie every phenomenon


the degree to which repeated measurements agree with each other


the study of objects smaller than can be seen with a microscope


the study of objects moving at speeds greater than about 1% of the speed of light, or of objects being affected by a
strong gravitational field


a method that typically begins with an observation and question that the scientist will research; next, the
scientist typically performs some research about the topic and then devises a hypothesis; then, the scientist will test the
hypothesis by performing an experiment; finally, the scientist analyzes the results of the experiment and draws a
conclusion


the SI unit for time, abbreviated (s)


the international system of units that scientists in most countries have agreed to use; includes units such as meters,
liters, and grams


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


an explanation for patterns in nature that is supported by scientific evidence and verified multiple times by various
groups of researchers


a quantitative measure of how much your measured values deviate from a standard or expected value


a standard used for expressing and comparing measurements


Section Summary


1.1 Physics: An Introduction
• Science seeks to discover and describe the underlying order and simplicity in nature.
• Physics is the most basic of the sciences, concerning itself with energy, matter, space and time, and their interactions.
• Scientific laws and theories express the general truths of nature and the body of knowledge they encompass. These laws


of nature are rules that all natural processes appear to follow.


1.2 Physical Quantities and Units
• Physical quantities are a characteristic or property of an object that can be measured or calculated from other


measurements.
• Units are standards for expressing and comparing the measurement of physical quantities. All units can be expressed as


combinations of four fundamental units.


Chapter 1 | Introduction: The Nature of Science and Physics 27




• The four fundamental units we will use in this text are the meter (for length), the kilogram (for mass), the second (for time),
and the ampere (for electric current). These units are part of the metric system, which uses powers of 10 to relate quantities
over the vast ranges encountered in nature.


• The four fundamental units are abbreviated as follows: meter, m; kilogram, kg; second, s; and ampere, A. The metric
system also uses a standard set of prefixes to denote each order of magnitude greater than or lesser than the fundamental
unit itself.


• Unit conversions involve changing a value expressed in one type of unit to another type of unit. This is done by using
conversion factors, which are ratios relating equal quantities of different units.


1.3 Accuracy, Precision, and Significant Figures
• Accuracy of a measured value refers to how close a measurement is to the correct value. The uncertainty in a


measurement is an estimate of the amount by which the measurement result may differ from this value.
• Precision of measured values refers to how close the agreement is between repeated measurements.
• The precision of a measuring tool is related to the size of its measurement increments. The smaller the measurement


increment, 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 the least


precise value.
• When adding or subtracting measured values, the final answer cannot contain more decimal places than the least precise


value.


1.4 Approximation
Scientists often approximate the values of quantities to perform calculations and analyze systems.


Conceptual Questions


1.1 Physics: An Introduction
1. Models are particularly useful in relativity and quantum mechanics, where conditions are outside those normally encountered
by humans. What is a model?


2. How does a model differ from a theory?


3. If two different theories describe experimental observations equally well, can one be said to be more valid than 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 or observation is to be believed. Will the criteria necessarily be as strict for
an expected result as for an unexpected result?


6. Can the validity of a model be limited, or must it be universally valid? How does this compare to the required validity of a
theory or a law?


7. Classical physics is a good approximation to modern physics under certain circumstances. What are they?


8.When is it necessary to use relativistic quantum mechanics?


9. Can classical physics be used to accurately describe a satellite moving at a speed of 7500 m/s? Explain why or why not.


1.2 Physical Quantities and Units
10. Identify some advantages of metric units.


1.3 Accuracy, Precision, and Significant Figures
11.What is the relationship between the accuracy and uncertainty of a measurement?


12. Prescriptions for vision correction are given in units called diopters (D). Determine the meaning of that unit. Obtain
information (perhaps by calling an optometrist or performing an internet search) on the minimum uncertainty with which
corrections in diopters are determined and the accuracy with which corrective lenses can be produced. Discuss the sources of
uncertainties in both the prescription and accuracy in the manufacture of lenses.


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Problems & Exercises


1.2 Physical Quantities and Units
1. The speed limit on some interstate highways is roughly 100
km/h. (a) What is this in meters per second? (b) How many
miles per hour is this?


2. A car is traveling at a speed of 33 m/s . (a) What is its
speed in kilometers per hour? (b) Is it exceeding the
90 km/h speed limit?


3. Show that 1.0 m/s = 3.6 km/h . Hint: Show the explicit
steps involved in converting 1.0 m/s = 3.6 km/h.
4. American football is played on a 100-yd-long field,
excluding the end zones. How long is the field in meters?
(Assume that 1 meter equals 3.281 feet.)


5. Soccer fields vary in size. A large soccer field is 115 m long
and 85 m wide. What are its dimensions in feet and inches?
(Assume that 1 meter equals 3.281 feet.)


6.What is the height in meters of a person who is 6 ft 1.0 in.
tall? (Assume that 1 meter equals 39.37 in.)


7. Mount Everest, at 29,028 feet, is the tallest mountain on
the Earth. What is its height in kilometers? (Assume that 1
kilometer equals 3,281 feet.)


8. The speed of sound is measured to be 342 m/s on a
certain day. What is this in km/h?


9. Tectonic plates are large segments of the Earth’s crust that
move slowly. Suppose that one such plate has an average
speed of 4.0 cm/year. (a) What distance does it move in 1 s at
this speed? (b) What is its speed in kilometers per million
years?


10. (a) Refer to Table 1.3 to determine the average distance
between the Earth and the Sun. Then calculate the average
speed of the Earth in its orbit in kilometers per second. (b)
What is this in meters per second?


1.3 Accuracy, Precision, and Significant
Figures
Express your answers to problems in this section to the
correct number of significant figures and proper units.


11. Suppose that your bathroom scale reads your mass as 65
kg with a 3% uncertainty. What is the uncertainty in your mass
(in kilograms)?


12. A good-quality measuring tape can be off by 0.50 cm over
a distance of 20 m. What is its percent uncertainty?


13. (a) A car speedometer has a 5.0% uncertainty. What is
the range of possible speeds when it reads 90 km/h ? (b)
Convert this range to miles per hour. (1 km = 0.6214 mi)


14. An infant’s pulse rate is measured to be 130 ± 5 beats/
min. What is the percent uncertainty in this measurement?


15. (a) Suppose that a person has an average heart rate of
72.0 beats/min. How many beats does he or she have in 2.0
y? (b) In 2.00 y? (c) In 2.000 y?


16. A can contains 375 mL of soda. How much is left after
308 mL is removed?


17. State how many significant figures are proper in the
results of the following calculations: (a)


(106.7)(98.2) / (46.210)(1.01) (b) (18.7)2 (c)

⎝1.60×10


−19⎞
⎠(3712) .


18. (a) How many significant figures are in the numbers 99
and 100? (b) If the uncertainty in each number is 1, what is
the percent uncertainty in each? (c) Which is a more
meaningful way to express the accuracy of these two
numbers, significant figures or percent uncertainties?


19. (a) If your speedometer has an uncertainty of 2.0 km/h
at 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?


20. (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 the
uncertainty in a blood pressure measurement of 80 mm Hg
?


21. A person measures his or her heart rate by counting the
number 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 per minute?


22.What is the area of a circle 3.102 cm in diameter?
23. If a marathon runner averages 9.5 mi/h, how long does it
take him or her to run a 26.22-mi marathon?


24. A marathon runner completes a 42.188-km course in
2 h , 30 min, and 12 s . There is an uncertainty of 25 m in
the distance traveled and an uncertainty of 1 s in the elapsed
time. (a) Calculate the percent uncertainty in the distance. (b)
Calculate the uncertainty in the elapsed time. (c) What is the
average speed in meters per second? (d) What is the
uncertainty in the average speed?


25. The sides of a small rectangular box are measured to be
1.80 ± 0.01 cm , 2.05 ± 0.02 cm, and 3.1 ± 0.1 cm
long. Calculate its volume and uncertainty in cubic
centimeters.


26.When non-metric units were used in the United Kingdom,
a unit of mass called the pound-mass (lbm) was employed,
where 1 lbm = 0.4539 kg . (a) If there is an uncertainty of
0.0001 kg in the pound-mass unit, what is its percent
uncertainty? (b) Based on that percent uncertainty, what
mass in pound-mass has an uncertainty of 1 kg when
converted to kilograms?


27. The length and width of a rectangular room are measured
to be 3.955 ± 0.005 m and 3.050 ± 0.005 m . Calculate
the area of the room and its uncertainty in square meters.


28. A car engine moves a piston with a circular cross section
of 7.500 ± 0.002 cm diameter a distance of
3.250 ± 0.001 cm to compress the gas in the cylinder. (a)
By what amount is the gas decreased in volume in cubic
centimeters? (b) Find the uncertainty in this volume.


1.4 Approximation
29. How many heartbeats are there in a lifetime?


Chapter 1 | Introduction: The Nature of Science and Physics 29




30. A generation is about one-third of a lifetime.
Approximately how many generations have passed since the
year 0 AD?


31. How many times longer than the mean life of an
extremely unstable atomic nucleus is the lifetime of a human?
(Hint: The lifetime of an unstable atomic nucleus is on the


order of 10−22 s .)
32. Calculate the approximate number of atoms in a
bacterium. Assume that the average mass of an atom in the
bacterium is ten times the mass of a hydrogen atom. (Hint:


The mass of a hydrogen atom is on the order of 10−27 kg


and the mass of a bacterium is on the order of 10−15 kg. )


Figure 1.28 This color-enhanced photo shows Salmonella typhimurium
(red) attacking human cells. These bacteria are commonly known for
causing foodborne illness. Can you estimate the number of atoms in
each bacterium? (credit: Rocky Mountain Laboratories, NIAID, NIH)


33. Approximately how many atoms thick is a cell membrane,
assuming all atoms there average about twice the size of a
hydrogen atom?


34. (a) What fraction of Earth’s diameter is the greatest ocean
depth? (b) The greatest mountain height?


35. (a) Calculate the number of cells in a hummingbird
assuming the mass of an average cell is ten times the mass
of a bacterium. (b) Making the same assumption, how many
cells are there in a human?


36. Assuming one nerve impulse must end before another
can begin, what is the maximum firing rate of a nerve in
impulses per second?


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2 KINEMATICS


Figure 2.1 The motion of an American kestrel through the air can be described by the bird’s displacement, speed, velocity, and acceleration. When it
flies in a straight line without any change in direction, its motion is said to be one dimensional. (credit: Vince Maidens, Wikimedia Commons)


Chapter Outline
2.1. Displacement


• Define position, displacement, distance, and distance traveled.
• Explain the relationship between position and displacement.
• Distinguish between displacement and distance traveled.
• Calculate displacement and distance given initial position, final position, and the path between the two.


2.2. Vectors, Scalars, and Coordinate Systems
• Define and distinguish between scalar and vector quantities.
• Assign a coordinate system for a scenario involving one-dimensional motion.


2.3. Time, Velocity, and Speed
• Explain the relationships between instantaneous velocity, average velocity, instantaneous speed, average speed,


displacement, and time.
• Calculate velocity and speed given initial position, initial time, final position, and final time.
• Derive a graph of velocity vs. time given a graph of position vs. time.
• Interpret a graph of velocity vs. time.


2.4. Acceleration
• Define and distinguish between instantaneous acceleration, average acceleration, and deceleration.
• Calculate acceleration given initial time, initial velocity, final time, and final velocity.


2.5. Motion Equations for Constant Acceleration in One Dimension
• Calculate displacement of an object that is not accelerating, given initial position and velocity.
• Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time.
• Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and


acceleration.
2.6. Problem-Solving Basics for One-Dimensional Kinematics


• Apply problem-solving steps and strategies to solve problems of one-dimensional kinematics.
• Apply strategies to determine whether or not the result of a problem is reasonable, and if not, determine the cause.


2.7. Falling Objects
• Describe the effects of gravity on objects in motion.
• Describe the motion of objects that are in free fall.
• Calculate the position and velocity of objects in free fall.


2.8. Graphical Analysis of One-Dimensional Motion
• Describe a straight-line graph in terms of its slope and y-intercept.
• Determine average velocity or instantaneous velocity from a graph of position vs. time.
• Determine average or instantaneous acceleration from a graph of velocity vs. time.
• Derive a graph of velocity vs. time from a graph of position vs. time.
• Derive a graph of acceleration vs. time from a graph of velocity vs. time.


Chapter 2 | Kinematics 31




Introduction to One-Dimensional Kinematics
Objects are in motion everywhere we look. Everything from a tennis game to a space-probe flyby of the planet Neptune involves
motion. When you are resting, your heart moves blood through your veins. And even in inanimate objects, there is continuous
motion in the vibrations of atoms and molecules. Questions about motion are interesting in and of themselves: How long will it
take for a space probe to get to Mars? Where will a football land if it is thrown at a certain angle? But an understanding of motion
is also key to understanding other concepts in physics. An understanding of acceleration, for example, is crucial to the study of
force.


Our formal study of physics begins with kinematics which is defined as the study of motion without considering its causes. The
word “kinematics” comes from a Greek term meaning motion and is related to other English words such as “cinema” (movies)
and “kinesiology” (the study of human motion). In one-dimensional kinematics and Two-Dimensional Kinematics we will study
only the motion of a football, for example, without worrying about what forces cause or change its motion. Such considerations
come in other chapters. In this chapter, we examine the simplest type of motion—namely, motion along a straight line, or one-
dimensional motion. In Two-Dimensional Kinematics, we apply concepts developed here to study motion along curved paths
(two- and three-dimensional motion); for example, that of a car rounding a curve.


2.1 Displacement


Figure 2.2 These cyclists in Vietnam can be described by their position relative to buildings and a canal. Their motion can be described by their change
in position, or displacement, in the frame of reference. (credit: Suzan Black, Fotopedia)


Position
In order to describe the motion of an object, you must first be able to describe its position—where it is at any particular time.
More precisely, you need to specify its position relative to a convenient reference frame. Earth is often used as a reference
frame, and we often describe the position of an object as it relates to stationary objects in that reference frame. For example, a
rocket launch would be described in terms of the position of the rocket with respect to the Earth as a whole, while a professor’s
position could be described in terms of where she is in relation to the nearby white board. (See Figure 2.3.) In other cases, we
use reference frames that are not stationary but are in motion relative to the Earth. To describe the position of a person in an
airplane, for example, we use the airplane, not the Earth, as the reference frame. (See Figure 2.4.)


Displacement
If an object moves relative to a reference frame (for example, if a professor moves to the right relative to a white board or a
passenger moves toward the rear of an airplane), then the object’s position changes. This change in position is known as
displacement. The word “displacement” implies that an object has moved, or has been displaced.


Displacement


Displacement is the change in position of an object:


(2.1)Δx = xf − x0,


where Δx is displacement, xf is the final position, and x0 is the initial position.


In this text the upper case Greek letter Δ (delta) always means “change in” whatever quantity follows it; thus, Δx means
change in position. Always solve for displacement by subtracting initial position x0 from final position xf .


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Note that the SI unit for displacement is the meter (m) (see Physical Quantities and Units), but sometimes kilometers, miles,
feet, and other units of length are used. Keep in mind that when units other than the meter are used in a problem, you may need
to convert them into meters to complete the calculation.


Figure 2.3 A professor paces left and right while lecturing. Her position relative to Earth is given by x . The +2.0 m displacement of the professor
relative to Earth is represented by an arrow pointing to the right.


Figure 2.4 A passenger moves from his seat to the back of the plane. His location relative to the airplane is given by x . The −4.0-m displacement
of the passenger relative to the plane is represented by an arrow toward the rear of the plane. Notice that the arrow representing his displacement is
twice as long as the arrow representing the displacement of the professor (he moves twice as far) in Figure 2.3.


Note that displacement has a direction as well as a magnitude. The professor’s displacement is 2.0 m to the right, and the airline
passenger’s displacement is 4.0 m toward the rear. In one-dimensional motion, direction can be specified with a plus or minus
sign. When you begin a problem, you should select which direction is positive (usually that will be to the right or up, but you are
free to select positive as being any direction). The professor’s initial position is x0 = 1.5 m and her final position is
xf = 3.5 m . Thus her displacement is


(2.2)Δx = xf−x0 = 3.5 m − 1.5 m = + 2.0 m.


In this coordinate system, motion to the right is positive, whereas motion to the left is negative. Similarly, the airplane
passenger’s initial position is x0 = 6.0 m and his final position is xf = 2.0 m , so his displacement is


(2.3)Δx = xf−x0 = 2.0 m − 6.0 m = −4.0 m.


Chapter 2 | Kinematics 33




His displacement is negative because his motion is toward the rear of the plane, or in the negative x direction in our coordinate
system.


Distance
Although displacement is described in terms of direction, distance is not. Distance is defined to be the magnitude or size of
displacement between two positions. Note that the distance between two positions is not the same as the distance traveled
between them. Distance traveled is the total length of the path traveled between two positions. Distance has no direction and,
thus, no sign. For example, the distance the professor walks is 2.0 m. The distance the airplane passenger walks is 4.0 m.


Misconception Alert: Distance Traveled vs. Magnitude of Displacement


It is important to note that the distance traveled, however, can be greater than the magnitude of the displacement (by
magnitude, we mean just the size of the displacement without regard to its direction; that is, just a number with a unit). For
example, the professor could pace back and forth many times, perhaps walking a distance of 150 m during a lecture, yet still
end up only 2.0 m to the right of her starting point. In this case her displacement would be +2.0 m, the magnitude of her
displacement would be 2.0 m, but the distance she traveled would be 150 m. In kinematics we nearly always deal with
displacement and magnitude of displacement, and almost never with distance traveled. One way to think about this is to
assume you marked the start of the motion and the end of the motion. The displacement is simply the difference in the
position of the two marks and is independent of the path taken in traveling between the two marks. The distance traveled,
however, is the total length of the path taken between the two marks.


Check Your Understanding


A cyclist rides 3 km west and then turns around and rides 2 km east. (a) What is her displacement? (b) What distance does
she ride? (c) What is the magnitude of her displacement?


Solution


Figure 2.5


(a) The rider’s displacement is Δx = xf − x0 = −1 km . (The displacement is negative because we take east to be
positive and west to be negative.)


(b) The distance traveled is 3 km + 2 km = 5 km .


(c) The magnitude of the displacement is 1 km .


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2.2 Vectors, Scalars, and Coordinate Systems


Figure 2.6 The motion of this Eclipse Concept jet can be described in terms of the distance it has traveled (a scalar quantity) or its displacement in a
specific direction (a vector quantity). In order to specify the direction of motion, its displacement must be described based on a coordinate system. In
this case, it may be convenient to choose motion toward the left as positive motion (it is the forward direction for the plane), although in many cases,
the x -coordinate runs from left to right, with motion to the right as positive and motion to the left as negative. (credit: Armchair Aviator, Flickr)


What is the difference between distance and displacement? Whereas displacement is defined by both direction and magnitude,
distance is defined only by magnitude. Displacement is an example of a vector quantity. Distance is an example of a scalar
quantity. A vector is any quantity with both magnitude and direction. Other examples of vectors include a velocity of 90 km/h east
and a force of 500 newtons straight down.


The direction of a vector in one-dimensional motion is given simply by a plus ( + ) or minus ( − ) sign. Vectors are
represented graphically by arrows. An arrow used to represent a vector has a length proportional to the vector’s magnitude (e.g.,
the larger the magnitude, the longer the length of the vector) and points in the same direction as the vector.


Some physical quantities, like distance, either have no direction or none is specified. A scalar is any quantity that has a
magnitude, but no direction. For example, a 20ºC temperature, the 250 kilocalories (250 Calories) of energy in a candy bar, a
90 km/h speed limit, a person’s 1.8 m height, and a distance of 2.0 m are all scalars—quantities with no specified direction. Note,
however, that a scalar can be negative, such as a −20ºC temperature. In this case, the minus sign indicates a point on a scale
rather than a direction. Scalars are never represented by arrows.


Coordinate Systems for One-Dimensional Motion
In order to describe the direction of a vector quantity, you must designate a coordinate system within the reference frame. For
one-dimensional motion, this is a simple coordinate system consisting of a one-dimensional coordinate line. In general, when
describing horizontal motion, motion to the right is usually considered positive, and motion to the left is considered negative. With
vertical motion, motion up is usually positive and motion down is negative. In some cases, however, as with the jet in Figure 2.6,
it can be more convenient to switch the positive and negative directions. For example, if you are analyzing the motion of falling
objects, it can be useful to define downwards as the positive direction. If people in a race are running to the left, it is useful to
define left as the positive direction. It does not matter as long as the system is clear and consistent. Once you assign a positive
direction and start solving a problem, you cannot change it.


Figure 2.7 It is usually convenient to consider motion upward or to the right as positive ( + ) and motion downward or to the left as negative ( − ) .


Chapter 2 | Kinematics 35




Check Your Understanding


A person’s speed can stay the same as he or she rounds a corner and changes direction. Given this information, is speed a
scalar or a vector quantity? Explain.


Solution
Speed is a scalar quantity. It does not change at all with direction changes; therefore, it has magnitude only. If it were a
vector quantity, it would change as direction changes (even if its magnitude remained constant).


2.3 Time, Velocity, and Speed


Figure 2.8 The motion of these racing snails can be described by their speeds and their velocities. (credit: tobitasflickr, Flickr)


There is more to motion than distance and displacement. Questions such as, “How long does a foot race take?” and “What was
the runner’s speed?” cannot be answered without an understanding of other concepts. In this section we add definitions of time,
velocity, and speed to expand our description of motion.


Time
As discussed in Physical Quantities and Units, the most fundamental physical quantities are defined by how they are
measured. This is the case with time. Every measurement of time involves measuring a change in some physical quantity. It may
be a number on a digital clock, a heartbeat, or the position of the Sun in the sky. In physics, the definition of time is simple— time
is change, or the interval over which change occurs. It is impossible to know that time has passed unless something changes.


The amount of time or change is calibrated by comparison with a standard. The SI unit for time is the second, abbreviated s. We
might, for example, observe that a certain pendulum makes one full swing every 0.75 s. We could then use the pendulum to
measure time by counting its swings or, of course, by connecting the pendulum to a clock mechanism that registers time on a
dial. This allows us to not only measure the amount of time, but also to determine a sequence of events.


How does time relate to motion? We are usually interested in elapsed time for a particular motion, such as how long it takes an
airplane passenger to get from his seat to the back of the plane. To find elapsed time, we note the time at the beginning and end
of the motion and subtract the two. For example, a lecture may start at 11:00 A.M. and end at 11:50 A.M., so that the elapsed
time would be 50 min. Elapsed time Δt is the difference between the ending time and beginning time,


(2.4)Δt = tf − t0,


where Δt is the change in time or elapsed time, tf is the time at the end of the motion, and t0 is the time at the beginning of
the motion. (As usual, the delta symbol, Δ , means the change in the quantity that follows it.)
Life is simpler if the beginning time t0 is taken to be zero, as when we use a stopwatch. If we were using a stopwatch, it would


simply read zero at the start of the lecture and 50 min at the end. If t0 = 0 , then Δt = tf ≡ t .


In this text, for simplicity’s sake,


• motion starts at time equal to zero (t0 = 0)
• the symbol t is used for elapsed time unless otherwise specified (Δt = tf ≡ t)


Velocity
Your notion of velocity is probably the same as its scientific definition. You know that if you have a large displacement in a small
amount of time you have a large velocity, and that velocity has units of distance divided by time, such as miles per hour or
kilometers per hour.


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Average Velocity


Average velocity is displacement (change in position) divided by the time of travel,


(2.5)v- = ΔxΔt =
xf − x0
tf − t0


,


where v- is the average (indicated by the bar over the v ) velocity, Δx is the change in position (or displacement), and xf
and x0 are the final and beginning positions at times tf and t0 , respectively. If the starting time t0 is taken to be zero,


then the average velocity is simply


(2.6)v- = Δxt .


Notice that this definition indicates that velocity is a vector because displacement is a vector. It has both magnitude and direction.
The SI unit for velocity is meters per second or m/s, but many other units, such as km/h, mi/h (also written as mph), and cm/s,
are in common use. Suppose, for example, an airplane passenger took 5 seconds to move −4 m (the negative sign indicates that
displacement is toward the back of the plane). His average velocity would be


(2.7)v- = Δxt =
−4 m
5 s = − 0.8 m/s.


The minus sign indicates the average velocity is also toward the rear of the plane.


The average velocity of an object does not tell us anything about what happens to it between the starting point and ending point,
however. For example, we cannot tell from average velocity whether the airplane passenger stops momentarily or backs up
before he goes to the back of the plane. To get more details, we must consider smaller segments of the trip over smaller time
intervals.


Figure 2.9 A more detailed record of an airplane passenger heading toward the back of the plane, showing smaller segments of his trip.


The smaller the time intervals considered in a motion, the more detailed the information. When we carry this process to its logical
conclusion, we are left with an infinitesimally small interval. Over such an interval, the average velocity becomes the
instantaneous velocity or the velocity at a specific instant. A car’s speedometer, for example, shows the magnitude (but not the
direction) of the instantaneous velocity of the car. (Police give tickets based on instantaneous velocity, but when calculating how
long it will take to get from one place to another on a road trip, you need to use average velocity.) Instantaneous velocity v is
the average velocity at a specific instant in time (or over an infinitesimally small time interval).


Mathematically, finding instantaneous velocity, v , at a precise instant t can involve taking a limit, a calculus operation beyond
the scope of this text. However, under many circumstances, we can find precise values for instantaneous velocity without
calculus.


Speed
In everyday language, most people use the terms “speed” and “velocity” interchangeably. In physics, however, they do not have
the same meaning and they are distinct concepts. One major difference is that speed has no direction. Thus speed is a scalar.
Just as we need to distinguish between instantaneous velocity and average velocity, we also need to distinguish between
instantaneous speed and average speed.


Instantaneous speed is the magnitude of instantaneous velocity. For example, suppose the airplane passenger at one instant
had an instantaneous velocity of −3.0 m/s (the minus meaning toward the rear of the plane). At that same time his instantaneous
speed was 3.0 m/s. Or suppose that at one time during a shopping trip your instantaneous velocity is 40 km/h due north. Your
instantaneous speed at that instant would be 40 km/h—the same magnitude but without a direction. Average speed, however, is
very different from average velocity. Average speed is the distance traveled divided by elapsed time.


Chapter 2 | Kinematics 37




We have noted that distance traveled can be greater than displacement. So average speed can be greater than average velocity,
which is displacement divided by time. For example, if you drive to a store and return home in half an hour, and your car’s
odometer shows the total distance traveled was 6 km, then your average speed was 12 km/h. Your average velocity, however,
was zero, because your displacement for the round trip is zero. (Displacement is change in position and, thus, is zero for a round
trip.) Thus average speed is not simply the magnitude of average velocity.


Figure 2.10 During a 30-minute round trip to the store, the total distance traveled is 6 km. The average speed is 12 km/h. The displacement for the
round trip is zero, since there was no net change in position. Thus the average velocity is zero.


Another way of visualizing the motion of an object is to use a graph. A plot of position or of velocity as a function of time can be
very useful. For example, for this trip to the store, the position, velocity, and speed-vs.-time graphs are displayed in Figure 2.11.
(Note that these graphs depict a very simplified model of the trip. We are assuming that speed is constant during the trip, which
is unrealistic given that we’ll probably stop at the store. But for simplicity’s sake, we will model it with no stops or changes in
speed. We are also assuming that the route between the store and the house is a perfectly straight line.)


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Figure 2.11 Position vs. time, velocity vs. time, and speed vs. time on a trip. Note that the velocity for the return trip is negative.


Making Connections: Take-Home Investigation—Getting a Sense of Speed


If you have spent much time driving, you probably have a good sense of speeds between about 10 and 70 miles per hour.
But what are these in meters per second? What do we mean when we say that something is moving at 10 m/s? To get a
better sense of what these values really mean, do some observations and calculations on your own:


• calculate typical car speeds in meters per second
• estimate jogging and walking speed by timing yourself; convert the measurements into both m/s and mi/h
• determine the speed of an ant, snail, or falling leaf


Chapter 2 | Kinematics 39




Check Your Understanding


A commuter train travels from Baltimore to Washington, DC, and back in 1 hour and 45 minutes. The distance between the
two stations is approximately 40 miles. What is (a) the average velocity of the train, and (b) the average speed of the train in
m/s?


Solution


(a) The average velocity of the train is zero because xf = x0 ; the train ends up at the same place it starts.


(b) The average speed of the train is calculated below. Note that the train travels 40 miles one way and 40 miles back, for a
total distance of 80 miles.


(2.8)distance
time =


80 miles
105 minutes


(2.9)80 miles
105 minutes×


5280 feet
1 mile ×


1 meter
3.28 feet×


1 minute
60 seconds = 20 m/s


2.4 Acceleration


Figure 2.12 A plane decelerates, or slows down, as it comes in for landing in St. Maarten. Its acceleration is opposite in direction to its velocity. (credit:
Steve Conry, Flickr)


In everyday conversation, to accelerate means to speed up. The accelerator in a car can in fact cause it to speed up. The greater
the acceleration, the greater the change in velocity over a given time. The formal definition of acceleration is consistent with
these notions, but more inclusive.


Average Acceleration


Average Acceleration is the rate at which velocity changes,


(2.10)
a- = ΔvΔt =


vf − v0
tf − t0


,


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 m/s divided by time in s, the SI units for acceleration are m/s2 , meters per second squared
or meters per second per second, which literally means by how many meters per second the velocity changes every second.


Recall that velocity is a vector—it has both magnitude and direction. This means that a change in velocity can be a change in
magnitude (or speed), but it can also be a change in direction. For example, if a car turns a corner at constant speed, it is
accelerating because its direction is changing. The quicker you turn, the greater the acceleration. So there is an acceleration
when velocity changes either 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 either
in magnitude or in direction. Acceleration is therefore a change in either 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 of motion.
When an object slows down, its acceleration is opposite to the direction of its motion. This is known as deceleration.


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Figure 2.13 A subway train in Sao Paulo, Brazil, decelerates as it comes into a station. It is accelerating in a direction opposite to its direction of
motion. (credit: Yusuke Kawasaki, Flickr)


Misconception Alert: Deceleration vs. Negative Acceleration


Deceleration always refers to acceleration in the direction opposite to the direction of the velocity. Deceleration always
reduces speed. Negative acceleration, however, is acceleration in the negative direction in the chosen coordinate system.
Negative acceleration may or may not be deceleration, and deceleration may or may not be considered negative
acceleration. For example, consider Figure 2.14.


Chapter 2 | Kinematics 41




Figure 2.14 (a) This car is speeding up as it moves toward the right. It therefore has positive acceleration in our coordinate system. (b) This car is
slowing down as it moves toward the right. Therefore, it has negative acceleration in our coordinate system, because its acceleration is toward the
left. The car is also decelerating: the direction of its acceleration is opposite to its direction of motion. (c) This car is moving toward the left, but
slowing down over time. Therefore, its acceleration is positive in our coordinate system because it is toward the right. However, the car is
decelerating because its acceleration is opposite to its motion. (d) This car is speeding up as it moves toward the left. It has negative acceleration
because it is accelerating toward the left. However, because its acceleration is in the same direction as its motion, it is speeding up (not
decelerating).


Example 2.1 Calculating 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 its average
acceleration?


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Figure 2.15 (credit: Jon Sullivan, PD Photo.org)


Strategy


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


Figure 2.16


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- = ΔvΔt =
vf − v0
tf − t0


.


Solution


1. Identify the knowns. v0 = 0 , vf = −15.0 m/s (the negative sign indicates direction toward the west), Δt = 1.80 s .


2. Find the change in velocity. Since the horse is going from zero to − 15.0 m/s , its change in velocity equals its final
velocity: Δv = vf = −15.0 m/s .


3. Plug in the known values (Δv and Δt ) and solve for the unknown a- .
(2.11)a- = ΔvΔt =


−15.0 m/s
1.80 s = −8.33 m/s


2.


Discussion


The negative sign for acceleration indicates that acceleration is toward the west. An acceleration of 8.33 m/s2 due west
means that the horse increases its velocity by 8.33 m/s due west each second, that is, 8.33 meters per second per second,


which we write as 8.33 m/s2 . This is truly an average acceleration, because the ride is not smooth. We shall see later that
an acceleration of this magnitude would require the rider to hang on with a force nearly equal to his weight.


Instantaneous Acceleration
Instantaneous acceleration a , or the acceleration at a specific instant in time, is obtained by the same process as discussed
for instantaneous velocity in Time, Velocity, and Speed—that is, by considering an infinitesimally small interval of time. How do
we find instantaneous acceleration using only algebra? The answer is that we choose an average acceleration that is
representative of the motion. Figure 2.17 shows graphs of instantaneous acceleration versus time for two very different motions.
In Figure 2.17(a), the acceleration varies slightly and the average over the entire interval is nearly the same as the
instantaneous acceleration at any time. In this case, we should treat this motion as if it had a constant acceleration equal to the


Chapter 2 | Kinematics 43




average (in this case about 1.8 m/s2 ). In Figure 2.17(b), the acceleration varies drastically over time. In such situations it is
best to consider smaller time intervals and choose an average acceleration for each. For example, we could consider motion


over the time intervals from 0 to 1.0 s and from 1.0 to 3.0 s as separate motions with accelerations of +3.0 m/s2 and
–2.0 m/s2 , respectively.


Figure 2.17 Graphs of instantaneous acceleration versus time for two different one-dimensional motions. (a) Here acceleration varies only slightly and
is always in the same direction, since it is positive. The average over the interval is nearly the same as the acceleration at any given time. (b) Here the
acceleration varies greatly, perhaps representing a package on a post office conveyor belt that is accelerated forward and backward as it bumps along.
It is necessary to consider small time intervals (such as from 0 to 1.0 s) with constant or nearly constant acceleration in such a situation.


The next several examples consider the motion of the subway train shown in Figure 2.18. In (a) the shuttle moves to the right,
and in (b) it moves to the left. The examples are designed to further illustrate aspects of motion and to illustrate some of the
reasoning that goes into solving problems.


44 Chapter 2 | Kinematics


Figure 2.18 One-dimensional motion of a subway train considered in Example 2.2, Example 2.3, Example 2.4, Example 2.5, Example 2.6, and
Example 2.7. Here we have chosen the x -axis so that + means to the right and − means to the left for displacements, velocities, and accelerations.
(a) The subway train moves to the right from x0 to xf . Its displacement Δx is +2.0 km. (b) The train moves to the left from x′0 to x′f . Its
displacement Δx′ is −1.5 km . (Note that the prime symbol (′) is used simply to distinguish between displacement in the two different situations.
The distances of travel and the size of the cars are on different scales to fit everything into the diagram.)


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Example 2.2 Calculating Displacement: A Subway Train


What are the magnitude and sign of displacements for the motions of the subway train shown in parts (a) and (b) of Figure
2.18?


Strategy


A drawing with a coordinate system is already provided, so we don’t need to make a sketch, but we should analyze it to
make sure we understand what it is showing. Pay particular attention to the coordinate system. To find displacement, we use
the equation Δx = xf − x0 . This is straightforward since the initial and final positions are given.


Solution


1. Identify the knowns. In the figure we see that xf = 6.70 km and x0 = 4.70 km for part (a), and x′f = 3.75 km and
x′0 = 5.25 km for part (b).


2. Solve for displacement in part (a).


(2.12)Δx = xf − x0 = 6.70 km − 4.70 km=+2.00 km


3. Solve for displacement in part (b).


(2.13)Δx′ = x′f − x′0 = 3.75 km − 5.25 km = − 1.50 km


Discussion


The direction of the motion in (a) is to the right and therefore its displacement has a positive sign, whereas motion in (b) is to
the left and thus has a negative sign.


Example 2.3 Comparing Distance Traveled with Displacement: A Subway Train


What are the distances traveled for the motions shown in parts (a) and (b) of the subway train in Figure 2.18?


Strategy


To answer this question, think about the definitions of distance and distance traveled, and how they are related to
displacement. Distance between two positions is defined to be the magnitude of displacement, which was found in Example
2.2. Distance traveled is the total length of the path traveled between the two positions. (See Displacement.) In the case of
the subway train shown in Figure 2.18, the distance traveled is the same as the distance between the initial and final
positions of the train.


Solution


1. The displacement for part (a) was +2.00 km. Therefore, the distance between the initial and final positions was 2.00 km,
and the distance traveled was 2.00 km.


2. The displacement for part (b) was −1.5 km. Therefore, the distance between the initial and final positions was 1.50 km,
and the distance traveled was 1.50 km.


Discussion


Distance is a scalar. It has magnitude but no sign to indicate direction.


Example 2.4 Calculating Acceleration: A Subway Train Speeding Up


Suppose the train in Figure 2.18(a) accelerates from rest to 30.0 km/h in the first 20.0 s of its motion. What is its average
acceleration during that time interval?


Strategy


It is worth it at this point to make a simple sketch:


Figure 2.19


Chapter 2 | Kinematics 45




This problem involves three steps. First we must determine the change in velocity, then we must determine the change in
time, and finally we use these values to calculate the acceleration.


Solution


1. Identify the knowns. v0 = 0 (the trains starts at rest), vf = 30.0 km/h , and Δt = 20.0 s .


2. Calculate Δv . Since the train starts from rest, its change in velocity is Δv=+30.0 km/h , where the plus sign means
velocity to the right.


3. Plug in known values and solve for the unknown, a- .
(2.14)a- = ΔvΔt =


+30.0 km/h
20.0 s


4. Since the units are mixed (we have both hours and seconds for time), we need to convert everything into SI units of
meters and seconds. (See Physical Quantities and Units for more guidance.)


(2.15)
a- = ⎛⎝


+30 km/h
20.0 s






103 m
1 km




1 h


3600 s

⎠ = 0.417 m/s


2


Discussion


The plus sign means that acceleration is to the right. This is reasonable because the train starts from rest and ends up with
a velocity to the right (also positive). So acceleration is in the same direction as the change in velocity, as is always the case.


Example 2.5 Calculate Acceleration: A Subway Train Slowing Down


Now suppose that at the end of its trip, the train in Figure 2.18(a) slows to a stop from a speed of 30.0 km/h in 8.00 s. What
is its average acceleration while stopping?


Strategy


Figure 2.20


In this case, the train is decelerating and its acceleration is negative because it is toward the left. As in the previous
example, we must find the change in velocity and the change in time and then solve for acceleration.


Solution


1. Identify the knowns. v0 = 30.0 km/h , vf = 0 km/h (the train is stopped, so its velocity is 0), and Δt = 8.00 s .


2. Solve for the change in velocity, Δv .
(2.16)Δv = vf − v0 = 0 − 30.0 km/h = −30.0 km/h


3. Plug in the knowns, Δv and Δt , and solve for a- .
(2.17)a- = ΔvΔt =


−30.0 km/h
8.00 s


4. Convert the units to meters and seconds.


(2.18)
a- = ΔvΔt =




−30.0 km/h
8.00 s






103 m
1 km




1 h


3600 s

⎠ = −1.04 m/s


2.


Discussion


The minus sign indicates that acceleration is to the left. This sign is reasonable because the train initially has a positive
velocity in this problem, and a negative acceleration would oppose the motion. Again, acceleration is in the same direction
as the change in velocity, which is negative here. This acceleration can be called a deceleration because it has a direction
opposite to the velocity.


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The graphs of position, velocity, and acceleration vs. time for the trains in Example 2.4 and Example 2.5 are displayed in Figure
2.21. (We have taken the velocity to remain constant from 20 to 40 s, after which the train decelerates.)


Figure 2.21 (a) Position of the train over time. Notice that the train’s position changes slowly at the beginning of the journey, then more and more
quickly as it picks up speed. Its position then changes more slowly as it slows down at the end of the journey. In the middle of the journey, while the
velocity remains constant, the position changes at a constant rate. (b) Velocity of the train over time. The train’s velocity increases as it accelerates at
the beginning of the journey. It remains the same in the middle of the journey (where there is no acceleration). It decreases as the train decelerates at
the end of the journey. (c) The acceleration of the train over time. The train has positive acceleration as it speeds up at the beginning of the journey. It
has no acceleration as it travels at constant velocity in the middle of the journey. Its acceleration is negative as it slows down at the end of the journey.


Example 2.6 Calculating Average Velocity: The Subway Train


What is the average velocity of the train in part b of Example 2.2, and shown again below, if it takes 5.00 min to make its
trip?


Chapter 2 | Kinematics 47




Figure 2.22


Strategy


Average velocity is displacement divided by time. It will be negative here, since the train moves to the left and has a negative
displacement.


Solution


1. Identify the knowns. x′f = 3.75 km , x′0 = 5.25 km , Δt = 5.00 min .


2. Determine displacement, Δx′ . We found Δx′ to be − 1.5 km in Example 2.2.
3. Solve for average velocity.


(2.19)v- = Δx′Δt =
−1.50 km
5.00 min


4. Convert units.


(2.20)v- = Δx′Δt =


−1.50 km
5.00 min






60 min
1 h

⎠ = −18.0 km/h


Discussion


The negative velocity indicates motion to the left.


Example 2.7 Calculating Deceleration: The Subway Train


Finally, suppose the train in Figure 2.22 slows to a stop from a velocity of 20.0 km/h in 10.0 s. What is its average
acceleration?


Strategy


Once again, let’s draw a sketch:


Figure 2.23


As before, we must find the change in velocity and the change in time to calculate average acceleration.


Solution


1. Identify the knowns. v0 = −20 km/h , vf = 0 km/h , Δt = 10.0 s .


2. Calculate Δv . The change in velocity here is actually positive, since
(2.21)Δv = vf − v0 = 0 − (−20 km/h)=+20 km/h.


3. Solve for a- .
(2.22)a- = ΔvΔt =


+20.0 km/h
10.0 s


4. Convert units.


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(2.23)
a- = ⎛⎝


+20.0 km/h
10.0 s






103 m
1 km




1 h


3600 s

⎠= +0.556 m/s


2


Discussion


The plus sign means that acceleration is to the right. This is reasonable because the train initially has a negative velocity (to
the left) in this problem and a positive acceleration opposes the motion (and so it is to the right). Again, acceleration is in the
same direction as the change in velocity, which is positive here. As in Example 2.5, this acceleration can be called a
deceleration since it is in the direction opposite to the velocity.


Sign and Direction
Perhaps the most important thing to note about these examples is the signs of the answers. In our chosen coordinate system,
plus means the quantity is to the right and minus means it is to the left. This is easy to imagine for displacement and velocity. But
it is a little less obvious for acceleration. Most people interpret negative acceleration as the slowing of an object. This was not the
case in Example 2.7, where a positive acceleration slowed a negative velocity. The crucial distinction was that the acceleration
was in the opposite direction from the velocity. In fact, a negative acceleration will increase a negative velocity. For example, the
train moving to the left in Figure 2.22 is sped up by an acceleration to the left. In that case, both v and a are negative. The
plus and minus signs give the directions of the accelerations. If acceleration has the same sign as the velocity, the object is
speeding up. If acceleration has the opposite sign as the velocity, the object is slowing down.


Check Your Understanding


An airplane lands on a runway traveling east. Describe its acceleration.


Solution
If we take east to be positive, then the airplane has negative acceleration, as it is accelerating toward the west. It is also
decelerating: its acceleration is opposite in direction to its velocity.


PhET Explorations: Moving Man Simulation


Learn about position, velocity, and acceleration graphs. Move the little man back and forth with the mouse and plot his
motion. Set the position, velocity, or acceleration and let the simulation move the man for you.


Figure 2.24 Moving Man (http://


/content/m42100/1.4/moving-man_en.jar)
2.5 Motion Equations for Constant Acceleration in One Dimension


Figure 2.25 Kinematic equations can help us describe and predict the motion of moving objects such as these kayaks racing in Newbury, England.
(credit: Barry Skeates, Flickr)


We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a
given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop
some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration
already covered.


Chapter 2 | Kinematics 49




Notation: t, x, v, a
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 that Δt = 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—now, Δt = t . It also simplifies the
expression for 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,


(2.24)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 is under
consideration.


We now make the important assumption that acceleration is constant. This assumption allows us to avoid using calculus to find
instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal. That is,


(2.25)a- = a = constant,


so we use the symbol a for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations
we can study nor degrade the accuracy of our treatment. For one thing, acceleration is constant in a great number of situations.
Furthermore, in many other situations we can accurately describe motion by assuming a constant acceleration equal to the
average acceleration for that motion. Finally, in motions where acceleration changes drastically, such as a car accelerating to top
speed and then braking to a stop, the motion can be considered in separate parts, each of which has its own constant
acceleration.


Solving for Displacement (Δx ) and Final Position ( x ) from Average Velocity when Acceleration ( a ) is Constant


To get our first two new equations, we start with the definition of average velocity:


(2.26)v- = ΔxΔt .


Substituting the simplified notation for Δx and Δt yields
(2.27)v- = x − x0t .


Solving for x yields


(2.28)x = x0 + v
- t,


where the average velocity is


(2.29)
v- = v0 + v2 (constant a).


The equation v- = v0 + v2 reflects the fact that, when acceleration is constant, v is just the simple average of the initial and


final velocities. For example, if you steadily increase your velocity (that is, with constant acceleration) from 30 to 60 km/h, then


your average velocity during this steady increase is 45 km/h. Using the equation v- = v0 + v2 to check this, we see that


(2.30)
v- = v0 + v2 =


30 km/h + 60 km/h
2 = 45 km/h,


which seems logical.


Example 2.8 Calculating Displacement: How Far does the Jogger Run?


A jogger runs down a straight stretch of road with an average velocity of 4.00 m/s for 2.00 min. What is his final position,
taking his initial position to be zero?


Strategy


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Draw a sketch.


Figure 2.26


The final position x is given by the equation


(2.31)x = x0 + v
- t.


To find x , we identify the values of x0 , v
- , and t from the statement of the problem and substitute them into the equation.


Solution


1. Identify the knowns. v- = 4.00 m/s , Δt = 2.00 min , and x0 = 0 m .


2. Enter the known values into the equation.


(2.32)x = x0 + v
- t = 0 + (4.00 m/s)(120 s) = 480 m


Discussion


Velocity and final displacement are both positive, which means they are in the same direction.


The equation x = x0 + v
- t gives insight into the relationship between displacement, average velocity, and time. It shows, for


example, that displacement is a linear function of average velocity. (By linear function, we mean that displacement depends on


v- rather than on v- raised to some other power, such as v- 2 . When graphed, linear functions look like straight lines with a
constant slope.) On a car trip, for example, we will get twice as far in a given time if we average 90 km/h than if we average 45
km/h.


Figure 2.27 There is a linear relationship between displacement and average velocity. For a given time t , an object moving twice as fast as another
object will move twice as far as the other object.


Solving for Final Velocity


We can derive another useful equation by manipulating the definition of acceleration.


(2.33)a = ΔvΔt


Chapter 2 | Kinematics 51




Substituting the simplified notation for Δv and Δt gives us
(2.34)a = v − v0t (constant a).


Solving for v yields


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


Example 2.9 Calculating Final Velocity: An Airplane Slowing Down after Landing


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 final velocity?
Strategy


Draw a sketch. We draw the acceleration vector in the direction opposite the velocity vector because the plane is
decelerating.


Figure 2.28


Solution


1. Identify the knowns. v0 = 70.0 m/s , a = −1.50 m/s
2 , t = 40.0 s .


2. Identify the unknown. In this case, it is final velocity, vf .


3. Determine which equation to use. We can calculate the final velocity using the equation v = v0 + at .


4. Plug in the known values and solve.


(2.36)v = v0 + at = 70.0 m/s +

⎝−1.50 m/s2



⎠(40.0 s) = 10.0 m/s


Discussion


The final velocity is much less than the initial velocity, as desired when slowing down, but still positive. With jet engines,
reverse thrust could be maintained long enough to stop the plane and start moving it backward. That would be indicated by
a negative final velocity, which is not the case here.


Figure 2.29 The airplane lands with an initial velocity of 70.0 m/s and slows to a final velocity of 10.0 m/s before heading for the terminal. Note
that the acceleration is negative because its direction is opposite to its velocity, which is positive.


In addition to being useful in problem solving, the equation v = v0 + at gives us insight into the relationships among velocity,
acceleration, and time. From it 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 (i.e., velocity is constant)
• if a is negative, then the final velocity is less than the initial velocity


(All of these observations fit our intuition, and it is always useful to examine basic equations in light of our intuition and
experiences to check that they do indeed describe nature accurately.)


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Making Connections: Real-World Connection


Figure 2.30 The Space Shuttle Endeavor blasts off from the Kennedy Space Center in February 2010. (credit: Matthew Simantov, Flickr)


An intercontinental ballistic missile (ICBM) has a larger average acceleration than the Space Shuttle and achieves a greater
velocity in the first minute or two of flight (actual ICBM burn times are classified—short-burn-time missiles are more difficult
for an enemy to destroy). But the Space Shuttle obtains a greater final velocity, so that it can orbit the earth rather than come
directly back down as an ICBM does. The Space Shuttle does this by accelerating for a longer time.


Solving for Final Position When Velocity is Not Constant ( a ≠ 0 )


We can combine the equations above to find a third equation that allows us to calculate the final position of an object
experiencing constant acceleration. We start with


(2.37)v = v0 + at.


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


(2.38)v0 + v
2 = v0 +


1
2at.


Since
v0 + v
2 = v


- for constant acceleration, then


(2.39)v- = v0 + 12at.


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


(2.40)x = x0 + v0t + 12at
2 (constant a).


Example 2.10 Calculating Displacement of an Accelerating Object: Dragsters


Dragsters can achieve average accelerations of 26.0 m/s2 . Suppose such a dragster accelerates from rest at this rate for
5.56 s. How far does it travel in this time?


Figure 2.31 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.)


Chapter 2 | Kinematics 53




Strategy


Draw a sketch.


Figure 2.32


We are asked to find displacement, which is x if we take x0 to be zero. (Think about it like the starting line of a race. It can


be anywhere, but we call it 0 and measure all other positions relative to it.) We can use the equation x = x0 + v0t + 12at
2


once we identify v0 , a , and t from the statement of the problem.


Solution


1. Identify the knowns. Starting from rest means that v0 = 0 , a is given as 26.0 m/s
2 and t is given as 5.56 s.


2. Plug the known values into the equation to solve for the unknown x :


(2.41)x = x0 + v0t + 12at
2.


Since the initial position and velocity are both zero, this simplifies to


(2.42)x = 12at
2.


Substituting the identified values of a and t gives


(2.43)x = 12

⎝26.0 m/s2



⎠(5.56 s)2 ,


yielding


(2.44)x = 402 m.
Discussion


If we convert 402 m to miles, we find that the distance covered is very close to one quarter of a mile, the standard distance
for drag racing. So the answer is reasonable. This is an impressive displacement in only 5.56 s, but top-notch dragsters can
do a quarter mile in even less time than this.


What else can we learn by examining the equation x = x0 + v0t + 12at
2? We see that:


• displacement depends on the square of the elapsed time when acceleration is not zero. In Example 2.10, the dragster
covers only one fourth of the total distance in the first half of the elapsed time


• if acceleration is zero, then the initial velocity equals average velocity ( v0 = v
- ) and x = x0 + v0t + 12at


2 becomes


x = x0 + v0t


Solving for Final Velocity when Velocity Is Not Constant ( a ≠ 0 )


A fourth useful equation can be obtained from another algebraic manipulation of previous equations.


If we solve v = v0 + at for t , we get


(2.45)t = v − v0a .


Substituting this and v- = v0 + v2 into x = x0 + v
- t , we get


(2.46)v2 = v0
2 + 2a(x − x0) (constanta).


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Example 2.11 Calculating Final Velocity: Dragsters


Calculate the final velocity of the dragster in Example 2.10 without using information about time.


Strategy


Draw a sketch.


Figure 2.33


The equation v2 = v0
2 + 2a(x − x0) is ideally suited to this task because it relates velocities, acceleration, and


displacement, and no time information is required.


Solution


1. Identify the known values. We know that v0 = 0 , since the dragster starts from rest. Then we note that
x − x0 = 402 m (this was the answer in Example 2.10). Finally, the average acceleration was given to be


a = 26.0 m/s2 .


2. Plug the knowns into the equation v2 = v0
2 + 2a(x − x0) and solve for v.


(2.47)v2 = 0 + 2⎛⎝26.0 m/s2

⎠(402 m).


Thus


(2.48)v2 = 2.09×104 m2 /s2.
To get v , we take the square root:


(2.49)v = 2.09×104 m2 /s2 = 145 m/s.
Discussion


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


An examination of the equation v2 = v0
2 + 2a(x − x0) can produce further 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 deceleration, a car that is going twice as fast doesn’t simply stop in twice the distance—it takes much further to


stop. (This is why we have reduced speed zones near schools.)


Putting Equations Together
In the following examples, we further explore one-dimensional motion, but in situations requiring slightly more algebraic
manipulation. The examples also give insight into problem-solving techniques. The box below provides easy reference to the
equations needed.


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


- t
(2.51)


v- = v0 + v2
(2.52)v = v0 + at
(2.53)x = x0 + v0t + 12at


2


Chapter 2 | Kinematics 55




(2.54)v2 = v0
2 + 2a(x − x0)


Example 2.12 Calculating Displacement: How Far Does a Car Go When Coming to a Halt?


On dry concrete, a car can decelerate at a rate of 7.00 m/s2 , whereas on wet concrete it can decelerate at only
5.00 m/s2 . Find the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h) (a) on dry concrete and (b) on
wet concrete. (c) Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn
red, taking into account his reaction time of 0.500 s to get his foot on the brake.


Strategy


Draw a sketch.


Figure 2.34


In order to determine which equations are best to use, we need to list all of the known values and identify exactly what we
need to solve for. We shall do this explicitly in the next several examples, using tables to set them off.


Solution for (a)


1. Identify the knowns and what we want to solve for. We know that v0 = 30.0 m/s ; v = 0 ; a = −7.00 m/s
2 ( a is


negative because it is in a direction opposite to velocity). We take x0 to be 0. We are looking for displacement Δx , or
x − x0 .


2. Identify the equation that will help up solve the problem. The best equation to use is


(2.55)v2 = v0
2 + 2a(x − x0).


This equation is best because it includes only one unknown, x . We know the values of all the other variables in this
equation. (There are other equations that would allow us to solve for x , but they require us to know the stopping time, t ,
which we do not know. We could use them but it would entail additional calculations.)


3. Rearrange the equation to solve for x .


(2.56)
x − x0 =


v2 − v0
2


2a
4. Enter known values.


(2.57)
x − 0 = 0


2 − (30.0 m/s)2


2⎛⎝−7.00 m/s2



Thus,


(2.58)x = 64.3 m on dry concrete.


Solution for (b)


This part can be solved in exactly the same manner as Part A. The only difference is that the deceleration is – 5.00 m/s2 .
The result is


(2.59)xwet = 90.0 m on wet concrete.


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Solution for (c)


Once the driver reacts, the stopping distance is the same as it is in Parts 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 add that to the stopping time. It
is reasonable to assume that the velocity remains constant during the driver’s reaction time.


1. Identify the knowns and what we want to solve for. We know that v- = 30.0 m/s ; treaction = 0.500 s ; areaction = 0 .
We take x0 − reaction to be 0. We are looking for xreaction .


2. Identify the best equation to use.


x = x0 + v
- t works well because the only unknown value is x , which is what we want to solve for.


3. Plug in the knowns to solve the equation.


(2.60)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 cases of dry and wet
concrete 15.0 m greater than if he reacted instantly.


4. Add the displacement during the reaction time to the displacement when braking.


(2.61)xbraking + xreaction = xtotal
a. 64.3 m + 15.0 m = 79.3 m when dry


b. 90.0 m + 15.0 m = 105 m when wet


Figure 2.35 The distance necessary to stop a car varies greatly, depending on road conditions and driver reaction time. Shown here are the
braking distances for dry and wet pavement, as calculated in this example, for a car initially traveling at 30.0 m/s. Also shown are the total
distances traveled from the point where the driver first sees a light turn red, assuming a 0.500 s reaction time.


Discussion


The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer to stop a car
on wet rather than dry pavement. 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 be determined and then
find an appropriate equation. There is often more than one way to solve a problem. The various parts of this example can in
fact be solved by other methods, but the solutions presented above are the shortest.


Example 2.13 Calculating Time: A Car Merges into Traffic


Suppose a car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is 10.0 m/s and it accelerates at


2.00 m/s2 , how long does it take to travel the 200 m up the ramp? (Such information might be useful to a traffic engineer.)
Strategy


Draw a sketch.


Chapter 2 | Kinematics 57




Figure 2.36


We are asked to solve for the time t . As before, we identify the known quantities in order to choose a convenient physical
relationship (that is, an equation with one unknown, t ).


Solution


1. Identify the knowns and what we want to solve for. We know that v0 = 10 m/s ; a = 2.00 m/s
2 ; and x = 200 m .


2. We need to solve for t . Choose the best equation. x = x0 + v0t + 12at
2 works best because the only unknown in the


equation is the variable t for which we need to solve.


3. We will need to rearrange the equation to solve for t . In this case, it will be easier to plug in the knowns first.


(2.62)200 m = 0 m + (10.0 m/s)t + 12

⎝2.00 m/s2



⎠ t
2


4. Simplify the equation. The units of meters (m) cancel because they are in each term. We can get the units of seconds (s)
to cancel by taking t = t s , where t is the magnitude of time and s is the unit. Doing so leaves


(2.63)200 = 10t + t2.
5. Use the quadratic formula to solve for t .


(a) Rearrange the equation to get 0 on one side of the equation.


(2.64)t2 + 10t − 200 = 0
This is a quadratic equation of the form


(2.65)at2 + bt + c = 0,


where the constants are a = 1.00, b = 10.0, and c = −200 .


(b) Its solutions are given by the quadratic formula:


(2.66)
t = −b ± b


2 − 4ac
2a .


This yields two solutions for t , which are


(2.67)t = 10.0 and−20.0.
In this case, then, the time is t = t in seconds, or


(2.68)t = 10.0 s and − 20.0 s.
A negative value for time is unreasonable, since it would mean that the event happened 20 s before the motion began. We
can discard that solution. Thus,


(2.69)t = 10.0 s.
Discussion


Whenever an equation contains an unknown squared, there will be two solutions. In some problems both solutions are
meaningful, but in others, such as the above, only one solution is reasonable. The 10.0 s answer seems reasonable for a
typical freeway on-ramp.


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With the basics of kinematics established, we can go on to many other interesting examples and applications. In the process of
developing kinematics, we have also glimpsed a general approach to problem solving that produces both correct answers and
insights into physical relationships. Problem-Solving Basics discusses problem-solving basics and outlines an approach that
will help you succeed in this invaluable task.


Making Connections: Take-Home Experiment—Breaking News


We have been using SI units of meters per second squared to describe some examples of acceleration or deceleration of
cars, runners, and trains. To achieve a better feel for these numbers, one can measure the braking deceleration of a car
doing a slow (and safe) stop. Recall that, for average acceleration, a- = Δv / Δt . While traveling in a car, slowly apply the
brakes as you come up to a stop sign. Have a passenger note the initial speed in miles per hour and the time taken (in
seconds) to stop. From this, calculate the deceleration in miles per hour per second. Convert this to meters per second
squared and compare with other decelerations mentioned in this chapter. Calculate the distance traveled in braking.


Check Your Understanding


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


Solution


To answer this, choose an equation that allows you to solve for time t , given only a , v0 , and v .


(2.70)v = v0 + at


Rearrange to solve for t .


(2.71)t = v − va =
400 m/s − 0 m/s


20 m/s2
= 20 s


2.6 Problem-Solving Basics for One-Dimensional Kinematics


Figure 2.37 Problem-solving skills are essential to your success in Physics. (credit: scui3asteveo, Flickr)


Problem-solving skills are obviously essential to success in a quantitative course in physics. More importantly, the ability to apply
broad physical principles, usually represented by equations, to specific situations is a very powerful form of knowledge. It is much
more powerful than memorizing a list of facts. Analytical skills and problem-solving abilities can be applied to new situations,
whereas a list of facts cannot be made long enough to contain every possible circumstance. Such analytical skills are useful both
for solving problems in this text and for applying physics in everyday and professional life.


Problem-Solving Steps
While there is no simple step-by-step method that works for every problem, the following general procedures facilitate problem
solving and make it more meaningful. A certain amount of creativity and insight is required as well.


Step 1


Examine the situation to determine which physical principles are involved. It often helps to draw a simple sketch at the outset.
You will also need to decide which direction is positive and note that on your sketch. Once you have identified 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 of nature, and relationships among physical quantities.
Without a conceptual understanding of a problem, a numerical solution is meaningless.


Chapter 2 | Kinematics 59




Step 2


Make a list of what is given or can be inferred from the problem as stated (identify the knowns). Many problems are stated very
succinctly and require some inspection to determine what is known. A sketch can also be very useful at this point. Formally
identifying the knowns is of particular importance in applying physics to real-world situations. Remember, “stopped” means
velocity is zero, and we often can take initial time and position as zero.


Step 3


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.


Step 4


Find an equation or set of equations that can help you solve the problem. Your list of knowns and unknowns can help here. It is
easiest if you can find equations that contain only one unknown—that is, all of the other variables are known, so you can easily
solve for the unknown. If the equation contains more than one unknown, then an additional equation is needed to solve the
problem. In some problems, several unknowns must be determined to get at the one needed most. In such problems it is
especially important to keep physical principles in mind to avoid going astray in a sea of equations. You may have to use two (or
more) different equations to get the final answer.


Step 5


Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units.
This step produces the numerical answer; it also provides a check on units that can help you find errors. If the units of the
answer are incorrect, then an error has been made. However, be warned that correct units do not guarantee that the numerical
part of the answer is also correct.


Step 6


Check the answer to see if it is reasonable: Does it make sense? This final step is extremely important—the goal of physics is to
accurately describe nature. To see if the answer is reasonable, check both its magnitude and its sign, in addition to its units. Your
judgment will improve as you solve more and more physics problems, and it will become possible for you to make finer and finer
judgments regarding whether nature is adequately described by the answer to a problem. This step brings the problem back to
its conceptual meaning. If you can judge whether the answer is reasonable, you have a deeper understanding of physics than
just being able to mechanically solve a problem.


When solving problems, we often perform these steps in different order, and we also tend to do several steps simultaneously.
There is no rigid procedure that will work every time. Creativity and insight grow with experience, and the basics of problem
solving become almost automatic. One way to get practice is to work out the text’s examples for yourself as you read. Another is
to work as many end-of-section problems as possible, starting with the easiest to build confidence and progressing to the more
difficult. Once you become involved in physics, you will see it all around you, and you can begin to apply it to situations you
encounter outside the classroom, just as is done in many of the applications in this text.


Unreasonable Results
Physics must describe nature accurately. Some problems have results that are unreasonable because one premise is
unreasonable or because certain premises are inconsistent with one another. The physical principle applied correctly then


produces an unreasonable result. For example, if a person starting a foot race accelerates at 0.40 m/s2 for 100 s, his final
speed will be 40 m/s (about 150 km/h)—clearly unreasonable because the time of 100 s is an unreasonable premise. The
physics is correct in a sense, but there is more to describing nature than just manipulating equations correctly. Checking the
result of a problem to see if it is reasonable does more than help uncover errors in problem solving—it also builds intuition in
judging whether nature is being accurately described.


Use the following strategies to determine whether an answer is reasonable and, if it is not, to determine what is the cause.


Step 1


Solve the problem using strategies as outlined and in the format followed in the worked examples in the text. In the example
given in the preceding paragraph, you would identify the givens as the acceleration and time and use the equation below to find
the unknown final velocity. That is,


(2.72)v = v0 + at = 0 +

⎝0.40 m/s2



⎠(100 s) = 40 m/s.


Step 2


Check to see if the answer is reasonable. Is it too large or too small, or does it have the wrong sign, improper units, …? In this
case, you may need to convert meters per second into a more familiar unit, such as miles per hour.


(2.73)⎛

40 m
s




3.28 ft
m




1 mi
5280 ft






60 s
min




60 min
1 h

⎠ = 89 mph


This velocity is about four times greater than a person can run—so it is too large.


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Step 3


If the answer is unreasonable, look for what specifically could cause the identified difficulty. In the example of the runner, there
are only two assumptions that are suspect. The acceleration could be too great or the time too long. First look at the acceleration


and think about what the number means. If someone accelerates at 0.40 m/s2 , their velocity is increasing by 0.4 m/s each
second. Does this seem reasonable? If so, the time must be too long. It is not possible for someone to accelerate at a constant


rate of 0.40 m/s2 for 100 s (almost two minutes).


2.7 Falling Objects
Falling objects form an interesting class of motion problems. For example, we can estimate the depth of a vertical mine shaft by
dropping a rock into it and listening for the rock to hit the bottom. By applying the kinematics developed so far to falling objects,
we can examine some interesting situations and learn much about gravity in the process.


Gravity
The most remarkable and unexpected fact about falling objects is that, if air resistance and friction are negligible, then in a 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 friction that we
expect light objects to fall slower than heavy ones.


Figure 2.38 A hammer and a feather will fall with the same constant acceleration if air resistance is considered negligible. This is a general
characteristic of gravity not unique to Earth, as astronaut David R. Scott demonstrated on the Moon in 1971, where the acceleration due to gravity is


only 1.67 m/s2 .


In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball will
reach the ground after a hard baseball dropped at the same time. (It might be difficult to observe the difference if the height is not
large.) Air resistance opposes the motion of an object through the air, while friction between objects—such as between clothes
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 in free-fall.


The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called
the acceleration due to gravity. The acceleration due to gravity is constant, which means we can apply the kinematics
equations to any falling object where air resistance and friction are negligible. This opens a broad class of interesting situations
to us. The acceleration due to gravity is so important that its magnitude is given its own symbol, g . It is constant at any given


location on Earth and has the average value


(2.74)g = 9.80 m/s2.


Although g varies from 9.78 m/s2 to 9.83 m/s2 , depending on latitude, altitude, underlying geological formations, and local


topography, the average value of 9.80 m/s2 will be used in this text unless otherwise specified. The direction of the
acceleration due to gravity is downward (towards the center of Earth). In fact, its direction defines what we call vertical. Note that
whether the acceleration a in the kinematic equations has the value +g or −g depends on how we define our coordinate


system. If we define the upward direction as positive, then a = −g = −9.80 m/s2 , and if we define the downward direction as


positive, then a = g = 9.80 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 progress toward
more complex ones. So we start by considering straight up and down motion with no air resistance or friction. These
assumptions mean that the velocity (if there is any) is vertical. If the object is dropped, we know the initial velocity is zero. Once
the object has left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is one-
dimensional and has constant acceleration of magnitude g . We will also represent vertical displacement with the symbol y and


use x for horizontal displacement.


Chapter 2 | Kinematics 61




Kinematic Equations for Objects in Free-Fall where Acceleration = -g
(2.75)v = v0 - gt
(2.76)y = y0 + v0t - 12gt


2


(2.77)v2 = v0
2 - 2g(y − y0)


Example 2.14 Calculating Position and Velocity of a Falling Object: A Rock Thrown Upward


A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of 13.0 m/s. The rock misses the
edge of the cliff as it falls back to earth. Calculate the position and velocity of the rock 1.00 s, 2.00 s, and 3.00 s after it is
thrown, neglecting the effects of air resistance.


Strategy


Draw a sketch.


Figure 2.39


We are asked to determine the position y at various times. It is reasonable to take the initial position y0 to be zero. This


problem involves one-dimensional motion in the vertical direction. We use plus and minus signs to indicate direction, with up
being positive and down negative. Since up is positive, and the rock is thrown upward, the initial velocity must be positive
too. The acceleration due to gravity is downward, so a is negative. It is crucial that the initial velocity and the acceleration
due to gravity have opposite signs. Opposite signs indicate that the acceleration due to gravity opposes the initial motion
and will slow and eventually reverse it.


Since we are asked for values of position and velocity at three times, we will refer to these as y1 and v1 ; y2 and v2 ; and


y3 and v3 .


Solution for Position y1


1. Identify the knowns. We know that y0 = 0 ; v0 = 13.0 m/s ; a = −g = −9.80 m/s
2 ; and t = 1.00 s .


2. Identify the best equation to use. We will use y = y0 + v0t + 12at
2 because it includes only one unknown, y (or y1 ,


here), which is the value we want to find.


3. Plug in the known values and solve for y1 .


(2.78)y = 0 + (13.0 m/s)(1.00 s) + 12

⎝−9.80 m/s2



⎠(1.00 s)2 = 8.10 m


Discussion


The rock is 8.10 m above its starting point at t = 1.00 s, since y1 > y0 . It could be moving up or down; the only way to
tell is to calculate v1 and find out if it is positive or negative.


Solution for Velocity v1


1. Identify the knowns. We know that y0 = 0 ; v0 = 13.0 m/s ; a = −g = −9.80 m/s
2 ; and t = 1.00 s . We also know


from the solution above that y1 = 8.10 m .


2. Identify the best equation to use. The most straightforward is v = v0 − gt (from v = v0 + at , where
a = gravitational acceleration = −g ).


3. Plug in the knowns and solve.


(2.79)v1 = v0 − gt = 13.0 m/s −

⎝9.80 m/s2



⎠(1.00 s) = 3.20 m/s


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Discussion


The positive value for v1 means that the rock is still heading upward at t = 1.00 s . However, it has slowed from its original
13.0 m/s, as expected.


Solution for Remaining Times


The procedures for calculating the position and velocity at t = 2.00 s and 3.00 s are the same as those above. The
results are summarized in Table 2.1 and illustrated in Figure 2.40.


Table 2.1 Results


Time, t Position, y Velocity, v Acceleration, a
1.00 s 8.10 m 3.20 m/s −9.80 m/s2


2.00 s 6.40 m −6.60 m/s −9.80 m/s2


3.00 s −5.10 m −16.4 m/s −9.80 m/s2


Graphing the data helps us understand it more clearly.


Chapter 2 | Kinematics 63




Figure 2.40 Vertical position, vertical velocity, and vertical acceleration vs. time for a rock thrown vertically up at the edge of a cliff. Notice that
velocity changes linearly with time and that acceleration is constant. Misconception Alert! Notice that the position vs. time graph shows vertical
position only. It is easy to get the impression that the graph shows some horizontal motion—the shape of the graph looks like the path of a
projectile. But this is not the case; the horizontal axis is time, not space. The actual path of the rock in space is straight up, and straight down.


Discussion


The interpretation of these results is important. At 1.00 s the rock is above its starting point and heading upward, since y1
and v1 are both positive. At 2.00 s, the rock is still above its starting point, but the negative velocity means it is moving


downward. At 3.00 s, both y3 and v3 are negative, meaning the rock is below its starting point and continuing to move


downward. Notice that when the rock is at its highest point (at 1.5 s), its velocity is zero, but its acceleration is still


−9.80 m/s2 . Its acceleration is −9.80 m/s2 for the whole trip—while it is moving up and while it is moving down. Note
that the values for y are the positions (or displacements) of the rock, not the total distances traveled. Finally, note that free-


fall applies to upward motion as well as downward. Both have the same acceleration—the acceleration due to gravity, which
remains constant the entire time. Astronauts training in the famous Vomit Comet, for example, experience free-fall while
arcing up as well as down, as we will discuss in more detail later.


Making Connections: Take-Home Experiment—Reaction Time


A simple experiment can be done to determine your reaction time. Have a friend hold a ruler between your thumb and index
finger, separated by about 1 cm. Note the mark on the ruler that is right between your fingers. Have your friend drop the ruler


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unexpectedly, and try to catch it between your two fingers. Note the new reading on the ruler. Assuming acceleration is that
due to gravity, calculate your reaction time. How far would you travel in a car (moving at 30 m/s) if the time it took your foot
to go from the gas pedal to the brake was twice this reaction time?


Example 2.15 Calculating Velocity of a Falling Object: A Rock Thrown Down


What happens if the person on the cliff throws the rock straight down, instead of straight up? To explore this question,
calculate the velocity of the rock when it is 5.10 m below the starting point, and has been thrown downward with an initial
speed of 13.0 m/s.


Strategy


Draw a sketch.


Figure 2.41


Since up is positive, the final position of the rock will be negative because it finishes below the starting point at y0 = 0 .
Similarly, the initial velocity is downward and therefore negative, as is the acceleration due to gravity. We expect the final
velocity to be negative since the rock will continue to move downward.


Solution


1. Identify the knowns. y0 = 0 ; y1 = − 5.10 m ; v0 = −13.0 m/s ; a = −g = −9.80 m/s
2 .


2. Choose the kinematic equation that makes it easiest to solve the problem. The equation v2 = v0
2 + 2a(y − y0) works


well because the only unknown in it is v . (We will plug y1 in for y .)


3. Enter the known values


(2.80)v2 = (−13.0 m/s)2 + 2⎛⎝−9.80 m/s2

⎠(−5.10 m − 0 m) = 268.96 m2 /s2,


where we have retained extra significant figures because this is an intermediate result.


Taking the square root, and noting that a square root can be positive or negative, gives


(2.81)v = ±16.4 m/s.


The negative root is chosen to indicate that the rock is still heading down. Thus,


(2.82)v = −16.4 m/s.
Discussion


Note that this is exactly the same velocity the rock had at this position when it was thrown straight upward with the same
initial speed. (See Example 2.14 and Figure 2.42(a).) This is not a coincidental result. Because we only consider the
acceleration due to gravity in this problem, the speed of a falling object depends only on its initial speed and its vertical
position relative to the starting point. For example, if the velocity of the rock is calculated at a height of 8.10 m above the
starting point (using the method from Example 2.14) when the initial velocity is 13.0 m/s straight up, a result of ±3.20 m/s
is obtained. Here both signs are meaningful; the positive value occurs when the rock is at 8.10 m and heading up, and the
negative value occurs when the rock is at 8.10 m and heading back down. It has the same speed but the opposite direction.


Chapter 2 | Kinematics 65




Figure 2.42 (a) A person throws a rock straight up, as explored in Example 2.14. The arrows are velocity vectors at 0, 1.00, 2.00, and 3.00 s. (b)
A person throws a rock straight down from a cliff with the same initial speed as before, as in Example 2.15. Note that at the same distance below
the point of release, the rock has the same velocity in both cases.


Another way to look at it is this: In Example 2.14, the rock is thrown up with an initial velocity of 13.0 m/s . It rises and then
falls back down. When its position is y = 0 on its way back down, its velocity is −13.0 m/s . That is, it has the same
speed on its way down as on its way up. We would then expect its velocity at a position of y = −5.10 m to be the same
whether we have thrown it upwards at +13.0 m/s or thrown it downwards at −13.0 m/s . The velocity of the rock on its
way down from y = 0 is the same whether we have thrown it up or down to start with, as long as the speed with which it
was initially thrown is the same.


Example 2.16 Find g from Data on a Falling Object
The acceleration due to gravity on Earth differs slightly from place to place, depending on topography (e.g., whether you are
on a hill or in a valley) and subsurface geology (whether there is dense rock like iron ore as opposed to light rock like salt
beneath you.) The precise acceleration due to gravity can be calculated from data taken in an introductory physics
laboratory course. An object, usually a metal ball for which air resistance is negligible, is dropped and the time it takes to fall
a known distance is measured. See, for example, Figure 2.43. Very precise results can be produced with this method if
sufficient care is taken in measuring the distance fallen and the elapsed time.


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Figure 2.43 Positions and velocities of a metal ball released from rest when air resistance is negligible. Velocity is seen to increase linearly with
time while displacement increases with time squared. Acceleration is a constant and is equal to gravitational acceleration.


Suppose the ball falls 1.0000 m in 0.45173 s. Assuming the ball is not affected by air resistance, what is the precise
acceleration due to gravity at this location?


Strategy


Draw a sketch.


Figure 2.44


Chapter 2 | Kinematics 67




We need to solve for acceleration a . Note that in this case, displacement is downward and therefore negative, as is
acceleration.


Solution


1. Identify the knowns. y0 = 0 ; y = –1.0000 m ; t = 0.45173 ; v0 = 0 .


2. Choose the equation that allows you to solve for a using the known values.


(2.83)y = y0 + v0t + 12at
2


3. Substitute 0 for v0 and rearrange the equation to solve for a . Substituting 0 for v0 yields


(2.84)y = y0 + 12at
2.


Solving for a gives


(2.85)
a = 2(y − y0)


t2
.


4. Substitute known values yields


(2.86)
a = 2( − 1.0000 m – 0)


(0.45173 s)2
= −9.8010 m/s2 ,


so, because a = −g with the directions we have chosen,


(2.87)g = 9.8010 m/s2.


Discussion


The negative value for a indicates that the gravitational acceleration is downward, as expected. We expect the value to be


somewhere around the average value of 9.80 m/s2 , so 9.8010 m/s2 makes sense. Since the data going into the
calculation are relatively precise, this value for g is more precise than the average value of 9.80 m/s2 ; it represents the
local value for the acceleration due to gravity.


Check Your Understanding


A chunk of ice breaks off a glacier and falls 30.0 meters before it hits the water. Assuming it falls freely (there is no air
resistance), how long does it take to hit the water?


Solution


We know that initial position y0 = 0 , final position y = −30.0 m , and a = −g = −9.80 m/s
2 . We can then use the


equation y = y0 + v0t + 12at
2 to solve for t . Inserting a = −g , we obtain


(2.88)y = 0 + 0 − 12gt
2


t2 = 2y−g


t = ± 2y−g = ±
2( − 30.0 m)
−9.80 m/s2


= ± 6.12 s2 = 2.47 s ≈ 2.5 s


where we take the positive value as the physically relevant answer. Thus, it takes about 2.5 seconds for the piece of ice to
hit the water.


PhET Explorations: Equation Grapher


Learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the
individual terms (e.g. y = bx ) to see how they add to generate the polynomial curve.


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Figure 2.45 Equation Grapher (http://


/content/m42102/1.5/equation-grapher_en.jar)


2.8 Graphical Analysis of One-Dimensional Motion
A graph, like a picture, is worth a thousand words. Graphs not only contain numerical information; they also reveal relationships
between physical quantities. This section uses graphs of displacement, velocity, and acceleration versus time to illustrate one-
dimensional kinematics.


Slopes and General Relationships
First note that graphs in this text have perpendicular axes, one horizontal and the other vertical. When two physical quantities are
plotted against one another in such a graph, the horizontal axis is usually considered to be an independent variable and the
vertical axis a dependent variable. If we call the horizontal axis the x -axis and the vertical axis the y -axis, as in Figure 2.46, a


straight-line graph has the general form


(2.89)y = mx + b.


Here m is the slope, defined to be the rise divided by the run (as seen in the figure) of the straight line. The letter b is used for
the y-intercept, which is the point at which the line crosses the vertical axis.


Figure 2.46 A straight-line graph. The equation for a straight line is y = mx + b .


Graph of Displacement vs. Time (a = 0, so v is constant)
Time is usually an independent variable that other quantities, such as displacement, depend upon. A graph of displacement
versus time would, thus, have x on the vertical axis and t on the horizontal axis. Figure 2.47 is just such a straight-line graph.
It shows a graph of displacement versus time for a jet-powered car on a very flat dry lake bed in Nevada.


Chapter 2 | Kinematics 69




Figure 2.47 Graph of displacement versus time for a jet-powered car on the Bonneville Salt Flats.


Using the relationship between dependent and independent variables, we see that the slope in the graph above is average
velocity v- and the intercept is displacement at time zero—that is, x0 . Substituting these symbols into y = mx + b gives


(2.90)x = v- t + x0
or


(2.91)x = x0 + v
- t.


Thus a graph of displacement versus time gives a general relationship among displacement, velocity, and time, as well as giving
detailed numerical information about a specific situation.


The Slope of x vs. t


The slope of the graph of displacement x vs. time t is velocity v .


(2.92)slope = ΔxΔt = v


Notice that this equation is the same as that derived algebraically from other motion equations in Motion Equations for
Constant Acceleration in One Dimension.


From the figure we can see that the car has a displacement of 400 m at time 0.650 m at t = 1.0 s, and so on. Its displacement at
times other than those listed in the table can be read from the graph; furthermore, information about its velocity and acceleration
can also be obtained from the graph.


Example 2.17 Determining Average Velocity from a Graph of Displacement versus Time: Jet
Car


Find the average velocity of the car whose position is graphed in Figure 2.47.


Strategy


The slope of a graph of x vs. t is average velocity, since slope equals rise over run. In this case, rise = change in
displacement and run = change in time, so that


(2.93)slope = ΔxΔt = v
- .


Since the slope is constant here, any two points on the graph can be used to find the slope. (Generally speaking, it is most
accurate to use two widely separated points on the straight line. This is because any error in reading data from the graph is
proportionally smaller if the interval is larger.)


Solution


1. Choose two points on the line. In this case, we choose the points labeled on the graph: (6.4 s, 2000 m) and (0.50 s, 525
m). (Note, however, that you could choose any two points.)


2. Substitute the x and t values of the chosen points into the equation. Remember in calculating change (Δ) we always
use final value minus initial value.


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(2.94)v- = ΔxΔt =
2000 m − 525 m
6.4 s − 0.50 s ,


yielding


(2.95)v- = 250 m/s.
Discussion


This is an impressively large land speed (900 km/h, or about 560 mi/h): much greater than the typical highway speed limit of
60 mi/h (27 m/s or 96 km/h), but considerably shy of the record of 343 m/s (1234 km/h or 766 mi/h) set in 1997.


Graphs of Motion when a is constant but a ≠ 0


The graphs in Figure 2.48 below represent the motion of the jet-powered car as it accelerates toward its top speed, but only
during the time when its acceleration is constant. Time starts at zero for this motion (as if measured with a stopwatch), and the
displacement and velocity are initially 200 m and 15 m/s, respectively.


Chapter 2 | Kinematics 71




Figure 2.48 Graphs of motion of a jet-powered car during the time span when its acceleration is constant. (a) The slope of an x vs. t graph is
velocity. This is shown at two points, and the instantaneous velocities obtained are plotted in the next graph. Instantaneous velocity at any point is the
slope of the tangent at that point. (b) The slope of the v vs. t graph is constant for this part of the motion, indicating constant acceleration. (c)


Acceleration has the constant value of 5.0 m/s2 over the time interval plotted.


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Figure 2.49 A U.S. Air Force jet car speeds down a track. (credit: Matt Trostle, Flickr)


The graph of displacement versus time in Figure 2.48(a) is a curve rather than a straight line. The slope of the curve becomes
steeper as time progresses, showing that the velocity is increasing over time. The slope at any point on a displacement-versus-
time graph is the instantaneous velocity at that point. It is found by drawing a straight line tangent to the curve at the point of
interest and taking the slope of this straight line. Tangent lines are shown for two points in Figure 2.48(a). If this is done at every
point on the curve and the values are plotted against time, then the graph of velocity versus time shown in Figure 2.48(b) is
obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in Figure 2.48(c).


Example 2.18 Determining Instantaneous Velocity from the Slope at a Point: Jet Car


Calculate the velocity of the jet car at a time of 25 s by finding the slope of the x vs. t graph in the graph below.


Figure 2.50 The slope of an x vs. t graph is velocity. This is shown at two points. Instantaneous velocity at any point is the slope of the tangent
at that point.


Strategy


The slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. This principle is
illustrated in Figure 2.50, where Q is the point at t = 25 s .
Solution


1. Find the tangent line to the curve at t = 25 s .
2. Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m
at time 32 s.


3. Plug these endpoints into the equation to solve for the slope, v .


(2.96)
slope = vQ =


ΔxQ
ΔtQ


= (3120 m − 1300 m)(32 s − 19 s)


Thus,


(2.97)vQ = 1820 m13 s = 140 m/s.


Discussion


Chapter 2 | Kinematics 73




This is the value given in this figure’s table for v at t = 25 s . The value of 140 m/s for vQ is plotted in Figure 2.50. The
entire graph of v vs. t can be obtained in this fashion.


Carrying this one step further, we note that the slope of a velocity versus time graph is acceleration. Slope is rise divided by run;
on a v vs. t graph, rise = change in velocity Δv and run = change in time Δt .


The Slope of v vs. t


The slope of a graph of velocity v vs. time t is acceleration a .


(2.98)slope = ΔvΔt = a


Since the velocity versus time graph in Figure 2.48(b) is a straight line, its slope is the same everywhere, implying that
acceleration is constant. Acceleration versus time is graphed in Figure 2.48(c).


Additional general information can be obtained from Figure 2.50 and the expression for a straight line, y = mx + b .


In this case, the vertical axis y is V , the intercept b is v0 , the slope m is a , and the horizontal axis x is t . Substituting


these symbols yields


(2.99)v = v0 + at.


A general relationship for velocity, acceleration, and time has again been obtained from a graph. Notice that this equation was
also derived algebraically from other motion equations in Motion Equations for Constant Acceleration in One Dimension.


It is not accidental that the same equations are obtained by graphical analysis as by algebraic techniques. In fact, an important
way to discover physical relationships is to measure various physical quantities and then make graphs of one quantity against
another to see if they are correlated in any way. Correlations imply physical relationships and might be shown by smooth graphs
such as those above. From such graphs, mathematical relationships can sometimes be postulated. Further experiments are then
performed to determine the validity of the hypothesized relationships.


Graphs of Motion Where Acceleration is Not Constant
Now consider the motion of the jet car as it goes from 165 m/s to its top velocity of 250 m/s, graphed in Figure 2.51. Time again
starts at zero, and the initial displacement and velocity are 2900 m and 165 m/s, respectively. (These were the final displacement


and velocity of the car in the motion graphed in Figure 2.48.) Acceleration gradually decreases from 5.0 m/s2 to zero when the
car hits 250 m/s. The slope of the x vs. t graph increases until t = 55 s , after which time the slope is constant. Similarly,
velocity increases until 55 s and then becomes constant, since acceleration decreases to zero at 55 s and remains zero
afterward.


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Figure 2.51 Graphs of motion of a jet-powered car as it reaches its top velocity. This motion begins where the motion in Figure 2.48 ends. (a) The
slope of this graph is velocity; it is plotted in the next graph. (b) The velocity gradually approaches its top value. The slope of this graph is acceleration;
it is plotted in the final graph. (c) Acceleration gradually declines to zero when velocity becomes constant.


Example 2.19 Calculating Acceleration from a Graph of Velocity versus Time


Calculate the acceleration of the jet car at a time of 25 s by finding the slope of the v vs. t graph in Figure 2.51(b).


Strategy


The slope of the curve at t = 25 s is equal to the slope of the line tangent at that point, as illustrated in Figure 2.51(b).
Solution


Determine endpoints of the tangent line from the figure, and then plug them into the equation to solve for slope, a .


(2.100)slope = ΔvΔt =
(260 m/s − 210 m/s)


(51 s − 1.0 s)


Chapter 2 | Kinematics 75




acceleration:


acceleration due to gravity:


average acceleration:


average speed:


average velocity:


deceleration:


dependent variable:


displacement:


distance:


distance traveled:


(2.101)a = 50 m/s50 s = 1.0 m/s
2.


Discussion


Note that this value for a is consistent with the value plotted in Figure 2.51(c) at t = 25 s .


A graph of displacement versus time can be used to generate a graph of velocity versus time, and a graph of velocity versus time
can be used to generate a graph of acceleration versus time. We do this by finding the slope of the graphs at every point. If the
graph is linear (i.e., a line with a constant slope), it is easy to find the slope at any point and you have the slope for every point.
Graphical analysis of motion can be used to describe both specific and general characteristics of kinematics. Graphs can also be
used for other topics in physics. An important aspect of exploring physical relationships is to graph them and look for underlying
relationships.


Check Your Understanding


A graph of velocity vs. time of a ship coming into a harbor is shown below. (a) Describe the motion of the ship based on the
graph. (b)What would a graph of the ship’s acceleration look like?


Figure 2.52


Solution


(a) The ship moves at constant velocity and then begins to decelerate at a constant rate. At some point, its deceleration rate
decreases. It maintains this lower deceleration rate until it stops moving.


(b) A graph of acceleration vs. time would show zero acceleration in the first leg, large and constant negative acceleration in
the second leg, and constant negative acceleration.


Figure 2.53


Glossary
the rate of change in velocity; the change in velocity over time


acceleration of an object as a result of gravity


the change in velocity divided by the time over which it changes


distance traveled divided by time during which motion occurs


displacement divided by time over which displacement occurs


acceleration in the direction opposite to velocity; acceleration that results in a decrease in velocity


the variable that is being measured; usually plotted along the y -axis


the change in position of an object


the magnitude of displacement between two positions


the total length of the path traveled between two positions


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elapsed time:


free-fall:


independent variable:


instantaneous acceleration:


instantaneous speed:


instantaneous velocity:


kinematics:


model:


position:


scalar:


slope:


time:


vector:


y-intercept:


the difference between the ending time and beginning time


the state of movement that results from gravitational force only


the variable that the dependent variable is measured with respect to; usually plotted along the x -axis


acceleration at a specific point in time


magnitude of the instantaneous velocity


velocity at a specific instant, or the average velocity over an infinitesimal time interval


the study of motion without considering its causes


simplified description that contains only those elements necessary to describe the physics of a physical situation


the location of an object at a particular time


a quantity that is described by magnitude, but not direction


the difference in y -value (the rise) divided by the difference in x -value (the run) of two points on a straight line


change, or the interval over which change occurs


a quantity that is described by both magnitude and direction


the y- value when x = 0, or when the graph crosses the y -axis


Section Summary


2.1 Displacement
• Kinematics is the study of motion without considering its causes. In this chapter, it is limited to motion along a straight line,


called one-dimensional motion.
• Displacement is the change in position of an object.
• In symbols, displacement Δx is defined to be


Δx = xf − x0,
where x0 is the initial position and xf is the final position. In this text, the Greek letter Δ (delta) always means “change
in” whatever quantity follows it. The SI unit for displacement is the meter (m). Displacement has a direction as well as a
magnitude.


• When you start a problem, assign which direction will be positive.
• Distance is the magnitude of displacement between two positions.
• Distance traveled is the total length of the path traveled between two positions.


2.2 Vectors, Scalars, and Coordinate Systems
• A vector is any quantity that has magnitude and direction.
• A scalar is any quantity that has magnitude but no direction.
• Displacement and velocity are vectors, whereas distance and speed are scalars.
• In one-dimensional motion, direction is specified by a plus or minus sign to signify left or right, up or down, and the like.


2.3 Time, Velocity, and Speed
• Time is measured in terms of change, and its SI unit is the second (s). 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, as if measured with a


stopwatch; the elapsed time is then just t .
• Average velocity v- is defined as displacement divided by the travel time. In symbols, average velocity is


v- = ΔxΔt =
xf − x0
tf − t0


.
• The SI unit for velocity is m/s.
• Velocity is a vector and thus has a direction.
• Instantaneous velocity v is the velocity at a specific instant or the average velocity for an infinitesimal interval.
• Instantaneous speed is the magnitude of the instantaneous velocity.
• Instantaneous speed is a scalar quantity, as it has no direction specified.


Chapter 2 | Kinematics 77




• Average speed is the total distance traveled divided by the elapsed time. (Average speed is not the magnitude of the
average velocity.) Speed is a scalar quantity; it has no direction associated with it.


2.4 Acceleration
• Acceleration is the rate at which velocity changes. In symbols, average acceleration a- is


a- = ΔvΔt =
vf − v0
tf − t0


.


• The SI unit for acceleration is m/s2 .
• Acceleration is a vector, and thus has a both a magnitude and direction.
• Acceleration can be caused by either a change in the magnitude or the direction of the velocity.
• Instantaneous acceleration a is the acceleration at a specific instant in time.
• Deceleration is an acceleration with a direction opposite to that of the velocity.


2.5 Motion Equations for Constant Acceleration in One Dimension
• To simplify calculations we take acceleration to be constant, so that a- = a at all times.
• We also take initial time to be zero.
• Initial position and velocity are given a subscript 0; final values have no subscript. Thus,


Δt = t
Δx = x − x0
Δv = v − v0







• The following kinematic equations for motion with constant a are useful:


x = x0 + v
- t


v- = v0 + v2
v = v0 + at


x = x0 + v0t + 12at
2


v2 = v0
2 + 2a(x − x0)


• In vertical motion, y is substituted for x .


2.6 Problem-Solving Basics for One-Dimensional Kinematics
• The six basic problem solving steps for physics are:


Step 1. Examine the situation to determine which physical principles are involved.


Step 2. Make a list of what is given or can be inferred from the problem as stated (identify the knowns).


Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns).


Step 4. Find an equation or set of equations that can help you solve the problem.


Step 5. Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete
with units.


Step 6. Check the answer to see if it is reasonable: Does it make sense?


2.7 Falling Objects
• An object in free-fall experiences constant acceleration if air resistance is negligible.
• On Earth, all free-falling objects have an acceleration due to gravity g , which averages


g = 9.80 m/s2.
• Whether the acceleration a should be taken as +g or −g is determined by your choice of coordinate system. If you


choose the upward direction as positive, a = −g = −9.80 m/s2 is negative. In the opposite case,


a = +g = 9.80 m/s2 is positive. Since acceleration is constant, the kinematic equations above can be applied with the
appropriate +g or −g substituted for a .


• For objects in free-fall, up is normally taken as positive for displacement, velocity, and acceleration.


2.8 Graphical Analysis of One-Dimensional Motion
• Graphs of motion can be used to analyze motion.


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• Graphical solutions yield identical solutions to mathematical methods for deriving motion equations.
• The slope of a graph of displacement x vs. time t is velocity v .
• The slope of a graph of velocity v vs. time t graph is acceleration a .
• Average velocity, instantaneous velocity, and acceleration can all be obtained by analyzing graphs.


Conceptual Questions


2.1 Displacement
1. Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement.
Specifically identify each quantity in your example.


2. Under what circumstances does distance traveled equal magnitude of displacement? What is the only case in which
magnitude of displacement and displacement are exactly the same?


3. Bacteria move back and forth by 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 distance traveled by a bacterium is large for its size, while its


displacement is small. Why is this?


2.2 Vectors, Scalars, and Coordinate Systems


4. A student writes, “A bird that is diving for prey has a speed of − 10 m / s .” What is wrong with the student’s statement? What
has the student actually described? Explain.


5.What is the speed of the bird in Exercise 2.4?


6. Acceleration is the change in velocity over time. Given this information, is acceleration a vector or a scalar quantity? Explain.


7. A weather forecast states that the temperature is predicted to be −5ºC the following day. Is this temperature a vector or a
scalar quantity? Explain.


2.3 Time, Velocity, and Speed
8. Give an example (but not one from the text) of a device used to measure time and identify what change in that device
indicates a change in time.


9. There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the
difference between these two quantities.


10. Does a car’s odometer measure position or displacement? Does its speedometer measure speed or velocity?


11. If you divide the total distance traveled on a car trip (as determined by the odometer) by the time for the trip, are you
calculating the average speed or the magnitude of the average velocity? Under what circumstances are these two quantities the
same?


12. How are instantaneous velocity and instantaneous speed related to one another? How do they differ?


2.4 Acceleration
13. Is it possible for speed to be constant while acceleration is not zero? Give an example of such a situation.


14. Is it possible for velocity to be constant while acceleration is not zero? Explain.


15. Give an example in which velocity is zero yet acceleration is not.


16. If a subway train is moving to the left (has a negative velocity) and then comes to a stop, what is the direction of its
acceleration? Is the acceleration positive or negative?


17. Plus and minus signs are used in one-dimensional motion to indicate direction. What is the sign of an acceleration that
reduces the magnitude of a negative velocity? Of a positive velocity?


2.6 Problem-Solving Basics for One-Dimensional Kinematics
18.What information do you need in order to choose which equation or equations to use to solve a problem? Explain.


19.What is the last thing you should do when solving a problem? Explain.


2.7 Falling Objects
20.What is the acceleration of a rock thrown straight upward on the way up? At the top of its flight? On the way down?


21. 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 the acceleration due to gravity have the same sign on the way up as on the way down?


22. Suppose you throw a rock nearly straight up at a coconut in a palm tree, and the rock misses on the way up but hits the
coconut on the way down. Neglecting air resistance, how does the speed of the rock when it hits the coconut on the way down
compare with what it would have been if it had hit the coconut on the way up? Is it more likely to dislodge the coconut on the way
up or down? Explain.


Chapter 2 | Kinematics 79




23. If an object is thrown straight up and air resistance is negligible, then its speed when it returns to the starting point is the
same as when it was released. If air resistance were not negligible, how would its speed upon return compare with its initial
speed? How would the maximum height to which it rises be affected?


24. The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration due to gravity being
the same, how many times higher could a safe fall on the Moon be than on Earth (gravitational acceleration on the Moon is about
1/6 that of the Earth)?


25. How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations
(gravitational acceleration on the Moon is about 1/6 of g on Earth)?


2.8 Graphical Analysis of One-Dimensional Motion
26. (a) Explain how you can use the graph of position versus time in Figure 2.54 to describe the change in velocity over time.
Identify (b) the time ( ta , tb , tc , td , or te ) at which the instantaneous velocity is greatest, (c) the time at which it is zero, and


(d) the time at which it is negative.


Figure 2.54


27. (a) Sketch a graph of velocity versus time corresponding to the graph of displacement versus time given in Figure 2.55. (b)
Identify the time or times ( ta , tb , tc , etc.) at which the instantaneous velocity is greatest. (c) At which times is it zero? (d) At


which times is it negative?


Figure 2.55


28. (a) Explain how you can determine the acceleration over time from a velocity versus time graph such as the one in Figure
2.56. (b) Based on the graph, how does acceleration change over time?


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Figure 2.56


29. (a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in Figure 2.57. (b)
Identify the time or times ( ta , tb , tc , etc.) at which the acceleration is greatest. (c) At which times is it zero? (d) At which times


is it negative?


Figure 2.57


30. Consider the velocity vs. time graph of a person in an elevator shown in Figure 2.58. Suppose the elevator is initially at rest.
It then accelerates for 3 seconds, maintains that velocity for 15 seconds, then decelerates for 5 seconds until it stops. The
acceleration for the entire trip is not constant so we cannot use the equations of motion from Motion Equations for Constant
Acceleration in One Dimension for the complete trip. (We could, however, use them in the three individual sections where
acceleration is a constant.) Sketch graphs of (a) position vs. time and (b) acceleration vs. time for this trip.


Figure 2.58


31. A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity,
and acceleration of the cylinder vs. time as it goes up and then down the plane.


Chapter 2 | Kinematics 81




Problems & Exercises


2.1 Displacement


Figure 2.59


1. Find the following for path A in Figure 2.59: (a) The
distance traveled. (b) The magnitude of the displacement
from start to finish. (c) The displacement from start to finish.


2. Find the following for path B in Figure 2.59: (a) The
distance traveled. (b) The magnitude of the displacement
from start to finish. (c) The displacement from start to finish.


3. Find the following for path C in Figure 2.59: (a) The
distance traveled. (b) The magnitude of the displacement
from start to finish. (c) The displacement from start to finish.


4. Find the following for path D in Figure 2.59: (a) The
distance traveled. (b) The magnitude of the displacement
from start to finish. (c) The displacement from start to finish.


2.3 Time, Velocity, and Speed
5. (a) Calculate Earth’s average speed relative to the Sun. (b)
What is its average velocity over a period of one year?


6. A helicopter blade spins at exactly 100 revolutions per
minute. Its tip is 5.00 m from the center of rotation. (a)
Calculate the average speed of the blade tip in the
helicopter’s frame of reference. (b) What is its average
velocity over one revolution?


7. The North American and European continents are moving
apart at a rate of about 3 cm/y. At this rate how long will it
take them to drift 500 km farther apart than they are at
present?


8. Land west of the San Andreas fault in southern California is
moving at an average velocity of about 6 cm/y northwest
relative to land east of the fault. Los Angeles is west of the
fault and may thus someday be at the same latitude as San
Francisco, which is east of the fault. How far in the future will
this occur if the displacement to be made is 590 km
northwest, assuming the motion remains constant?


9. On May 26, 1934, a streamlined, stainless steel diesel train
called the Zephyr set the world’s nonstop long-distance speed
record for trains. Its run from Denver to Chicago took 13
hours, 4 minutes, 58 seconds, and was witnessed by more
than a million people along the route. The total distance
traveled was 1633.8 km. What was its average speed in km/h
and m/s?


10. Tidal friction is slowing the rotation of the Earth. As a
result, the orbit of the Moon is increasing in radius at a rate of
approximately 4 cm/year. Assuming this to be a constant rate,
how many years will pass before the radius of the Moon’s


orbit increases by 3.84×106 m (1%)?


11. A student drove to the university from her home and
noted that the odometer reading of her car increased by 12.0
km. The trip took 18.0 min. (a) What was her average speed?
(b) If the straight-line distance from her home to the university
is 10.3 km in a direction 25.0º south of east, what was her
average velocity? (c) If she returned home by the same path
7 h 30 min after she left, what were her average speed and
velocity for the entire trip?


12. The speed of propagation of the action potential (an
electrical signal) in a nerve cell depends (inversely) on the
diameter of the axon (nerve fiber). If the nerve cell connecting
the spinal cord to your feet is 1.1 m long, and the nerve
impulse speed is 18 m/s, how long does it take for the nerve
signal to travel this distance?


13. Conversations with astronauts on the lunar surface were
characterized by a kind of echo in which the earthbound
person’s voice was so loud in the astronaut’s space helmet
that it was picked up by the astronaut’s microphone and
transmitted back to Earth. It is reasonable to assume that the
echo time equals the time necessary for the radio wave to
travel from the Earth to the Moon and back (that is, neglecting
any time delays in the electronic equipment). Calculate the
distance from Earth to the Moon given that the echo time was
2.56 s and that radio waves travel at the speed of light


(3.00×108 m/s) .


14. A football quarterback runs 15.0 m straight down the
playing field in 2.50 s. He is then hit and pushed 3.00 m
straight backward in 1.75 s. He breaks the tackle and runs
straight forward another 21.0 m in 5.20 s. Calculate his
average velocity (a) for each of the three intervals and (b) for
the entire motion.


15. The planetary model of the atom pictures electrons
orbiting the atomic nucleus much as planets orbit the Sun. In
this model you can view hydrogen, the simplest atom, as


having a single electron in a circular orbit 1.06×10−10 m in
diameter. (a) If the average speed of the electron in this orbit


is known to be 2.20×106 m/s , calculate the number of
revolutions per second it makes about the nucleus. (b) What
is the electron’s average velocity?


2.4 Acceleration
16. A cheetah can accelerate from rest to a speed of 30.0 m/s
in 7.00 s. What is its acceleration?


17. Professional Application


Dr. John Paul Stapp was U.S. Air Force officer who studied
the effects of extreme deceleration on the human body. On
December 10, 1954, Stapp rode a rocket sled, accelerating
from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and
was brought jarringly back to rest in only 1.40 s! Calculate his
(a) acceleration and (b) deceleration. Express each in


multiples of g (9.80 m/s2) by taking its ratio to the
acceleration of gravity.


18. A commuter backs her car out of her garage with an


acceleration of 1.40 m/s2 . (a) How long does it take her to
reach a speed of 2.00 m/s? (b) If she then brakes to a stop in
0.800 s, what is her deceleration?


19. Assume that an intercontinental ballistic missile goes from
rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual
speed and time are classified). What is its average


acceleration in m/s2 and in multiples of g (9.80 m/s2)?


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2.5 Motion Equations for Constant
Acceleration in One Dimension
20. An Olympic-class sprinter starts a race with an


acceleration of 4.50 m/s2 . (a) What is her speed 2.40 s
later? (b) Sketch a graph of her position vs. time for this
period.


21. A well-thrown ball is caught in a well-padded mitt. If the


deceleration 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 was the initial velocity of
the ball?


22. A bullet in a gun is accelerated from the firing chamber 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)?


23. (a) A light-rail commuter train accelerates at a rate of


1.35 m/s2 . How long does it take to reach its top speed of
80.0 km/h, starting from rest? (b) The same train ordinarily


decelerates at a rate of 1.65 m/s2 . How long does it take to
come to a stop from its top speed? (c) In emergencies the
train can decelerate more rapidly, coming to rest from 80.0


km/h in 8.30 s. What is its emergency deceleration in m/s2 ?
24.While entering a freeway, a car accelerates from rest at a


rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the
situation. (b) List the knowns in this problem. (c) How far does
the car travel in those 12.0 s? To solve this part, first identify
the unknown, and then discuss how you chose the
appropriate equation to solve for it. After choosing the
equation, show your steps in solving for the unknown, check
your units, and discuss whether the answer is reasonable. (d)
What is the car’s final velocity? Solve for this unknown in the
same manner as in part (c), showing all steps explicitly.


25. At the end of a race, a runner decelerates 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) What is her final velocity? (c)
Evaluate the result. Does it make sense?


26. Professional Application:


Blood is accelerated from rest to 30.0 cm/s in a distance of
1.80 cm by the left ventricle of the heart. (a) Make a sketch of
the situation. (b) List the knowns in this problem. (c) How long
does the acceleration take? To solve this part, first identify the
unknown, and then discuss how you chose the appropriate
equation to solve for it. After choosing the equation, show
your steps in solving for the unknown, checking your units. (d)
Is the answer reasonable when compared with the time for a
heartbeat?


27. In a slap shot, a hockey player accelerates the puck 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 , calculate the distance over which
the puck accelerates.


28. A powerful motorcycle can accelerate from rest to 26.8 m/
s (100 km/h) in only 3.90 s. (a) What is its average
acceleration? (b) How far does it travel in that time?


29. Freight trains can produce only relatively small
accelerations and decelerations. (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 will
it travel in each case?


30. A fireworks shell is accelerated from rest to a velocity of
65.0 m/s over a distance of 0.250 m. (a) How long did the
acceleration last? (b) Calculate the acceleration.


31. A swan on a lake gets airborne by flapping its wings and
running on top of the water. (a) If the swan must reach a
velocity of 6.00 m/s to take off and it accelerates from rest at


an average rate of 0.350 m/s2 , how far will it travel before
becoming airborne? (b) How long does this take?


32. Professional Application:


A woodpecker’s brain is specially protected from large
decelerations by tendon-like attachments inside the skull.
While pecking on a tree, the woodpecker’s head comes to a
stop from an initial velocity of 0.600 m/s in a distance of only


2.00 mm. (a) Find the acceleration in m/s2 and in multiples
of g ⎛⎝g = 9.80 m/s2



⎠ . (b) Calculate the stopping time. (c)


The tendons cradling the brain stretch, making its stopping
distance 4.50 mm (greater than the head and, hence, less
deceleration of the brain). What is the brain’s deceleration,
expressed in multiples of g ?


33. An unwary football player collides with a padded goalpost
while running at a velocity of 7.50 m/s and comes to a full
stop after compressing the padding and his body 0.350 m. (a)
What is his deceleration? (b) How long does the collision
last?


34. In World War II, there were several reported cases of
airmen who jumped from their flaming airplanes with no
parachute to escape certain death. Some fell about 20,000
feet (6000 m), and some of them survived, with few life-
threatening injuries. For these lucky pilots, the tree branches
and snow drifts on the ground allowed their deceleration to be
relatively small. If we assume that a pilot’s speed upon impact
was 123 mph (54 m/s), then what was his deceleration?
Assume that the trees and snow stopped him over a distance
of 3.0 m.


35. Consider a grey squirrel falling out of a tree to the ground.
(a) If we ignore air resistance in this case (only for the sake of
this problem), determine a squirrel’s velocity just before hitting
the ground, assuming it fell from a height of 3.0 m. (b) If the
squirrel stops in a distance of 2.0 cm through bending its
limbs, compare its deceleration with that of the airman in the
previous problem.


36. An express train passes through a station. It enters with
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 m long.
(a) How long is the nose of the train in the station? (b) How
fast is it going when the nose leaves the station? (c) If the
train is 130 m long, when does the end of the train leave the
station? (d) What is the velocity of the end of the train as it
leaves?


37. Dragsters can actually reach a top speed of 145 m/s in
only 4.45 s—considerably less time than given in Example
2.10 and Example 2.11. (a) Calculate the average
acceleration for such a dragster. (b) Find the final velocity of
this dragster starting from rest and accelerating at the rate
found in (a) for 402 m (a quarter mile) without using any


Chapter 2 | Kinematics 83




information on time. (c) Why is the final velocity greater than
that used to find the average acceleration? Hint: Consider
whether the assumption of constant acceleration is valid for a
dragster. If not, discuss whether the acceleration would be
greater at the beginning or end of the run and what effect that
would have on the final velocity.


38. A bicycle racer sprints at the end of a race to clinch a
victory. The racer has an initial velocity of 11.5 m/s and


accelerates at the rate of 0.500 m/s2 for 7.00 s. (a) What is
his final velocity? (b) The racer continues at this velocity to
the finish line. If he was 300 m from the finish line when he
started to accelerate, how much time did he save? (c) One
other racer was 5.00 m ahead when the winner started to
accelerate, but he was unable to accelerate, and traveled at
11.8 m/s until the finish line. How far ahead of him (in meters
and in seconds) did the winner finish?


39. In 1967, New Zealander Burt Munro set the world record
for an Indian motorcycle, on the Bonneville Salt Flats in Utah,
with a maximum speed of 183.58 mi/h. The one-way course
was 5.00 mi long. Acceleration rates are often described by
the time it takes to reach 60.0 mi/h from rest. If this time was
4.00 s, and Burt accelerated at this rate until he reached his
maximum speed, how long did it take Burt to complete the
course?


40. (a) A world record was set for the men’s 100-m dash in
the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica.
Bolt “coasted” across the finish line with a time of 9.69 s. If we
assume that Bolt accelerated for 3.00 s to reach his maximum
speed, and maintained that speed for the rest of the race,
calculate his maximum speed and his acceleration. (b) During
the same Olympics, Bolt also set 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, what was his maximum
speed for this race?


2.7 Falling Objects
Assume air resistance is negligible unless otherwise stated.


41. Calculate the displacement and velocity at times of (a)
0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown
straight up with an initial velocity of 15.0 m/s. Take the point of
release to be y0 = 0 .


42. Calculate the displacement and velocity at times of (a)
0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock
thrown straight down with an initial velocity of 14.0 m/s from
the Verrazano Narrows Bridge in New York City. The roadway
of this bridge is 70.0 m above the water.


43. A basketball referee tosses the ball straight up for the
starting tip-off. At what velocity must a basketball player leave
the ground to rise 1.25 m above the floor in an attempt to get
the ball?


44. A rescue helicopter is hovering over a person whose boat
has sunk. One of the rescuers throws a life preserver straight
down to the victim with an initial velocity of 1.40 m/s and
observes that it takes 1.8 s to reach the water. (a) List the
knowns in this problem. (b) How high above the water was
the preserver released? Note that the downdraft of the
helicopter reduces the effects of air resistance on the falling
life preserver, so that an acceleration equal to that of gravity
is reasonable.


45. A dolphin in an aquatic show jumps straight up out of the
water at a velocity of 13.0 m/s. (a) List the knowns in this
problem. (b) How high does his body rise above the water?
To solve this part, first note that the final velocity is now a


known and identify its value. Then identify the unknown, and
discuss how you chose the appropriate equation to solve for
it. After choosing the equation, show your steps in solving for
the unknown, checking units, and discuss whether the answer
is reasonable. (c) How long is the dolphin in the air? Neglect
any effects due to his size or orientation.


46. A swimmer bounces straight up from a diving board and
falls feet first into a pool. She starts with a velocity of 4.00 m/
s, and her takeoff point is 1.80 m above the pool. (a) How
long are her feet in the air? (b) What is her highest point
above the board? (c) What is her velocity when her feet hit
the water?


47. (a) Calculate the height of a cliff if it takes 2.35 s for a rock
to hit the ground when it is thrown straight up from the cliff
with an initial velocity of 8.00 m/s. (b) How long would it take
to reach the ground if it is thrown straight down with the same
speed?


48. A very strong, but inept, shot putter puts the shot straight
up vertically with an initial velocity of 11.0 m/s. How long does
he have to get out of the way if the shot was released at a
height of 2.20 m, and he is 1.80 m tall?


49. You throw a ball straight up with an initial velocity of 15.0
m/s. It passes a tree branch on the way up at a height of 7.00
m. How much additional time will pass before the ball passes
the tree branch on the way back down?


50. A kangaroo can jump over an object 2.50 m high. (a)
Calculate its vertical speed when it leaves the ground. (b)
How long is it in the air?


51. Standing at the base of one of the cliffs of Mt. Arapiles in
Victoria, Australia, a hiker hears a rock break loose from a
height of 105 m. He can’t see the rock right away but then
does, 1.50 s later. (a) How far above the hiker is the rock
when he can see it? (b) How much time does he have to
move before the rock hits his head?


52. An object is dropped from a height of 75.0 m above
ground level. (a) Determine the distance traveled during the
first second. (b) Determine the final velocity at which the
object hits the ground. (c) Determine the distance traveled
during the last second of motion before hitting the ground.


53. There is a 250-m-high cliff at Half Dome in Yosemite
National Park in California. Suppose a boulder breaks loose
from the top of this cliff. (a) How fast will it be going when it
strikes the ground? (b) Assuming a reaction time of 0.300 s,
how long will a tourist at the bottom have to get out of the way
after hearing the sound of the rock breaking loose (neglecting
the height of the tourist, which would become negligible
anyway if hit)? The speed of sound is 335 m/s on this day.


54. A ball is thrown straight up. It passes a 2.00-m-high
window 7.50 m off the ground on its path up and takes 1.30 s
to go past the window. What was the ball’s initial velocity?


55. Suppose you drop a rock into a dark well and, using
precision equipment, you measure the time for the sound of a
splash to return. (a) Neglecting the time required for sound to
travel up the well, calculate the distance to the water if the
sound returns in 2.0000 s. (b) Now calculate the distance
taking into account the time for sound to travel up the well.
The speed of sound is 332.00 m/s in this well.


56. A steel ball is dropped onto a hard floor from a height of
1.50 m and rebounds to a height of 1.45 m. (a) Calculate its
velocity just before it strikes the floor. (b) Calculate its velocity
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


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compress during its collision with the floor, assuming the floor
is absolutely rigid?


57. A coin is dropped from a hot-air balloon that is 300 m
above the ground and rising at 10.0 m/s upward. For the coin,
find (a) the maximum height reached, (b) its position and
velocity 4.00 s after being released, and (c) the time before it
hits the ground.


58. A soft tennis ball is dropped onto a hard floor from a
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 way
back 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 with the floor,
assuming the floor is absolutely rigid?


2.8 Graphical Analysis of One-Dimensional
Motion
Note: There is always uncertainty in numbers taken from
graphs. If your answers differ from expected values, examine
them to see if they are within data extraction uncertainties
estimated by you.


59. (a) By taking the slope of the curve in Figure 2.60, verify
that the velocity of the jet car is 115 m/s at t = 20 s . (b) By
taking the slope of the curve at any point in Figure 2.61,


verify that the jet car’s acceleration is 5.0 m/s2 .


Figure 2.60


Figure 2.61


60. Using approximate values, calculate the slope of the
curve in Figure 2.62 to verify that the velocity at t = 10.0 s
is 0.208 m/s. Assume all values are known to 3 significant
figures.


Figure 2.62


61. Using approximate values, calculate the slope of the
curve in Figure 2.62 to verify that the velocity at t = 30.0 s
is 0.238 m/s. Assume all values are known to 3 significant
figures.


62. By taking the slope of the curve in Figure 2.63, verify that


the acceleration is 3.2 m/s2 at t = 10 s .


Figure 2.63


63. Construct the displacement graph for the subway shuttle
train as shown in Figure 2.18(a). Your graph should show the
position of the train, in kilometers, from t = 0 to 20 s. You will
need to use the information on acceleration and velocity given
in the examples for this figure.


64. (a) Take the slope of the curve in Figure 2.64 to find the
jogger’s velocity at t = 2.5 s . (b) Repeat at 7.5 s. These
values must be consistent with the graph in Figure 2.65.


Figure 2.64


Chapter 2 | Kinematics 85




Figure 2.65


Figure 2.66


65. A graph of v(t) is shown for a world-class track sprinter
in a 100-m race. (See Figure 2.67). (a) What is his average
velocity for the first 4 s? (b) What is his instantaneous velocity
at t = 5 s ? (c) What is his average acceleration between 0
and 4 s? (d) What is his time for the race?


Figure 2.67


66. Figure 2.68 shows the displacement graph for a particle
for 5 s. Draw the corresponding velocity and acceleration
graphs.


Figure 2.68


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3 TWO-DIMENSIONAL KINEMATICS


Figure 3.1 Everyday motion that we experience is, thankfully, rarely as tortuous as a rollercoaster ride like this—the Dragon Khan in Spain’s Universal
Port Aventura Amusement Park. However, most motion is in curved, rather than straight-line, paths. Motion along a curved path is two- or three-
dimensional motion, and can be described in a similar fashion to one-dimensional motion. (credit: Boris23/Wikimedia Commons)


Chapter Outline
3.1. Kinematics in Two Dimensions: An Introduction


• Observe that motion in two dimensions consists of horizontal and vertical components.
• Understand the independence of horizontal and vertical vectors in two-dimensional motion.


3.2. Vector Addition and Subtraction: Graphical Methods
• Understand the rules of vector addition, subtraction, and multiplication.
• Apply graphical methods of vector addition and subtraction to determine the displacement of moving objects.


3.3. Vector Addition and Subtraction: Analytical Methods
• Understand the rules of vector addition and subtraction using analytical methods.
• Apply analytical methods to determine vertical and horizontal component vectors.
• Apply analytical methods to determine the magnitude and direction of a resultant vector.


3.4. Projectile Motion
• Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and


trajectory.
• Determine the location and velocity of a projectile at different points in its trajectory.
• Apply the principle of independence of motion to solve projectile motion problems.


3.5. Addition of Velocities
• Apply principles of vector addition to determine relative velocity.
• Explain the significance of the observer in the measurement of velocity.


Chapter 3 | Two-Dimensional Kinematics 87




Introduction to Two-Dimensional Kinematics
The arc of a basketball, the orbit of a satellite, a bicycle rounding a curve, a swimmer diving into a pool, blood gushing out of a
wound, and a puppy chasing its tail are but a few examples of motions along curved paths. In fact, most motions in nature follow
curved paths rather than straight lines. Motion along a curved path on a flat surface or a plane (such as that of a ball on a pool
table or a skater on an ice rink) is two-dimensional, and thus described by two-dimensional kinematics. Motion not confined to a
plane, such as a car following a winding mountain road, is described by three-dimensional kinematics. Both two- and three-
dimensional kinematics are simple extensions of the one-dimensional kinematics developed for straight-line motion in the
previous chapter. This simple extension will allow us to apply physics to many more situations, and it will also yield unexpected
insights about nature.


3.1 Kinematics in Two Dimensions: An Introduction


Figure 3.2Walkers and drivers in a city like New York are rarely able to travel in straight lines to reach their destinations. Instead, they must follow
roads and sidewalks, making two-dimensional, zigzagged paths. (credit: Margaret W. Carruthers)


Two-Dimensional Motion: Walking in a City
Suppose you want to walk from one point to another in a city with uniform square blocks, as pictured in Figure 3.3.


Figure 3.3 A pedestrian walks a two-dimensional path between two points in a city. In this scene, all blocks are square and are the same size.


The straight-line path that a helicopter might fly is blocked to you as a pedestrian, and so you are forced to take a two-
dimensional path, such as the one shown. You walk 14 blocks in all, 9 east followed by 5 north. What is the straight-line
distance?


An old adage states that the shortest distance between two points is a straight line. The two legs of the trip and the straight-line


path form a right triangle, and so the Pythagorean theorem, a2 + b2 = c2 , can be used to find the straight-line distance.


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Figure 3.4 The Pythagorean theorem relates the length of the legs of a right triangle, labeled a and b , with the hypotenuse, labeled c . The


relationship is given by: a2+ b2= c2 . This can be rewritten, solving for c : c = a2+ b2 .


The hypotenuse of the triangle is the straight-line path, and so in this case its length in units of city blocks is


(9 blocks)2+ (5 blocks)2= 10.3 blocks , considerably shorter than the 14 blocks you walked. (Note that we are using three
significant figures in the answer. Although it appears that “9” and “5” have only one significant digit, they are discrete numbers. In
this case “9 blocks” is the same as “9.0 or 9.00 blocks.” We have decided to use three significant figures in the answer in order to
show the result more precisely.)


Figure 3.5 The straight-line path followed by a helicopter between the two points is shorter than the 14 blocks walked by the pedestrian. All blocks are
square and the same size.


The fact that the straight-line distance (10.3 blocks) in Figure 3.5 is less than the total distance walked (14 blocks) is one
example of a general characteristic of vectors. (Recall that vectors are quantities that have both magnitude and direction.)


As for one-dimensional kinematics, we use arrows to represent vectors. The length of the arrow is proportional to the vector’s
magnitude. The arrow’s length is indicated by hash marks in Figure 3.3 and Figure 3.5. The arrow points in the same direction
as the vector. For two-dimensional motion, the path of an object can be represented with three vectors: one vector shows the
straight-line path between the initial and final points of the motion, one vector shows the horizontal component of the motion, and
one vector shows the vertical component of the motion. The horizontal and vertical components of the motion add together to
give the straight-line path. For example, observe the three vectors in Figure 3.5. The first represents a 9-block displacement
east. The second represents a 5-block displacement north. These vectors are added to give the third vector, with a 10.3-block
total displacement. The third vector is the straight-line path between the two points. Note that in this example, the vectors that we
are adding are perpendicular to each other and thus form a right triangle. This means that we can use the Pythagorean theorem
to calculate the magnitude of the total displacement. (Note that we cannot use the Pythagorean theorem to add vectors that are
not perpendicular. We will develop techniques for adding vectors having any direction, not just those perpendicular to one
another, in Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical
Methods.)


The Independence of Perpendicular Motions
The person taking the path shown in Figure 3.5 walks east and then north (two perpendicular directions). How far he or she
walks east is only affected by his or her motion eastward. Similarly, how far he or she walks north is only affected by his or her
motion northward.


Independence of Motion


The horizontal and vertical components of two-dimensional motion are independent of each other. Any motion in the
horizontal direction does not affect motion in the vertical direction, and vice versa.


This is true in a simple scenario like that of walking in one direction first, followed by another. It is also true of more complicated
motion involving movement in two directions at once. For example, let’s compare the motions of two baseballs. One baseball is
dropped from rest. At the same instant, another is thrown horizontally from the same height and follows a curved path. A
stroboscope has captured the positions of the balls at fixed time intervals as they fall.


Chapter 3 | Two-Dimensional Kinematics 89




Figure 3.6 This shows the motions of two identical balls—one falls from rest, the other has an initial horizontal velocity. Each subsequent position is an
equal time interval. Arrows represent horizontal and vertical velocities at each position. The ball on the right has an initial horizontal velocity, while the
ball on the left has no horizontal velocity. Despite the difference in horizontal velocities, the vertical velocities and positions are identical for both balls.
This shows that the vertical and 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 implies that
the vertical motion is independent of whether or not the ball is moving horizontally. (Assuming no air resistance, the vertical
motion of a falling object is influenced by gravity only, and not by any horizontal forces.) Careful examination of the ball thrown
horizontally shows that it travels the same horizontal distance between flashes. This is due to the fact that there are no additional
forces on the ball in the horizontal direction after it is thrown. This result means that the horizontal velocity is constant, and
affected neither by vertical motion nor by gravity (which is vertical). Note that this case is true only for ideal conditions. In the real
world, air resistance will affect the speed of the balls in both directions.


The two-dimensional curved path of the horizontally thrown ball is composed of two independent one-dimensional motions
(horizontal and vertical). The key to analyzing such motion, called projectile motion, is to resolve (break) it into motions along
perpendicular directions. Resolving two-dimensional motion into perpendicular components is possible because the components
are independent. We shall see how to resolve vectors in Vector Addition and Subtraction: Graphical Methods and Vector
Addition and Subtraction: Analytical Methods. We will find such techniques to be useful in many areas of physics.


PhET Explorations: Ladybug Motion 2D


Learn about position, velocity and acceleration vectors. Move the ladybug by setting the position, velocity or acceleration,
and see how the vectors change. Choose linear, circular or elliptical motion, and record and playback the motion to analyze
the behavior.


Figure 3.7 Ladybug Motion 2D (http://cnx.org/content/m42104/1.4/ladybug-motion-2d_en.jar)


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3.2 Vector Addition and Subtraction: Graphical Methods


Figure 3.8 Displacement can be determined graphically using a scale map, such as this one of the Hawaiian Islands. A journey from Hawai’i to
Moloka’i has a number of legs, or journey segments. These segments can be added graphically with a ruler to determine the total two-dimensional
displacement of the journey. (credit: US Geological Survey)


Vectors in Two Dimensions
A vector is a quantity that has magnitude and direction. Displacement, velocity, acceleration, and force, for example, are all
vectors. In one-dimensional, or straight-line, motion, the direction of a vector can be given simply by a plus or minus sign. In two
dimensions (2-d), however, we specify the direction of a vector relative to some reference frame (i.e., coordinate system), using
an arrow having length proportional to the vector’s magnitude and pointing in the direction of the vector.


Figure 3.9 shows such a graphical representation of a vector, using as an example the total displacement for the person walking
in a city considered in Kinematics in Two Dimensions: An Introduction. We shall use the notation that a boldface symbol,
such as D , stands for a vector. Its magnitude is represented by the symbol in italics, D , and its direction by θ .


Vectors in this Text


In this text, we will represent a vector with a boldface variable. For example, we will represent the quantity force with the
vector F , which has both magnitude and direction. The magnitude of the vector will be represented by a variable in italics,
such as F , and the direction of the variable will be given by an angle θ .


Figure 3.9 A person walks 9 blocks east and 5 blocks north. The displacement is 10.3 blocks at an angle 29.1º north of east.


Chapter 3 | Two-Dimensional Kinematics 91




Figure 3.10 To describe the resultant vector for the person walking in a city considered in Figure 3.9 graphically, draw an arrow to represent the total
displacement vector D . Using a protractor, draw a line at an angle θ relative to the east-west axis. The length D of the arrow is proportional to the
vector’s magnitude and is measured along the line with a ruler. In this example, the magnitude D of the vector is 10.3 units, and the direction θ is
29.1º north of east.


Vector Addition: Head-to-Tail Method
The head-to-tail method is a graphical way to add vectors, described in Figure 3.11 below and in the steps following. The tail
of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow.


Figure 3.11 Head-to-Tail Method: The head-to-tail method of graphically adding vectors is illustrated for the two displacements of the person walking
in a city considered in Figure 3.9. (a) Draw a vector representing the displacement to the east. (b) Draw a vector representing the displacement to the
north. The tail of this vector should originate from the head of the first, east-pointing vector. (c) Draw a line from the tail of the east-pointing vector to the
head of the north-pointing vector to form the sum or resultant vectorD . The length of the arrow D is proportional to the vector’s magnitude and is
measured to be 10.3 units . Its direction, described as the angle with respect to the east (or horizontal axis) θ is measured with a protractor to be
29.1º .


Step 1. Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor.


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Figure 3.12


Step 2. Now draw an arrow to represent the second vector (5 blocks to the north). Place the tail of the second vector at the head
of the first vector.


Figure 3.13


Step 3. If there are more than two vectors, continue this process for each vector to be added. Note that in our example, we have
only two vectors, so we have finished placing arrows tip to tail.


Step 4. Draw an arrow from the tail of the first vector to the head of the last vector. This is the resultant, or the sum, of the other
vectors.


Figure 3.14


Step 5. To get the magnitude of the resultant, measure its length with a ruler. (Note that in most calculations, we will use the
Pythagorean theorem to determine this length.)


Step 6. To get the direction of the resultant, measure the angle it makes with the reference frame using a protractor. (Note that
in most calculations, we will use trigonometric relationships to determine this angle.)


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The graphical addition of vectors is limited in accuracy only by the precision with which the drawings can be made and the
precision of the measuring tools. It is valid for any number of vectors.


Example 3.1 Adding Vectors Graphically Using the Head-to-Tail Method: A Woman Takes a
Walk


Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths
(displacements) on a flat field. First, she walks 25.0 m in a direction 49.0º north of east. Then, she walks 23.0 m heading
15.0º north of east. Finally, she turns and walks 32.0 m in a direction 68.0° south of east.
Strategy


Represent each displacement vector graphically with an arrow, labeling the first A , the second B , and the third C ,
making the lengths proportional to the distance and the directions as specified relative to an east-west line. The head-to-tail
method outlined above will give a way to determine the magnitude and direction of the resultant displacement, denoted R .
Solution


(1) Draw the three displacement vectors.


Figure 3.15


(2) Place the vectors head to tail retaining both their initial magnitude and direction.


Figure 3.16


(3) Draw the resultant vector, R .


Figure 3.17


(4) Use a ruler to measure the magnitude of R , and a protractor to measure the direction of R . While the direction of the
vector can be specified in many ways, the easiest way is to measure the angle between the vector and the nearest


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horizontal or vertical axis. Since the resultant vector is south of the eastward pointing axis, we flip the protractor upside down
and measure the angle between the eastward axis and the vector.


Figure 3.18


In this case, the total displacement R is seen to have a magnitude of 50.0 m and to lie in a direction 7.0º south of east. By
using its magnitude and direction, this vector can be expressed as R = 50.0 m and θ = 7.0º south of east.
Discussion


The head-to-tail graphical method of vector addition works for any number of vectors. It is also important to note that the
resultant is independent of the order in which the vectors are added. Therefore, we could add the vectors in any order as
illustrated in Figure 3.19 and we will still get the same solution.


Figure 3.19


Here, we see that when the same vectors are added in a different order, the result is the same. This characteristic is true in
every case and is an important characteristic of vectors. Vector addition is commutative. Vectors can be added in any order.


(3.1)A + B = B + A.


(This is true for the addition of ordinary numbers as well—you get the same result whether you add 2 + 3 or 3 + 2 , for
example).


Vector Subtraction


Vector subtraction is a straightforward extension of vector addition. To define subtraction (say we want to subtract B from A ,
written A – B , we must first define what we mean by subtraction. The negative of a vector B is defined to be –B ; that is,
graphically the negative of any vector has the same magnitude but the opposite direction, as shown in Figure 3.20. In other
words, B has the same length as –B , but points in the opposite direction. Essentially, we just flip the vector so it points in the
opposite direction.


Chapter 3 | Two-Dimensional Kinematics 95




Figure 3.20 The negative of a vector is just another vector of the same magnitude but pointing in the opposite direction. So B is the negative of –B ;
it has the same length but opposite direction.


The subtraction of vector B from vector A is then simply defined to be the addition of –B to A . Note that vector subtraction
is the addition of a negative vector. The order of subtraction does not affect the results.


(3.2)A – B = A + (–B).


This is analogous to the subtraction of scalars (where, for example, 5 – 2 = 5 + (–2) ). Again, the result is independent of the
order in which the subtraction is made. When vectors are subtracted graphically, the techniques outlined above are used, as the
following example illustrates.


Example 3.2 Subtracting Vectors Graphically: A Woman Sailing a Boat


A woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction
66.0º north of east from her current location, and then travel 30.0 m in a direction 112º north of east (or 22.0º west of
north). If the woman makes a mistake and travels in the opposite direction for the second leg of the trip, where will she end
up? Compare this location with the location of the dock.


Figure 3.21


Strategy


We can represent the first leg of the trip with a vector A , and the second leg of the trip with a vector B . The dock is
located at a location A + B . If the woman mistakenly travels in the opposite direction for the second leg of the journey, she
will travel a distance B (30.0 m) in the direction 180º – 112º = 68º south of east. We represent this as –B , as shown
below. The vector –B has the same magnitude as B but is in the opposite direction. Thus, she will end up at a location
A + (–B) , or A – B .


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Figure 3.22


We will perform vector addition to compare the location of the dock, A + B , with the location at which the woman
mistakenly arrives, A + (–B) .


Solution


(1) To determine the location at which the woman arrives by accident, draw vectors A and –B .
(2) Place the vectors head to tail.


(3) Draw the resultant vector R .


(4) Use a ruler and protractor to measure the magnitude and direction of R .


Figure 3.23


In this case, R = 23.0 m and θ = 7.5º south of east.


(5) To determine the location of the dock, we repeat this method to add vectors A and B . We obtain the resultant vector
R ' :


Figure 3.24


In this case R = 52.9 m and θ = 90.1º north of east.
We can see that the woman will end up a significant distance from the dock if she travels in the opposite direction for the
second leg of the trip.


Chapter 3 | Two-Dimensional Kinematics 97




Discussion


Because subtraction of a vector is the same as addition of a vector with the opposite direction, the graphical method of
subtracting vectors works the same as for addition.


Multiplication of Vectors and Scalars
If we decided to walk three times as far on the first leg of the trip considered in the preceding example, then we would walk
3 × 27.5 m , or 82.5 m, in a direction 66.0º north of east. This is an example of multiplying a vector by a positive scalar.
Notice that the magnitude changes, but the direction stays the same.


If the scalar is negative, then multiplying a vector by it changes the vector’s magnitude and gives the new vector the opposite
direction. For example, if you multiply by –2, the magnitude doubles but the direction changes. We can summarize these rules in
the following way: When vector A is multiplied by a scalar c ,


• the magnitude of the vector becomes the absolute value of c A ,
• if c is positive, the direction of the vector does not change,
• if c is negative, the direction is reversed.


In our case, c = 3 and A = 27.5 m . Vectors are multiplied by scalars in many situations. Note that division is the inverse of
multiplication. For example, dividing by 2 is the same as multiplying by the value (1/2). The rules for multiplication of vectors by
scalars are the same for division; simply treat the divisor as a scalar between 0 and 1.


Resolving a Vector into Components
In the examples above, we have been adding vectors to determine the resultant vector. In many cases, however, we will need to
do the opposite. We will need to take a single vector and find what other vectors added together produce it. In most cases, this
involves determining the perpendicular components of a single vector, for example the x- and y-components, or the north-south
and east-west components.


For example, we may know that the total displacement of a person walking in a city is 10.3 blocks in a direction 29.0º north of
east and want to find out how many blocks east and north had to be walked. This method is called finding the components (or
parts) of the displacement in the east and north directions, and it is the inverse of the process followed to find the total
displacement. It is one example of finding the components of a vector. There are many applications in physics where this is a
useful thing to do. We will see this soon in Projectile Motion, and much more when we cover forces in Dynamics: Newton’s
Laws of Motion. Most of these involve finding components along perpendicular axes (such as north and east), so that right
triangles are involved. The analytical techniques presented in Vector Addition and Subtraction: Analytical Methods are ideal
for finding vector components.


PhET Explorations: Maze Game


Learn about position, velocity, and acceleration in the "Arena of Pain". Use the green arrow to move the ball. Add more walls
to the arena to make the game more difficult. Try to make a goal as fast as you can.


Figure 3.25 Maze Game (http://
/content/m42127/1.7/maze-game_en.jar)


98 Chapter 3 | Two-Dimensional Kinematics


3.3 Vector Addition and Subtraction: Analytical Methods
Analytical methods of vector addition and subtraction employ geometry and simple trigonometry rather than the ruler and
protractor of graphical methods. Part of the graphical technique is retained, because vectors are still represented by arrows for
easy visualization. However, analytical methods are more concise, accurate, and precise than graphical methods, which are
limited by the accuracy with which a drawing can be made. Analytical methods are limited only by the accuracy and precision
with which physical quantities are known.


Resolving a Vector into Perpendicular Components
Analytical techniques and right triangles go hand-in-hand in physics because (among other things) motions along perpendicular
directions are independent. We very often need to separate a vector into perpendicular components. For example, given a vector
like A in Figure 3.26, we may wish to find which two perpendicular vectors, Ax and Ay , add to produce it.


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Figure 3.26 The vector A , with its tail at the origin of an x, y-coordinate system, is shown together with its x- and y-components, Ax and Ay .
These vectors form a right triangle. The analytical relationships among these vectors are summarized below.


Ax and Ay are defined to be the components of A along the x- and y-axes. The three vectors A , Ax , and Ay form a right
triangle:


(3.3)Ax + Ay = A.


Note that this relationship between vector components and the resultant vector holds only for vector quantities (which include
both magnitude and direction). The relationship does not apply for the magnitudes alone. For example, if Ax = 3 m east,
Ay = 4 m north, and A = 5 m north-east, then it is true that the vectors Ax + Ay = A . However, it is not true that the sum
of the magnitudes of the vectors is also equal. That is,


(3.4)3 m + 4 m ≠ 5 m


Thus,


(3.5)Ax + Ay ≠ A


If the vector A is known, then its magnitude A (its length) and its angle θ (its direction) are known. To find Ax and Ay , its x-
and y-components, we use the following relationships for a right triangle.


(3.6)Ax = A cos θ


and


(3.7)Ay = A sin θ.


Figure 3.27 The magnitudes of the vector components Ax and Ay can be related to the resultant vector A and the angle θ with trigonometric
identities. Here we see that Ax = A cos θ and Ay = A sin θ .


Suppose, for example, that A is the vector representing the total displacement of the person walking in a city considered in
Kinematics in Two Dimensions: An Introduction and Vector Addition and Subtraction: Graphical Methods.


Chapter 3 | Two-Dimensional Kinematics 99




Figure 3.28We can use the relationships Ax = A cos θ and Ay = A sin θ to determine the magnitude of the horizontal and vertical
component vectors in this example.


Then A = 10.3 blocks and θ = 29.1º , so that
(3.8)Ax = A cos θ = ⎛⎝10.3 blocks⎞⎠⎛⎝cos 29.1º⎞⎠ = 9.0 blocks
(3.9)Ay = A sin θ = ⎛⎝10.3 blocks⎞⎠⎛⎝sin 29.1º⎞⎠ = 5.0 blocks.


Calculating a Resultant Vector


If the perpendicular components Ax and Ay of a vector A are known, then A can also be found analytically. To find the
magnitude A and direction θ of a vector from its perpendicular components Ax and Ay , we use the following relationships:


(3.10)A = Ax2 + Ay2


(3.11)θ = tan−1(Ay / Ax).


Figure 3.29 The magnitude and direction of the resultant vector can be determined once the horizontal and vertical components Ax and Ay have


been determined.


Note that the equation A = Ax2 + Ay2 is just the Pythagorean theorem relating the legs of a right triangle to the length of the


hypotenuse. For example, if Ax and Ay are 9 and 5 blocks, respectively, then A = 92+52=10.3 blocks, again consistent


with the example of the person walking in a city. Finally, the direction is θ = tan–1(5/9)=29.1º , as before.


Determining Vectors and Vector Components with Analytical Methods


Equations Ax = A cos θ and Ay = A sin θ are used to find the perpendicular components of a vector—that is, to go


from A and θ to Ax and Ay . Equations A = Ax2 + Ay2 and θ = tan–1(Ay / Ax) are used to find a vector from its


perpendicular components—that is, to go from Ax and Ay to A and θ . Both processes are crucial to analytical methods


of vector addition and subtraction.


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Adding Vectors Using Analytical Methods


To see how to add vectors using perpendicular components, consider Figure 3.30, in which the vectors A and B are added to
produce the resultant R .


Figure 3.30 Vectors A and B are two legs of a walk, and R is the resultant or total displacement. You can use analytical methods to determine
the magnitude and direction of R .


If A and B represent two legs of a walk (two displacements), then R is the total displacement. The person taking the walk
ends up at the tip of R. There are many ways to arrive at the same point. In particular, the person could have walked first in the
x-direction and then in the y-direction. Those paths are the x- and y-components of the resultant, Rx and Ry . If we know Rx


and Ry , we can find R and θ using the equations A = Ax2 + Ay2 and θ = tan–1(Ay / Ax) . When you use the analytical
method of vector addition, you can determine the components or the magnitude and direction of a vector.


Step 1. Identify the x- and y-axes that will be used in the problem. Then, find the components of each vector to be added along
the chosen perpendicular axes. Use the equations Ax = A cos θ and Ay = A sin θ to find the components. In Figure 3.31,
these components are Ax , Ay , Bx , and By . The angles that vectors A and B make with the x-axis are θA and θB ,
respectively.


Figure 3.31 To add vectors A and B , first determine the horizontal and vertical components of each vector. These are the dotted vectors Ax ,
Ay , Bx and By shown in the image.


Step 2. Find the components of the resultant along each axis by adding the components of the individual vectors along that axis.
That is, as shown in Figure 3.32,


(3.12)Rx = Ax + Bx
and


(3.13)Ry = Ay + By.


Chapter 3 | Two-Dimensional Kinematics 101




Figure 3.32 The magnitude of the vectors Ax and Bx add to give the magnitude Rx of the resultant vector in the horizontal direction. Similarly,
the magnitudes of the vectors Ay and By add to give the magnitude Ry of the resultant vector in the vertical direction.


Components along the same axis, say the x-axis, are vectors along the same line and, thus, can be added to one another like
ordinary numbers. The same is true for components along the y-axis. (For example, a 9-block eastward walk could be taken in
two legs, the first 3 blocks east and the second 6 blocks east, for a total of 9, because they are along the same direction.) So
resolving vectors into components along common axes makes it easier to add them. Now that the components of R are known,
its magnitude and direction can be found.


Step 3. To get the magnitude R of the resultant, use the Pythagorean theorem:
(3.14)R = Rx2 + Ry2.


Step 4. To get the direction of the resultant:
(3.15)θ = tan−1(Ry / Rx).


The following example illustrates this technique for adding vectors using perpendicular components.


Example 3.3 Adding Vectors Using Analytical Methods


Add the vector A to the vector B shown in Figure 3.33, using perpendicular components along the x- and y-axes. The x-
and y-axes are along the east–west and north–south directions, respectively. Vector A represents the first leg of a walk in
which a person walks 53.0 m in a direction 20.0º north of east. Vector B represents the second leg, a displacement of
34.0 m in a direction 63.0º north of east.


Figure 3.33 Vector A has magnitude 53.0 m and direction 20.0 º north of the x-axis. Vector B has magnitude 34.0 m and direction
63.0º north of the x-axis. You can use analytical methods to determine the magnitude and direction of R .


Strategy


The components of A and B along the x- and y-axes represent walking due east and due north to get to the same ending
point. Once found, they are combined to produce the resultant.


Solution


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Following the method outlined above, we first find the components of A and B along the x- and y-axes. Note that
A = 53.0 m , θA = 20.0º , B = 34.0 m , and θB = 63.0º . We find the x-components by using Ax = A cos θ , which
gives


(3.16)Ax = A cos θA = (53.0 m)(cos 20.0º)
= (53.0 m)(0.940) = 49.8 m


and


(3.17)Bx = B cos θB = (34.0 m)(cos 63.0º)
= (34.0 m)(0.454) = 15.4 m.


Similarly, the y-components are found using Ay = A sin θA :


(3.18)Ay = A sin θA = (53.0 m)(sin 20.0º)
= (53.0 m)(0.342) = 18.1 m


and


(3.19)By = B sin θB = (34.0 m)(sin 63.0 º )
= (34.0 m)(0.891) = 30.3 m.


The x- and y-components of the resultant are thus


(3.20)Rx = Ax + Bx = 49.8 m + 15.4 m = 65.2 m


and


(3.21)Ry = Ay + By = 18.1 m+30.3 m = 48.4 m.


Now we can find the magnitude of the resultant by using the Pythagorean theorem:


(3.22)R = Rx2 + Ry2 = (65.2)2 + (48.4)2 m


so that


(3.23)R = 81.2 m.
Finally, we find the direction of the resultant:


(3.24)θ = tan−1(Ry / Rx)=+tan−1(48.4 / 65.2).


Thus,


(3.25)θ = tan−1(0.742) = 36.6 º .


Figure 3.34 Using analytical methods, we see that the magnitude of R is 81.2 m and its direction is 36.6º north of east.


Discussion


This example illustrates the addition of vectors using perpendicular components. Vector subtraction using perpendicular
components is very similar—it is just the addition of a negative vector.


Chapter 3 | Two-Dimensional Kinematics 103




Subtraction of vectors is accomplished by the addition of a negative vector. That is, A − B ≡ A + (–B) . Thus, the method
for the subtraction of vectors using perpendicular components is identical to that for addition. The components of –B are
the negatives of the components of B . The x- and y-components of the resultant A − B = R are thus


(3.26)Rx = Ax + ⎛⎝ – Bx⎞⎠


and


(3.27)Ry = Ay + ⎛⎝ – By⎞⎠


and the rest of the method outlined above is identical to that for addition. (See Figure 3.35.)


Analyzing vectors using perpendicular components is very useful in many areas of physics, because perpendicular quantities are
often independent of one another. The next module, Projectile Motion, is one of many in which using perpendicular components
helps make the picture clear and simplifies the physics.


Figure 3.35 The subtraction of the two vectors shown in Figure 3.30. The components of –B are the negatives of the components of B . The
method of subtraction is the same as that for addition.


PhET Explorations: Vector Addition


Learn how to add vectors. 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.


Figure 3.36 Vector Addition (http://
/content/m42128/1.10/vector-addition_en.jar)


104 Chapter 3 | Two-Dimensional Kinematics


3.4 Projectile Motion
Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The
object is called a projectile, and its path is called its trajectory. The motion of falling objects, as covered in Problem-Solving
Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal
movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air
resistance is negligible.


The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed
separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction, where vertical and horizontal motions
were seen to be 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 acceleration due to gravity
is vertical—thus, there will be 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. Figure 3.37 illustrates the notation for displacement, where s is
defined to be the total displacement and x and y are its components along the horizontal and vertical axes, respectively. The
magnitudes of these vectors are s, x, and y. (Note that in the last section we used the notation A to represent a vector with


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components Ax and Ay . If we continued this format, we would call displacement s with components sx and sy . However, to
simplify the notation, we will simply represent the component vectors as x and y .)


Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their
components along the x- and y-axes, too. We will assume all forces except gravity (such as air resistance and friction, for


example) are negligible. The components of acceleration are then very simple: ay = – g = – 9.80 m/s2 . (Note that this
definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead
such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical,
ax = 0 . Both accelerations are constant, so the kinematic equations can be used.


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


- t
(3.29)


v- = v0 + v2
(3.30)v = v0 + at
(3.31)x = x0 + v0t + 12at


2


(3.32)v2 = v0
2 + 2a(x − x0).


Figure 3.37 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.


Given these assumptions, the following steps are then used to analyze projectile motion:


Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are
perpendicular, so Ax = A cos θ and Ay = A sin θ are used. The magnitude of the components of displacement s along
these axes are x and y. The magnitudes of the components of the velocity v are vx = v cos θ and vy = v sin θ, where v
is the magnitude of the velocity and θ is its direction, as shown in Figure 3.38. Initial values are denoted with a subscript 0, as
usual.


Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic
equations for horizontal and vertical motion take the following forms:


(3.33)Horizontal Motion(ax = 0)
(3.34)x = x0 + vxt
(3.35)vx = v0x = vx = velocity is a constant.
(3.36)Vertical Motion(assuming positive is up ay = −g = −9.80m/s2)
(3.37)y = y0 + 12(v0y + vy)t


(3.38)vy = v0y − gt
(3.39)y = y0 + v0yt − 12gt


2


Chapter 3 | Two-Dimensional Kinematics 105




(3.40)vy
2 = v0y


2 − 2g(y − y0).


Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common
variable between the motions is time t . The problem solving procedures here are the same as for one-dimensional kinematics
and are illustrated in the solved examples below.


Step 4. Recombine the two motions to find the total displacement s and velocity v . Because the x - and y -motions are
perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical


Methods and employing A = Ax2 + Ay2 and θ = tan−1(Ay / Ax) in the following form, where θ is the direction of the


displacement s and θv is the direction of the velocity v :


Total displacement and velocity


(3.41)s = x2 + y2


(3.42)θ = tan−1(y / x)
(3.43)v = vx2 + vy2


(3.44)θv = tan−1(vy / vx).


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Figure 3.38 (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and
horizontal axes. (b) The horizontal motion is simple, because ax = 0 and vx is thus 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 the object falls towards the Earth again, the vertical velocity increases
again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x - and y -motions are recombined to give the total velocity
at any given point on the trajectory.


Example 3.4 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 75.0º above the
horizontal, as illustrated in Figure 3.39. The fuse is timed to ignite the shell just as it reaches its highest point above the
ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and
the explosion? (c) What is the horizontal displacement of the shell when it explodes?


Strategy


Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. 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 for (a)


Chapter 3 | Two-Dimensional Kinematics 107




By “height” we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, 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 :


(3.45)vy
2 = v0y


2 − 2g(y − y0).


Figure 3.39 The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a
height of 233 m and 125 m away horizontally.


Because y0 and vy are both zero, the equation simplifies to


(3.46)0 = v0y
2 − 2gy.


Solving for y gives


(3.47)
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 sin θ , where v0y is


the initial velocity of 70.0 m/s, and θ0 = 75.0º is the initial angle. Thus,


(3.48)v0y = v0 sin θ0 = (70.0 m/s)(sin 75º) = 67.6 m/s.


and y is


(3.49)
y = (67.6 m/s)


2


2(9.80 m/s2)
,


so that


(3.50)y = 233m.


Discussion for (a)


Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity
is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any
projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air
resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such
heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be
somewhat larger than that given to reach the same height.


Solution for (b)


As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest


method is to use y = y0 + 12(v0y + vy)t . Because y0 is zero, this equation reduces to simply


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(3.51)y = 12(v0y + vy)t.


Note that the final vertical velocity, vy , at the highest point is zero. Thus,


(3.52)
t = 2y(v0y + vy)


= 2(233 m)(67.6 m/s)
= 6.90 s.


Discussion for (b)


This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several


seconds pass before the shell explodes. (Another way of finding the time is by using y = y0 + v0yt − 12gt
2 , and solving


the quadratic equation for t .)


Solution for (c)


Because air resistance is negligible, ax = 0 and the horizontal velocity is constant, as discussed above. The horizontal
displacement is horizontal velocity multiplied by time as given by x = x0 + vxt , where x0 is equal to zero:


(3.53)x = vxt,


where vx is the x-component of the velocity, which is given by vx = v0 cos θ0 . Now,


(3.54)vx = v0 cos θ0 = (70.0 m/s)(cos 75.0º) = 18.1 m/s.


The time t for both motions is the same, and so x is


(3.55)x = (18.1 m/s)(6.90 s) = 125 m.


Discussion for (c)


The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could
be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major
effect, and many fragments will land directly below.


In solving part (a) of the preceding example, the expression we found for y is valid for any projectile motion where air resistance


is negligible. Call the maximum height y = h ; then,


(3.56)
h =


v0y
2


2g .


This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity.


Defining a Coordinate System


It is important to set up a coordinate system when analyzing projectile motion. One part of defining the coordinate system is
to define an origin for the x and y positions. Often, it is convenient to choose the initial position of the object as the origin


such that x0 = 0 and y0 = 0 . It is also important to define the positive and negative directions in the x and y directions.
Typically, we define the positive vertical direction as upwards, and the positive horizontal direction is usually the direction of
the object’s motion. When this is the case, the vertical acceleration, g , takes a negative value (since it is directed


downwards towards the Earth). However, it is occasionally useful to define the coordinates differently. For example, if you
are analyzing the motion of a ball thrown downwards from the top of a cliff, it may make sense to define the positive direction
downwards since the motion of the ball is solely in the downwards direction. If this is the case, g takes a positive value.


Example 3.5 Calculating Projectile Motion: Hot Rock Projectile


Kilauea in Hawaii is the world’s most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks
and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an
angle 35.0º above the horizontal, as shown in Figure 3.40. The rock strikes the side of the volcano at an altitude 20.0 m
lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and
direction of the rock’s velocity at impact?


Chapter 3 | Two-Dimensional Kinematics 109




Figure 3.40 The trajectory of a rock ejected from the Kilauea volcano.


Strategy


Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the
desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for t first. While
the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final
velocity. Thus, the vertical and horizontal results will be recombined to obtain v and θv at the final time t determined in


the first part of the example.


Solution for (a)


While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time
for this by using


(3.57)y = y0 + v0yt − 12gt
2.


If we take the initial position y0 to be zero, then the final position is y = −20.0 m. Now the initial vertical velocity is the
vertical component of the initial velocity, found from v0y = v0 sin θ0 = ( 25.0 m/s )( sin 35.0º ) = 14.3 m/s . Substituting
known values yields


(3.58)−20.0 m = (14.3 m/s)t − ⎛⎝4.90 m/s2

⎠t
2.


Rearranging terms gives a quadratic equation in t :


(3.59)⎛
⎝4.90 m/s2



⎠t
2 − (14.3 m/s)t − (20.0 m) = 0.


This expression is a quadratic equation of the form at2 + bt + c = 0 , where the constants are a = 4.90 , b = – 14.3 ,
and c = – 20.0. Its solutions are given by the quadratic formula:


(3.60)
t = −b ± b


2 − 4ac
2a .


This equation yields two solutions: t = 3.96 and t = – 1.03 . (It is left as an exercise for the reader to verify these
solutions.) The time is t = 3.96 s or – 1.03 s . The negative value of time implies an event before the start of motion, and
so we discard it. Thus,


(3.61)t = 3.96 s.
Discussion for (a)


The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical
velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.


Solution for (b)


From the information now in hand, we can find the final horizontal and vertical velocities vx and vy and combine them to


find the total velocity v and the angle θ0 it makes with the horizontal. Of course, vx is constant so we can solve for it at


any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle.
Therefore:


(3.62)vx = v0 cos θ0 = (25.0 m/s)(cos 35º) = 20.5 m/s.


The final vertical velocity is given by the following equation:


(3.63)vy = v0y − gt,


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where v0y was found in part (a) to be 14.3 m/s . Thus,


(3.64)vy = 14.3 m/s − (9.80 m/s2)(3.96 s)


so that


(3.65)vy = −24.5 m/s.


To find the magnitude of the final velocity v we combine its perpendicular components, using the following equation:


(3.66)v = vx2 + vy2 = (20.5 m/s)2 + ( − 24.5 m/s)2,


which gives


(3.67)v = 31.9 m/s.


The direction θv is found from the equation:


(3.68)θv = tan−1(vy / vx)


so that


(3.69)θv = tan−1( − 24.5 / 20.5) = tan−1( − 1.19).


Thus,


(3.70)θv = −50.1 º .


Discussion for (b)


The negative angle means that the velocity is 50.1º below the horizontal. This result is consistent with the fact that the final
vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the
initial altitude. (See Figure 3.40.)


One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each
other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level
ground, we define range to be the horizontal distance R traveled by a projectile. Galileo and many others were interested in the
range of projectiles primarily for military purposes—such as aiming cannons. However, investigating the range of projectiles can
shed light on other interesting phenomena, such as the orbits of satellites around the Earth. Let us consider projectile range
further.


Figure 3.41 Trajectories of projectiles on level ground. (a) The greater 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 same for 15º and 75º , although the
maximum heights of those paths are different.


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How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed v0 , the greater the range, as


shown in Figure 3.41(a). The initial angle θ0 also has a dramatic effect on the range, as illustrated in Figure 3.41(b). For a fixed


initial speed, such as might be produced by a cannon, the maximum range is obtained with θ0 = 45º . This is true only for
conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately 38º . Interestingly, for
every initial angle except 45º , there are two angles that give the same range—the sum of those angles is 90º . The range also
depends on the value of the acceleration of gravity g . The lunar astronaut Alan Shepherd was able to drive a golf ball a great


distance on the Moon because gravity is weaker there. The range R of a projectile on level ground for which air resistance is
negligible is given by


(3.71)
R =


v0
2 sin 2θ0


g ,


where v0 is the initial speed and θ0 is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-


chapter problem (hints are given), but it does fit the major features of projectile range as described.


When we speak of the range of a projectile on level ground, we assume that R is very small compared with the circumference of
the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes
direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther
to fall than it would on level ground. (See Figure 3.42.) If the initial speed is great enough, the projectile goes into orbit. This
possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from
underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other
aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text.


Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth
orbits. In Addition of Velocities, we will examine the addition of velocities, which is another important aspect of two-dimensional
kinematics and will also yield insights beyond the immediate topic.


Figure 3.42 Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing
initial speed, the range increases and becomes longer than it would be on level ground because the Earth curves away underneath its path. With a
large enough initial speed, orbit is achieved.


PhET Explorations: Projectile Motion


Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass.
Add air resistance. Make a game out of this simulation by trying to hit a target.


Figure 3.43 Projectile Motion (http://
/content/m42042/1.12/projectile-motion_en.jar)


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3.5 Addition of Velocities


Relative Velocity
If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves
diagonally relative to the shore, as in Figure 3.44. The boat does not move in the direction in which it is pointed. The reason, of
course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can
sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure 3.45. The plane is
moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways.


Figure 3.44 A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow)
relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore.


Figure 3.45 An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground
in the direction it points; rather, it moves in the direction of its total velocity (solid arrow).


In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity
relative to an observer on solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as
indicated in Figure 3.44 and Figure 3.45. These situations are only two of many in which it is useful to add velocities. In this
module, we first re-examine how to add velocities and then consider certain aspects of what relative velocity means.


How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in
Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods apply to
the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is
simple—they add like ordinary numbers. For example, if a field hockey player is moving at 5 m/s straight toward the goal and
drives the ball in the same direction with a velocity of 30 m/s relative to her body, then the velocity of the ball is 35 m/s relative
to the stationary, profusely sweating goalkeeper standing in front of the goal.


Chapter 3 | Two-Dimensional Kinematics 113




In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on
analytical techniques. The following equations give the relationships between the magnitude and direction of velocity ( v and θ )
and its components ( vx and vy ) along the x- and y-axes of an appropriately chosen coordinate system:


(3.72)vx = v cos θ
(3.73)vy = v sin θ
(3.74)v = vx2 + vy2


(3.75)θ = tan−1(vy / vx).


Figure 3.46 The velocity, v , of an object traveling at an angle θ to the horizontal axis is the sum of component vectors vx and vy .


These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the
components of a velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction
of velocity when its components are known.


Take-Home Experiment: Relative Velocity of a Boat


Fill a bathtub half-full of water. Take a toy boat or some other object that floats in water. Unplug the drain so water starts to
drain. Try pushing the boat from one side of the tub to the other and perpendicular to the flow of water. Which way do you
need to push the boat so that it ends up immediately opposite? Compare the directions of the flow of water, heading of the
boat, and actual velocity of the boat.


Example 3.6 Adding Velocities: A Boat on a River


Figure 3.47 A boat attempts to travel straight across a river at a speed 0.75 m/s. The current in the river, however, flows at a speed of 1.20 m/s to
the right. What is the total displacement of the boat relative to the shore?


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Refer to Figure 3.47, which shows a boat trying to go straight across the river. Let us calculate the magnitude and direction
of the boat’s velocity relative to an observer on the shore, vtot . The velocity of the boat, vboat , is 0.75 m/s in the y -
direction relative to the river and the velocity of the river, vriver , is 1.20 m/s to the right.


Strategy


We start by choosing a coordinate system with its x -axis parallel to the velocity of the river, as shown in Figure 3.47.
Because the boat is directed straight toward the other shore, its velocity relative to the water is parallel to the y -axis and


perpendicular to the velocity of the river. Thus, we can add the two velocities by using the equations vtot = vx2 + vy2 and


θ = tan−1(vy / vx) directly.


Solution


The magnitude of the total velocity is


(3.76)vtot = vx2 + vy2,


where


(3.77)vx = vriver = 1.20 m/s


and


(3.78)vy = vboat = 0.750 m/s.


Thus,


(3.79)vtot = (1.20 m/s)2 + (0.750 m/s)2


yielding


(3.80)vtot = 1.42 m/s.


The direction of the total velocity θ is given by:


(3.81)θ = tan−1(vy / vx) = tan−1(0.750 / 1.20).


This equation gives


(3.82)θ = 32.0º.
Discussion


Both the magnitude v and the direction θ of the total velocity are consistent with Figure 3.47. Note that because the
velocity of the river is large compared with the velocity of the boat, it is swept rapidly downstream. This result is evidenced
by the small angle (only 32.0º ) the total velocity has relative to the riverbank.


Example 3.7 Calculating Velocity: Wind Velocity Causes an Airplane to Drift


Calculate the wind velocity for the situation shown in Figure 3.48. The plane is known to be moving at 45.0 m/s due north
relative to the air mass, while its velocity relative to the ground (its total velocity) is 38.0 m/s in a direction 20.0º west of
north.


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Figure 3.48 An airplane is known to be heading north at 45.0 m/s, though its velocity relative to the ground is 38.0 m/s at an angle west of north.
What is the speed and direction of the wind?


Strategy


In this problem, somewhat different from the previous example, we know the total velocity vtot and that it is the sum of two
other velocities, vw (the wind) and vp (the plane relative to the air mass). The quantity vp is known, and we are asked to
find vw . None of the velocities are perpendicular, but it is possible to find their components along a common set of
perpendicular axes. If we can find the components of vw , then we can combine them to solve for its magnitude and
direction. As shown in Figure 3.48, we choose a coordinate system with its x-axis due east and its y-axis due north (parallel
to vp ). (You may wish to look back at the discussion of the addition of vectors using perpendicular components in Vector
Addition and Subtraction: Analytical Methods.)


Solution


Because vtot is the vector sum of the vw and vp , its x- and y-components are the sums of the x- and y-components of
the wind and plane velocities. Note that the plane only has vertical component of velocity so vpx = 0 and vpy = vp . That
is,


(3.83)vtotx = vwx
and


(3.84)vtoty = vwy + vp.


We can use the first of these two equations to find vwx :


(3.85)vwy = vtotx = vtotcos 110º.


Because vtot = 38.0 m / s and cos 110º = – 0.342 we have


(3.86)vwy = (38.0 m/s)(–0.342) = –13 m/s.


The minus sign indicates motion west which is consistent with the diagram.


Now, to find vwy we note that


(3.87)vtoty = vwy + vp


Here vtoty = vtotsin 110º ; thus,


(3.88)vwy = (38.0 m/s)(0.940) − 45.0 m/s = −9.29 m/s.


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This minus sign indicates motion south which is consistent with the diagram.


Now that the perpendicular components of the wind velocity vwx and vwy are known, we can find the magnitude and


direction of vw . First, the magnitude is


(3.89)vw = vwx2 + vwy2


= ( − 13.0 m/s)2 + ( − 9.29 m/s)2


so that


(3.90)vw = 16.0 m/s.


The direction is:


(3.91)θ = tan−1(vwy / vwx) = tan−1( − 9.29 / −13.0)


giving


(3.92)θ = 35.6º.
Discussion


The wind’s speed and direction are consistent with the significant effect the wind has on the total velocity of the plane, as
seen in Figure 3.48. Because the plane is fighting a strong combination of crosswind and head-wind, it ends up with a total
velocity significantly less than its velocity relative to the air mass as well as heading in a different direction.


Note that in both of the last two examples, we were able to make the mathematics easier by choosing a coordinate system with
one axis parallel to one of the velocities. We will repeatedly find that choosing an appropriate coordinate system makes problem
solving easier. For example, in projectile motion we always use a coordinate system with one axis parallel to gravity.


Relative Velocities and Classical Relativity
When adding velocities, we have been careful to specify that the velocity is relative to some reference frame. These velocities
are called relative velocities. For example, the velocity of an airplane relative to an air mass is different from its velocity relative
to the ground. Both are quite different from the velocity of an airplane relative to its passengers (which should be close to zero).
Relative velocities are one aspect of relativity, which is defined to be the study of how different observers moving relative to
each other measure the same phenomenon.


Nearly everyone has heard of relativity and immediately associates it with Albert Einstein (1879–1955), the greatest physicist of
the 20th century. Einstein revolutionized our view of nature with his modern theory of relativity, which we shall study in later
chapters. The relative velocities in this section are actually aspects of classical relativity, first discussed correctly by Galileo and
Isaac Newton. Classical relativity is limited to situations where speeds are less than about 1% of the speed of light—that is,
less than 3,000 km/s . Most things we encounter in daily life move slower than this speed.


Let us consider an example of what two different observers see in a situation analyzed long ago by Galileo. Suppose a sailor at
the top of a mast on a moving ship drops his binoculars. Where will it hit the deck? Will it hit at the base of the mast, or will it hit
behind the mast because the ship is moving forward? The answer is that if air resistance is negligible, the binoculars will hit at
the base of the mast at a point directly below its point of release. Now let us consider what two different observers see when the
binoculars drop. One observer is on the ship and the other on shore. The binoculars have no horizontal velocity relative to the
observer on the ship, and so he sees them fall straight down the mast. (See Figure 3.49.) To the observer on shore, the
binoculars and the ship have the same horizontal velocity, so both move the same distance forward while the binoculars are
falling. This observer sees the curved path shown in Figure 3.49. Although the paths look different to the different observers,
each sees the same result—the binoculars hit at the base of the mast and not behind it. To get the correct description, it is crucial
to correctly specify the velocities relative to the observer.


Chapter 3 | Two-Dimensional Kinematics 117




Figure 3.49 Classical relativity. The same motion as viewed by two different observers. An observer on the moving ship sees the binoculars dropped
from the top of its mast fall straight down. An observer on shore sees the binoculars take the curved path, moving forward with the ship. Both observers
see the binoculars strike the deck at the base of the mast. The initial horizontal velocity is different relative to the two observers. (The ship is shown
moving rather fast to emphasize the effect.)


Example 3.8 Calculating Relative Velocity: An Airline Passenger Drops a Coin


An airline passenger drops a coin while the plane is moving at 260 m/s. What is the velocity of the coin when it strikes the
floor 1.50 m below its point of release: (a) Measured relative to the plane? (b) Measured relative to the Earth?


Figure 3.50 The motion of a coin dropped inside an airplane as viewed by two different observers. (a) An observer in the plane sees the coin fall
straight down. (b) An observer on the ground sees the coin move almost horizontally.


Strategy


Both problems can be solved with the techniques for falling objects and projectiles. In part (a), the initial velocity of the coin
is zero relative to the plane, so the motion is that of a falling object (one-dimensional). In part (b), the initial velocity is 260 m/
s horizontal relative to the Earth and gravity is vertical, so this motion is a projectile motion. In both parts, it is best to use a
coordinate system with vertical and horizontal axes.


Solution for (a)


Using the given information, we note that the initial velocity and position are zero, and the final position is 1.50 m. The final
velocity can be found using the equation:


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(3.93)vy
2 = v0y


2 − 2g(y − y0).


Substituting known values into the equation, we get


(3.94)vy
2 = 02 − 2(9.80 m/s2)( − 1.50 m − 0 m) = 29.4 m2 /s2


yielding


(3.95)vy = −5.42 m/s.


We know that the square root of 29.4 has two roots: 5.42 and -5.42. We choose the negative root because we know that the
velocity is directed downwards, and we have defined the positive direction to be upwards. There is no initial horizontal
velocity relative to the plane and no horizontal acceleration, and so the motion is straight down relative to the plane.


Solution for (b)


Because the initial vertical velocity is zero relative to the ground and vertical motion is independent of horizontal motion, the
final vertical velocity for the coin relative to the ground is vy = − 5.42 m/s , the same as found in part (a). In contrast to
part (a), there now is a horizontal component of the velocity. However, since there is no horizontal acceleration, the initial
and final horizontal velocities are the same and vx = 260 m/s . The x- and y-components of velocity can be combined to
find the magnitude of the final velocity:


(3.96)v = vx2 + vy2.


Thus,


(3.97)v = (260 m/s)2 + ( − 5.42 m/s)2


yielding


(3.98)v = 260.06 m/s.
The direction is given by:


(3.99)θ = tan−1(vy / vx) = tan−1( − 5.42 / 260)


so that


(3.100)θ = tan−1( − 0.0208) = −1.19º.


Discussion


In part (a), the final velocity relative to the plane is the same as it would be if the coin were dropped from rest on the Earth
and fell 1.50 m. This result fits our experience; objects in a plane fall the same way when the plane is flying horizontally as
when it is at rest on the ground. This result is also true in moving cars. In part (b), an observer on the ground sees a much
different motion for the coin. The plane is moving so fast horizontally to begin with that its final velocity is barely greater than
the initial velocity. Once again, we see that in two dimensions, vectors do not add like ordinary numbers—the final velocity v
in part (b) is not (260 – 5.42) m/s ; rather, it is 260.06 m/s . The velocity’s magnitude had to be calculated to five digits to
see any difference from that of the airplane. The motions as seen by different observers (one in the plane and one on the
ground) in this example are analogous to those discussed for the binoculars dropped from the mast of a moving ship, except
that the velocity of the plane is much larger, so that the two observers see very different paths. (See Figure 3.50.) In
addition, both observers see the coin fall 1.50 m vertically, but the one on the ground also sees it move forward 144 m (this
calculation is left for the reader). Thus, one observer sees a vertical path, the other a nearly horizontal path.


Making Connections: Relativity and Einstein


Because Einstein was able to clearly define how measurements are made (some involve light) and because the speed
of light is the same for all observers, the outcomes are spectacularly unexpected. Time varies with observer, energy is
stored as increased mass, and more surprises await.


PhET Explorations: Motion in 2D


Try the new "Ladybug Motion 2D" simulation for the latest updated version. Learn about position, velocity, and acceleration
vectors. Move the ball with the mouse or let the simulation move the ball in four types of motion (2 types of linear, simple
harmonic, circle).


Chapter 3 | Two-Dimensional Kinematics 119




air resistance:


analytical method:


classical relativity:


commutative:


component (of a 2-d vector):


direction (of a vector):


head (of a vector):


head-to-tail method:


kinematics:


magnitude (of a vector):


motion:


projectile:


projectile motion:


range:


relative velocity:


relativity:


resultant:


resultant vector:


scalar:


tail:


trajectory:


vector:


vector addition:


velocity:


Figure 3.51 Motion in 2D (http://


/content/m42045/1.12/motion-2d_en.jar) Glossary


a frictional force that slows the motion of objects as they travel through the air; when solving basic physics
problems, air resistance is assumed to be zero


the method of determining the magnitude and direction of a resultant vector using the Pythagorean
theorem and trigonometric identities


the study of relative velocities in situations where speeds are less than about 1% of the speed of
light—that is, less than 3000 km/s


refers to the interchangeability of order in a function; vector addition is commutative because the order in which
vectors are added together does not affect the final sum


a piece of a vector that points in either the vertical or the horizontal direction; every 2-d vector
can be expressed as a sum of two vertical and horizontal vector components


the orientation of a vector in space


the end point of a vector; the location of the tip of the vector’s arrowhead; also referred to as the “tip”


a method of adding vectors in which the tail of each vector is placed at the head of the previous vector


the study of motion without regard to mass or force


the length or size of a vector; magnitude is a scalar quantity


displacement of an object as a function of time


an object that travels through the air and experiences only acceleration due to gravity


the motion of an object that is subject only to the acceleration of gravity


the maximum horizontal distance that a projectile travels


the velocity of an object as observed from a particular reference frame


the study of how different observers moving relative to each other measure the same phenomenon


the sum of two or more vectors


the vector sum of two or more vectors


a quantity with magnitude but no direction


the start point of a vector; opposite to the head or tip of the arrow


the path of a projectile through the air


a quantity that has both magnitude and direction; an arrow used to represent quantities with both magnitude and
direction


the rules that apply to adding vectors together


speed in a given direction


Section Summary


3.1 Kinematics in Two Dimensions: An Introduction
• The shortest path between any two points is a straight line. In two dimensions, this path can be represented by a vector


with horizontal and vertical components.


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• The horizontal and vertical components of a vector are independent of one another. Motion in the horizontal direction does
not affect motion in the vertical direction, and vice versa.


3.2 Vector Addition and Subtraction: Graphical Methods
• The graphical method of adding vectors A and B involves drawing vectors on a graph and adding them using the


head-to-tail method. The resultant vector R is defined such that A + B = R . The magnitude and direction of R are
then determined with a ruler and protractor, respectively.


• The graphical method of subtracting vector B from A involves adding the opposite of vector B , which is defined as
−B . In this case, A – B = A + (–B) = R . Then, the head-to-tail method of addition is followed in the usual way to
obtain the resultant vector R .


• Addition of vectors is commutative such that A + B = B + A .
• The head-to-tail method of adding vectors involves drawing the first vector on a graph and then placing the tail of each


subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to
the head of the final vector.


• If a vector A is multiplied by a scalar quantity c , the magnitude of the product is given by cA . If c is positive, the
direction of the product points in the same direction as A ; if c is negative, the direction of the product points in the
opposite direction as A .


3.3 Vector Addition and Subtraction: Analytical Methods
• The analytical method of vector addition and subtraction involves using the Pythagorean theorem and trigonometric


identities to determine the magnitude and direction of a resultant vector.
• The steps to add vectors A and B using the analytical method are as follows:


Step 1: Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each
vector using the equations


Ax = A cos θ
Bx = B cos θ


and


Ay = A sin θ
By = B sin θ.


Step 2: Add the horizontal and vertical components of each vector to determine the components Rx and Ry of the


resultant vector, R :
Rx = Ax + Bx


and


Ry = Ay + By.
Step 3: Use the Pythagorean theorem to determine the magnitude, R , of the resultant vector R :


R = Rx2 + Ry2.


Step 4: Use a trigonometric identity to determine the direction, θ , of R :


θ = tan−1(Ry / Rx).


3.4 Projectile Motion
• Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.
• To solve projectile motion problems, perform the following steps:


1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical
components. The components of position s are given by the quantities x and y , and the components of the velocity
v are given by vx = v cos θ and vy = v sin θ , where v is the magnitude of the velocity and θ is its direction.


2. Analyze the motion of the projectile in the horizontal direction using the following equations:


Horizontal motion(ax = 0)


x = x0 + vxt


vx = v0x = vx = velocity is a constant.
3. Analyze the motion of the projectile in the vertical direction using the following equations:


Chapter 3 | Two-Dimensional Kinematics 121




Vertical motion(Assuming positive direction is up; ay = −g = −9.80 m/s2)


y = y0 + 12(v0y + vy)t


vy = v0y − gt


y = y0 + v0yt − 12gt
2


vy
2 = v0y


2 − 2g(y − y0).


4. Recombine the horizontal and vertical components of location and/or velocity using the following equations:


s = x2 + y2


θ = tan−1(y / x)


v = vx2 + vy2


θv = tan−1(vy / vx).
• The maximum height h of a projectile launched with initial vertical velocity v0y is given by


h =
v0y
2


2g .


• The maximum horizontal distance traveled by a projectile is called the range. The range R of a projectile on level ground
launched at an angle θ0 above the horizontal with initial speed v0 is given by


R =
v0
2 sin 2θ0


g .


3.5 Addition of Velocities
• Velocities in two dimensions are added using the same analytical vector techniques, which are rewritten as


vx = v cos θ
vy = v sin θ


v = vx2 + vy2


θ = tan−1(vy / vx).
• Relative velocity is the velocity of an object as observed from a particular reference frame, and it varies dramatically with


reference frame.
• Relativity is the study of how different observers measure the same phenomenon, particularly when the observers move


relative to one another. Classical relativity is limited to situations where speed is less than about 1% of the speed of light
(3000 km/s).


Conceptual Questions


3.2 Vector Addition and Subtraction: Graphical Methods
1.Which of the following is a vector: a person’s height, the altitude on Mt. Everest, the age of the Earth, the boiling point of water,
the cost of this book, the Earth’s population, the acceleration of gravity?


2. Give a specific example of a vector, stating its magnitude, units, and direction.


3.What do vectors and scalars have in common? How do they differ?


4. Two campers in a national park hike from their cabin to the same spot on a lake, each taking a different path, as illustrated
below. The total distance traveled along Path 1 is 7.5 km, and that along Path 2 is 8.2 km. What is the final displacement of each
camper?


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Figure 3.52


5. If an airplane pilot is told to fly 123 km in a straight line to get from San Francisco to Sacramento, explain why he could end up
anywhere on the circle shown in Figure 3.53. What other information would he need to get to Sacramento?


Figure 3.53


6. Suppose you take two steps A and B (that is, two nonzero displacements). Under what circumstances can you end up at
your starting point? More generally, under what circumstances can two nonzero vectors add to give zero? Is the maximum
distance you can end up from the starting point A + B the sum of the lengths of the two steps?
7. Explain why it is not possible to add a scalar to a vector.


8. If you take two steps of different sizes, can you end up at your starting point? More generally, can two vectors with different
magnitudes ever add to zero? Can three or more?


3.3 Vector Addition and Subtraction: Analytical Methods


9. Suppose you add two vectors A and B . What relative direction between them produces the resultant with the greatest
magnitude? What is the maximum magnitude? What relative direction between them produces the resultant with the smallest
magnitude? What is the minimum magnitude?


10. Give an example of a nonzero vector that has a component of zero.


11. Explain why a vector cannot have a component greater than its own magnitude.


12. If the vectors A and B are perpendicular, what is the component of A along the direction of B ? What is the component
of B along the direction of A ?


3.4 Projectile Motion
13. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being
neither 0º nor 90º ): (a) Is the velocity ever zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever
be the same as the initial velocity at a time other than at t = 0 ? (d) Can the speed ever be the same as the initial speed at a
time other than at t = 0 ?


Chapter 3 | Two-Dimensional Kinematics 123




14. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being
neither 0º nor 90º ): (a) Is the acceleration ever zero? (b) Is the acceleration ever in the same direction as a component of
velocity? (c) Is the acceleration ever opposite in direction to a component of velocity?


15. For a fixed initial speed, the range of a projectile is determined by the angle at which it is fired. For all but the maximum, there
are two angles that give the same range. Considering factors that might affect the ability of an archer to hit a target, such as
wind, explain why the smaller angle (closer to the horizontal) is preferable. When would it be necessary for the archer to use the
larger angle? Why does the punter in a football game use the higher trajectory?


16. During a lecture demonstration, a professor places two coins on the edge of a table. She then flicks one of the coins
horizontally off the table, simultaneously nudging the other over the edge. Describe the subsequent motion of the two coins, in
particular discussing whether they hit the floor at the same time.


3.5 Addition of Velocities
17.What frame or frames of reference do you instinctively use when driving a car? When flying in a commercial jet airplane?


18. A basketball player dribbling down the court usually keeps his eyes fixed on the players around him. He is moving fast. Why
doesn’t he need to keep his eyes on the ball?


19. If someone is riding in the back of a pickup truck and throws a softball straight backward, is it possible for the ball to fall
straight down as viewed by a person standing at the side of the road? Under what condition would this occur? How would the
motion of the ball appear to the person who threw it?


20. The hat of a jogger running at constant velocity falls off the back of his head. Draw a sketch showing the path of the hat in the
jogger’s frame of reference. Draw its path as viewed by a stationary observer.


21. A clod of dirt falls from the bed of a moving truck. It strikes the ground directly below the end of the truck. What is the
direction of its velocity relative to the truck just before it hits? Is this the same as the direction of its velocity relative to ground just
before it hits? Explain your answers.


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Problems & Exercises


3.2 Vector Addition and Subtraction: Graphical
Methods
Use graphical methods to solve these problems. You may
assume data taken from graphs is accurate to three
digits.


1. Find the following for path A in Figure 3.54: (a) the total
distance traveled, and (b) the magnitude and direction of the
displacement from start to finish.


Figure 3.54 The various lines represent paths taken by different people
walking in a city. All blocks are 120 m on a side.


2. Find the following for path B in Figure 3.54: (a) the total
distance traveled, and (b) the magnitude and direction of the
displacement from start to finish.


3. Find the north and east components of the displacement
for the hikers shown in Figure 3.52.


4. Suppose you walk 18.0 m straight west and then 25.0 m
straight north. How far are you from your starting point, and
what is the compass direction of a line connecting your
starting point to your final position? (If you represent the two
legs of the walk as vector displacements A and B , as in
Figure 3.55, then this problem asks you to find their sum
R = A + B .)


Figure 3.55 The two displacements A and B add to give a total
displacement R having magnitude R and direction θ .
5. Suppose you first walk 12.0 m in a direction 20º west of
north and then 20.0 m in a direction 40.0º south of west.
How far are you from your starting point, and what is the
compass direction of a line connecting your starting point to
your final position? (If you represent the two legs of the walk
as vector displacements A and B , as in Figure 3.56, then
this problem finds their sum R = A + B .)


Figure 3.56


6. Repeat the problem above, but reverse the order of the two
legs of the walk; show that you get the same final result. That
is, you first walk leg B , which is 20.0 m in a direction exactly
40º south of west, and then leg A , which is 12.0 m in a
direction exactly 20º west of north. (This problem shows that
A + B = B + A .)
7. (a) Repeat the problem two problems prior, but for the
second leg you walk 20.0 m in a direction 40.0º north of east
(which is equivalent to subtracting B from A —that is, to
finding R′ = A − B ). (b) Repeat the problem two problems
prior, but now you first walk 20.0 m in a direction 40.0º south
of west and then 12.0 m in a direction 20.0º east of south
(which is equivalent to subtracting A from B —that is, to
finding R′′ = B - A = - R′ ). Show that this is the case.
8. Show that the order of addition of three vectors does not
affect their sum. Show this property by choosing any three
vectors A , B , and C , all having different lengths and
directions. Find the sum A + B + C then find their sum
when added in a different order and show the result is the
same. (There are five other orders in which A , B , and C
can be added; choose only one.)


9. Show that the sum of the vectors discussed in Example
3.2 gives the result shown in Figure 3.24.


10. Find the magnitudes of velocities vA and vB in Figure


3.57


Figure 3.57 The two velocities vA and vB add to give a total vtot .


11. Find the components of vtot along the x- and y-axes in


Figure 3.57.


Chapter 3 | Two-Dimensional Kinematics 125




12. Find the components of vtot along a set of perpendicular


axes rotated 30º counterclockwise relative to those in
Figure 3.57.


3.3 Vector Addition and Subtraction: Analytical
Methods
13. Find the following for path C in Figure 3.58: (a) the total
distance traveled and (b) the magnitude and direction of the
displacement from start to finish. In this part of the problem,
explicitly show how you follow the steps of the analytical
method of vector addition.


Figure 3.58 The various lines represent paths taken by different people
walking in a city. All blocks are 120 m on a side.


14. Find the following for path D in Figure 3.58: (a) the total
distance traveled and (b) the magnitude and direction of the
displacement from start to finish. In this part of the problem,
explicitly show how you follow the steps of the analytical
method of vector addition.


15. Find the north and east components of the displacement
from San Francisco to Sacramento shown in Figure 3.59.


Figure 3.60 The two displacements A and B add to give a total
displacement R having magnitude R and direction θ .
Note that you can also solve this graphically. Discuss why the
analytical technique for solving this problem is potentially
more accurate than the graphical technique.


17. Repeat Exercise 3.16 using analytical techniques, but
reverse the order of the two legs of the walk and show that
you get the same final result. (This problem shows that
adding them in reverse order gives the same result—that is,
B + A = A + B .) Discuss how taking another path to
reach the same point might help to overcome an obstacle
blocking you other path.


18. You drive 7.50 km in a straight line in a direction 15º
east of north. (a) Find the distances you would have to drive
straight east and then straight north to arrive at the same
point. (This determination is equivalent to find the
components of the displacement along the east and north
directions.) (b) Show that you still arrive at the same point if
the east and north legs are reversed in order.


19. Do Exercise 3.16 again using analytical techniques and
change the second leg of the walk to 25.0 m straight south.
(This is equivalent to subtracting B from A —that is, finding
R′ = A – B ) (b) Repeat again, but now you first walk
25.0 m north and then 18.0 m east. (This is equivalent to
subtract A from B —that is, to find A = B + C . Is that
consistent with your result?)


20. A new landowner has a triangular piece of flat land she
wishes to fence. Starting at the west corner, she measures
the first side to be 80.0 m long and the next to be 105 m.
These sides are represented as displacement vectors A
from B in Figure 3.61. She then correctly calculates the
length and orientation of the third side C . What is her
result?


126 Chapter 3 | Two-Dimensional Kinematics


Figure 3.59


16. Solve the following problem using analytical techniques:
Suppose you walk 18.0 m straight west and then 25.0 m
straight north. How far are you from your starting point, and
what is the compass direction of a line connecting your
starting point to your final position? (If you represent the two
legs of the walk as vector displacements A and B , as in
Figure 3.60, then this problem asks you to find their sum
R = A + B .)


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Figure 3.61


21. You fly 32.0 km in a straight line in still air in the
direction 35.0º south of west. (a) Find the distances you
would have to fly straight south and then straight west to
arrive at the same point. (This determination is equivalent to
finding the components of the displacement along the south
and west directions.) (b) Find the distances you would have to
fly first in a direction 45.0º south of west and then in a
direction 45.0º west of north. These are the components of
the displacement along a different set of axes—one rotated
45º .
22. A farmer wants to fence off his four-sided plot of flat land.
He measures the first three sides, shown as A, B, and C
in Figure 3.62, and then correctly calculates the length and
orientation of the fourth side D . What is his result?


Figure 3.62


23. In an attempt to escape his island, Gilligan builds a raft
and sets to sea. The wind shifts a great deal during the day,
and he is blown along the following straight lines: 2.50 km
45.0º north of west; then 4.70 km 60.0º south of east;
then 1.30 km 25.0º south of west; then 5.10 km straight
east; then 1.70 km 5.00º east of north; then 7.20 km
55.0º south of west; and finally 2.80 km 10.0º north of
east. What is his final position relative to the island?


24. Suppose a pilot flies 40.0 km in a direction 60º north of
east and then flies 30.0 km in a direction 15º north of east
as shown in Figure 3.63. Find her total distance R from the
starting point and the direction θ of the straight-line path to
the final position. Discuss qualitatively how this flight would be
altered by a wind from the north and how the effect of the
wind would depend on both wind speed and the speed of the
plane relative to the air mass.


Figure 3.63


3.4 Projectile Motion
25. A projectile is launched at ground level with an initial
speed of 50.0 m/s at an angle of 30.0º above the horizontal.
It strikes a target above the ground 3.00 seconds later. What
are the x and y distances from where the projectile was


launched to where it lands?


26. A ball is kicked with an initial velocity of 16 m/s in the
horizontal direction and 12 m/s in the vertical direction. (a) At
what speed does the ball hit the ground? (b) For how long
does the ball remain in the air? (c)What maximum height is
attained by the ball?


27. A ball is thrown horizontally from the top of a 60.0-m
building and lands 100.0 m from the base of the building.
Ignore air resistance. (a) How long is the ball in the air? (b)
What must have been the initial horizontal component of the
velocity? (c) What is the vertical component of the velocity
just before the ball hits the ground? (d) What is the velocity
(including both the horizontal and vertical components) of the
ball just before it hits the ground?


28. (a) A daredevil is attempting to jump his motorcycle over a
line of buses parked end to end by driving up a 32º ramp at
a speed of 40.0 m/s (144 km/h) . How many buses can he
clear if the top of the takeoff ramp is at the same height as the
bus tops and the buses are 20.0 m long? (b) Discuss what
your answer implies about the margin of error in this act—that
is, consider how much greater the range is than the horizontal
distance he must travel to miss the end of the last bus.
(Neglect air resistance.)


29. An archer shoots an arrow at a 75.0 m distant target; the
bull’s-eye of the target is at same height as the release height
of the arrow. (a) At what angle must the arrow be released to
hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of
the problem, explicitly show how you follow the steps involved
in solving projectile motion problems. (b) There is a large tree
halfway between the archer and the target with an
overhanging horizontal branch 3.50 m above the release
height of the arrow. Will the arrow go over or under the
branch?


30. A rugby player passes the ball 7.00 m across the field,
where it is caught at the same height as it left his hand. (a) At
what angle was the ball thrown if its initial speed was 12.0 m/
s, assuming that the smaller of the two possible angles was
used? (b) What other angle gives the same range, and why
would it not be used? (c) How long did this pass take?


31. Verify the ranges for the projectiles in Figure 3.41(a) for
θ = 45º and the given initial velocities.
32. Verify the ranges shown for the projectiles in Figure
3.41(b) for an initial velocity of 50 m/s at the given initial
angles.


33. The cannon on a battleship can fire a shell a maximum
distance of 32.0 km. (a) Calculate the initial velocity of the
shell. (b) What maximum height does it reach? (At its highest,
the shell is above 60% of the atmosphere—but air resistance
is not really negligible as assumed to make this problem
easier.) (c) The ocean is not flat, because the Earth is curved.


Assume that the radius of the Earth is 6.37×103 km . How
many meters lower will its surface be 32.0 km from the ship
along a horizontal line parallel to the surface at the ship?
Does your answer imply that error introduced by the
assumption of a flat Earth in projectile motion is significant
here?


Chapter 3 | Two-Dimensional Kinematics 127




34. An arrow is shot from a height of 1.5 m toward a cliff of
height H . It is shot with a velocity of 30 m/s at an angle of
60º above the horizontal. It lands on the top edge of the cliff
4.0 s later. (a) What is the height of the cliff? (b) What is the
maximum height reached by the arrow along its trajectory? (c)
What is the arrow’s impact speed just before hitting the cliff?


35. In the standing broad jump, one squats and then pushes
off with the legs to see how far one can jump. Suppose the
extension of the legs from the crouch position is 0.600 m and
the acceleration achieved from this position is 1.25 times the
acceleration due to gravity, g . How far can they jump? State


your assumptions. (Increased range can be achieved by
swinging the arms in the direction of the jump.)


36. The world long jump record is 8.95 m (Mike Powell, USA,
1991). Treated as a projectile, what is the maximum range
obtainable by a person if he has a take-off speed of 9.5 m/s?
State your assumptions.


37. Serving at a speed of 170 km/h, a tennis player hits the
ball at a height of 2.5 m and an angle θ below the horizontal.
The service line is 11.9 m from the net, which is 0.91 m high.
What is the angle θ such that the ball just crosses the net?
Will the ball land in the service box, whose out line is 6.40 m
from the net?


38. A football quarterback is moving straight backward at a
speed of 2.00 m/s when he throws a pass to a player 18.0 m
straight downfield. (a) If the ball is thrown at an angle of 25º
relative to the ground and is caught at the same height as it is
released, what is its initial speed relative to the ground? (b)
How long does it take to get to the receiver? (c) What is its
maximum height above its point of release?


39. Gun sights are adjusted to aim high to compensate for the
effect of gravity, effectively making the gun accurate only for a
specific range. (a) If a gun is sighted to hit targets that are at
the same height as the gun and 100.0 m away, how low will
the bullet hit if aimed directly at a target 150.0 m away? The
muzzle velocity of the bullet is 275 m/s. (b) Discuss
qualitatively how a larger muzzle velocity would affect this
problem and what would be the effect of air resistance.


40. An eagle is flying horizontally at a speed of 3.00 m/s when
the fish in her talons wiggles loose and falls into the lake 5.00
m below. Calculate the velocity of the fish relative to the water
when it hits the water.


41. An owl is carrying a mouse to the chicks in its nest. Its
position at that time is 4.00 m west and 12.0 m above the
center of the 30.0 cm diameter nest. The owl is flying east at
3.50 m/s at an angle 30.0º below the horizontal when it
accidentally drops the mouse. Is the owl lucky enough to
have the mouse hit the nest? To answer this question,
calculate the horizontal position of the mouse when it has
fallen 12.0 m.


42. Suppose a soccer player kicks the ball from a distance 30
m toward the goal. Find the initial speed of the ball if it just
passes over the goal, 2.4 m above the ground, given the
initial direction to be 40º above the horizontal.
43. Can a goalkeeper at her/ his goal kick a soccer ball into
the opponent’s goal without the ball touching the ground? The
distance will be about 95 m. A goalkeeper can give the ball a
speed of 30 m/s.


44. The free throw line in basketball is 4.57 m (15 ft) from the
basket, which is 3.05 m (10 ft) above the floor. A player
standing on the free throw line throws the ball with an initial


speed of 7.15 m/s, releasing it at a height of 2.44 m (8 ft)
above the floor. At what angle above the horizontal must the
ball be thrown to exactly hit the basket? Note that most
players will use a large initial angle rather than a flat shot
because it allows for a larger margin of error. Explicitly show
how you follow the steps involved in solving projectile motion
problems.


45. In 2007, Michael Carter (U.S.) set a world record in the
shot put with a throw of 24.77 m. What was the initial speed
of the shot if he released it at a height of 2.10 m and threw it
at an angle of 38.0º above the horizontal? (Although the
maximum distance for a projectile on level ground is achieved
at 45º when air resistance is neglected, the actual angle to
achieve maximum range is smaller; thus, 38º will give a
longer range than 45º in the shot put.)


46. A basketball player is running at 5.00 m/s directly
toward the basket when he jumps into the air to dunk the ball.
He maintains his horizontal velocity. (a) What vertical velocity
does he need to rise 0.750 m above the floor? (b) How far
from the basket (measured in the horizontal direction) must
he start his jump to reach his maximum height at the same
time as he reaches the basket?


47. A football player punts the ball at a 45.0º angle. Without
an effect from the wind, the ball would travel 60.0 m
horizontally. (a) What is the initial speed of the ball? (b) When
the ball is near its maximum height it experiences a brief gust
of wind that reduces its horizontal velocity by 1.50 m/s. What
distance does the ball travel horizontally?


48. Prove that the trajectory of a projectile is parabolic, having


the form y = ax + bx2 . To obtain this expression, solve the
equation x = v0x t for t and substitute it into the expression


for y = v0yt – (1 / 2)gt
2 (These equations describe the x


and y positions of a projectile that starts at the origin.) You


should obtain an equation of the form y = ax + bx2 where
a and b are constants.


49. Derive R =
v0
2 sin 2θ0


g for the range of a projectile on


level ground by finding the time t at which y becomes zero


and substituting this value of t into the expression for
x − x0 , noting that R = x − x0
50. Unreasonable Results (a) Find the maximum range of a
super cannon that has a muzzle velocity of 4.0 km/s. (b) What
is unreasonable about the range you found? (c) Is the
premise unreasonable or is the available equation
inapplicable? Explain your answer. (d) If such a muzzle
velocity could be obtained, discuss the effects of air
resistance, thinning air with altitude, and the curvature of the
Earth on the range of the super cannon.


51. Construct Your Own Problem Consider a ball tossed
over a fence. Construct a problem in which you calculate the
ball’s needed initial velocity to just clear the fence. Among the
things to determine are; the height of the fence, the distance
to the fence from the point of release of the ball, and the
height at which the ball is released. You should also consider
whether it is possible to choose the initial speed for the ball
and just calculate the angle at which it is thrown. Also
examine the possibility of multiple solutions given the
distances and heights you have chosen.


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3.5 Addition of Velocities
52. Bryan Allen pedaled a human-powered aircraft across the
English Channel from the cliffs of Dover to Cap Gris-Nez on
June 12, 1979. (a) He flew for 169 min at an average velocity
of 3.53 m/s in a direction 45º south of east. What was his
total displacement? (b) Allen encountered a headwind
averaging 2.00 m/s almost precisely in the opposite direction
of his motion relative to the Earth. What was his average
velocity relative to the air? (c) What was his total
displacement relative to the air mass?


53. A seagull flies at a velocity of 9.00 m/s straight into the
wind. (a) If it takes the bird 20.0 min to travel 6.00 km relative
to the Earth, what is the velocity of the wind? (b) If the bird
turns around and flies with the wind, how long will he take to
return 6.00 km? (c) Discuss how the wind affects the total
round-trip time compared to what it would be with no wind.


54. Near the end of a marathon race, the first two runners are
separated by a distance of 45.0 m. The front runner has a
velocity of 3.50 m/s, and the second a velocity of 4.20 m/s. (a)
What is the velocity of the second runner relative to the first?
(b) If the front runner is 250 m from the finish line, who will
win the race, assuming they run at constant velocity? (c)
What distance ahead will the winner be when she crosses the
finish line?


55. Verify that the coin dropped by the airline passenger in
the Example 3.8 travels 144 m horizontally while falling 1.50
m in the frame of reference of the Earth.


56. A football quarterback is moving straight backward at a
speed of 2.00 m/s when he throws a pass to a player 18.0 m
straight downfield. The ball is thrown at an angle of 25.0º
relative to the ground and is caught at the same height as it is
released. What is the initial velocity of the ball relative to the
quarterback ?


57. A ship sets sail from Rotterdam, The Netherlands,
heading due north at 7.00 m/s relative to the water. The local
ocean current is 1.50 m/s in a direction 40.0º north of east.
What is the velocity of the ship relative to the Earth?


58. (a) A jet airplane flying from Darwin, Australia, has an air
speed of 260 m/s in a direction 5.0º south of west. It is in the
jet stream, which is blowing at 35.0 m/s in a direction 15º
south of east. What is the velocity of the airplane relative to
the Earth? (b) Discuss whether your answers are consistent
with your expectations for the effect of the wind on the plane’s
path.


59. (a) In what direction would the ship in Exercise 3.57 have
to travel in order to have a velocity straight north relative to
the Earth, assuming its speed relative to the water remains
7.00 m/s ? (b) What would its speed be relative to the Earth?
60. (a) Another airplane is flying in a jet stream that is blowing
at 45.0 m/s in a direction 20º south of east (as in Exercise
3.58). Its direction of motion relative to the Earth is 45.0º
south of west, while its direction of travel relative to the air is
5.00º south of west. What is the airplane’s speed relative to
the air mass? (b) What is the airplane’s speed relative to the
Earth?


61. A sandal is dropped from the top of a 15.0-m-high mast
on a ship moving at 1.75 m/s due south. Calculate the
velocity of the sandal when it hits the deck of the ship: (a)
relative to the ship and (b) relative to a stationary observer on


shore. (c) Discuss how the answers give a consistent result
for the position at which the sandal hits the deck.


62. The velocity of the wind relative to the water is crucial to
sailboats. Suppose a sailboat is in an ocean current that has
a velocity of 2.20 m/s in a direction 30.0º east of north
relative to the Earth. It encounters a wind that has a velocity
of 4.50 m/s in a direction of 50.0º south of west relative to
the Earth. What is the velocity of the wind relative to the
water?


63. The great astronomer Edwin Hubble discovered that all
distant galaxies are receding from our Milky Way Galaxy with
velocities proportional to their distances. It appears to an
observer on the Earth that we are at the center of an
expanding universe. Figure 3.64 illustrates this for five
galaxies lying along a straight line, with the Milky Way Galaxy
at the center. Using the data from the figure, calculate the
velocities: (a) relative to galaxy 2 and (b) relative to galaxy 5.
The results mean that observers on all galaxies will see
themselves at the center of the expanding universe, and they
would likely be aware of relative velocities, concluding that it
is not possible to locate the center of expansion with the
given information.


Figure 3.64 Five galaxies on a straight line, showing their distances and
velocities relative to the Milky Way (MW) Galaxy. The distances are in
millions of light years (Mly), where a light year is the distance light
travels in one year. The velocities are nearly proportional to the
distances. The sizes of the galaxies are greatly exaggerated; an
average galaxy is about 0.1 Mly across.


64. (a) Use the distance and velocity data in Figure 3.64 to
find the rate of expansion as a function of distance.


(b) If you extrapolate back in time, how long ago would all of
the galaxies have been at approximately the same position?
The two parts of this problem give you some idea of how the
Hubble constant for universal expansion and the time back to
the Big Bang are determined, respectively.


65. An athlete crosses a 25-m-wide river by swimming
perpendicular to the water current at a speed of 0.5 m/s
relative to the water. He reaches the opposite side at a
distance 40 m downstream from his starting point. How fast is
the water in the river flowing with respect to the ground? What
is the speed of the swimmer with respect to a friend at rest on
the ground?


66. A ship sailing in the Gulf Stream is heading 25.0º west
of north at a speed of 4.00 m/s relative to the water. Its
velocity relative to the Earth is 4.80 m/s 5.00º west of
north. What is the velocity of the Gulf Stream? (The velocity
obtained is typical for the Gulf Stream a few hundred
kilometers off the east coast of the United States.)


67. An ice hockey player is moving at 8.00 m/s when he hits
the puck toward the goal. The speed of the puck relative to
the player is 29.0 m/s. The line between the center of the goal
and the player makes a 90.0º angle relative to his path as
shown in Figure 3.65. What angle must the puck’s velocity
make relative to the player (in his frame of reference) to hit
the center of the goal?


Chapter 3 | Two-Dimensional Kinematics 129




Figure 3.65 An ice hockey player moving across the rink must shoot
backward to give the puck a velocity toward the goal.


68. Unreasonable Results Suppose you wish to shoot
supplies straight up to astronauts in an orbit 36,000 km above
the surface of the Earth. (a) At what velocity must the
supplies be launched? (b) What is unreasonable about this
velocity? (c) Is there a problem with the relative velocity
between the supplies and the astronauts when the supplies
reach their maximum height? (d) Is the premise unreasonable
or is the available equation inapplicable? Explain your
answer.


69. Unreasonable Results A commercial airplane has an air
speed of 280 m/s due east and flies with a strong tailwind. It
travels 3000 km in a direction 5º south of east in 1.50 h. (a)
What was the velocity of the plane relative to the ground? (b)
Calculate the magnitude and direction of the tailwind’s
velocity. (c) What is unreasonable about both of these
velocities? (d) Which premise is unreasonable?


70. Construct Your Own Problem Consider an airplane
headed for a runway in a cross wind. Construct a problem in
which you calculate the angle the airplane must fly relative to
the air mass in order to have a velocity parallel to the runway.
Among the things to consider are the direction of the runway,
the wind speed and direction (its velocity) and the speed of
the plane relative to the air mass. Also calculate the speed of
the airplane relative to the ground. Discuss any last minute
maneuvers the pilot might have to perform in order for the
plane to land with its wheels pointing straight down the
runway.


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4 DYNAMICS: FORCE AND NEWTON'S
LAWS OF MOTION


Figure 4.1 Newton’s laws of motion describe the motion of the dolphin’s path. (credit: Jin Jang)


Chapter Outline
4.1. Development of Force Concept


• Understand the definition of force.
4.2. Newton’s First Law of Motion: Inertia


• Define mass and inertia.
• Understand Newton's first law of motion.


4.3. Newton’s Second Law of Motion: Concept of a System
• Define net force, external force, and system.
• Understand Newton’s second law of motion.
• Apply Newton’s second law to determine the weight of an object.


4.4. Newton’s Third Law of Motion: Symmetry in Forces
• Understand Newton's third law of motion.
• Apply Newton's third law to define systems and solve problems of motion.


4.5. Normal, Tension, and Other Examples of Forces
• Define normal and tension forces.
• Apply Newton's laws of motion to solve problems involving a variety of forces.
• Use trigonometric identities to resolve weight into components.


4.6. Problem-Solving Strategies
• Understand and apply a problem-solving procedure to solve problems using Newton's laws of motion.


4.7. Further Applications of Newton’s Laws of Motion
• Apply problem-solving techniques to solve for quantities in more complex systems of forces.
• Integrate concepts from kinematics to solve problems using Newton's laws of motion.


4.8. Extended Topic: The Four Basic Forces—An Introduction
• Understand the four basic forces that underlie the processes in nature.


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 131




Introduction to Dynamics: Newton’s Laws of Motion
Motion draws our attention. Motion itself can be beautiful, causing us to marvel at the forces needed to achieve spectacular
motion, such as that of a dolphin jumping out of the water, or a pole vaulter, or the flight of a bird, or the orbit of a satellite. The
study of motion is kinematics, but kinematics only describes the way objects move—their velocity and their acceleration.
Dynamics considers the forces that affect the motion of moving objects and systems. Newton’s laws of motion are the
foundation of dynamics. These laws provide an example of the breadth and simplicity of principles under which nature functions.
They are also universal laws in that they apply to similar situations on Earth as well as in space.


Issac Newton’s (1642–1727) laws of motion were just one part of the monumental work that has made him legendary. The
development of Newton’s laws marks the transition from the Renaissance into the modern era. This transition was characterized
by a revolutionary change in the way people thought about the physical universe. For many centuries natural philosophers had
debated the nature of the universe based largely on certain rules of logic with great weight given to the thoughts of earlier
classical philosophers such as Aristotle (384–322 BC). Among the many great thinkers who contributed to this change were
Newton and Galileo.


Figure 4.2 Issac Newton’s monumental work, Philosophiae Naturalis Principia Mathematica, was published in 1687. It proposed scientific laws that are
still used today to describe the motion of objects. (credit: Service commun de la documentation de l'Université de Strasbourg)


Galileo was instrumental in establishing observation as the absolute determinant of truth, rather than “logical” argument. Galileo’s
use of the telescope was his most notable achievement in demonstrating the importance of observation. He discovered moons
orbiting Jupiter and made other observations that were inconsistent with certain ancient ideas and religious dogma. For this
reason, and because of the manner in which he dealt with those in authority, Galileo was tried by the Inquisition and punished.
He spent the final years of his life under a form of house arrest. Because others before Galileo had also made discoveries by
observing the nature of the universe, and because repeated observations verified those of Galileo, his work could not be
suppressed or denied. After his death, his work was verified by others, and his ideas were eventually accepted by the church and
scientific communities.


Galileo also contributed to the formation of what is now called Newton’s first law of motion. Newton made use of the work of his
predecessors, which enabled him to develop laws of motion, discover the law of gravity, invent calculus, and make great
contributions to the theories of light and color. It is amazing that many of these developments were made with Newton working
alone, without the benefit of the usual interactions that take place among scientists today.


It was not until the advent of modern physics early in the 20th century that it was discovered that Newton’s laws of motion
produce a good approximation to motion only when the objects are moving at speeds much, 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 classical mechanics, as discussed in Introduction to the Nature of Science and Physics. At the beginning of the 20th


century, Albert Einstein (1879–1955) developed the theory of relativity and, along with many other scientists, developed quantum
theory. This theory does not have the constraints present in classical physics. All of the situations we consider in this chapter,
and all those preceding the introduction of relativity in Special Relativity, are in the realm of classical physics.


Making Connections: Past and Present Philosophy


The importance of observation and the concept of cause and effect were not always so entrenched in human thinking. This
realization was a part of the evolution of modern physics from natural philosophy. The achievements of Galileo, Newton,
Einstein, and others were key milestones in the history of scientific thought. Most of the scientific theories that are described
in this book descended from the work of these scientists.


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4.1 Development of Force Concept
Dynamics is the study of the forces that cause objects and systems to move. To understand this, we need a working definition of
force. Our intuitive definition of force—that is, a push or a pull—is a good place to start. We know that a push or pull has both
magnitude and direction (therefore, it is a vector quantity) and can vary considerably in each regard. For example, a cannon
exerts a strong force on a cannonball that is launched into the air. In contrast, Earth exerts only a tiny downward pull on a flea.
Our everyday experiences 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 4.3, we might expect the total force to be in the direction shown. Since force is a vector, it
adds just like other vectors, as illustrated in Figure 4.3(a) for two ice skaters. Forces, like other vectors, are represented by
arrows and can be added using the familiar head-to-tail method or by trigonometric methods. These ideas were developed in
Two-Dimensional Kinematics.


Figure 4.3 Part (a) shows an overhead view of two ice skaters pushing on a third. Forces are vectors and add like other vectors, so the total force on
the third skater is in the direction shown. In part (b), we see a free-body diagram representing the forces acting on the third skater.


Figure 4.3(b) is our first example of a free-body diagram, which is a technique used to illustrate all the external forces acting
on a body. The body is represented by a single isolated point (or free body), and only those forces acting on the body from the
outside (external forces) are shown. (These forces are the only ones shown, because only external forces acting on the body
affect its motion. We can ignore any internal forces within the body.) Free-body diagrams are very useful in analyzing forces
acting on a system and are employed extensively in the study and application of Newton’s laws of motion.


A more quantitative definition of force can be based on some standard force, just as distance is measured in units relative to a
standard distance. One possibility is to stretch a spring a certain fixed distance, as illustrated in Figure 4.4, and use the force it
exerts to pull itself back to its relaxed shape—called a restoring force—as a standard. The magnitude of all other forces can be
stated as multiples of this standard unit of force. Many other possibilities exist for standard forces. (One that we will encounter in
Magnetism is the magnetic force between two wires carrying electric current.) Some alternative definitions of force will be given
later in this chapter.


Figure 4.4 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, Frestore , which is reproducible. (c) A spring scale is one device that uses a spring to
measure force. The force Frestore is exerted on whatever is attached to the hook. Here Frestore has a magnitude of 6 units in the force standard
being employed.


Take-Home Experiment: Force Standards


To investigate force standards and cause and effect, get two identical rubber bands. Hang one rubber band vertically on a
hook. Find a small household item that could be attached to the rubber band using a paper clip, and use this item as a
weight to investigate the stretch of the rubber band. Measure the amount of stretch produced in the rubber band with one,
two, and four of these (identical) items suspended from the rubber band. What is the relationship between the number of
items and the amount of stretch? How large a stretch would you expect for the same number of items suspended from two


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 133




rubber bands? What happens to the amount of stretch of the rubber band (with the weights attached) if the weights are also
pushed to the side with a pencil?


4.2 Newton’s First Law of Motion: Inertia
Experience suggests that an object at rest will remain at rest if left alone, and that an object in motion tends to slow down and
stop unless some effort is made to keep it moving. What Newton’s first law of motion states, however, is the following:


Newton’s First Law of Motion


A body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external
force.


Note the repeated use of the verb “remains.” We can think of this law as preserving the status quo of motion.


Rather than contradicting our experience, Newton’s first law of motion states that there must be a cause (which is a net
external force) for there to be any change in velocity (either a change in magnitude or direction). We will define net external force
in the next section. An object sliding across a table or floor slows down due to the net force of friction acting on the object. If
friction disappeared, would the object still slow down?


The idea of cause and effect is crucial in accurately describing what happens in various situations. For example, consider what
happens to an object sliding along a rough horizontal surface. The object quickly grinds to a halt. If we spray the surface with
talcum powder to make the surface smoother, the object slides farther. If we make the surface even smoother by rubbing
lubricating oil on it, the object slides farther yet. Extrapolating to a frictionless surface, we can imagine the object sliding in a
straight line indefinitely. Friction is thus the cause of the slowing (consistent with Newton’s first law). The object would not slow
down at all if friction were completely eliminated. Consider an air hockey table. When the air is turned off, the puck slides only a
short distance before friction slows it to a stop. However, when the air is turned on, it creates a nearly frictionless surface, and
the puck glides long distances without slowing down. Additionally, if we know enough about the friction, we can accurately predict
how quickly the object will slow down. Friction is an external force.


Newton’s first law is completely general and can be applied to anything from an object sliding on a table to a satellite in orbit to
blood pumped from the heart. Experiments have thoroughly verified that any change in velocity (speed or direction) must be
caused by an external force. The idea of generally applicable or universal laws is important not only here—it is a basic feature of
all laws of physics. Identifying these laws is like recognizing patterns in nature from which further patterns can be discovered.
The genius of Galileo, who first developed the idea for the first law, and Newton, who clarified it, was to ask the fundamental
question, “What is the cause?” Thinking in terms of cause and effect is a worldview fundamentally different from the typical
ancient Greek approach when questions such as “Why does a tiger have stripes?” would have been answered in Aristotelian
fashion, “That is the nature of the beast.” True perhaps, but not a useful insight.


Mass
The property of a body to remain at rest or to remain in motion with constant velocity is called inertia. Newton’s first law is often
called the law of inertia. As we know from experience, some objects have more inertia than others. It is obviously more difficult
to change the motion of a large boulder than that of a basketball, for example. The inertia of an object is measured by its mass.
Roughly speaking, mass is a measure of the amount of “stuff” (or matter) in something. The quantity or amount of matter in an
object is determined by the numbers of atoms and molecules of various types it contains. Unlike weight, mass does not vary with
location. The mass of an object is the same on Earth, in orbit, or on the surface of the Moon. In practice, it is very difficult to count
and identify all of the atoms and molecules in an object, so masses are not often determined in this manner. Operationally, the
masses of objects are determined by comparison with the standard kilogram.


Check Your Understanding


Which has more mass: a kilogram of cotton balls or a kilogram of gold?


Solution


They are equal. A kilogram of one substance is equal in mass to a kilogram of another substance. The quantities that might
differ between them are volume and density.


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4.3 Newton’s Second Law of Motion: Concept of a System
Newton’s second law of motion is closely related to Newton’s first law of motion. It mathematically states the cause and effect
relationship between force and changes in motion. Newton’s second law of motion is more quantitative and is used extensively to
calculate what happens in situations involving a force. Before we can write down Newton’s second law as a simple equation
giving the exact relationship of force, mass, and acceleration, we need to sharpen some ideas that have already been
mentioned.


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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 an acceleration. Newton’s first law says that a net external force causes a
change in motion; thus, we see that a net external force causes acceleration.


Another question immediately arises. What do we mean by an external force? An intuitive notion of external is correct—an
external force acts from outside the system of interest. For example, in Figure 4.5(a) the system of interest is the wagon plus
the child in it. The two forces exerted by the other children are external forces. An internal force acts between elements of the
system. Again looking at Figure 4.5(a), the force the child in the wagon exerts to hang onto the wagon is an internal force
between elements of the system of interest. Only external forces affect the motion of a system, according to Newton’s first law.
(The internal forces actually cancel, as we shall see in the next section.) You must define the boundaries of the system before
you can determine which forces are external. Sometimes the system is obvious, whereas other times identifying the boundaries
of a system is more subtle. The concept of a system is fundamental to many areas of physics, as is the correct application of
Newton’s laws. This concept will be revisited many times on our journey through physics.


Figure 4.5 Different forces exerted on the same mass produce different accelerations. (a) Two children push a wagon with a child in it. Arrows
representing all external forces are shown. The system of interest is the wagon and its rider. The weight w of the system and the support of the
ground N are also shown for completeness and are assumed to cancel. The vector f represents the friction acting on the wagon, and it acts to the
left, opposing the motion of the wagon. (b) All of the external forces acting on the system add together to produce a net force, Fnet . 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
this dot. Because there are two forces acting to the right, we draw the vectors collinearly. (c) A larger net external force produces a larger acceleration (


a′ > a ) when an adult pushes the child.


Now, it seems reasonable that acceleration should be directly proportional to and in the same direction as the net (total) external
force acting on a system. This assumption has been verified experimentally and is illustrated in Figure 4.5. In part (a), a smaller
force causes a smaller acceleration than the larger force illustrated in part (c). For completeness, the vertical forces are also
shown; they are assumed to cancel since there is no acceleration in the vertical direction. The vertical forces are the weight w
and the support of the ground N , and the horizontal force f represents the force of friction. These will be discussed in more
detail in later sections. For now, we will define friction as a force that opposes the motion past each other of objects that are
touching. Figure 4.5(b) shows how vectors representing the external forces add together to produce a net force, Fnet .


To obtain an equation for Newton’s second law, we first write the relationship of acceleration and net external force as the
proportionality


(4.1)a ∝ Fnet ,


where the symbol ∝ means “proportional to,” and Fnet is the net external force. (The net external force is the vector sum of
all external forces and can be determined graphically, using the head-to-tail method, or analytically, using components. The
techniques are the same as for the addition of other vectors, and are covered in Two-Dimensional Kinematics.) This
proportionality states what we have said in words—acceleration is directly proportional to the net external force. Once the system
of interest is chosen, it is important to identify the external forces and ignore the internal ones. It is a tremendous simplification
not to have to consider the numerous internal forces acting between objects within the system, such as muscular forces within


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 135




the child’s body, let alone the myriad of forces between atoms in the objects, but by doing so, we can easily solve some very
complex problems with only minimal error due to our simplification


Now, it also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the
larger the mass (the inertia), the smaller the acceleration produced by a given force. And indeed, as illustrated in Figure 4.6, the
same net external force applied to a car produces a much smaller acceleration than when applied to a basketball. The
proportionality is written as


(4.2)a ∝ 1m
where m is the mass of the system. Experiments have shown that acceleration is exactly inversely proportional to mass, just as
it is exactly linearly proportional to the net external force.


Figure 4.6 The same force exerted on systems of different masses produces different accelerations. (a) A basketball player pushes on a basketball to
make a pass. (The effect of gravity on the ball is ignored.) (b) The same player exerts an identical force on a stalled SUV and produces a far smaller
acceleration (even if friction is negligible). (c) The free-body diagrams are identical, permitting direct comparison of the two situations. A series of
patterns for the free-body diagram will emerge as you do more problems.


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 of motion.


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 the
system, and inversely proportional to its mass.


In equation form, Newton’s second law of motion is


(4.3)
a = Fnetm .


This is often written in the more familiar form


(4.4)Fnet = ma.


When only the magnitude of force and acceleration are considered, this equation is simply


(4.5)Fnet = ma.


Although these last two equations are really the same, the first gives more insight into what Newton’s second law means. The
law is a cause and effect relationship among three quantities that is not simply based on their definitions. The validity of the
second law is completely based on experimental verification.


Units of Force


Fnet = ma is used to define the units of force in terms of the three basic units for mass, length, and time. The SI unit of force is


called the newton (abbreviated N) and is the force needed to accelerate a 1-kg system at the rate of 1m/s2 . That is, since
Fnet = ma ,


(4.6)1 N = 1 kg ⋅ m/s2.


While almost the entire world uses the newton for the unit of force, in the United States the most familiar unit of force is the
pound (lb), where 1 N = 0.225 lb.


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Weight and the Gravitational Force
When an object is dropped, it accelerates toward the center of Earth. Newton’s second law states that a net force on an object is
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 . Weight can be denoted as a vector w because it has a direction; down is, by definition, the direction of
gravity, and hence weight is a downward force. The magnitude of weight is denoted as w . Galileo was instrumental in showing
that, in the absence of air resistance, all objects fall 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 downward toward Earth. It experiences only the downward force of gravity, which has
magnitude w . Newton’s second law states that the magnitude of the net external force on an object is Fnet = ma .


Since the object experiences only the downward force of gravity, Fnet = w . We know that the acceleration of an object due to
gravity is g , or a = g . Substituting these into Newton’s second law gives


Weight


This is the equation for weight—the gravitational force on a mass m :


(4.7)w = mg.


Since g = 9.80 m/s2 on Earth, the weight of a 1.0 kg object on Earth is 9.8 N, as we see:


(4.8)w = mg = (1.0 kg)(9.80 m/s2 ) = 9.8 N.


Recall that g can take a positive or negative value, depending on the positive direction in the coordinate system. Be sure to


take this into consideration when solving problems with weight.


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 object is
the force of gravity. In the real world, when objects fall downward toward Earth, they are never truly in free-fall because there is
always some upward force from the air acting on the object.


The acceleration due to gravity g varies slightly over the surface of Earth, so that the weight of an object depends on location


and is not an intrinsic property of the object. Weight varies dramatically if one leaves Earth’s surface. On the Moon, for example,


the 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 nearest large
body, such as Earth, the Moon, the Sun, and so on. This is the most common and useful definition of weight in physics. It differs
dramatically, however, from the definition of weight used by NASA and the popular media in relation to space travel and
exploration. When they speak of “weightlessness” and “microgravity,” they are really referring to the phenomenon we call “free-
fall” in physics. We shall use the above definition of weight, and we will make careful distinctions between free-fall and actual
weightlessness.


It is important to be aware that weight and mass are very different physical quantities, although they are closely related. Mass is
the quantity of matter (how much “stuff”) and does not vary in classical physics, whereas weight is the gravitational force and
does vary depending on gravity. It is tempting to equate the two, since most of our examples take place on Earth, where the
weight of an object only varies a little with the location of the object. Furthermore, the terms mass and weight are used
interchangeably in everyday language; for example, our medical records often show our “weight” in kilograms, but never in the
correct units of newtons.


Common Misconceptions: Mass vs. Weight


Mass and weight are often used interchangeably in everyday language. However, in science, these terms are distinctly
different from one another. Mass is a measure of how much matter is in an object. The typical measure of mass is the
kilogram (or the “slug” in English units). Weight, on the other hand, is a measure of the force of gravity acting on an object.
Weight is equal to the mass of an object (m ) multiplied by the acceleration due to gravity ( g ). Like any other force, weight


is measured in terms of newtons (or pounds in English units).


Assuming the mass of an object is kept intact, it will remain the same, regardless of its location. However, because weight
depends on the acceleration due to gravity, the weight of an object can change when the object enters into a region with


stronger or weaker gravity. For example, the acceleration due to gravity on the Moon is 1.67 m/s2 (which is much less than
the acceleration due to gravity on Earth, 9.80 m/s2 ). If you measured your weight on Earth and then measured your weight
on the Moon, you would find that you “weigh” much less, even though you do not look any skinnier. This is because the force


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 137




of gravity is weaker on the Moon. In fact, when people say that they are “losing weight,” they really mean that they are losing
“mass” (which in turn causes them to weigh less).


Take-Home Experiment: Mass and Weight


What do bathroom scales measure? When you stand on a bathroom scale, what happens to the scale? It depresses slightly.
The scale contains springs that compress in proportion to your weight—similar to rubber bands expanding when pulled. The
springs provide a measure of your weight (for an object which is not accelerating). This is a force in newtons (or pounds). In
most countries, the measurement is divided by 9.80 to give a reading in mass units of kilograms. The scale measures weight
but is calibrated to provide information about mass. While standing on a bathroom scale, push down on a table next to you.
What happens to the reading? Why? Would your scale measure the same “mass” on Earth as on the Moon?


Example 4.1 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) parallel to the
ground. The mass of the mower is 24 kg. What is its acceleration?


Figure 4.7 The net force on a lawn mower is 51 N to the right. At what rate does the lawn mower accelerate to the right?


Strategy


Since Fnet and m are given, the acceleration can be calculated directly from Newton’s second law as stated in
Fnet = ma .


Solution


The magnitude of the acceleration a is a = Fnetm . Entering known values gives


(4.9)a = 51 N24 kg


Substituting the units kg ⋅ m/s2 for N yields


(4.10)
a = 51 kg ⋅ m/s


2


24 kg = 2.1 m/s
2.


Discussion


The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. There is no
information given in this example about the individual external forces acting on the system, but we can say something about
their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction
opposing the motion (since we know the mower moves forward), and the vertical forces must cancel if there is to be no
acceleration in the vertical direction (the mower is moving only horizontally). The acceleration found is small enough to be
reasonable for a person pushing a mower. Such an effort would not last too long because the person’s top speed would
soon be reached.


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Example 4.2 What Rocket Thrust Accelerates This Sled?


Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human
subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled 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 4.8. 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 known to be 650 N.


Figure 4.8 A sled experiences a rocket thrust that accelerates it to the right. Each rocket creates an identical thrust T . As in other situations
where there is only horizontal acceleration, the vertical forces cancel. The ground exerts an upward force N on the system that is equal in
magnitude and opposite in direction to its weight, w . The system here is the sled, its rockets, and rider, so none of the forces between these
objects are considered. The arrow representing friction ( f ) is drawn larger than scale.


Strategy


Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is 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-body diagram in the figure.


Solution


Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the
thrust of the engines. Since we have defined the direction of the force and acceleration as acting “to the right,” we need to
consider only the magnitudes of these quantities in the calculations. Hence we begin with


(4.11)Fnet = ma,


where Fnet is the net force along the horizontal direction. We can see from Figure 4.8 that the engine thrusts add, while


friction opposes the thrust. In equation form, the net external force is


(4.12)Fnet = 4T − f .


Substituting this into Newton’s second law gives


(4.13)Fnet = ma = 4T − f .


Using a little algebra, we solve for the total thrust 4T:


(4.14)4T = ma + f .


Substituting known values yields


(4.15)4T = ma + f = (2100 kg)(49 m/s2 ) + 650 N.


So the total thrust is


(4.16)4T = 1.0×105 N,


and the individual thrusts are


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 139




(4.17)
T = 1.0×10


5 N
4 = 2.6×10


4 N.


Discussion


The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s
to test the limits of human endurance and the setup designed to protect human subjects in jet fighter emergency ejections.
Speeds of 1000 km/h were obtained, with accelerations of 45 g 's. (Recall that g , the acceleration due to gravity, is


9.80 m/s2 . When we say that an acceleration is 45 g 's, it is 45×9.80 m/s2 , which is approximately 440 m/s2 .) While
living subjects are not used any more, land speeds of 10,000 km/h have been obtained with rocket sleds. In this example, as
in the preceding one, the system of interest is obvious. We will see in later examples that choosing the system of interest is
crucial—and the choice is not always obvious.


Newton’s second law of motion is more than a definition; it is a relationship among acceleration, force, and mass. It can help
us make predictions. Each of those physical quantities can be defined independently, so the second law tells us something
basic and universal about nature. The next section introduces the third and final law of motion.


4.4 Newton’s Third Law of Motion: Symmetry in Forces
There is a passage in the musical Man of la Mancha that relates to Newton’s third law of motion. Sancho, in describing a fight
with his wife to Don Quixote, says, “Of course I hit her back, Your Grace, but she’s a lot harder than me and you know what they
say, ‘Whether the stone hits the pitcher or the pitcher hits the stone, it’s going to be bad for the pitcher.’” This is exactly what
happens whenever one body exerts a force on another—the first also experiences a force (equal in magnitude and opposite in
direction). Numerous common experiences, such as stubbing a toe or throwing a ball, confirm this. It is precisely stated in
Newton’s third law of motion.


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.


This law represents a certain symmetry in nature: Forces always occur in pairs, and one body cannot exert a force on another
without experiencing a force itself. We sometimes refer to this law loosely as “action-reaction,” where the force exerted is the
action and the force experienced as a consequence is the reaction. Newton’s third law has practical uses in analyzing 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 pushing off from
the side of a pool, as illustrated in Figure 4.9. She pushes against the pool wall with her feet and accelerates in the direction
opposite to that of her push. The wall has exerted an equal and opposite force back on the swimmer. You might think that two
equal and opposite forces would cancel, but they do not because they act on different systems. In this case, there are two
systems that we could investigate: the swimmer or 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
Fwall on feet . In contrast, the force Ffeet on wall acts on the wall and 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 . Note that the swimmer pushes in the direction
opposite to that in which she wishes to move. The reaction to her push is thus in the desired direction.


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Figure 4.9When the swimmer exerts a force Ffeet on wall on the wall, she accelerates in the direction opposite to that of her push. This means the
net external force on her is in the direction opposite to Ffeet on wall . This opposition occurs because, in accordance with Newton’s third law of
motion, the wall exerts a force Fwall on feet on her, equal in magnitude but in the direction opposite to the one she exerts on it. The line around the
swimmer indicates the system of interest. Note that Ffeet on wall does not act on this system (the swimmer) and, thus, does not cancel
Fwall on feet . Thus the free-body diagram shows only Fwall on feet , w , the gravitational force, and BF , the buoyant force of the water
supporting the swimmer’s weight. The vertical forces w and BF cancel since there is no vertical motion.


Other examples of Newton’s third law are easy to find. As a professor paces in front of a whiteboard, she exerts a force
backward on the floor. The floor exerts a reaction force forward on the professor that causes her to accelerate forward. Similarly,
a car accelerates because the ground pushes forward on the drive wheels in reaction to the drive wheels pushing backward on
the ground. You can see evidence of the wheels pushing backward when tires spin on a gravel road and throw rocks backward.
In another example, rockets move forward by expelling gas backward at high velocity. This means the rocket exerts a large
backward force on the gas in the rocket combustion chamber, and the gas therefore exerts a large reaction force forward on the
rocket. This reaction force is called thrust. It is a common misconception that rockets propel themselves by pushing on the
ground or on the air behind them. They actually work better in a vacuum, where they can more readily expel the exhaust gases.
Helicopters similarly create lift by pushing air down, thereby experiencing an upward reaction force. Birds and airplanes also fly
by exerting force on air in a direction opposite to that of whatever force they need. For example, the wings of a bird force air
downward and backward in order 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. In a situation similar to Sancho’s, professional cage fighters experience reaction forces
when they punch, sometimes breaking their hand by hitting an opponent’s body.


Example 4.3 Getting Up To Speed: Choosing the Correct System


A physics professor pushes a cart of demonstration equipment to a lecture hall, as seen in Figure 4.10. Her mass is 65.0
kg, the cart’s is 12.0 kg, and the equipment’s is 7.0 kg. Calculate the acceleration produced when the professor exerts a
backward force of 150 N on the floor. All forces opposing the motion, such as friction on the cart’s wheels and air resistance,
total 24.0 N.


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 141




Figure 4.10 A professor pushes a cart of demonstration equipment. The lengths of the arrows are proportional to the magnitudes of the forces


(except for f , since it is too small to draw to scale). Different questions are asked in each example; thus, the system of interest must be defined
differently for each. System 1 is appropriate for Example 4.4, since it asks for the acceleration of the entire group of objects. Only Ffloor 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
this example so that Fprof will be an external force and enter into Newton’s second law. Note that the free-body diagrams, which allow us to
apply Newton’s second law, vary with the system chosen.


Strategy


Since they accelerate as a unit, we define the system to be the professor, cart, and equipment. This is System 1 in Figure
4.10. The professor pushes backward with a force Ffoot of 150 N. According to Newton’s third law, the floor exerts a
forward reaction force Ffloor of 150 N on System 1. Because all motion is horizontal, we can assume there is no net force
in the vertical direction. The problem is therefore one-dimensional along the horizontal direction. As noted, f opposes the
motion and is thus in the opposite direction of Ffloor . Note that 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 net external force can be found from all this information, we can use Newton’s
second law to find the acceleration as requested. See the free-body diagram in the figure.


Solution


Newton’s second law is given by


(4.18)
a = Fnetm .


The net external force on System 1 is deduced from Figure 4.10 and the discussion above to be


(4.19)Fnet = Ffloor − f = 150 N − 24.0 N = 126 N.


The mass of System 1 is


(4.20)m = (65.0 + 12.0 + 7.0) kg = 84 kg.


These values of Fnet and m produce an acceleration of


(4.21)
a = Fnetm ,


a = 126 N84 kg = 1.5 m/s
2 .


Discussion


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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 to note that forces 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 her. In this case both forces act on the
same system and, therefore, cancel. Thus internal forces (between components of a system) cancel. Choosing System 1
was crucial to solving this problem.


Example 4.4 Force on the Cart—Choosing a New System


Calculate the force the professor exerts on the cart in Figure 4.10 using data from the previous example if needed.


Strategy


If we now define the system of interest to be the cart plus equipment (System 2 in Figure 4.10), then the net external force
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 will enter Newton’s
second law for System 2.


Solution


Newton’s second law can be used to find Fprof . Starting with


(4.22)
a = Fnetm


and noting that the magnitude of the net external force on System 2 is


(4.23)Fnet = Fprof − f ,


we solve for Fprof , the desired quantity:


(4.24)Fprof = Fnet + f .


The value of f is given, so we must calculate net Fnet . That can be done since both the acceleration and mass of System


2 are known. Using Newton’s second law we see that


(4.25)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,


(4.26)Fnet = ma,


(4.27)Fnet = (19.0 kg)(1.5 m/s2 ) = 29 N.


Now we can find the desired force:


(4.28)Fprof = Fnet + f ,


(4.29)Fprof = 29 N+24.0 N = 53 N.


Discussion


It is interesting that this force is significantly less than the 150-N force the professor exerted backward on the floor. Not all of
that 150-N force is transmitted to the cart; some of it accelerates the professor.


The choice of a system is an important analytical step both in solving problems and in thoroughly understanding the physics
of the situation (which is not necessarily the same thing).


PhET Explorations: Gravity Force Lab


Visualize the gravitational force that two objects exert on each other. Change properties of the objects in order to see how it
changes the gravity force.


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 143




Figure 4.11 Gravity Force Lab (http://cnx.org/content/m42074/1.5/gravity-force-lab_en.jar)


4.5 Normal, Tension, and Other Examples of Forces
Forces are given many names, such as push, pull, thrust, lift, weight, friction, and tension. Traditionally, forces have been
grouped into several categories and given names relating to their source, how they are transmitted, or their effects. The most
important of these categories are discussed in this section, together with some interesting applications. Further examples of
forces are discussed later in this text.


Normal Force
Weight (also called force of gravity) is a pervasive force that acts at all times and must be counteracted to keep an object from
falling. You definitely notice that you must support the weight of a heavy object by pushing up on it when you hold it stationary, as
illustrated in Figure 4.12(a). But how do inanimate objects like a table support the weight of a mass placed on them, such as
shown in Figure 4.12(b)? When the bag of dog food is placed on the table, the table actually sags slightly under the load. This
would be noticeable if the load were placed on a card table, but even rigid objects deform when a force is applied to them.
Unless the object is deformed beyond its limit, it will exert a restoring force much like a deformed spring (or trampoline or diving
board). The greater the deformation, the greater the restoring force. So when the load is placed on the table, the table sags until
the restoring force becomes as large as the weight of the load. At this point the net external force on the load is zero. That is the
situation when the load is stationary on the table. The table sags quickly, and the sag is slight so we do not notice it. But it is
similar to the sagging of a trampoline when you climb onto it.


144 Chapter 4 | Dynamics: Force and Newton's Laws of Motion


Figure 4.12 (a) The person holding the bag of dog food must supply an upward force Fhand equal in magnitude and opposite in direction to the
weight of the food w . (b) The card table sags when the dog food is placed on it, much like a stiff trampoline. 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 of the
load, as we assumed in a few of the previous examples. If the force supporting a load is perpendicular to the surface of contact
between the load and its support, this force is defined to be a normal force and here is given the symbol N . (This is not the unit


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for force N.) The word normal means perpendicular to a surface. The normal force can be less than the object’s weight if the
object is on an incline, as you will see in the next example.


Common Misconception: Normal Force (N) vs. Newton (N)


In this section we have introduced the quantity normal force, which is represented by the variable N . This should not be
confused with the symbol for the newton, which is also represented by the letter N. These symbols are particularly important
to distinguish because the units of a normal force (N ) happen to be newtons (N). For example, the normal force N that the
floor exerts on a chair might be N = 100 N . One important difference is that normal force is a vector, while the newton is
simply a unit. Be careful not to confuse these letters in your calculations! You will encounter more similarities among
variables and units as you proceed in physics. Another example of this is the quantity work (W ) and the unit watts (W).


Example 4.5 Weight on an Incline, a Two-Dimensional Problem


Consider the skier on a slope shown in Figure 4.13. Her mass including equipment is 60.0 kg. (a) What is her acceleration if
friction is negligible? (b) What is her acceleration if friction is known to be 45.0 N?


Figure 4.13 Since motion and friction are parallel to the slope, it is most convenient to project all forces onto a coordinate system where one axis
is parallel to the slope and the other is perpendicular (axes shown to left of skier). N is perpendicular to the slope and f is parallel to the slope,
but w has components along both axes, namely w⊥ and w ∥ . N is equal in magnitude to w⊥ , so that there is no motion
perpendicular to the slope, but f is less than w ∥ , so that there is a downslope acceleration (along the parallel axis).


Strategy


This is a two-dimensional problem, since the forces on the skier (the system of interest) are not parallel. The approach we
have used in two-dimensional kinematics also works very well here. Choose a convenient coordinate system and project the
vectors onto its axes, creating two connected one-dimensional problems to solve. The most convenient coordinate system
for motion on an incline is one that has one coordinate parallel to the slope and one perpendicular to the slope. (Remember
that motions along mutually perpendicular axes are independent.) We use the symbols ⊥ and ∥ to represent
perpendicular and parallel, respectively. This choice of axes simplifies this type of problem, because there is no motion
perpendicular to the slope and because friction is always parallel to the surface between two objects. The only external
forces acting on the system are the skier’s weight, friction, and the support of the slope, respectively labeled w , f , and N
in Figure 4.13. N is always perpendicular to the slope, and f is parallel to it. But w is not in the direction of either axis,
and so the first step we take is to project it into components along the chosen axes, defining w ∥ to be the component of


weight parallel to the slope and w⊥ the component of weight perpendicular to the slope. Once this is done, we can


consider the two separate problems of forces parallel to the slope and forces perpendicular to the slope.


Solution


The magnitude of the component of the weight parallel to the slope is w ∥ = w sin (25º) = mg sin (25º) , and the


magnitude of the component of the weight perpendicular to the slope is w⊥ = w cos (25º) = mg cos (25º) .


(a) Neglecting friction. Since the acceleration is parallel to the slope, we need only consider forces parallel to the slope.
(Forces perpendicular to the slope add to zero, since there is no acceleration in that direction.) The forces parallel to the
slope are the amount of the skier’s weight parallel to the slope w ∥ and friction f . Using Newton’s second law, with


subscripts to denote quantities parallel to the slope,


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 145




(4.30)
a ∥ =


Fnet ∥
m


where Fnet ∥ = w ∥ = mg sin (25º) , assuming no friction for this part, so that


(4.31)
a ∥ =


Fnet ∥
m =


mg sin (25º)
m = g sin (25º)


(4.32)(9.80 m/s2)(0.4226) = 4.14 m/s2


is the acceleration.


(b) Including friction. We now have a given value for friction, and we know its direction is parallel to the slope and it opposes
motion between surfaces in contact. So the net external force is now


(4.33)Fnet ∥ = w ∥ − f ,


and substituting this into Newton’s second law, a ∥ =
Fnet ∥
m , gives


(4.34)
a ∥ =


Fnet ∣ ∣
m =


w ∥ − f
m =


mg sin (25º) − f
m .


We substitute known values to obtain


(4.35)
a ∥ =


(60.0 kg)(9.80 m/s2)(0.4226) − 45.0 N
60.0 kg ,


which yields


(4.36)a ∥ = 3.39 m/s
2,


which is the acceleration parallel to the incline when there is 45.0 N of opposing friction.


Discussion


Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is
none. In fact, it is a general result that if friction on an incline is negligible, then the acceleration down the incline is
a = g sinθ , regardless of mass. This is related to the previously discussed fact that 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
the same acceleration (if the angle is the same).


Resolving Weight into Components


Figure 4.14 An object rests on an incline that makes an angle θ with the horizontal.


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, w⊥ , and a force acting parallel to the plane, w ∥
. The perpendicular force of weight, w⊥ , is typically equal in magnitude and opposite in direction to the normal force, N .
The force acting parallel to the plane, w ∥ , causes the object to accelerate down the incline. The force of friction, f ,
opposes the motion of the object, so it acts upward along the plane.


It is important to be careful when resolving the weight of the object into components. If the angle of the incline is at an angle
θ to the horizontal, then the magnitudes of the weight components are


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(4.37)w ∥ = w sin (θ) = mg sin (θ)


and


(4.38)w⊥ = w cos (θ) = mg cos (θ).


Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, draw the right
triangle formed by the three weight vectors. Notice that the angle θ of the incline is the same as the angle formed between
w and w⊥ . Knowing this property, you can use trigonometry to determine the magnitude of the weight components:


(4.39)cos (θ) = w⊥w
w⊥ = w cos (θ) = mg cos (θ)


(4.40)sin (θ) =
w ∥
w


w ∥ = w sin (θ) = mg sin (θ)


Take-Home Experiment: Force Parallel


To investigate how a force parallel to an inclined plane changes, find a rubber band, some objects to hang from the end of
the rubber band, and a board you can position at different angles. How much does the rubber band stretch when you hang
the object from the end of the board? Now place the board at an angle so that the object slides off when placed on the
board. How much does the rubber band extend if it is lined up parallel to the board and used to hold the object stationary on
the board? Try two more angles. What does this show?


Tension
A tension is a force along the length of a medium, especially a force carried by a flexible medium, 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 exert pulls
only parallel to its length; thus, a force carried by a flexible connector is a tension with direction parallel to the connector. It is
important to understand that tension is a pull in a connector. In contrast, consider the phrase: “You can’t push a rope.” The
tension force pulls outward along the two ends of a rope.


Consider a person holding a mass on a rope as shown in Figure 4.15.


Figure 4.15When a perfectly flexible connector (one requiring 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. The pull such a flexible connector exerts is a tension. Note that the rope pulls with equal force but in opposite
directions on the hand and the supported mass (neglecting the weight of the rope). This is an example of Newton’s third law. The rope is the medium
that carries the equal and opposite forces between the two objects. The tension anywhere in the rope between the hand and the mass is equal. Once
you have determined the tension in one location, you have determined the tension at all locations along the rope.


Tension in the rope must equal the weight of the supported mass, as we can prove using Newton’s second law. If the 5.00-kg
mass in the figure is stationary, then its acceleration is zero, and thus Fnet = 0 . The only external forces acting on the mass are
its weight w and the tension T supplied by the rope. Thus,


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 147




(4.41)Fnet = T − w = 0,


where T and w are the magnitudes of the tension and weight and their signs indicate direction, with up being positive here.
Thus, just as you would expect, the tension equals the weight of the supported mass:


(4.42)T = w = mg.


For a 5.00-kg mass, then (neglecting the mass of the rope) we see that


(4.43)T = mg = (5.00 kg)(9.80 m/s2 ) = 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 direct
observation and measure of the tension force in the rope.


Flexible connectors are often used to transmit forces around corners, such as in a hospital traction system, a finger joint, or a
bicycle brake cable. If there is no friction, the tension is transmitted undiminished. Only its direction changes, and it is always
parallel to the flexible connector. This is illustrated in Figure 4.16 (a) and (b).


Figure 4.16 (a) Tendons in the finger carry force T from the muscles to other parts of the finger, usually changing the force’s direction, but not its
magnitude (the tendons are relatively friction free). (b) The brake cable on a bicycle carries the tension T from the handlebars to the brake
mechanism. Again, the direction but not the magnitude of T is changed.


Example 4.6 What Is the Tension in a Tightrope?


Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in Figure 4.17.


Figure 4.17 The weight of a tightrope walker causes a wire to sag by 5.0 degrees. The system of interest here is the point in the wire at which the
tightrope walker is standing.


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Strategy


As you can see in the figure, the wire is not perfectly horizontal (it cannot be!), but is bent under the person’s weight. Thus,
the tension on either side of the person has an upward component that can support his weight. As usual, forces are vectors
represented pictorially by arrows having the same directions 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 TL (left
tension) and TR (right tension), as illustrated. It is reasonable to neglect the weight of the wire itself. The net external force
is zero since the system is stationary. A little trigonometry can now be used to find the tensions. One conclusion is possible
at the outset—we can see from part (b) of the figure that the magnitudes of the tensions TL and TR must be equal. This is


because there is no horizontal acceleration in the rope, and the only forces acting to the left and right are TL and TR .


Thus, the magnitude of those 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 method of solution is
to pick a convenient coordinate system and project the vectors onto its axes. In this case the best coordinate system has
one axis horizontal and the other vertical. We call the horizontal the x -axis and the vertical the y -axis.


Solution


First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to draw a new free-body
diagram showing all of the horizontal and vertical components of each force acting on the system.


Figure 4.18When the vectors are projected onto vertical and horizontal axes, their components along those axes must add to zero, since the
tightrope walker is stationary. The small angle results in T being much greater than w .


Consider the horizontal components of the forces (denoted with a subscript x ):


(4.44)Fnetx = TLx − TRx.


The net external horizontal force Fnetx = 0 , since the person is stationary. Thus,


(4.45)Fnetx = 0 = TLx − TRx
TLx = TRx .


Now, observe Figure 4.18. You can use trigonometry to determine the magnitude of TL and TR . Notice that:


(4.46)
cos (5.0º) = TLx


TL
TLx = TL cos (5.0º)


cos (5.0º) = TRx
TR


TRx = TR cos (5.0º).


Equating TLx and TRx :


(4.47)TL cos (5.0º) = TR cos (5.0º).


Thus,


(4.48)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 net Fy = 0 . Thus, as illustrated in the free-body diagram in Figure
4.18,


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 149




(4.49)Fnety = TLy + TRy − w = 0.


Observing Figure 4.18, we can use trigonometry to determine the relationship between TLy , TRy , and T . As we


determined from the analysis in the horizontal direction, TL = TR = T :


(4.50)
sin (5.0º) =


TLy
TL


TLy = TL sin (5.0º) = T sin (5.0º)


sin (5.0º) =
TRy
TR


TRy = TR sin (5.0º) = T sin (5.0º).


Now, we can substitute the values for TLy and TRy , into the net force equation in the vertical direction:


(4.51)Fnety = TLy + TRy − w = 0
Fnety = T sin (5.0º) + T sin (5.0º) − w = 0
2 T sin (5.0º) − w = 0
2 T sin (5.0º) = w


and


(4.52)T = w2 sin (5.0º) =
mg


2 sin (5.0º),


so that


(4.53)
T = (70.0 kg)(9.80 m/s


2)
2(0.0872) ,


and the tension is


(4.54)T = 3900 N.
Discussion


Note that the vertical tension in the wire acts as a normal force that supports the weight of the tightrope walker. The tension
is almost six times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its
tension is only a small fraction of the tension in the wire. The large horizontal components are in opposite directions and
cancel, and so most of the tension in the wire is not used to support the weight of the tightrope walker.


If we wish to create a very large tension, all we have to do is exert a force perpendicular to a flexible connector, as illustrated in
Figure 4.19. As we saw in the last example, the weight of the tightrope walker acted as a force perpendicular to the rope. We
saw that the tension in the roped related to the weight of the tightrope walker in the following way:


(4.55)T = w2 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:


(4.56)
T = F⊥2 sin (θ).


Note that θ is the angle between the horizontal and the bent connector. In this case, T becomes very large as θ approaches
zero. Even the relatively small weight of any flexible connector will cause it to sag, since an infinite tension would result if it were
horizontal (i.e., θ = 0 and sin θ = 0 ). (See Figure 4.19.)


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Figure 4.19We can create a very large tension in the chain by pushing on it perpendicular to its length, as shown. Suppose 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 nearly straight as possible. The tension in


the chain is given by T = F⊥2 sin (θ) ; since θ is small, T is very large. This situation is analogous to the tightrope walker shown in Figure 4.17,


except that the tensions shown here are those transmitted to the car and the tree rather than those acting at the point where F⊥ is applied.


Figure 4.20 Unless an infinite tension is exerted, any flexible connector—such as the chain at the bottom of the picture—will sag under its own weight,
giving a characteristic curve when the weight is evenly distributed along the length. Suspension bridges—such as the Golden Gate Bridge shown in
this image—are essentially very heavy flexible connectors. The weight of the bridge is evenly distributed along the length of flexible connectors, usually
cables, which take on the characteristic shape. (credit: Leaflet, Wikimedia Commons)


Extended Topic: Real Forces and Inertial Frames
There is another distinction among forces in addition to the types already mentioned. Some forces are real, whereas others are
not. Real forces are those that have some physical origin, such as the gravitational pull. Contrastingly, fictitious forces are those
that arise simply because an observer is in an accelerating frame of reference, such as one that rotates (like a merry-go-round)
or undergoes linear acceleration (like a car slowing down). For example, if a satellite is heading due north above Earth’s northern
hemisphere, then to an observer on Earth it will appear to experience a force to the west that has no physical origin. Of course,
what is happening here is that Earth is rotating toward the east and moves east under the satellite. In Earth’s frame this looks like
a westward force on the satellite, or it can be interpreted as a violation of Newton’s first law (the law of inertia). An inertial frame
of reference is one in 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 to observe
fictitious forces and the slight departures from Newton’s laws, such as the effect just described. On the large scale, such as for
the rotation of weather systems and ocean currents, the effects can be easily observed.


The crucial factor in determining whether a frame of reference is inertial is whether it accelerates or rotates relative to a known
inertial frame. Unless stated otherwise, all phenomena discussed in this text are considered in inertial frames.


All the forces discussed in this section are real forces, but there are a number of other real forces, such as lift and thrust, that are
not discussed in this section. They are more specialized, and it is not necessary to discuss every type of force. It is natural,
however, to ask where the basic simplicity we seek to find in physics is in the long list of forces. Are some more basic than
others? Are some different manifestations of the same underlying force? The answer to both questions is yes, as will be seen in
the next (extended) section and in the treatment of modern physics later in the text.


PhET Explorations: Forces in 1 Dimension


Explore the forces at work when you try to push a filing cabinet. Create an applied force and see the resulting friction force
and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body
diagram of all the forces (including gravitational and normal forces).


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 151




Figure 4.21 Forces in 1 Dimension (http://


/content/m42075/1.8/forces-1d_en.jar)


4.6 Problem-Solving Strategies
Success in problem solving is obviously necessary to understand and apply physical principles, not to mention the more
immediate need of passing exams. The basics of problem solving, presented earlier in this text, are followed here, but specific
strategies useful in applying Newton’s laws of motion are emphasized. These techniques also reinforce concepts that are useful
in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, and so the following
techniques should reinforce skills you have already begun to develop.


Problem-Solving Strategy for Newton’s Laws of Motion
Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton’s laws of
motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a
sketch is shown in Figure 4.22(a). Then, as in Figure 4.22(b), use arrows to represent all forces, label them carefully, and make
their lengths and directions correspond to the forces they represent (whenever sufficient information exists).


152 Chapter 4 | Dynamics: Force and Newton's Laws of Motion


Figure 4.22 (a) A sketch of Tarzan hanging from a vine. (b) Arrows are used to represent all forces. T is the tension in the vine above Tarzan, FT is
the force he exerts on the vine, and w is his weight. All other forces, such as the nudge of a breeze, are assumed negligible. (c) Suppose we are
given the ape man’s mass and asked to find the tension in the vine. We then define the system of interest as shown and draw a free-body diagram.
FT is no longer shown, because it is not a force acting on the system of interest; rather, FT acts on the outside world. (d) Showing only the arrows,
the head-to-tail method of addition is used. It is apparent that T = - w , if Tarzan is stationary.


Step 2. Identify what needs to be determined and what is known or can be inferred from the problem as stated. That is, make a
list of knowns and unknowns. Then carefully determine the system of interest. This decision is a crucial step, since Newton’s
second law involves only external forces. Once the system of interest has been identified, it becomes possible to determine
which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 4.22(c).)
Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between
the system and something outside (external). As illustrated earlier in this chapter, the system of interest depends on what
question we need to answer. This choice becomes easier with practice, eventually developing into an almost unconscious
process. Skill in clearly defining systems will be beneficial in later chapters as well.


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A diagram showing the system of interest and all of the external forces is called a free-body diagram. Only forces are shown on
free-body diagrams, not acceleration or velocity. We have drawn several of these in worked examples. Figure 4.22(c) shows a
free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.


Step 3. Once a free-body diagram is drawn, Newton’s second law can be applied to solve the problem. This is done in Figure
4.22(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 them into equation form and solve for the unknown, as done in all previous examples. If the problem is
one-dimensional—that is, if all forces are parallel—then they add like scalars. If the problem is two-dimensional, then it must be
broken down into a pair of one-dimensional problems. This is done by projecting the force vectors onto a set of axes chosen for
convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is
involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always
convenient to make one axis parallel to the direction of motion, if this is known.


Applying Newton’s Second Law


Before you write net force equations, it is critical to determine whether the system is accelerating in a particular direction. If
the acceleration is zero in a particular direction, then the net force is zero in that direction. Similarly, if the acceleration is
nonzero in a particular direction, then the net force is described by the equation: Fnet = ma .


For example, if the system is accelerating in the horizontal direction, but it is not accelerating in the vertical direction, then
you will have the following conclusions:


(4.57)Fnet x = ma,


(4.58)Fnet y = 0.


You will need this information in order to determine unknown forces acting in a system.


Step 4. As always, check the solution to see whether it is reasonable. In some cases, this is obvious. For example, it is
reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice,
intuition develops gradually through problem solving, and with experience it becomes progressively easier to judge whether an
answer is reasonable. Another way to check your solution is to check the units. If you are solving for force and end up with units
of m/s, then you have made a mistake.


4.7 Further Applications of Newton’s Laws of Motion
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.


Example 4.7 Drag Force on a Barge


Suppose two tugboats push on a barge at different angles, as shown in Figure 4.23. 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.


Figure 4.23 (a) A view from above of two tugboats pushing on a barge. (b) The free-body diagram for the ship contains 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. Since the applied forces are perpendicular, the x- and y-axes are in the same direction as Fx and Fy . The problem quickly becomes a
one-dimensional problem along the direction of Fapp , since friction is in the direction opposite to Fapp .


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 153




If 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,
such as air or water. The drag force opposes the motion of the object.)


Strategy


The directions and magnitudes of acceleration and the applied forces are given in Figure 4.23(a). We will define the total
force of the tugboats on the barge as Fapp so that:


(4.59)Fapp=Fx + Fy


Since the barge is flat bottomed, the drag of the water FD will be in the direction opposite to Fapp , as shown in the free-
body diagram in Figure 4.23(b). The system of interest here is the barge, since the forces on it are given as well as its
acceleration. Our strategy is to find the magnitude and direction of the net applied force Fapp , and then apply Newton’s
second law to solve for the drag force FD .


Solution


Since Fx and Fy are perpendicular, the magnitude and direction of Fapp are easily found. First, the resultant magnitude
is given by the Pythagorean theorem:


(4.60)Fapp = Fx2 + Fy2


Fapp = (2.7×105 N)2 + (3.6×105 N)2 = 4.5×105 N.


The angle is given by


(4.61)
θ = tan−1⎛⎝


Fy
Fx



θ = tan−1⎛⎝
3.6×105 N
2.7×105 N



⎠ = 53º,


which we know, because of Newton’s first law, is the same direction as the acceleration. FD is in the opposite direction of
Fapp , since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as Fapp , but its
magnitude is slightly less than Fapp . The problem is now one-dimensional. From Figure 4.23(b), we can see that


(4.62)Fnet = Fapp − FD.


But Newton’s second law states that


(4.63)Fnet = ma.


Thus,


(4.64)Fapp − FD = ma.


This can be solved for the magnitude of the drag force of the water FD in terms of known quantities:


(4.65)FD = Fapp − ma.


Substituting known values gives


(4.66)FD = (4.5×10
5 N) − (5.0×106 kg)(7.5×10–2 m/s2 ) = 7.5×104 N.


The direction of FD has already been determined to be in the direction opposite to Fapp , or at an angle of 53º south of
west.


Discussion


The numbers used in this example are reasonable for a moderately large barge. It is certainly difficult to obtain larger
accelerations with tugboats, and small speeds are desirable to avoid running the barge into the docks. Drag is 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.


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In the earlier example of a tightrope walker we noted that the tensions in wires supporting a mass were equal only because the
angles on either side were equal. Consider the following example, where the angles are not equal; slightly more trigonometry is
involved.


Example 4.8 Different Tensions at Different Angles


Consider the traffic light (mass 15.0 kg) suspended from two wires as shown in Figure 4.24. Find the tension in each wire,
neglecting the masses of the wires.


Figure 4.24 A traffic light is suspended from two wires. (b) Some of the forces involved. (c) Only forces acting on the system are shown here. The
free-body diagram for the traffic light is also shown. (d) The forces projected onto vertical (y) and horizontal (x) axes. The horizontal components
of the tensions must cancel, and the sum of the vertical components of the tensions must equal the weight of the traffic light. (e) The free-body
diagram shows the vertical and horizontal forces acting on the traffic light.


Strategy


The system of interest is the traffic light, and its free-body diagram is shown in Figure 4.24(c). The three forces involved are
not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis
vertical and one horizontal, and the vector projections on it are shown in part (d) of the figure. 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


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 155




second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because
acceleration is zero.


Solution


First consider the horizontal or x-axis:


(4.67)Fnetx = T2x − T1x = 0.


Thus, as you might expect,


(4.68)T1x = T2x.


This gives us the following relationship between T1 and T2 :


(4.69)T1 cos (30º) = T2 cos (45º).


Thus,


(4.70)T2 = (1.225)T1.


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:


(4.71)Fnet y = T1y + T2y − w = 0.


This implies


(4.72)T1y + T2y = w.


Substituting the expressions for the vertical components gives


(4.73)T1 sin (30º) + T2 sin (45º) = 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:


(4.74)T1(0.500) + (1.225T1)(0.707) = w = mg,


which yields


(4.75)(1.366)T1 = (15.0 kg)(9.80 m/s
2).


Solving this last equation gives the magnitude of T1 to be


(4.76)T1 = 108 N.


Finally, the magnitude of T2 is determined using the relationship between them, T2 = 1.225 T1 , found above. Thus we


obtain


(4.77)T2 = 132 N.


Discussion


Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either
side are the same (as they were in the earlier example of a tightrope walker).


The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it
must push upward to support the weight of an 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 at a constant speed: will the scale still read more than your weight at rest? Consider the
following example.


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Example 4.9 What Does the Bathroom Scale Read in an Elevator?


Figure 4.25 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 4.25 (a) The various forces acting when a person stands on a bathroom scale in an elevator. The arrows 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, ws is the weight of the scale, we is the weight of the elevator, Fs is the force of the scale on the
person, Fp is the force of the person on the scale, Ft 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.


Strategy


If the scale is accurate, its reading will equal Fp , the magnitude of the force the person exerts downward on it. Figure


4.25(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 a free-body diagram is drawn as in
Figure 4.25(b). Analysis of the free-body diagram using Newton’s laws can produce answers to both parts (a) and (b) of this
example, as well as some other questions that might arise. The only forces acting on the person are his weight w and the
upward force of the scale Fs . According to Newton’s third law Fp and Fs 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,


(4.78)Fnet = ma.


From the free-body diagram we see that Fnet = Fs − w , so that


(4.79)Fs − w = ma.


Solving for Fs gives an equation with only one unknown:


(4.80)Fs = ma + w,


or, because w = mg , simply


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 157




(4.81)Fs = ma + mg.


No assumptions were made about the acceleration, and so this solution should be valid for a variety of accelerations in
addition to the ones in this exercise.


Solution for (a)


In this part of the problem, a = 1.20 m/s2 , so that
(4.82)Fs = (75.0 kg)(1.20 m/s2 ) + (75.0 kg)(9.80 m/s2),


yielding


(4.83)Fs = 825 N.


Discussion for (a)


This is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force
of the scale would be equal to his weight:


(4.84)Fnet = ma = 0 = Fs − w
Fs = w = mg


Fs = (75.0 kg)(9.80 m/s2)
Fs = 735 N.


So, the scale reading in the elevator is greater than his 735-N (165 lb) weight. This means that the scale is pushing 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 feel in rapidly accelerating versus slowly
accelerating elevators.


Solution for (b)


Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight?


For any constant velocity—up, down, or stationary—acceleration is zero because a = ΔvΔt , and Δv = 0 .


Thus,


(4.85)Fs = ma + mg = 0 + mg.


Now


(4.86)Fs = (75.0 kg)(9.80 m/s2),


which gives


(4.87)Fs = 735 N.


Discussion for (b)


The scale reading is 735 N, which equals the person’s weight. This will be the case whenever the elevator has a constant
velocity—moving up, moving down, or stationary.


158 Chapter 4 | Dynamics: Force and Newton's Laws of Motion


The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator
accelerates downward, a is negative, and the scale reading is less than the weight of the person, until a constant downward
velocity is reached, at which time the scale reading again becomes equal to the person’s weight. If the elevator is in free-fall and
accelerating downward at g , then the scale reading will be zero and the person will appear to be weightless.


Integrating Concepts: 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 of physical
principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to
solve problems of motion. For example, forces produce accelerations, a topic of kinematics, and hence the relevance of earlier
chapters. When approaching problems that involve various types of forces, acceleration, velocity, and/or position, use the
following steps to approach the problem:


Problem-Solving Strategy


Step 1. Identify which physical principles are involved. Listing the givens and the quantities to be calculated will allow you to
identify the principles involved.
Step 2. Solve the problem using strategies outlined in the text. If these are available for the specific topic, you should refer to
them. You should also refer to the sections of the text that deal with a particular topic. The following worked example illustrates
how these strategies are applied to an integrated concept problem.


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Example 4.10 What Force Must a Soccer Player Exert to Reach Top Speed?


A soccer player starts from rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What was his average
acceleration? (b) What average force did he exert backward on the ground to achieve this acceleration? The player’s mass
is 70.0 kg, and air resistance is negligible.


Strategy


1. To solve an integrated concept problem, we must first identify the physical principles involved and identify the chapters
in which they are found. Part (a) of this example considers acceleration along a straight line. This is a topic of
kinematics. Part (b) deals with force, a topic of dynamics found in this chapter.


2. The following solutions to each part of the example illustrate how the specific problem-solving strategies are applied.
These involve identifying knowns and unknowns, checking to see if the answer is reasonable, and so forth.


Solution for (a)


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, and so Δt = 2.50 s . The unknown is acceleration, which can be found from its definition:


(4.88)a = ΔvΔt .


Substituting the known values yields


(4.89)a = 8.00 m/s2.50 s
= 3.20 m/s2.


Discussion for (a)


This is an attainable acceleration for an athlete in good condition.


Solution for (b)


Here we are asked to find the average force the player exerts backward to achieve this forward acceleration. Neglecting air
resistance, this would be equal in magnitude to the net external force on the player, since this force causes his acceleration.
Since we now know the player’s acceleration and are given his mass, we can use Newton’s second law to find the force
exerted. That is,


(4.90)Fnet = ma.


Substituting the known values of m and a gives


(4.91)Fnet = (70.0 kg)(3.20 m/s2)
= 224 N.


Discussion for (b)


This is about 50 pounds, a reasonable average force.


This worked example illustrates how to apply problem-solving strategies to situations that include topics from different
chapters. The first step is to identify the physical principles involved in the problem. The second step is to solve for the
unknown using familiar problem-solving strategies. These strategies are found throughout the text, and many worked
examples show how to use them for single topics. You will find these techniques for integrated concept problems useful in
applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday
life. The following problems will build your skills in the broad application of physical principles.


4.8 Extended Topic: The Four Basic Forces—An Introduction
One of the most remarkable simplifications in physics is that only four distinct forces account for all known phenomena. In fact,
nearly all of the forces we experience directly are due to only one basic force, called the electromagnetic force. (The gravitational
force is the only force we experience directly that is not electromagnetic.) This is a tremendous simplification of the myriad of
apparently different forces we can list, only a few of which were discussed in the previous section. As we will see, the basic
forces are all thought to act through the exchange of microscopic carrier particles, and the characteristics of the basic forces are
determined by the types of particles exchanged. Action at a distance, such as the gravitational force of Earth on the Moon, is
explained by the existence of a force field rather than by “physical contact.”


The four basic forces are the gravitational force, the electromagnetic force, the weak nuclear force, and the strong nuclear force.
Their properties are summarized in Table 4.1. Since the weak and strong nuclear forces act over an extremely short range, the
size of a nucleus or less, we do not experience them directly, although they are crucial to the very structure of matter. These
forces determine which nuclei are stable and which decay, and they are the basis of the release of energy in certain nuclear
reactions. Nuclear forces determine not only the stability of nuclei, but also the relative abundance of elements in nature. The


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 159




properties of the nucleus of an atom determine the number of electrons it has and, thus, indirectly determine the chemistry of the
atom. More will be said of all of these topics in later chapters.


Concept Connections: The Four Basic Forces


The four basic forces will be encountered in more detail as you progress through the text. The gravitational force is defined
in Uniform Circular Motion and Gravitation, electric force in Electric Charge and Electric Field, magnetic force in
Magnetism, and nuclear forces in Radioactivity and Nuclear Physics. On a macroscopic scale, electromagnetism and
gravity are the basis for all forces. The nuclear forces are vital to the substructure of matter, but they are not directly
experienced on the macroscopic scale.


Table 4.1 Properties of the Four Basic Forces[1]


Force Approximate Relative Strengths Range Attraction/Repulsion Carrier Particle


Gravitational 10−38 ∞ attractive only Graviton


Electromagnetic 10 – 2 ∞ attractive and repulsive Photon


Weak nuclear 10 – 13 < 10–18m attractive and repulsive W+ , W – , Z0


Strong nuclear 1 < 10–15m attractive and repulsive gluons


The gravitational force is surprisingly weak—it is only because gravity is always attractive that we notice it at all. Our weight is
the gravitational force due to the entire Earth acting on us. On the very large scale, as in astronomical systems, the gravitational
force is the dominant force determining the motions of moons, planets, stars, and galaxies. The gravitational force also affects
the nature of space and time. As we shall see later in the study of general relativity, space is curved in the vicinity of very
massive bodies, such as the Sun, and time actually slows down near massive bodies.


Electromagnetic forces can be either attractive or repulsive. They are long-range forces, which act over extremely large
distances, and they nearly cancel for macroscopic objects. (Remember that it is the net external force that is important.) If they
did not cancel, electromagnetic forces would completely overwhelm the gravitational force. The electromagnetic force is a
combination of electrical forces (such as those that cause static electricity) and magnetic forces (such as those that affect a
compass needle). These two forces were thought to be quite distinct until early in the 19th century, when scientists began to
discover that they are different manifestations of the same force. This discovery is a classical case of the unification of forces.
Similarly, friction, tension, and all of the other classes of forces we experience directly (except gravity, of course) are due to
electromagnetic interactions of atoms and molecules. It is still convenient to consider these forces separately in specific
applications, however, because of the ways they manifest themselves.


Concept Connections: Unifying Forces


Attempts to unify the four basic forces are discussed in relation to elementary particles later in this text. By “unify” we mean
finding connections between the forces that show that they are different manifestations of a single force. Even if such
unification is achieved, the forces will retain their separate characteristics on the macroscopic scale and may be identical
only under extreme conditions such as those existing in the early universe.


Physicists are now exploring whether the four basic forces are in some way related. Attempts to unify all forces into one come
under the rubric of Grand Unified Theories (GUTs), with which there has been some success in recent years. It is now known
that under conditions of extremely high density and temperature, such as existed in the early universe, the electromagnetic and
weak nuclear forces are indistinguishable. They can now be considered to be different manifestations of one force, called the
electroweak force. So the list of four has been reduced in a sense to only three. Further progress in unifying all forces is proving
difficult—especially the inclusion of the gravitational force, which has the special characteristics of affecting the space and time in
which the other forces exist.


While the unification of forces will not affect how we discuss forces in this text, it is fascinating that such underlying simplicity
exists in the face of the overt complexity of the universe. There is no reason that nature must be simple—it simply is.


Action at a Distance: Concept of a Field
All forces act at a distance. This is obvious for the gravitational force. Earth and the Moon, for example, interact without coming
into contact. It is also true for all other forces. Friction, for example, is an electromagnetic force between atoms that may not
actually touch. What is it that carries forces between objects? One way to answer this question is to imagine that a force field
surrounds whatever object creates the force. A second object (often called a test object) placed in this field will experience a


160 Chapter 4 | Dynamics: Force and Newton's Laws of Motion


1. The graviton is a proposed particle, though it has not yet been observed by scientists. See the discussion of gravitational


waves later in this section. The particles W+ , W− , and Z0 are called vector bosons; these were predicted by theory and first
observed in 1983. There are eight types of gluons proposed by scientists, and their existence is indicated by meson exchange in
the nuclei of atoms.


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force that is a function of location and other variables. The field itself is the “thing” that carries the force from one object to
another. The field is defined so as to be a characteristic of the object creating it; the field does not depend on the test object
placed in it. Earth’s gravitational field, for example, is a function of the mass of Earth and the distance from its center,
independent of the presence of other masses. The concept of a field is useful because equations can be written for force fields
surrounding objects (for gravity, this yields w = mg at Earth’s surface), and motions can be calculated from these equations.
(See Figure 4.26.)


Figure 4.26 The electric force field between a positively charged particle and a negatively charged particle. When a positive test charge is placed in the
field, the charge will experience a force in the direction of the force field lines.


Concept Connections: Force Fields


The concept of a force field is also used in connection with electric charge and is presented in Electric Charge and Electric
Field. It is also a useful idea for all the basic forces, as will be seen in Particle Physics. Fields help us to visualize forces
and how they are transmitted, as well as to describe them with precision and to link forces with subatomic carrier particles.


The field concept has been applied very successfully; we can calculate motions and describe nature to high precision using field
equations. As useful as the field concept is, however, it leaves unanswered the question of what carries the force. It has been
proposed in recent decades, starting in 1935 with Hideki Yukawa’s (1907–1981) work on the strong nuclear force, that all forces
are transmitted by the exchange of elementary particles. We can visualize particle exchange as analogous to macroscopic
phenomena such as two people passing a basketball back and forth, thereby exerting a repulsive force without touching one
another. (See Figure 4.27.)


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 161




Figure 4.27 The exchange of masses resulting in repulsive forces. (a) The person throwing the basketball exerts a force Fp1 on it toward the other


person and feels a reaction force FB away from the second person. (b) The person catching the basketball exerts a force Fp2 on it to stop the ball


and feels a reaction force F′B away from the first person. (c) The analogous exchange of a meson between a proton and a neutron carries the
strong nuclear forces Fexch and F′exch between them. An attractive force can also be exerted by the exchange of a mass—if person 2 pulled the
basketball away from the first person as he tried to retain it, then the force between them would be attractive.


This idea of particle exchange deepens rather than contradicts field concepts. It is more satisfying philosophically to think of
something physical actually moving between objects acting at a distance. Table 4.1 lists the exchange or carrier particles, both
observed and proposed, that carry the four forces. But the real fruit of the particle-exchange proposal is that searches for
Yukawa’s proposed particle found it and a number of others that were completely unexpected, stimulating yet more research. All
of this research eventually led to the proposal of quarks as the underlying substructure of matter, which is a basic tenet of GUTs.
If successful, these theories will explain not only forces, but also the structure of matter itself. Yet physics is an experimental
science, so the test of these theories must lie in the domain of the real world. As of this writing, scientists at the CERN laboratory
in Switzerland are starting to test these theories using the world’s largest particle accelerator: the Large Hadron Collider. This
accelerator (27 km in circumference) allows two high-energy proton beams, traveling in opposite directions, to collide. An energy
of 14 trillion electron volts will be available. It is anticipated that some new particles, possibly force carrier particles, will be found.
(See Figure 4.28.) One of the force carriers of high interest that researchers hope to detect is the Higgs boson. The observation
of its properties might tell us why different particles have different masses.


162 Chapter 4 | Dynamics: Force and Newton's Laws of Motion


Figure 4.28 The world’s largest particle accelerator spans the border between Switzerland and France. Two beams, traveling in opposite directions
close to the speed of light, collide in a tube similar to the central tube shown here. External magnets determine the beam’s path. Special detectors will
analyze particles created in these collisions. Questions as broad as what is the origin of mass and what was matter like the first few seconds of our
universe will be explored. This accelerator began preliminary operation in 2008. (credit: Frank Hommes)


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acceleration:


carrier particle:


dynamics:


external force:


force:


force field:


free-body diagram:


free-fall:


friction:


Tiny particles also have wave-like behavior, something we will explore more in a later chapter. To better understand force-carrier
particles from another perspective, let us consider gravity. The search for gravitational waves has been going on for a number of
years. Almost 100 years ago, Einstein predicted the existence of these waves as part of his general theory of relativity.
Gravitational waves are created during the collision of massive stars, in black holes, or in supernova explosions—like shock
waves. These gravitational waves will travel through space from such sites much like a pebble dropped into a pond sends out
ripples—except these waves move at the speed of light. A detector apparatus has been built in the U.S., consisting of two large
installations nearly 3000 km apart—one in Washington state and one in Louisiana! The facility is called the Laser Interferometer
Gravitational-Wave Observatory (LIGO). Each installation is designed to use optical lasers to examine any slight shift in the
relative positions of two masses due to the effect of gravity waves. The two sites allow simultaneous measurements of these
small effects to be separated from other natural phenomena, such as earthquakes. Initial operation of the detectors began in
2002, and work is proceeding on increasing their sensitivity. Similar installations have been built in Italy (VIRGO), Germany
(GEO600), and Japan (TAMA300) to provide a worldwide network of gravitational wave detectors.


International collaboration in this area is moving into space with the joint EU/US project LISA (Laser Interferometer Space
Antenna). Earthquakes and other Earthly noises will be no problem for these monitoring spacecraft. LISA will complement LIGO
by looking at much more massive black holes through the observation of gravitational-wave sources emitting much larger
wavelengths. Three satellites will be placed in space above Earth in an equilateral triangle (with 5,000,000-km sides) (Figure
4.29). The system will measure the relative positions of each satellite to detect passing gravitational waves. Accuracy to within
10% of the size of an atom will be needed to detect any waves. The launch of this project might be as early as 2018.


“I’m sure LIGO will tell us something about the universe that we didn’t know before. The history of science tells us that any time
you go where you haven’t been before, you usually find something that really shakes the scientific paradigms of the day. Whether
gravitational wave astrophysics will do that, only time will tell.”—David Reitze, LIGO Input Optics Manager, University of Florida


Figure 4.29 Space-based future experiments for the measurement of gravitational waves. Shown here is a drawing of LISA’s orbit. Each satellite of
LISA will consist of a laser source and a mass. The lasers will transmit a signal to measure the distance between each satellite’s test mass. The
relative motion of these masses will provide information about passing gravitational waves. (credit: NASA)


The ideas presented in this section are but a glimpse into topics of modern physics that will be covered in much greater depth in
later chapters.


Glossary
the rate at which an object’s velocity changes over a period of time


a fundamental particle of nature that is surrounded by a characteristic force field; photons are carrier particles
of the electromagnetic force


the study of how forces affect the motion of objects and systems


a force acting on an object or system that originates outside of the object or system


a push or pull on an object with a specific magnitude and direction; can be represented by vectors; can be expressed
as a multiple of a standard force


a region in which a test particle will experience a force


a sketch showing all of the external forces acting on an object or system; the system is represented by a
dot, and the forces are represented by vectors extending outward from the dot


a situation in which the only force acting on an object is the force due to gravity


a force past each other of objects that are touching; examples include rough surfaces and air resistance


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 163




inertia:


inertial frame of reference:


law of inertia:


mass:


net external force:


Newton’s first law of motion:


Newton’s second law of motion:


Newton’s third law of motion:


normal force:


system:


tension:


thrust:


weight:


the tendency of an object to remain at rest or remain in motion


a coordinate system that is not accelerating; all forces acting in an inertial frame of reference are
real forces, as opposed to fictitious forces that are observed due to an accelerating frame of reference


see Newton’s first law of motion


the quantity of matter in a substance; measured in kilograms


the vector sum of all external forces acting on an object or system; causes a mass to accelerate


a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless
acted on by a net external force; also known as the law of inertia


the net external force Fnet on an object with mass m is proportional to and in the same
direction as the acceleration of the object, a , and inversely proportional to the mass; defined mathematically as


a = Fnetm


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 the first body exerts


the force that a surface applies to an object to support the weight of the object; acts perpendicular to the
surface on which the object rests


defined by the boundaries of an object or collection of objects being observed; all forces originating from outside of
the system are considered external forces


the pulling force that acts along a medium, especially a stretched flexible connector, such as a rope or cable; when a
rope supports the weight of an object, the force on the object due to the rope is called a tension force


a reaction force that pushes a body forward in response to a backward force; rockets, airplanes, and cars are pushed
forward by a thrust reaction force


the force w due to gravity acting on an object of mass m ; defined mathematically as: w = mg , where g is the
magnitude and direction of the acceleration due to gravity


Section Summary


4.1 Development of Force Concept
• Dynamics is the study of how forces affect the motion of objects.
• Force is a push or pull that can be defined in terms of various standards, and it is a vector having both magnitude and


direction.
• External forces are any outside forces that act on a body. A free-body diagram is a drawing of all external forces acting


on a body.


4.2 Newton’s First Law of Motion: Inertia
• Newton’s first law of motion states that a body at rest remains at rest, or, if in motion, remains in motion at a constant


velocity unless acted on by a net external force. This is also known as the law of inertia.
• Inertia is the tendency of an object to remain at rest or remain in motion. Inertia is related to an object’s mass.
• Mass is the quantity of matter in a substance.


4.3 Newton’s Second Law of Motion: Concept of a System
• Acceleration, a , is defined as a change in velocity, meaning a change in its magnitude or direction, or both.
• An external force is one acting on a system from outside the system, as opposed to internal forces, which act between


components within the system.
• Newton’s second law of motion states that the acceleration of a system is directly proportional to and in the same direction


as the net external force acting on the system, and inversely proportional to its mass.


• In equation form, Newton’s second law of motion is a = Fnetm .
• This is often written in the more familiar form: Fnet = ma .
• The weight w of an object is defined as the force of gravity acting on an object of mass m . The object experiences an


acceleration due to gravity g :


w = mg.
• If the only force acting on an object is due to gravity, the object is in free fall.


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• Friction is a force that opposes the motion past each other of objects that are touching.


4.4 Newton’s Third Law of Motion: Symmetry in Forces
• Newton’s third law of motion represents a basic symmetry in nature. It states: 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 the first
body exerts.


• A thrust is a reaction force that pushes a body forward in response to a backward force. Rockets, airplanes, and cars are
pushed forward by a thrust reaction force.


4.5 Normal, Tension, and Other Examples of Forces
• When objects rest 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, N .
• When objects rest on a non-accelerating horizontal surface, the magnitude of the normal force is equal to the weight of the


object:


N = mg.
• When objects rest 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 (w⊥ ) and parallel (w ∥ ) to the surface of the plane. These
components can be calculated using:


w ∥ = w sin (θ) = mg sin (θ)


w⊥ = w cos (θ) = mg cos (θ).
• The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension, T . When a rope


supports the weight of an object that is at rest, the tension in the rope is equal to the weight of the object:


T = mg.
• In any inertial frame of reference (one that is not accelerated or rotated), Newton’s laws have the simple forms given in this


chapter and all forces are real forces having a physical origin.


4.6 Problem-Solving Strategies
• To solve problems involving Newton’s laws of motion, follow the procedure described:


1. Draw a sketch of the problem.
2. Identify known and unknown quantities, and identify the system of interest. Draw a free-body diagram, which is a


sketch showing all of the forces acting on an object. The object is represented by a dot, and the forces are
represented by vectors extending in different directions from the dot. If vectors act in directions that are not horizontal
or vertical, resolve the vectors into horizontal and vertical components and draw them on the free-body diagram.


3. Write Newton’s second law in the horizontal and vertical directions and add the forces acting on the object. If the
object does not accelerate in a particular direction (for example, the x -direction) then Fnet x = 0 . If the object does
accelerate in that direction, Fnet x = ma .


4. Check your answer. Is the answer reasonable? Are the units correct?


4.7 Further Applications of Newton’s Laws of Motion
• Newton’s laws of motion can be applied in numerous situations to solve problems of motion.
• Some problems will 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 analyze the
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 is accelerating, the
normal force will be less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal
force will always be less than the full weight of the object.


• Some problems will contain various physical quantities, such as forces, acceleration, velocity, or position. You can apply
concepts from kinematics and dynamics in order to solve these problems of motion.


4.8 Extended Topic: The Four Basic Forces—An Introduction
• The various types of forces that are categorized for use in many applications are all manifestations of the four basic forces


in nature.
• The properties of these forces are summarized in Table 4.1.
• Everything we experience directly without sensitive instruments is due to either electromagnetic forces or gravitational


forces. The nuclear forces are responsible for the submicroscopic structure of matter, but they are not directly sensed
because of their short ranges. Attempts are being made to show all four forces are different manifestations of a single
unified force.


• A force field surrounds an object creating a force and is the carrier of that force.


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 165




Conceptual Questions


4.1 Development of Force Concept
1. Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be
capable of producing the same force repeatedly.


2.What properties do forces have that allow us to classify them as vectors?


4.2 Newton’s First Law of Motion: Inertia
3. How are inertia and mass related?


4.What is the relationship between weight and mass? Which is an intrinsic, unchanging property of a body?


4.3 Newton’s Second Law of Motion: Concept of a System
5.Which statement is correct? (a) Net force causes motion. (b) Net force causes change in motion. Explain your answer and
give an example.


6.Why can we neglect forces such as those holding a body together when we apply Newton’s second law of motion?


7. Explain how the choice of the “system of interest” affects which forces must be considered when applying Newton’s second
law of motion.


8. Describe a situation in which the net external force on a system is not zero, yet its speed remains constant.


9. A system can have a nonzero velocity while the net external force on it is zero. Describe such a situation.


10. A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory?


11. (a) Give an example of different net external forces acting on the same system to produce different accelerations. (b) Give an
example of the same net external force acting on systems of different masses, producing different accelerations. (c) What law
accurately describes both effects? State it in words and as an equation.


12. If the acceleration of a system is zero, are no external forces acting on it? What about internal forces? Explain your answers.


13. If a constant, nonzero force is applied to an object, what can you say about the velocity and acceleration of the object?


14. The gravitational force on the basketball in Figure 4.6 is ignored. When gravity is taken into account, what is the direction of
the net external force on the basketball—above horizontal, below horizontal, or still horizontal?


4.4 Newton’s Third Law of Motion: Symmetry in Forces
15.When you take off in a jet aircraft, there is a sensation of being pushed back into the seat. Explain why you move backward
in the seat—is there really a force backward on you? (The same reasoning explains whiplash injuries, in which the head is
apparently thrown backward.)


16. A device used since the 1940s to measure the kick or recoil of the body due to heart beats is the “ballistocardiograph.” What
physics principle(s) are involved here to measure the force of cardiac contraction? How might we construct such a device?


17. Describe a situation in which one system exerts a force on another and, as a consequence, experiences a force that is equal
in magnitude and opposite in direction. Which of Newton’s laws of motion apply?


18.Why does an ordinary rifle recoil (kick backward) when fired? The barrel of a recoilless rifle is open at both ends. Describe
how Newton’s third law applies when one is fired. Can you safely stand close behind one when it is fired?


19. An American football lineman reasons that it is senseless to try to out-push the opposing player, since no matter how hard he
pushes he will experience an equal and opposite force from the other player. Use Newton’s laws and draw a free-body diagram
of an appropriate system to explain how he can still out-push the opposition if he is strong enough.


20. Newton’s third law of motion tells us that forces always occur in pairs of equal and opposite magnitude. Explain how the
choice of the “system of interest” affects whether one such pair of forces cancels.


4.5 Normal, Tension, and Other Examples of Forces
21. If a leg is suspended by a traction setup as shown in Figure 4.30, what is the tension in the rope?


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Figure 4.30 A leg is suspended by a traction system in which wires are used to transmit forces. Frictionless pulleys change the direction of the force T
without changing its magnitude.


22. In a traction setup for a broken bone, with pulleys and rope available, how might we be able to increase the force along the
tibia using the same weight? (See Figure 4.30.) (Note that the tibia is the shin bone shown in this image.)


4.7 Further Applications of Newton’s Laws of Motion
23. To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is
accelerating downward at g . Why will they appear to be weightless, as measured by standing on a bathroom scale, in this


accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?


24. A cartoon shows the toupee coming off the head of an elevator passenger when the elevator rapidly stops during an upward
ride. Can this really happen without the person being tied to the floor of the elevator? Explain your answer.


4.8 Extended Topic: The Four Basic Forces—An Introduction
25. Explain, in terms of the properties of the four basic forces, why people notice the gravitational force acting on their bodies if it
is such a comparatively weak force.


26.What is the dominant force between astronomical objects? Why are the other three basic forces less significant over these
very large distances?


27. Give a detailed example of how the exchange of a particle can result in an attractive force. (For example, consider one child
pulling a toy out of the hands of another.)


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 167




Problems & Exercises


4.3 Newton’s Second Law of Motion: Concept
of a System
You may assume data taken from illustrations is accurate
to three digits.


1. A 63.0-kg sprinter starts a race with an acceleration of


4.20 m/s2 . What is the net external force on him?
2. If the sprinter from the previous problem accelerates at that
rate for 20 m, and then maintains that velocity for the
remainder of the 100-m dash, what will be his time for the
race?


3. A cleaner pushes a 4.50-kg laundry cart in such a way that
the net external force on it is 60.0 N. Calculate the magnitude
of its acceleration.


4. Since astronauts in orbit are apparently weightless, a
clever method of measuring their masses is needed to
monitor their mass gains or losses to adjust diets. One way to
do this is to exert a known force on an astronaut and measure
the acceleration produced. Suppose a net external force of
50.0 N is exerted and the 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 they
orbit experiences an equal and opposite force. Discuss how
this would affect the measurement of the astronaut’s
acceleration. Propose a method in which recoil of the vehicle
is avoided.


5. In Figure 4.7, the net external force on the 24-kg mower is
stated to be 51 N. If the force of friction opposing the motion
is 24 N, what force F (in newtons) is the person exerting on
the mower? Suppose the mower is moving at 1.5 m/s when
the force F is removed. How far will the mower go before
stopping?


6. The same rocket sled drawn in Figure 4.31 is decelerated


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 2100 kg.


Figure 4.31


7. (a) If the rocket sled shown in Figure 4.32 starts with only
one rocket burning, what is the magnitude of its acceleration?
Assume that the mass of the system is 2100 kg, the thrust T


is 2.4×104 N, and the force of friction opposing the motion
is known to be 650 N. (b) Why is the acceleration not one-
fourth of what it is with all rockets burning?


Figure 4.32


8.What is the deceleration of the rocket sled if it comes to
rest in 1.1 s from a speed of 1000 km/h? (Such deceleration
caused one test subject to black out and have temporary
blindness.)


9. Suppose two children push horizontally, but in exactly
opposite directions, on a third child in a wagon. The first child
exerts a force of 75.0 N, the second a force of 90.0 N, friction
is 12.0 N, and the mass of the third child plus wagon is 23.0
kg. (a) What is the system of interest if the acceleration of the
child in the wagon is to be calculated? (b) Draw a free-body
diagram, including all forces acting on the system. (c)
Calculate the acceleration. (d) What would the acceleration
be if friction were 15.0 N?


10. 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 air resistance,
total 400 N. (Air resistance is analogous to air friction. It
always opposes the motion of an object.) What is the
magnitude of the force the motorcycle exerts backward on the
ground to produce its acceleration if the mass of the
motorcycle with rider is 245 kg?


11. The rocket sled shown in Figure 4.33 accelerates at a


rate of 49.0 m/s2 . Its passenger has a mass of 75.0 kg. (a)
Calculate the horizontal component of the force the seat
exerts against his body. Compare this with his weight by
using a ratio. (b) Calculate the direction and magnitude of the
total force the seat exerts against his body.


Figure 4.33


12. Repeat the previous problem for the 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 restraining
belts.


13. The weight of an astronaut plus his space suit on the
Moon is only 250 N. How much do they weigh on Earth?
What is the mass on the Moon? On Earth?


14. Suppose the mass of a fully loaded module in which
astronauts take off from the Moon is 10,000 kg. The thrust of
its engines is 30,000 N. (a) Calculate its the magnitude of
acceleration in a vertical takeoff from the Moon. (b) Could it lift
off from Earth? If not, why not? If it could, calculate the
magnitude of its acceleration.


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4.4 Newton’s Third Law of Motion: Symmetry in
Forces
15.What net external force is exerted on a 1100-kg artillery
shell fired from a battleship if the shell is accelerated at


2.40×104 m/s2 ? What is the magnitude of the force
exerted on the ship by the artillery shell?


16. A brave but inadequate rugby player is being pushed
backward by an opposing player who is exerting a force of
800 N on him. The mass of the losing player plus equipment


is 90.0 kg, and he is accelerating at 1.20 m/s2 backward.
(a) What is the force of friction between the losing player’s
feet and the grass? (b) What force does the winning player
exert on the ground to move forward if his mass plus
equipment is 110 kg? (c) Draw a sketch of the situation
showing the system of interest used to solve each part. For
this situation, draw a free-body diagram and write the net
force equation.


4.5 Normal, Tension, and Other Examples of
Forces
17. Two teams of nine members each engage in a tug of war.
Each of the first team’s members has an average mass of 68
kg and exerts an average force of 1350 N horizontally. Each
of the second team’s members has an average mass of 73 kg
and exerts an average force of 1365 N horizontally. (a) What
is magnitude of the acceleration of the two teams? (b) What is
the tension in the section of rope between the teams?


18.What force does a trampoline have to apply to a 45.0-kg


gymnast to accelerate her straight up at 7.50 m/s2 ? Note
that the answer is independent of the velocity of the
gymnast—she can be moving either up or down, or be
stationary.


19. (a) Calculate the tension in a vertical strand of spider web


if a spider of mass 8.00×10−5 kg hangs motionless on it.
(b) Calculate the tension in a horizontal strand of spider web if
the same spider sits motionless in the middle of it much like
the tightrope walker in Figure 4.17. The strand sags at an
angle of 12º below the horizontal. Compare this with the
tension in the vertical strand (find their ratio).


20. Suppose a 60.0-kg gymnast climbs a rope. (a) What is the
tension in the rope if he climbs at a constant speed? (b) What
is the tension in the rope if he accelerates upward at a rate of


1.50 m/s2 ?


21. Show that, as stated in the text, a force F⊥ exerted on
a flexible medium at its center and perpendicular to its length
(such as on the tightrope wire in Figure 4.17) gives rise to a


tension of magnitude T = F⊥2 sin (θ) .


22. Consider the baby being weighed in Figure 4.34. (a)
What is the mass of the child and basket if a scale reading of
55 N is observed? (b) What is the tension T1 in the cord


attaching the baby to the scale? (c) What is the tension T2 in


the cord attaching the scale to the ceiling, if the scale has a
mass of 0.500 kg? (d) Draw a sketch of the situation
indicating the system of interest used to solve each part. The
masses of the cords are negligible.


Figure 4.34 A baby is weighed using a spring scale.


4.6 Problem-Solving Strategies


23. A 5.00×105-kg rocket is accelerating straight up. Its


engines produce 1.250×107 N of thrust, and air resistance
is 4.50×106 N . What is the rocket’s acceleration? Explicitly
show how you follow the steps in the Problem-Solving
Strategy for Newton’s laws of motion.


24. The wheels of a midsize car exert a force of 2100 N
backward on the road to accelerate the car in the forward
direction. If the force of friction including air resistance is 250


N and the acceleration of the car is 1.80 m/s2 , what is the
mass of the car plus its occupants? Explicitly show how you
follow the steps in the Problem-Solving Strategy for Newton’s
laws of motion. For this situation, draw a free-body diagram
and write the net force equation.


25. Calculate the force a 70.0-kg high jumper must exert on
the ground to produce an upward acceleration 4.00 times the
acceleration due to gravity. Explicitly show how you follow the
steps in the Problem-Solving Strategy for Newton’s laws of
motion.


26.When landing after a spectacular somersault, a 40.0-kg
gymnast decelerates by pushing straight down on the mat.
Calculate the force she must exert if her deceleration is 7.00
times the acceleration due to gravity. Explicitly show how you
follow the steps in the Problem-Solving Strategy for Newton’s
laws of motion.


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 169




27. A freight train consists of two 8.00×104 -kg engines


and 45 cars with average masses of 5.50×104 kg . (a)
What force must each engine exert backward on the track to


accelerate the train at a rate of 5.00×10–2 m/s2 if the force
of friction is 7.50×105 N , assuming the engines exert
identical forces? This is not a large frictional force for such a
massive system. Rolling friction for trains is small, and
consequently trains are very energy-efficient transportation
systems. (b) What is the force in the coupling between the
37th and 38th cars (this is the force each exerts on the other),
assuming all cars have the same mass and that friction is
evenly distributed among all of the cars and engines?


28. Commercial airplanes are sometimes pushed out of the
passenger loading area by a tractor. (a) An 1800-kg tractor


exerts a force of 1.75×104 N backward on the pavement,
and the system experiences forces resisting motion that total


2400 N. If the acceleration is 0.150 m/s2 , what is the mass
of the airplane? (b) Calculate the force exerted by the tractor
on the airplane, assuming 2200 N of the friction is
experienced by the airplane. (c) Draw two sketches showing
the systems of interest used to solve each part, including the
free-body diagrams for each.


29. A 1100-kg car pulls a boat on a trailer. (a) What total force
resists the motion of the car, boat, and trailer, if the car exerts
a 1900-N force on the road and produces an acceleration of


0.550 m/s2 ? The mass of the boat plus trailer is 700 kg. (b)
What is the force in the hitch between the car and the trailer if
80% of the resisting forces are experienced by the boat and
trailer?


30. (a) Find the magnitudes of the forces F1 and F2 that
add to give the total force Ftot shown in Figure 4.35. This
may be done either graphically or by using trigonometry. (b)
Show graphically that the same total force is obtained
independent of the order of addition of F1 and F2 . (c) Find
the direction and magnitude of some other pair of vectors that
add to give Ftot . Draw these to scale on the same drawing
used in part (b) or a similar picture.


Figure 4.35


31. Two children pull a third child on a snow saucer sled
exerting forces F1 and F2 as shown from above in Figure
4.36. Find the acceleration of the 49.00-kg sled and child
system. Note that the direction of the frictional force is
unspecified; it will be in the opposite direction of the sum of
F1 and F2 .


Figure 4.36 An overhead view of the horizontal forces acting on a child’s
snow saucer sled.


32. Suppose your car was mired deeply in the mud and you
wanted to use the method illustrated in Figure 4.37 to pull it
out. (a) What force would you have to exert perpendicular to
the center of the rope to produce a force of 12,000 N on the
car if the angle is 2.00°? In this part, explicitly show how you
follow the steps in the Problem-Solving Strategy for Newton’s
laws of motion. (b) Real ropes stretch under such forces.
What force would be exerted on the car if the angle increases
to 7.00° and you still apply the force found in part (a) to its
center?


Figure 4.37


33.What force is exerted on the tooth in Figure 4.38 if the
tension in the wire is 25.0 N? Note that the force applied to
the tooth is smaller than the tension in the wire, but this is
necessitated by practical considerations of how force can be
applied in the mouth. Explicitly show how you follow steps in
the Problem-Solving Strategy for Newton’s laws of motion.


Figure 4.38 Braces are used to apply forces to teeth to realign them.
Shown in this figure are the tensions applied by the wire to the
protruding tooth. The total force applied to the tooth by the wire, Fapp ,
points straight toward the back of the mouth.


34. Figure 4.39 shows Superhero and Trusty Sidekick
hanging motionless from a rope. Superhero’s mass is 90.0
kg, while Trusty Sidekick’s is 55.0 kg, and the mass of the
rope is negligible. (a) Draw a free-body diagram of the
situation showing all forces acting on Superhero, Trusty
Sidekick, and the rope. (b) Find the tension in the rope above
Superhero. (c) Find the tension in the rope between


170 Chapter 4 | Dynamics: Force and Newton's Laws of Motion


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Superhero and Trusty Sidekick. Indicate on your free-body
diagram the system of interest used to solve each part.


Figure 4.39 Superhero and Trusty Sidekick hang motionless on a rope
as they try to figure out what to do next. Will the tension be the same
everywhere in the rope?


35. A nurse pushes a cart by exerting a force on the handle at
a downward angle 35.0º below the horizontal. The loaded
cart has a mass of 28.0 kg, and the force of friction is 60.0 N.
(a) Draw a free-body diagram for the system of interest. (b)
What force must the nurse exert to move at a constant
velocity?


36. Construct Your Own Problem Consider the tension in
an elevator cable during the time the elevator starts from rest
and accelerates its load upward to some cruising velocity.
Taking the elevator and its load to be the system of interest,
draw a free-body diagram. Then calculate the tension in the
cable. Among the things to consider are the mass of the
elevator and its load, the final velocity, and the time taken to
reach that velocity.


37. Construct Your Own Problem Consider two people
pushing a toboggan with four children on it up a snow-
covered slope. Construct a problem in which you calculate
the acceleration of the toboggan and its load. Include a free-
body diagram of the appropriate system of interest as the
basis for your analysis. Show vector forces and their
components and explain the choice of coordinates. Among
the things to be considered are the forces exerted by those
pushing, the angle of the slope, and the masses of the
toboggan and children.


38. Unreasonable Results (a) Repeat Exercise 4.29, but


assume an acceleration of 1.20 m/s2 is produced. (b) What


is unreasonable about the result? (c) Which premise is
unreasonable, and why is it unreasonable?


39. Unreasonable Results (a) What is the initial acceleration


of a rocket that has a mass of 1.50×106 kg at takeoff, the


engines of which produce a thrust of 2.00×106 N ? Do not
neglect gravity. (b) What is unreasonable about the result?
(This result has been unintentionally achieved by several real
rockets.) (c) Which premise is unreasonable, or which
premises are inconsistent? (You may find it useful to compare
this problem to the rocket problem earlier in this section.)


4.7 Further Applications of Newton’s Laws of
Motion


40. 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. 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.


41. Two muscles in the back of the leg pull upward on the
Achilles tendon, as shown in Figure 4.40. (These muscles
are called the medial and lateral heads of the gastrocnemius
muscle.) Find the magnitude and direction of the total force
on the Achilles tendon. What type of movement could be
caused by this force?


Figure 4.40 Achilles tendon


42. A 76.0-kg person is being pulled away from a burning
building as shown in Figure 4.41. Calculate the tension in the
two ropes if the person is momentarily motionless. Include a
free-body diagram in your solution.


Chapter 4 | Dynamics: Force and Newton's Laws of Motion 171




Figure 4.41 The force T2 needed to hold steady the person being
rescued from the fire is less than her weight and less than the force
T1 in the other rope, since the more vertical rope supports a greater
part of her weight (a vertical force).


43. Integrated Concepts 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
average force was exerted to slow him if he was moving
horizontally? (The gravitational force is balanced by the
buoyant force of the water.)


44. Integrated ConceptsWhen starting a foot race, a
70.0-kg sprinter exerts an average force of 650 N backward
on the ground for 0.800 s. (a) What is his final speed? (b)
How far does he travel?


45. Integrated Concepts A large rocket has a mass of


average force on the shell in the mortar? Express your
answer in newtons and as a ratio to the weight of the shell.


48. Integrated Concepts Repeat Exercise 4.47 for a shell
fired at an angle 10.0º from the vertical.
49. Integrated Concepts An elevator filled with passengers
has a mass of 1700 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 elevator
continues upward at constant velocity for 8.50 s. What is the
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 has
the elevator moved above its original starting point, and what
is its final velocity?


50. Unreasonable Results (a) What is the final velocity of a
car originally traveling at 50.0 km/h that decelerates at a rate


of 0.400 m/s2 for 50.0 s? (b) What is unreasonable about
the result? (c) Which premise is unreasonable, or which
premises are inconsistent?


51. Unreasonable Results A 75.0-kg man stands on a
bathroom scale in an elevator that accelerates from rest to
30.0 m/s in 2.00 s. (a) Calculate the scale reading in newtons
and compare it with his weight. (The scale exerts an upward
force on him equal to its reading.) (b) What is unreasonable
about the result? (c) Which premise is unreasonable, or which
premises are inconsistent?


4.8 Extended Topic: The Four Basic
Forces—An Introduction
52. (a) What is the strength of the weak nuclear force relative
to the strong nuclear force? (b) What is the strength of the
weak nuclear force relative to the electromagnetic force?
Since the weak nuclear force acts at only very short
distances, such as inside nuclei, where the strong and
electromagnetic forces also act, it might seem surprising that
we have any knowledge of it at all. We have such knowledge
because the weak nuclear force is responsible for beta decay,
a type of nuclear decay not explained by other forces.


53. (a) What is the ratio of the strength of the gravitational
force to that of the strong nuclear force? (b) What is the ratio
of the strength of the gravitational force to that of the weak
nuclear force? (c) What is the ratio of the strength of the
gravitational force to that of the electromagnetic force? What
do your answers imply about the influence of the gravitational
force on atomic nuclei?


54.What is the ratio of the strength of the strong nuclear
force to that of the electromagnetic force? Based on this ratio,
you might expect that the strong force dominates the nucleus,
which is true for small nuclei. Large nuclei, however, have
sizes greater than the range of the strong nuclear force. At
these sizes, the electromagnetic force begins to affect nuclear
stability. These facts will be used to explain nuclear fusion
and fission later in this text.


172 Chapter 4 | Dynamics: Force and Newton's Laws of Motion


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/h straight up, assuming constant mass and thrust? (c) In
reality, the mass of a rocket decreases significantly as its fuel
is consumed. Describe qualitatively how this affects the
acceleration and time for this motion.


46. Integrated Concepts A basketball player jumps straight
up for a ball. To do this, he lowers his body 0.300 m and then
accelerates through this distance by forcefully straightening
his legs. This player leaves the floor with a vertical velocity
sufficient 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 part (a) in a distance of 0.300 m.
(c) Calculate the force he exerts on the floor to do this, given
that his mass is 110 kg.


47. Integrated Concepts A 2.50-kg fireworks shell is fired
straight up from a mortar and reaches a height of 110 m. (a)
Neglecting air resistance (a poor assumption, but we will
make it for this example), calculate the shell’s velocity when it
leaves the mortar. (b) The mortar itself is a tube 0.450 m long.
Calculate the average acceleration of the shell in the tube as
it goes from zero to the velocity found in (a). (c) What is the


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5 FURTHER APPLICATIONS OF NEWTON'S
LAWS: FRICTION, DRAG, AND ELASTICITY


Figure 5.1 Total hip replacement surgery has become a common procedure. The head (or ball) of the patient’s femur fits into a cup that has a hard
plastic-like inner lining. (credit: National Institutes of Health, via Wikimedia Commons)


Chapter Outline
5.1. Friction


• Discuss the general characteristics of friction.
• Describe the various types of friction.
• Calculate the magnitude of static and kinetic friction.


5.2. Drag Forces
• Express mathematically the drag force.
• Discuss the applications of drag force.
• Define terminal velocity.
• Determine the terminal velocity given mass.


5.3. Elasticity: Stress and Strain
• State Hooke’s law.
• Explain Hooke’s law using graphical representation between deformation and applied force.
• Discuss the three types of deformations such as changes in length, sideways shear and changes in volume.
• Describe with examples the young’s modulus, shear modulus and bulk modulus.
• Determine the change in length given mass, length and radius.


Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 173




Introduction: Further Applications of Newton’s Laws
Describe the forces on the hip joint. What means are taken to ensure that this will be a good movable joint? From the photograph
(for an adult) in Figure 5.1, estimate the dimensions of the artificial device.


It is difficult to categorize forces into various types (aside from the four basic forces discussed in previous chapter). We know that
a net force affects the motion, position, and shape of an object. It is useful at this point to look at some particularly interesting and
common forces that will provide further applications of Newton’s laws of motion. We have in mind the forces of friction, air or
liquid drag, and deformation.


5.1 Friction
Friction is a force that is around us all the time that opposes relative motion between systems in contact but also allows us to
move (which you have discovered if you have ever tried to walk on ice). While a common force, the behavior of friction is actually
very complicated and is still not completely understood. We have to rely heavily on observations for whatever understandings we
can gain. However, we can still deal with its more elementary general characteristics and understand the circumstances in which
it behaves.


Friction


Friction is a force that opposes relative motion between systems in contact.


One of the simpler characteristics of friction is that it is parallel to the contact surface between systems and always in a direction
that opposes motion or attempted motion of the systems relative to each other. If two systems are in contact and moving relative
to one another, then the friction between them is called kinetic friction. For example, friction slows a hockey puck sliding on ice.
But when objects are stationary, static friction can act between them; the static friction is usually greater than the kinetic friction
between the objects.


Kinetic Friction


If two systems are in contact and moving relative to one another, then the friction between them is called kinetic friction.


Imagine, for example, trying to slide a heavy crate across a concrete floor—you may push harder and harder on the crate and
not move it at all. This means that the static friction responds to what you do—it increases to be equal to and in the opposite
direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion it
is easier to keep it in motion than it was to get it started, indicating that the kinetic friction force is less than the static friction
force. If you add mass to the crate, say by placing a box on top of it, you need to push even harder to get it started and also to
keep it moving. Furthermore, if you oiled the concrete you would find it to be easier to get the crate started and keep it going (as
you might expect).


Figure 5.2 is a crude pictorial representation of how friction occurs at the interface between two objects. Close-up inspection of
these surfaces shows them to be rough. So 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, break off the points, or do both. A considerable force can be
resisted by friction with no apparent motion. The harder the surfaces are pushed together (such as if another box is placed on
the crate), the more force is needed to move them. Part of the friction is due to adhesive forces between the surface molecules
of the two objects, which explain the dependence of friction on the nature of the substances. Adhesion varies with substances in
contact and is a complicated aspect of surface physics. Once an object is moving, there are fewer points of contact (fewer
molecules adhering), so less force is required to keep the object moving. At small but nonzero speeds, friction is nearly
independent of speed.


174 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity


Figure 5.2 Frictional forces, such as f , always oppose motion or attempted motion between objects in contact. Friction arises in part because of the
roughness of the surfaces in contact, as seen in the expanded view. In order for the object to move, it must rise to where the peaks can skip along the
bottom surface. Thus a force is required just to set the object in motion. Some of the peaks will be broken off, also requiring a force to maintain motion.
Much of the friction is actually due to attractive forces between molecules making up the two objects, so that even perfectly smooth surfaces are not
friction-free. Such adhesive forces also depend on the substances the surfaces are made of, explaining, for example, why rubber-soled shoes slip less
than those with leather soles.


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The magnitude of the frictional force has two forms: one for static situations (static friction), the other for when there is motion
(kinetic friction).


When there is no motion between the objects, the magnitude of static friction fs is


(5.1)fs ≤ µsN,


where µs is the coefficient of static friction and N is the magnitude of the normal force (the force perpendicular to the surface).


Magnitude of Static Friction


Magnitude of static friction fs is


(5.2)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 minimum and 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 will move. Thus


(5.3)fs(max) = µsN.


Once an object is moving, the magnitude of kinetic friction fk is given by


(5.4)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.


Magnitude of Kinetic Friction


The magnitude of kinetic friction fk is given by


(5.5)fk = µkN,


where µk is the coefficient of kinetic friction.


As seen in Table 5.1, the coefficients of kinetic friction are less than their static counterparts. That values of µ in Table 5.1 are


stated to only one or, at most, two digits is an indication of the approximate description of friction given by the above two
equations.


Table 5.1 Coefficients of Static and Kinetic Friction


System Static friction μs Kinetic friction μk


Rubber on dry concrete 1.0 0.7


Rubber on wet concrete 0.7 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


Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 175




The equations given earlier include the dependence of friction on materials and the normal force. The direction of friction 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 would be equal to its


weight, W = mg = (100 kg)(9.80 m/s2) = 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 floor greater 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 force of only 290 N (
fk = µkN = (0.30)(980 N) = 290 N ) would keep it moving at a constant speed. If the floor is lubricated, both coefficients
are considerably less than they would be without lubrication. Coefficient of friction is a unit less quantity with a magnitude usually
between 0 and 1.0. The coefficient of the friction depends on the two surfaces that are in contact.


Take-Home Experiment


Find a small plastic object (such as a food container) and slide it on a kitchen table by giving it a gentle tap. Now spray water
on the table, simulating a light shower of rain. What happens now when you give the object the same-sized tap? Now add a
few drops of (vegetable or olive) oil on the surface of the water and give the same tap. What happens now? This latter
situation is particularly important for drivers to note, especially after a light rain shower. Why?


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 two bones, 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 the joint are covered by cartilage,
which provides a smooth, almost glassy surface. The joints also produce a fluid (synovial fluid) that reduces friction and wear. A
damaged or arthritic joint can be replaced by an artificial joint (Figure 5.3). These replacements can be made of metals
(stainless steel or titanium) or plastic (polyethylene), also with very small coefficients of friction.


176 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity


Figure 5.3 Artificial knee replacement is a procedure that has been performed for more than 20 years. In this figure, we see the post-op x rays of the
right knee joint replacement. (credit: Mike Baird, Flickr)


Other natural lubricants include saliva produced in our mouths to aid in the swallowing process, and the slippery mucus found
between organs in the body, allowing them to move freely past each other during heartbeats, during breathing, and when a
person moves. Artificial lubricants are also common in hospitals and doctor’s clinics. For example, when ultrasonic imaging is
carried out, the gel that couples the transducer to the skin also serves to to lubricate the surface between the transducer and the
skin—thereby reducing the coefficient of friction between the two surfaces. This allows the transducer to mover freely over the
skin.


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Example 5.1 Skiing Exercise


A skier with a mass of 62 kg is sliding down a snowy slope. Find the coefficient of kinetic friction for the skier if friction is
known to be 45.0 N.


Strategy


The magnitude of kinetic friction was given in to be 45.0 N. Kinetic friction is related to the normal force N as fk = µkN ;
thus, the coefficient of kinetic friction can be found if we can find the normal force of the skier on a slope. 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 the skier and free-body diagram in Figure 5.4.)


Figure 5.4 The motion of the skier and friction are parallel to the slope and so it is most convenient to project all forces onto a coordinate system
where one axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). N (the normal force) is perpendicular to the
slope, and f (the friction) is parallel to the slope, but w (the skier’s weight) has components along both axes, namely w⊥ and W // . N is
equal in magnitude to w⊥ , so there is no motion perpendicular to the slope. However, f is less than W // in magnitude, so there is
acceleration down the slope (along the x-axis).


That is,


(5.6)N = w⊥ = w cos 25º = mg cos 25º.


Substituting this into our expression for kinetic friction, we get


(5.7)fk = µkmg cos 25º,


which can now be solved for the coefficient of kinetic friction µk .


Solution


Solving for µk gives


(5.8)
µk =


fk
N
= fk


w cos 25º
= fk


mg cos 25º.


Substituting known values on the right-hand side of the equation,


(5.9)µk = 45.0 N(62 kg)(9.80 m/s2)(0.906)
= 0.082.


Discussion


This result is a little smaller than the coefficient listed in Table 5.1 for waxed wood on snow, but it is still reasonable since
values of the coefficients of friction can vary greatly. In situations like this, where an object of mass m slides down a slope
that makes an angle θ with the horizontal, friction is given by fk = µkmg cos θ . All objects will slide down a slope with
constant acceleration under these circumstances. Proof of this is left for this chapter’s Problems and Exercises.


Take-Home Experiment


An object will slide down an inclined plane at a constant velocity if the net force on the object is zero. We can use this fact to
measure the coefficient of kinetic friction between two objects. As shown in Example 5.1, the kinetic friction on a slope
fk = µkmg cos θ . The component of the weight down the slope is equal to mg sin θ (see the free-body diagram in


Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 177




Figure 5.4). These forces act in opposite directions, so when they have equal magnitude, the acceleration is zero. Writing
these out:


(5.10)fk = Fgx
(5.11)µkmg cos θ = mg sin θ.


Solving for µk , we find that


(5.12)
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 book
lightly to get the coin to move. Measure the angle of tilt relative to the horizontal and find µk . Note that the coin will 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. Discuss how this may affect the value for µk and its uncertainty.


We have discussed that when an object rests on a horizontal surface, there is a normal force supporting it equal in magnitude to
its weight. Furthermore, simple friction is always proportional to the normal force.


Making Connections: Submicroscopic Explanations of Friction


The simpler aspects of friction dealt with so far are its macroscopic (large-scale) characteristics. Great strides have been
made in the atomic-scale explanation of friction during the past several decades. Researchers are finding that the atomic
nature of friction seems to have several fundamental characteristics. These characteristics not only explain some of the
simpler aspects of friction—they also hold the potential for the development of nearly friction-free environments that could
save hundreds of billions of dollars in energy which is currently being converted (unnecessarily) to heat.


Figure 5.5 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 area in contact, a somewhat counterintuitive notion. When two
rough surfaces are in contact, the actual contact area is a tiny fraction of the total area since only high spots touch. When a
greater normal force is exerted, the actual contact area increases, and it is found that the friction is proportional to this area.


178 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity


Figure 5.5 Two rough surfaces in contact have a much smaller area of actual contact than their total area. When there is a greater normal force as a
result of a greater applied force, the area of actual contact increases as does friction.


But the atomic-scale view promises to explain far more than the simpler features of friction. The mechanism for how 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 lattices to vibrate—essentially
creating sound waves that penetrate the material. The sound waves diminish with distance and their energy is converted into
heat. Chemical reactions that are related to frictional wear can also occur between atoms and molecules on the surfaces. Figure
5.6 shows how the tip of a probe drawn across another material is deformed by atomic-scale friction. The force needed to drag
the tip can be measured and is found to be related to shear stress, which will be discussed later in this chapter. 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 phenomenon known since ancient times—friction.


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Figure 5.6 The tip of a probe is deformed sideways by frictional force as the probe is dragged across a surface. Measurements of how the force varies
for different materials are yielding fundamental insights into the atomic nature of friction.


PhET Explorations: Forces and Motion


Explore the forces at work when you try to push a filing cabinet. Create an applied force and see the resulting friction force
and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. Draw a free-body
diagram of all the forces (including gravitational and normal forces).


Figure 5.7 Forces and Motion (http://cnx.org/content/m42139/1.7/forces-and-motion_en.jar)


5.2 Drag Forces
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 a liquid).
You feel the drag force when you move your hand through water. You might also feel it if you move your hand during a strong
wind. The faster you move your hand, the harder it is to move. You feel a smaller drag force when you tilt your hand so only the
side goes through the air—you have decreased the area of your hand that faces the direction of motion. Like friction, the drag
force always opposes the motion of an object. Unlike simple friction, the drag force is proportional to some function of the
velocity of the object in that fluid. This functionality is complicated and depends upon the shape of the object, its size, its velocity,
and the fluid it is in. For most large objects such as bicyclists, cars, and baseballs not moving too slowly, the magnitude of the
drag force FD is found to be proportional to the square of the speed of the object. We can write this relationship mathematically


as FD ∝ v
2 . When taking into account other factors, this relationship becomes


(5.13)FD = 12CρAv
2,


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 = bv
2 , where b is a constant


equivalent to 0.5CρA . We have set the exponent 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 in a few pages on fluid
dynamics, for small particles moving at low speeds in a fluid, the exponent is equal to 1.


Drag Force


Drag force FD is found to be proportional to the square of the speed of the object. Mathematically


(5.14)FD ∝ v
2


Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 179




(5.15)FD = 12CρAv
2,


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. (See Figure 5.8). “Aerodynamic”
shaping of an automobile can reduce the drag force and so increase a car’s gas mileage.


Figure 5.8 From racing cars to bobsled racers, aerodynamic shaping is crucial to achieving top speeds. Bobsleds are designed for speed. They are
shaped like a bullet with tapered fins. (credit: U.S. Army, via Wikimedia Commons)


The value of the drag coefficient, C , is determined empirically, usually with the use of a wind tunnel. (See Figure 5.9).


Figure 5.9 NASA researchers test a model plane in a wind tunnel. (credit: NASA/Ames)


The drag coefficient can depend upon velocity, but we will assume that it is a constant here. Table 5.2 lists some typical drag
coefficients 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 at about 90 km/h (55 mi/
h).


180 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity


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Table 5.2 Drag Coefficient
Values Typical values of
drag coefficient C .
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


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 Cathy Freeman
wore a full body suit in the 2000 Sydney Olympics, and won the gold medal for the 400 m race. Many swimmers in the 2008
Beijing Olympics wore (Speedo) body suits; it might have made a difference in breaking many world records (See Figure 5.10).
Most elite swimmers (and cyclists) shave their body hair. Such innovations can have the effect of slicing away milliseconds in a
race, sometimes making the difference between a gold and a silver medal. One consequence is that careful and precise
guidelines must be continuously developed to maintain the integrity of the sport.


Figure 5.10 Body suits, such as this LZR Racer Suit, have been credited with many world records after their release in 2008. Smoother “skin” and
more compression forces on a swimmer’s body provide at least 10% less drag. (credit: NASA/Kathy Barnstorff)


Some interesting situations connected to Newton’s second law occur when considering the effects of drag forces upon a moving
object. For instance, consider a skydiver falling through air under the influence of gravity. The two forces acting on him are the
force of gravity and the drag force (ignoring the buoyant force). The downward force of gravity remains constant regardless of the
velocity at which the person is moving. However, as the person’s velocity increases, the magnitude of the drag force increases
until the magnitude of the drag force is equal to the gravitational force, thus producing a net force of zero. A zero net force means
that there is no acceleration, as given 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, 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,


(5.16)Fnet = mg − FD = ma = 0.


Thus,


(5.17)mg = FD.


Using the equation for drag force, we have


Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 181




(5.18)mg = 12ρCAv
2.


Solving for the velocity, we obtain


(5.19)
v = 2mg


ρCA
.


Assume the density of air is ρ = 1.21 kg/m3 . A 75-kg skydiver descending head first will have an area approximately


A = 0.18 m2 and a drag coefficient of approximately C = 0.70 . We find that
(5.20)


v = 2(75 kg)(9.80 m/s
2)


(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 maximum terminal velocity of about 350 km/h while traveling in a pike
(head first) position, minimizing the area and his drag. In a spread-eagle position, that terminal velocity may decrease to about
200 km/h as the area increases. This terminal velocity becomes much smaller after the parachute opens.


Take-Home Experiment


This interesting activity examines the effect of weight upon terminal velocity. Gather together some nested coffee filters.
Leaving them in their original shape, measure the time it takes for one, two, three, four, and five nested filters to fall to the
floor from the same height (roughly 2 m). (Note that, due to the way the filters are nested, drag is constant and only mass
varies.) They obtain terminal velocity quite quickly, so find this velocity as a function of mass. Plot the terminal velocity v


versus mass. Also plot v2 versus mass. Which of these relationships is more linear? What can you conclude from these
graphs?


Example 5.2 A Terminal Velocity


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 = 12ρCAv
2 .


Thus the terminal velocity vt can be written as


(5.21)
vt =


2mg
ρCA


.


Solution


All quantities are known except the person’s projected area. This is an adult (82 kg) falling spread eagle. We can estimate
the frontal area as


(5.22)A = (2 m)(0.35 m) = 0.70 m2.


Using our equation for vt , we find that


(5.23)
vt =


2(85 kg)(9.80 m/s2)
(1.21 kg/m3)(1.0)(0.70 m2)


= 44 m/s.
Discussion


This result is consistent with the value for vt mentioned earlier. The 75-kg skydiver going feet first had a v = 98 m / s . He
weighed less but had a smaller frontal area and so a smaller drag due to the air.


182 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity


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 don’t reach a terminal velocity in such a short distance, but the squirrel does.


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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 proportional to the surface
of the moving object. Divide an animal’s length, breadth, and height each by ten; its weight is reduced to a thousandth, but its
surface only to a hundredth. So the resistance to falling in the case of the small animal is relatively ten times 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 is in a
denser medium than air. Then we find that the drag force is proportional just to the velocity. This relationship is given by Stokes’
law, which states that


(5.24)Fs = 6πrηv,


where r is the radius of the object, η is the viscosity of the fluid, and v is the object’s velocity.


Stokes’ Law
(5.25)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 this law are provided by microorganisms, pollen, and dust particles. Because each of these objects is so
small, we find that many of these objects travel unaided only at a constant (terminal) velocity. Terminal velocities for bacteria
(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 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. Fishes, dolphins, and
even massive whales are streamlined in shape to reduce drag forces. Birds are streamlined and migratory species that fly large
distances often have particular features such as long necks. Flocks of birds fly in the shape of a spear head as the flock forms a
streamlined pattern (see Figure 5.11). In humans, one important example of streamlining is the shape of sperm, which need to
be efficient in their use of energy.


Figure 5.11 Geese fly in a V formation during their long migratory travels. This shape reduces drag and energy consumption for individual birds, and
also allows them a better way to communicate. (credit: Julo, Wikimedia Commons)


Galileo’s Experiment


Galileo is said to have dropped two objects of different masses from the Tower of Pisa. He measured how long it took each
to reach the ground. Since stopwatches weren’t readily available, how do you think he measured their fall time? If the
objects were the same size, but with different masses, what do you think he should have observed? Would this result be
different if done on the Moon?


PhET Explorations: Masses & Springs


A realistic mass and spring laboratory. Hang masses from springs and adjust the spring stiffness and damping. You can
even slow time. Transport the lab to different planets. A chart shows the kinetic, potential, and thermal energy for each
spring.


Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 183




Figure 5.12 Masses & Springs (http://cnx.org/content/m42080/1.9/mass-spring-lab_en.jar)


5.3 Elasticity: Stress and Strain
We now move from consideration of forces that affect the motion of an object (such as friction and drag) to those that affect an
object’s shape. If a bulldozer pushes a car into a wall, the car will not move but it will noticeably change shape. A change in
shape due to the application of a force is a deformation. Even very small forces are known to cause some deformation. For
small deformations, two important characteristics are observed. First, the object returns to its original shape when the force is
removed—that is, the deformation is elastic for small deformations. Second, the size of the deformation is proportional to the
force—that is, for small deformations, Hooke’s law is obeyed. In equation form, Hooke’s law is given by


(5.26)F = kΔL,


where ΔL is the amount of deformation (the change in length, for example) produced by the force F , and k is a
proportionality constant that depends on the shape and composition of the object and the direction of the force. Note that this
force is a function of the deformation ΔL —it is not constant as a kinetic friction force is. Rearranging this to


(5.27)ΔL = F
k


makes it clear that the deformation is proportional to the applied force. Figure 5.13 shows the Hooke’s law relationship between
the extension ΔL of a spring or of a human bone. For metals or springs, the straight line region in which Hooke’s law pertains is
much larger. Bones are brittle and the elastic region is small and the fracture abrupt. Eventually a large enough stress to the
material will cause it to break or fracture. Tensile strength is the breaking stress that will cause permanent deformation or
fracture of a material.


Hooke’s Law
(5.28)F = kΔL,


where ΔL is the amount of deformation (the change in length, for example) produced by the force F , and k is a
proportionality constant that depends on the shape and composition of the object and the direction of the force.


(5.29)ΔL = F
k


Figure 5.13 A graph of deformation ΔL versus applied force F . The straight segment is the linear region where Hooke’s law is obeyed. The slope
of the straight region is 1


k
. For larger forces, the graph is curved but the deformation is still elastic— ΔL will return to zero if the force is removed.


184 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity


Still greater forces permanently deform the object until it finally fractures. The shape of the curve near fracture depends on several factors, including
how the force F is applied. Note that in this graph the slope increases just before fracture, indicating that a small increase in F is producing a large
increase in L near the fracture.


The proportionality constant k depends upon a number of factors for the material. For example, a guitar string made of nylon
stretches when it is tightened, and the elongation ΔL is proportional to the force applied (at least for small deformations).
Thicker nylon strings and ones made of steel stretch less for the same applied force, implying they have a larger k (see Figure


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5.14). Finally, all three strings return to their normal lengths when the force is removed, provided the deformation is small. Most


materials will behave in this manner if the deformation is less than about 0.1% or about 1 part in 103 .


Figure 5.14 The same force, in this case a weight ( w ), applied to three different guitar strings of identical length produces the three different
deformations shown as shaded segments. The string on the left is thin nylon, the one in the middle is thicker nylon, and the one on the right is steel.


Stretch Yourself a Little


How would you go about measuring the proportionality constant k of a rubber band? If a rubber band stretched 3 cm when
a 100-g mass was attached to it, then how much would it stretch if two similar rubber bands were attached to the same
mass—even if put together in parallel or alternatively if tied together in series?


We now consider three specific types of deformations: changes in length (tension and compression), sideways shear (stress),
and changes in volume. All deformations are assumed to be small unless otherwise stated.


Changes in Length—Tension and Compression: Elastic Modulus


A change in length ΔL is produced when a force is applied to a wire or rod parallel to its length L0 , either stretching it (a
tension) or compressing it. (See Figure 5.15.)


Figure 5.15 (a) Tension. The rod is stretched a length ΔL when a force is applied parallel to its length. (b) Compression. The same rod is
compressed by forces with the same magnitude in the opposite direction. For very small deformations and uniform materials, ΔL is approximately
the same for the same magnitude of tension or compression. For larger deformations, the cross-sectional area changes as the rod is compressed or
stretched.


Experiments have shown that the change in length (ΔL ) depends on only a few variables. As already noted, ΔL is
proportional to the force F and depends on the substance from which the object is made. Additionally, the change in length is


Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 185




proportional to the original length L0 and inversely proportional to the cross-sectional area of the wire or rod. For example, a


long guitar string will stretch more than a short one, and a thick string will stretch less than a thin one. We can combine all these
factors into one equation for ΔL :


(5.30)ΔL = 1
Y
F
A
L0,


where ΔL is the change in length, F the applied force, Y is a factor, called the elastic modulus or Young’s modulus, that
depends on the substance, A is the cross-sectional area, and L0 is the original length. Table 5.3 lists values of Y for several


materials—those with a large Y are said to have a large tensile stifness because they deform less for a given tension or
compression.


Table 5.3 Elastic Moduli[1]


Material
Young’s modulus (tension–compression)Y


(109 N/m2)
Shear modulus S
(109 N/m2)


Bulk modulus B
(109 N/m2)


Aluminum 70 25 75


Bone – tension 16 80 8


Bone –
compression 9


Brass 90 35 75


Brick 15


Concrete 20


Glass 70 20 30


Granite 45 20 45


Hair (human) 10


Hardwood 15 10


Iron, cast 100 40 90


Lead 16 5 50


Marble 60 20 70


Nylon 5


Polystyrene 3


Silk 6


Spider thread 3


Steel 210 80 130


Tendon 1


Acetone 0.7


Ethanol 0.9


Glycerin 4.5


Mercury 25


Water 2.2


Young’s moduli are not listed for liquids and gases in Table 5.3 because they cannot be stretched or compressed in only one
direction. Note that there is an assumption that the object does not accelerate, so that there are actually two applied forces of
magnitude F acting in opposite directions. For example, the strings in Figure 5.15 are being pulled down by a force of
magnitude w and held up by the ceiling, which also exerts a force of magnitude w .


186 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity


1. Approximate and average values. Young’s moduli Y for tension and compression sometimes differ but are averaged here.
Bone has significantly different Young’s moduli for tension and compression.


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Example 5.3 The Stretch of a Long Cable


Suspension cables are used to carry gondolas at ski resorts. (See Figure 5.16) Consider a suspension cable that includes
an unsupported span of 3 km. Calculate the amount of stretch in the steel cable. Assume that the cable has a diameter of


5.6 cm and the maximum tension it can withstand is 3.0×106 N .


Figure 5.16 Gondolas travel along suspension cables at the Gala Yuzawa ski resort in Japan. (credit: Rudy Herman, Flickr)


Strategy


The force is equal to the maximum tension, or F = 3.0×106 N . The cross-sectional area is πr2 = 2.46×10−3 m2 . The


equation ΔL = 1
Y
F
A
L0 can be used to find the change in length.


Solution


All quantities are known. Thus,


(5.31)
ΔL = ⎛⎝


1
210×109 N/m2






3.0×106 N


2.46×10–3 m2

⎠(3020 m)


= 18 m.
Discussion


This is quite a stretch, but only about 0.6% of the unsupported length. Effects of temperature upon length might be important
in these environments.


Bones, on the whole, do not fracture due to tension or compression. Rather they generally fracture due to sideways impact or
bending, resulting in the bone shearing or snapping. The behavior of bones under tension and compression is important because
it determines the load the bones can carry. Bones are classified as weight-bearing structures such as columns in buildings and
trees. Weight-bearing structures have special features; columns in building have steel-reinforcing rods while trees and bones are
fibrous. The bones in different parts of the body serve different structural functions and are prone to different stresses. Thus the
bone in the top of the femur is arranged in thin sheets separated by marrow while in other places the bones can be cylindrical
and filled with marrow or just solid. Overweight people have a tendency toward bone damage due to sustained compressions in
bone joints and tendons.


Another biological example of Hooke’s law occurs in tendons. Functionally, the tendon (the tissue connecting muscle to bone)
must stretch easily at first when a force is applied, but offer a much greater restoring force for a greater strain. Figure 5.17 shows
a stress-strain relationship for a human tendon. Some tendons have a high collagen content so there is relatively little strain, or
length change; others, like support tendons (as in the leg) can change length up to 10%. Note that this stress-strain curve is
nonlinear, since the slope of the line changes in different regions. In the first part of the stretch called the toe region, the fibers in
the tendon begin to align in the direction of the stress—this is called uncrimping. In the linear region, the fibrils will be stretched,
and in the failure region individual fibers begin to break. A simple model of this relationship can be illustrated by springs in
parallel: different springs are activated at different lengths of stretch. Examples of this are given in the problems at end of this
chapter. Ligaments (tissue connecting bone to bone) behave in a similar way.


Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 187




Figure 5.17 Typical stress-strain curve for mammalian tendon. Three regions are shown: (1) toe region (2) linear region, and (3) failure region.


Unlike bones and tendons, which need to be strong as well as elastic, the arteries and lungs need to be very stretchable. The
elastic properties of the arteries are essential for blood flow. The pressure in the arteries increases and arterial walls stretch
when the blood is pumped out of the heart. When the aortic valve shuts, the pressure in the arteries drops and the arterial walls
relax to maintain the blood flow. When you feel your pulse, you are feeling exactly this—the elastic behavior of the arteries as the
blood gushes through with each pump of the heart. If the arteries were rigid, you would not feel a pulse. The heart is also an
organ with special elastic properties. The lungs expand with muscular effort when we breathe in but relax freely and elastically
when we breathe out. Our skins are particularly elastic, especially for the young. A young person can go from 100 kg to 60 kg
with no visible sag in their skins. The elasticity of all organs reduces with age. Gradual physiological aging through reduction in
elasticity starts in the early 20s.


Example 5.4 Calculating Deformation: How Much Does Your Leg Shorten When You Stand on
It?


Calculate the change in length of the upper leg bone (the femur) when a 70.0 kg man supports 62.0 kg of his mass on it,
assuming the bone to be equivalent to a uniform rod that is 40.0 cm long and 2.00 cm in radius.


Strategy


The force is equal to the weight supported, or


(5.32)F = mg = ⎛⎝62.0 kg⎞⎠⎛⎝9.80 m/s2

⎠ = 607.6 N,


and the cross-sectional area is πr2 = 1.257×10−3 m2 . The equation ΔL = 1
Y
F
A
L0 can be used to find the change in


length.


Solution


All quantities except ΔL are known. Note that the compression value for Young’s modulus for bone must be used here.
Thus,


(5.33)
ΔL = ⎛⎝


1
9×109 N/m2







607.6 N
1.257×10−3 m2



⎠(0.400 m)


= 2×10−5 m.
Discussion


This small change in length seems reasonable, consistent with our experience that bones are rigid. In fact, even the rather
large forces encountered during strenuous physical activity do not compress or bend bones by large amounts. Although
bone is rigid compared with fat or muscle, several of the substances listed in Table 5.3 have larger values of Young’s
modulus Y . In other words, they are more rigid.


The equation for change in length is traditionally rearranged and written in the following form:


(5.34)F
A
= YΔL


L0
.


The ratio of force to area, F
A
, is defined as stress (measured in N/m2 ), and the ratio of the change in length to length, ΔL


L0
, is


defined as strain (a unitless quantity). In other words,


(5.35)stress = Y×strain.
In this form, the equation is analogous to Hooke’s law, with stress analogous to force and strain analogous to deformation. If we
again rearrange this equation to the form


188 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity


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(5.36)F = YAΔL
L0
,


we see that it is the same as Hooke’s law with a proportionality constant


(5.37)k = YA
L0
.


This general idea—that force and the deformation it causes are proportional for small deformations—applies to changes in
length, sideways bending, and changes in volume.


Stress


The ratio of force to area, F
A
, is defined as stress measured in N/m2.


Strain


The ratio of the change in length to length, ΔL
L0


, is defined as strain (a unitless quantity). In other words,


(5.38)stress = Y×strain.


Sideways Stress: Shear Modulus


Figure 5.18 illustrates what is meant by a sideways stress or a shearing force. Here the deformation is called Δx and it is
perpendicular to L0 , rather than parallel as with tension and compression. Shear deformation behaves similarly to tension and


compression and can be described with similar equations. The expression for shear deformation is


(5.39)Δx = 1
S
F
A
L0,


where S is the shear modulus (see Table 5.3) and F is the force applied perpendicular to L0 and parallel to the cross-


sectional area A . Again, to keep the object from accelerating, there are actually two equal and opposite forces F applied
across opposite faces, as illustrated in Figure 5.18. The equation is logical—for example, it is easier to bend a long thin pencil
(small A ) than a short thick one, and both are more easily bent than similar steel rods (large S ).


Shear Deformation
(5.40)Δx = 1


S
F
A
L0,


where S is the shear modulus and F is the force applied perpendicular to L0 and parallel to the cross-sectional area A .


Figure 5.18 Shearing forces are applied perpendicular to the length L0 and parallel to the area A , producing a deformation Δx . Vertical forces
are not shown, but it should be kept in mind that in addition to the two shearing forces, F , there must be supporting forces to keep the object from
rotating. The distorting effects of these supporting forces are ignored in this treatment. The weight of the object also is not shown, since it is usually
negligible compared with forces large enough to cause significant deformations.


Examination of the shear moduli in Table 5.3 reveals some telling patterns. For example, shear moduli are less than Young’s
moduli for most materials. Bone is a remarkable exception. Its shear modulus is not only greater than its Young’s modulus, but it
is as large as that of steel. This is why bones are so rigid.


The spinal column (consisting of 26 vertebral segments separated by discs) provides the main support for the head and upper
part of the body. The spinal column has normal curvature for stability, but this curvature can be increased, leading to increased


Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 189




shearing forces on the lower vertebrae. Discs are better at withstanding compressional forces than shear forces. Because the
spine is not vertical, the weight of the upper body exerts some of both. Pregnant women and people that are overweight (with
large abdomens) need to move their shoulders back to maintain balance, thereby increasing the curvature in their spine and so
increasing the shear component of the stress. An increased angle due to more curvature increases the shear forces along the
plane. These higher shear forces increase the risk of back injury through ruptured discs. The lumbosacral disc (the wedge
shaped disc below the last vertebrae) is particularly at risk because of its location.


The shear moduli for concrete and brick are very small; they are too highly variable to be listed. Concrete used in buildings can
withstand compression, as in pillars and arches, but is very poor against shear, as might be encountered in heavily loaded floors
or during earthquakes. Modern structures were made possible by the use of steel and steel-reinforced concrete. Almost by
definition, liquids and gases have shear moduli near zero, because they flow in response to shearing forces.


Example 5.5 Calculating Force Required to Deform: That Nail Does Not Bend Much Under a
Load


Find the mass of the picture hanging from a steel nail as shown in Figure 5.19, given that the nail bends only 1.80 µm .
(Assume the shear modulus is known to two significant figures.)


Figure 5.19 Side view of a nail with a picture hung from it. The nail flexes very slightly (shown much larger than actual) because of the shearing
effect of the supported weight. Also shown is the upward force of the wall on the nail, illustrating that there are equal and opposite forces applied
across opposite cross sections of the nail. See Example 5.5 for a calculation of the mass of the picture.


Strategy


The force F on the nail (neglecting the nail’s own weight) is the weight of the picture w . If we can find w , then the mass


of the picture is just wg . The equation Δx =
1
S
F
A
L0 can be solved for F .


Solution


Solving the equation Δx = 1
S
F
A
L0 for F , we see that all other quantities can be found:


(5.41)F = SA
L0
Δx.


S is found in Table 5.3 and is S = 80×109 N/m2 . The radius r is 0.750 mm (as seen in the figure), so the cross-
sectional area is


(5.42)A = πr2 = 1.77×10−6 m2.


The value for L0 is also shown in the figure. Thus,


(5.43)
F = (80×10


9 N/m2)(1.77×10−6 m2)
(5.00×10−3 m)


(1.80×10−6 m) = 51 N.


This 51 N force is the weight w of the picture, so the picture’s mass is


(5.44)m = wg =
F
g = 5.2 kg.


Discussion


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This is a fairly massive picture, and it is impressive that the nail flexes only 1.80 µm—an amount undetectable to the
unaided eye.


Changes in Volume: Bulk Modulus
An object will be compressed in all directions if inward forces are applied evenly on all its surfaces as in Figure 5.20. It is
relatively easy to compress gases and extremely difficult to compress liquids and solids. For example, air in a wine bottle is
compressed when it is corked. But if you try corking a brim-full bottle, you cannot compress the wine—some must be removed if
the cork is to be inserted. The reason for these different compressibilities is that atoms and molecules are separated by large
empty spaces in gases but packed close together in liquids and solids. To compress a gas, you must force its atoms and
molecules closer together. To compress liquids and solids, you must actually compress their atoms and molecules, and very
strong electromagnetic forces in them oppose this compression.


Figure 5.20 An inward force on all surfaces compresses this cube. Its change in volume is proportional to the force per unit area and its original
volume, and is related to the compressibility of the substance.


We can describe the compression or volume deformation of an object with an equation. First, we note that a force “applied


evenly” is defined to have the same stress, or ratio of force to area F
A


on all surfaces. The deformation produced is a change in


volume ΔV , which is found to behave very similarly to the shear, tension, and compression previously discussed. (This is not
surprising, since a compression of the entire object is equivalent to compressing each of its three dimensions.) The relationship
of the change in volume to other physical quantities is given by


(5.45)ΔV = 1
B
F
A
V0,


where B is the bulk modulus (see Table 5.3), V0 is the original volume, and
F
A


is the force per unit area applied uniformly


inward on all surfaces. Note that no bulk moduli are given for gases.


What are some examples of bulk compression of solids and liquids? One practical example is the manufacture of industrial-
grade diamonds by compressing carbon with an extremely large force per unit area. The carbon atoms rearrange their crystalline
structure into the more tightly packed pattern of diamonds. In nature, a similar process occurs deep underground, where
extremely large forces result from the weight of overlying material. Another natural source of large compressive forces is the
pressure created by the weight of water, especially in deep parts of the oceans. Water exerts an inward force on all surfaces of a
submerged object, and even on the water itself. At great depths, water is measurably compressed, as the following example
illustrates.


Example 5.6 Calculating Change in Volume with Deformation: How Much Is Water Compressed
at Great Ocean Depths?


Calculate the fractional decrease in volume ( ΔV
V0


) for seawater at 5.00 km depth, where the force per unit area is


5.00×107 N /m2 .
Strategy


Equation ΔV = 1
B
F
A
V0 is the correct physical relationship. All quantities in the equation except


ΔV
V0


are known.


Solution


Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 191




deformation:


drag force:


friction:


Hooke’s law:


kinetic friction:


magnitude of kinetic friction:


magnitude of static friction:


shear deformation:


static friction:


Stokes’ law:


strain:


stress:


tensile strength:


Solving for the unknown ΔV
V0


gives


(5.46)ΔV
V0


= 1
B
F
A
.


Substituting known values with the value for the bulk modulus B from Table 5.3,


(5.47)ΔV
V0


= 5.00×10
7 N/m2


2.2×109 N/m2
= 0.023 = 2.3%.


Discussion


Although measurable, this is not a significant decrease in volume considering that the force per unit area is about 500
atmospheres (1 million pounds per square foot). Liquids and solids are extraordinarily difficult to compress.


Conversely, very large forces are created by liquids and solids when they try to expand but are constrained from doing so—which
is equivalent to compressing them to less than their normal volume. This often occurs when a contained material warms up,
since most materials expand when their temperature increases. If the materials are tightly constrained, they deform or break their
container. Another very common example occurs when water freezes. Water, unlike most materials, expands when it freezes,
and it can easily fracture a boulder, rupture a biological cell, or crack an engine block that gets in its way.


Other types of deformations, such as torsion or twisting, behave analogously to the tension, shear, and bulk deformations
considered here.


Glossary
change in shape due to the application of force


FD , found to be proportional to the square of the speed of the object; mathematically


FD ∝ v
2


FD = 12CρAv
2,


where C is the drag coefficient, A is the area of the object facing the fluid, and ρ is the density of the fluid


a force that opposes relative motion or attempts at motion between systems in contact


proportional relationship between the force F on a material and the deformation ΔL it causes, F = kΔL


a force that opposes the motion of two systems that are in contact and moving relative to one another


fk = µkN , where µk is the coefficient of kinetic friction


fs ≤ µs N , where µs is the coefficient of static friction and N is the magnitude of the normal
force


deformation perpendicular to the original length of an object


a force that opposes the motion of two systems that are in contact and are not moving relative to one another


Fs = 6πrηv , where r is the radius of the object, η is the viscosity of the fluid, and v is the object’s velocity


ratio of change in length to original length


ratio of force to area


the breaking stress that will cause permanent deformation or fraction of a material


Section Summary


5.1 Friction
• Friction is a contact force between systems that opposes the motion or attempted motion between them. Simple friction is


proportional to the normal force N pushing the systems together. (A normal force is always perpendicular to the contact


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surface between systems.) Friction depends on both of the materials involved. The magnitude of static friction fs between


systems stationary relative to one another is given by


fs ≤ µsN,
where µs is the coefficient of static friction, which depends on both of the materials.


• The kinetic friction force fk between systems moving relative to one another is given by


fk = µkN,
where µk is the coefficient of kinetic friction, which also depends on both materials.


5.2 Drag Forces
• Drag forces acting on an object moving in a fluid oppose the motion. For larger objects (such as a baseball) moving at a


velocity v in air, the drag force is given by


FD = 12CρAv
2,


where C is the drag coefficient (typical values are given in Table 5.2), A is the area of the object facing the fluid, and ρ


is the fluid density.
• For small objects (such as a bacterium) moving in a denser medium (such as water), the drag force is given by Stokes’ law,


Fs = 6πηrv,
where r is the radius of the object, η is the fluid viscosity, and v is the object’s velocity.


5.3 Elasticity: Stress and Strain
• Hooke’s law is given by


F = kΔL,
where ΔL is the amount of deformation (the change in length), F is the applied force, and k is a proportionality constant
that depends on the shape and composition of the object and the direction of the force. The relationship between the
deformation and the applied force can also be written as


ΔL = 1
Y
F
A
L0,


where Y is Young’s modulus, which depends on the substance, A is the cross-sectional area, and L0 is the original


length.


• The ratio of force to area, F
A
, is defined as stress, measured in N/m2.


• The ratio of the change in length to length, ΔL
L0


, is defined as strain (a unitless quantity). In other words,


stress = Y×strain.
• The expression for shear deformation is


Δx = 1
S
F
A
L0,


where S is the shear modulus and F is the force applied perpendicular to L0 and parallel to the cross-sectional area A .


• The relationship of the change in volume to other physical quantities is given by


ΔV = 1
B
F
A
V0,


where B is the bulk modulus, V0 is the original volume, and
F
A


is the force per unit area applied uniformly inward on all


surfaces.


Conceptual Questions


5.1 Friction
1. Define normal force. What is its relationship to friction when friction behaves simply?


2. The glue on a piece of tape can exert forces. Can these forces be a type of simple friction? Explain, considering especially that
tape can stick to vertical walls and even to ceilings.


3.When you learn to drive, you discover that you need to let up slightly on the brake pedal as you come to a stop or the car will
stop with a jerk. Explain this in terms of the relationship between static and kinetic friction.


Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 193




4.When you push a piece of chalk across a chalkboard, it sometimes screeches because it rapidly alternates between slipping
and sticking to the board. Describe this process in more detail, in particular explaining how it is related to the fact that kinetic
friction is less than static friction. (The same slip-grab process occurs when tires screech on pavement.)


5.2 Drag Forces
5. Athletes such as swimmers and bicyclists wear body suits in competition. Formulate a list of pros and cons of such suits.


6. Two expressions were used for the drag force experienced by a moving object in a liquid. One depended upon the speed,
while the other was proportional to the square of the speed. In which types of motion would each of these expressions be more
applicable than the other one?


7. As cars travel, oil and gasoline leaks onto the road surface. If a light rain falls, what does this do to the control of the car?
Does a heavy rain make any difference?


8.Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in
such a fall?


5.3 Elasticity: Stress and Strain
9. The elastic properties of the arteries are essential for blood flow. Explain the importance of this in terms of the characteristics
of the flow of blood (pulsating or continuous).


10.What are you feeling when you feel your pulse? Measure your pulse rate for 10 s and for 1 min. Is there a factor of 6
difference?


11. Examine different types of shoes, including sports shoes and thongs. In terms of physics, why are the bottom surfaces
designed as they are? What differences will dry and wet conditions make for these surfaces?


12.Would you expect your height to be different depending upon the time of day? Why or why not?


13.Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in
such a fall?


14. Explain why pregnant women often suffer from back strain late in their pregnancy.


15. An old carpenter’s trick to keep nails from bending when they are pounded into hard materials is to grip the center of the nail
firmly with pliers. Why does this help?


16.When a glass bottle full of vinegar warms up, both the vinegar and the glass expand, but vinegar expands significantly more
with temperature than glass. The bottle will break if it was filled to its tightly capped lid. Explain why, and also explain how a
pocket of air above the vinegar would prevent the break. (This is the function of the air above liquids in glass containers.)


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Problems & Exercises


5.1 Friction
1. A physics major is cooking breakfast when he notices that
the frictional force between his steel spatula and his Teflon
frying pan is only 0.200 N. Knowing the coefficient of kinetic
friction between the two materials, he quickly calculates the
normal force. What is it?


2. (a) When rebuilding her car’s engine, a physics major must
exert 300 N of force to insert a dry steel piston into a steel
cylinder. What is the magnitude of the normal force between
the piston and cylinder? (b) What is the magnitude of the
force would she have to exert if the steel parts were oiled?


3. (a) What is the maximum frictional force in the knee joint of
a person who supports 66.0 kg of her mass on that knee? (b)
During strenuous exercise it is possible to exert forces to the
joints that are easily ten times greater than the weight being
supported. What is the maximum force of friction under such
conditions? The frictional forces in joints are relatively small in
all circumstances except when the joints deteriorate, such as
from injury or arthritis. Increased frictional forces can cause
further damage and pain.


4. Suppose you have a 120-kg wooden crate resting on a
wood floor. (a) What maximum force can you exert
horizontally on the crate without moving it? (b) If you continue
to exert this force once the crate starts to slip, what will the
magnitude of its acceleration then be?


5. (a) If half of the weight of a small 1.00×103 kg utility
truck is supported by its two drive wheels, what is the
magnitude of the maximum acceleration it can achieve on dry
concrete? (b) Will a metal cabinet lying on the wooden bed of
the truck slip if it accelerates at this rate? (c) Solve both
problems assuming the truck has four-wheel drive.


6. A team of eight dogs pulls a sled with waxed wood runners
on wet snow (mush!). The dogs have average masses of 19.0
kg, and the loaded sled with its rider has a mass of 210 kg.
(a) Calculate the magnitude of the acceleration starting from
rest if each dog exerts an average force of 185 N backward
on the snow. (b) What is the magnitude of the acceleration
once the sled starts to move? (c) For both situations,
calculate the magnitude of the force in the coupling between
the dogs and the sled.


7. Consider the 65.0-kg ice skater being pushed by two
others shown in Figure 5.21. (a) Find the direction and
magnitude of 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
she is initially stationary and wearing steel-bladed skates that
point 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 direction opposite
that of motion or attempted motion between surfaces in
contact.)


Figure 5.21


8. Show that the acceleration of any object down a frictionless
incline that makes an angle θ with the horizontal is
a = g sin θ . (Note that this acceleration is independent of
mass.)


9. Show that the acceleration of any object down an incline
where friction behaves simply (that is, where fk = µkN ) is
a = g( sin θ − µkcos θ). Note that the acceleration is
independent of mass and reduces to the expression found in
the previous problem when friction becomes negligibly small
(µk = 0).


10. Calculate the deceleration of a snow boarder going up a
5.0º , slope assuming the coefficient of friction for waxed
wood on wet snow. The result of Exercise 5.9 may be useful,
but be careful to consider the fact that the snow boarder is
going uphill. Explicitly show how you follow the steps in
Problem-Solving Strategies.


11. (a) Calculate the acceleration of a skier heading down a
10.0º slope, assuming the coefficient of friction for waxed
wood on wet snow. (b) Find the angle of the slope down
which this skier could coast at a constant velocity. You can
neglect air resistance in both parts, and you will find the result
of Exercise 5.9 to be useful. Explicitly show how you follow
the steps in the Problem-Solving Strategies.


12. If an object is to rest on an incline without slipping, then
friction must equal the component of the weight of the object
parallel to the incline. This requires greater and greater
friction for steeper slopes. Show that the maximum angle 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.


13. Calculate the maximum deceleration of a car that is
heading down a 6º slope (one that makes an angle of 6º
with the horizontal) under the following road conditions. You
may assume that the weight of the car is evenly distributed on
all four tires and that the coefficient of static friction is
involved—that is, the tires are not allowed to slip during the
deceleration. (Ignore rolling.) Calculate for a car: (a) On dry
concrete. (b) On wet concrete. (c) On ice, assuming that
µs = 0.100 , the same as for shoes on ice.


14. Calculate the maximum acceleration of a car that is
heading up a 4º slope (one that makes an angle of 4º with
the horizontal) under the following road conditions. Assume
that only half the weight of the car is supported by the two
drive wheels and that the coefficient of static friction is
involved—that is, the tires are not allowed to slip during the
acceleration. (Ignore rolling.) (a) On dry concrete. (b) On wet


Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 195




concrete. (c) On ice, assuming that μs = 0.100 , the same
as for shoes on ice.


15. Repeat Exercise 5.14 for a car with four-wheel drive.


16. 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 5.00×10−2 m / s2 if the
force of friction is 7.50×105 N , assuming the engines exert
identical forces? This is not a large frictional force for such a
massive system. Rolling friction for trains is small, and
consequently trains are very energy-efficient transportation
systems. (b) What is the magnitude of the force in the
coupling between the 37th and 38th cars (this is the force
each exerts on the other), assuming all cars have the same
mass and that friction is evenly distributed among all of the
cars and engines?


17. Consider the 52.0-kg mountain climber in Figure 5.22. (a)
Find the tension in the rope and the force that the mountain
climber must exert with her feet on the vertical rock face to
remain stationary. Assume that the force is exerted parallel to
her legs. Also, assume negligible force exerted by her arms.
(b) What is the minimum coefficient of friction between her
shoes and the cliff? Figure 5.23Which method of sliding a block of ice requires less


force—(a) pushing or (b) pulling at the same angle above the horizontal?


5.2 Drag Forces
20. The terminal velocity of a person falling in air depends
upon the weight and the area of the person facing the fluid.
Find the terminal velocity (in meters per second and
kilometers per hour) of an 80.0-kg skydiver falling in a pike


(headfirst) position with a surface area of 0.140 m2 .
21. A 60-kg and a 90-kg skydiver jump from an airplane at an
altitude of 6000 m, both falling in the pike position. Make
some assumption on their frontal areas and calculate their
terminal velocities. How long will it take for each skydiver to
reach the ground (assuming the time to reach terminal
velocity is small)? Assume all values are accurate to three
significant digits.


22. A 560-g squirrel with a surface area of 930 cm2 falls
from a 5.0-m tree to the ground. Estimate its terminal velocity.
(Use a drag coefficient for a horizontal skydiver.) What will be
the velocity of a 56-kg person hitting the ground, assuming no
drag contribution in such a short distance?


23. To maintain a constant speed, the force provided by a
car’s engine must equal the drag force plus the force of
friction of the road (the rolling resistance). (a) What are the
magnitudes of drag forces at 70 km/h and 100 km/h for a


Toyota Camry? (Drag area is 0.70 m2 ) (b) What is the
magnitude of 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.


24. By what factor does the drag force on a car increase as it
goes from 65 to 110 km/h?


25. Calculate the speed a spherical rain drop would achieve
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 .


196 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity


Figure 5.22 Part of the climber’s weight is supported by her rope and
part by friction between her feet and the rock face.


18. A contestant in a winter sporting event pushes a 45.0-kg
block of ice across a frozen lake as shown in Figure 5.23(a).
(a) Calculate the minimum force F he must exert to get the
block moving. (b) What is the magnitude of its acceleration
once it starts to move, if that force is maintained?


19. Repeat Exercise 5.18 with the contestant pulling the
block of ice with a rope over his shoulder at the same angle
above the horizontal as shown in Figure 5.23(b).


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26. Using Stokes’ law, verify that the units for viscosity are
kilograms per meter per second.


27. 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 terminal
velocity. Take the density of the bacterium to be


1.10×103 kg/m3 .


28. Stokes’ law describes sedimentation of particles in liquids
and can be used to measure viscosity. Particles in liquids
achieve terminal velocity quickly. One can measure the time it
takes for a particle to fall a certain distance and then 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 the viscosity of the oil.


5.3 Elasticity: Stress and Strain
29. During a circus act, one performer swings upside down
hanging from a trapeze holding another, also upside-down,
performer by the legs. If the upward force on the lower
performer is three times her weight, how much do the bones
(the femurs) in her upper legs stretch? You may assume each
is equivalent to a uniform rod 35.0 cm long and 1.80 cm in
radius. Her mass is 60.0 kg.


30. During a wrestling match, a 150 kg wrestler briefly stands
on one hand during a maneuver designed to perplex his
already moribund adversary. By how much does the upper
arm bone shorten in length? The bone can be represented by
a uniform rod 38.0 cm in length and 2.10 cm in radius.


31. (a) The “lead” in pencils is a graphite composition with a


Young’s modulus of about 1×109 N /m2 . Calculate the
change in length of the lead in an automatic pencil if you tap it
straight into the pencil with a force of 4.0 N. The lead is 0.50
mm in diameter and 60 mm long. (b) Is the answer
reasonable? That is, does it seem to be consistent with what
you have observed when using pencils?


32. TV broadcast antennas are the tallest artificial structures
on Earth. In 1987, a 72.0-kg physicist placed himself and 400
kg of equipment at the top of one 610-m high antenna to
perform gravity experiments. By how much was the antenna
compressed, if we consider it to be equivalent to a steel
cylinder 0.150 m in radius?


33. (a) By how much does a 65.0-kg mountain climber stretch
her 0.800-cm diameter nylon rope when she hangs 35.0 m
below a rock outcropping? (b) Does the answer seem to be
consistent with what you have observed for nylon ropes?
Would it make sense if the rope were actually a bungee
cord?


34. A 20.0-m tall hollow aluminum flagpole is equivalent in
stiffness to a solid cylinder 4.00 cm in diameter. A strong wind
bends the pole much as a horizontal force of 900 N exerted at
the top would. How far to the side does the top of the pole
flex?


35. As an oil well is drilled, each new section of drill pipe
supports its own weight and that of the pipe and drill bit
beneath it. Calculate the stretch in a new 6.00 m length of
steel pipe that supports 3.00 km of pipe having a mass of
20.0 kg/m and a 100-kg drill bit. The pipe is equivalent in
stiffness to a solid cylinder 5.00 cm in diameter.


36. Calculate the force a piano tuner applies to stretch a steel
piano wire 8.00 mm, if the wire is originally 0.850 mm in
diameter and 1.35 m long.


37. A vertebra is subjected to a shearing force of 500 N. Find
the shear deformation, taking the vertebra to be a cylinder
3.00 cm high and 4.00 cm in diameter.


38. A disk between vertebrae in the spine is subjected to a
shearing force of 600 N. Find its shear deformation, taking it


to have the shear modulus of 1×109 N /m2 . The disk is
equivalent to a solid cylinder 0.700 cm high and 4.00 cm in
diameter.


39.When using a pencil eraser, you exert a vertical force of
6.00 N at a distance of 2.00 cm from the hardwood-eraser
joint. The pencil is 6.00 mm in diameter and is held at an
angle of 20.0º to the horizontal. (a) By how much does the
wood flex perpendicular to its length? (b) How much is it
compressed lengthwise?


40. To consider the effect of wires hung on poles, we take
data from Example 4.8, in which tensions in wires supporting
a traffic light were calculated. The left wire made an angle
30.0º below the horizontal with the top of its pole and carried
a tension of 108 N. The 12.0 m tall hollow aluminum pole is
equivalent in stiffness to a 4.50 cm diameter solid cylinder. (a)
How far is it bent to the side? (b) By how much is it
compressed?


41. A farmer making grape juice fills a glass bottle to the brim
and caps it tightly. The juice expands more than the glass
when it warms up, in such a way that the volume increases by


0.2% (that is, ΔV /V0 = 2×10
−3 ) relative to the space


available. Calculate the magnitude of the normal force
exerted by the juice per square centimeter if its bulk modulus


is 1.8×109 N/m2 , assuming the bottle does not break. In
view of your answer, do you think the bottle will survive?


42. (a) When water freezes, its volume increases by 9.05%


(that is, ΔV /V0 = 9.05×10
−2 ). What force per unit area


is water capable of exerting on a container when it freezes?
(It is acceptable to use the bulk modulus of water in this
problem.) (b) Is it surprising that such forces can fracture
engine blocks, boulders, and the like?


43. This problem returns to the tightrope walker studied in


Example 4.6, who created a tension of 3.94×103 N in a
wire making an angle 5.0º below the horizontal with each
supporting pole. Calculate how much this tension stretches
the steel wire if it was originally 15 m long and 0.50 cm in
diameter.


44. The pole in Figure 5.24 is at a 90.0º bend in a power
line and is therefore subjected to more shear force than poles
in straight parts of the line. The tension in each line is


4.00×104 N , at the angles shown. The pole is 15.0 m tall,
has an 18.0 cm diameter, and can be considered to have half
the stiffness of hardwood. (a) Calculate the compression of
the pole. (b) Find how much it bends and in what direction. (c)
Find the tension in a guy wire used to keep the pole straight if
it is attached to the top of the pole at an angle of 30.0º with
the vertical. (Clearly, the guy wire must be in the opposite
direction of the bend.)


Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 197




Figure 5.24 This telephone pole is at a 90º bend in a power line. A
guy wire is attached to the top of the pole at an angle of 30º with the
vertical.


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6 UNIFORM CIRCULAR MOTION AND
GRAVITATION


Figure 6.1 This Australian Grand Prix Formula 1 race car moves in a circular path as it makes the turn. Its wheels also spin rapidly—the latter
completing many revolutions, the former only part of one (a circular arc). The same physical principles are involved in each. (credit: Richard Munckton)


Chapter Outline
6.1. Rotation Angle and Angular Velocity


• Define arc length, rotation angle, radius of curvature and angular velocity.
• Calculate the angular velocity of a car wheel spin.


6.2. Centripetal Acceleration
• Establish the expression for centripetal acceleration.
• Explain the centrifuge.


6.3. Centripetal Force
• Calculate coefficient of friction on a car tire.
• Calculate ideal speed and angle of a car on a turn.


6.4. Fictitious Forces and Non-inertial Frames: The Coriolis Force
• Discuss the inertial frame of reference.
• Discuss the non-inertial frame of reference.
• Describe the effects of the Coriolis force.


6.5. Newton’s Universal Law of Gravitation
• Explain Earth’s gravitational force.
• Describe the gravitational effect of the Moon on Earth.
• Discuss weightlessness in space.
• Examine the Cavendish experiment


6.6. Satellites and Kepler’s Laws: An Argument for Simplicity
• State Kepler’s laws of planetary motion.
• Derive the third Kepler’s law for circular orbits.
• Discuss the Ptolemaic model of the universe.


Chapter 6 | Uniform Circular Motion and Gravitation 199




Introduction to Uniform Circular Motion and Gravitation
Many motions, such as the arc of a bird’s flight or Earth’s path around the Sun, are curved. Recall that Newton’s first law tells us
that motion is along a straight line at constant speed unless there is a net external force. We will therefore study not only motion
along curves, but also the forces that cause it, including gravitational forces. In some ways, this chapter is a continuation of
Dynamics: Newton's Laws of Motion as we study more applications of Newton’s laws of motion.


This chapter deals with the simplest form of curved motion, uniform circular motion, motion in a circular path at constant
speed. Studying this topic illustrates most concepts associated with rotational motion and leads to the study of many new topics
we group under the name rotation. Pure rotational motion occurs when points in an object move in circular paths centered on
one point. Pure translational motion is motion with no rotation. Some motion combines both types, such as a rotating hockey
puck moving along ice.


6.1 Rotation Angle and Angular Velocity
In Kinematics, we studied motion along a straight line and introduced such concepts as displacement, velocity, and acceleration.
Two-Dimensional Kinematics dealt with motion in two dimensions. Projectile motion is a special case of two-dimensional
kinematics in which the object is projected into the air, while being subject to the gravitational force, and lands a distance away.
In this chapter, we consider situations where the object does not land but moves in a curve. We begin the study of uniform
circular motion by defining two angular quantities needed to describe rotational motion.


Rotation Angle
When objects rotate about some axis—for example, when the CD (compact disc) in Figure 6.2 rotates about its center—each
point in the object follows a circular arc. Consider a line from the center of the CD to its edge. Each pit used to record sound
along this line moves through the same angle in the same amount of time. The rotation angle is the amount of rotation and is
analogous to linear distance. We define the rotation angle Δθ to be the ratio of the arc length to the radius of curvature:


(6.1)Δθ = Δsr .


Figure 6.2 All points on a CD travel in circular arcs. The pits along a line from the center to the edge all move through the same angle Δθ in a time
Δt .


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Figure 6.3 The radius of a circle is rotated through an angle Δθ . The arc length Δs is described on the circumference.


The arc length Δs is the distance traveled along a circular path as shown in Figure 6.3 Note that r is the radius of curvature
of the circular path.


We know that for one complete revolution, the arc length is the circumference of a circle of radius r . The circumference of a
circle is 2πr . Thus for one complete revolution the rotation angle is


(6.2)Δθ = 2πrr = 2π.


This result is the basis for defining the units used to measure rotation angles, Δθ to be radians (rad), defined so that
(6.3)2π rad = 1 revolution.


A comparison of some useful angles expressed in both degrees and radians is shown in Table 6.1.


Table 6.1 Comparison of Angular Units


Degree Measures Radian Measure


30º π6


60º π3


90º π2


120º 2π3


135º 3π4


180º π


Chapter 6 | Uniform Circular Motion and Gravitation 201




Figure 6.4 Points 1 and 2 rotate through the same angle (Δθ ), but point 2 moves through a greater arc length (Δs) because it is at a greater
distance from the center of rotation (r) .


If Δθ = 2π rad, then the CD has made one complete revolution, and every point on the CD is back at its original position.
Because there are 360º in a circle or one revolution, the relationship between radians and degrees is thus


(6.4)2π rad = 360º
so that


(6.5)1 rad = 360º2π ≈ 57.3º.


Angular Velocity
How fast is an object rotating? We define angular velocity ω as the rate of change of an angle. In symbols, this is


(6.6)ω = ΔθΔt ,


where an angular rotation Δθ takes place in a time Δt . The greater the rotation angle in a given amount of time, the greater
the angular velocity. The units for angular velocity are radians per second (rad/s).


Angular velocity ω is analogous to linear velocity v . To get the precise relationship between angular and linear velocity, we
again consider a pit on the rotating CD. This pit moves an arc length Δs in a time Δt , and so it has a linear velocity


(6.7)v = ΔsΔt .


From Δθ = Δsr we see that Δs = rΔθ . Substituting this into the expression for v gives


(6.8)v = rΔθΔt = rω.


We write this relationship in two different ways and gain two different insights:


(6.9)v = rω or ω = vr .


The first relationship in v = rω or ω = vr states that the linear velocity v is proportional to the distance from the center of
rotation, thus, it is largest for a point on the rim (largest r ), as you might expect. We can also call this linear speed v of a point


on the rim the tangential speed. The second relationship in v = rω or ω = vr can be illustrated by considering the tire of a
moving car. Note that the speed of a point on the rim of the tire is the same as the speed v of the car. See Figure 6.5. So the
faster the car moves, the faster the tire spins—large v means a large ω , because v = rω . Similarly, a larger-radius tire
rotating at the same angular velocity (ω ) will produce a greater linear speed ( v ) for the car.


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Figure 6.5 A car moving at a velocity v to the right has a tire rotating with an angular velocity ω .The speed of the tread of the tire relative to the axle
is v , the same as if the car were jacked up. Thus the car moves forward at linear velocity v = rω , where r is the tire radius. A larger angular
velocity for the tire means a greater velocity for the car.


Example 6.1 How Fast Does a Car Tire Spin?


Calculate the angular velocity of a 0.300 m radius car tire when the car travels at 15.0 m/s (about 54 km/h ). See Figure
6.5.


Strategy


Because the linear speed of the tire rim is the same as the speed of the car, we have v = 15.0 m/s. The radius of the tire
is given to be r = 0.300 m. Knowing v and r , we can use the second relationship in v = rω, ω = vr to calculate the
angular velocity.


Solution


To calculate the angular velocity, we will use the following relationship:


(6.10)ω = vr .


Substituting the knowns,


(6.11)ω = 15.0 m/s0.300 m = 50.0 rad/s.


Discussion


When we cancel units in the above calculation, we get 50.0/s. But the angular velocity must have units of rad/s. Because
radians are actually unitless (radians are defined as a ratio of distance), we can simply insert them into the answer for the
angular velocity. Also note that if an earth mover with much larger tires, say 1.20 m in radius, were moving at the same
speed of 15.0 m/s, its tires would rotate more slowly. They would have an angular velocity


(6.12)ω = (15.0 m/s) / (1.20 m) = 12.5 rad/s.


Both ω and v have directions (hence they are angular and linear velocities, respectively). Angular velocity has only two
directions with respect to the axis of rotation—it is either clockwise or counterclockwise. Linear velocity is tangent to the path, as
illustrated in Figure 6.6.


Take-Home Experiment


Tie an object to the end of a string and swing it around in a horizontal circle above your head (swing at your wrist). Maintain
uniform speed as the object swings and measure the angular velocity of the motion. What is the approximate speed of the
object? Identify a point close to your hand and take appropriate measurements to calculate the linear speed at this point.
Identify other circular motions and measure their angular velocities.


Chapter 6 | Uniform Circular Motion and Gravitation 203




Figure 6.6 As an object moves in a circle, here a fly on the edge of an old-fashioned vinyl record, its instantaneous velocity is always tangent to the
circle. The direction of the angular velocity is clockwise in this case.


PhET Explorations: Ladybug Revolution


Figure 6.7 Ladybug Revolution (http://cnx.org/content/m42083/1.7/rotation_en.jar)


Join the ladybug in an exploration of rotational motion. Rotate the merry-go-round to change its angle, or choose a constant
angular velocity or angular acceleration. Explore how circular motion relates to the bug's x,y position, velocity, and
acceleration using vectors or graphs.


6.2 Centripetal Acceleration
We know from kinematics that acceleration is a change in velocity, either in its magnitude or in its direction, or both. In uniform
circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the
magnitude of the velocity might be constant. You experience this acceleration yourself when you turn a corner in your car. (If you
hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion.) What you notice is a
sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the
more noticeable this acceleration will become. In this section we examine the direction and magnitude of that acceleration.


Figure 6.8 shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at
two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of
rotation (the center of the circular path). This pointing is shown with the vector diagram in the figure. We call the acceleration of
an object moving in uniform circular motion (resulting from a net external force) the centripetal acceleration( ac ); centripetal


means “toward the center” or “center seeking.”


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Figure 6.8 The directions of the velocity of an object at two different points are shown, and the change in velocity Δv is seen to point directly toward
the center of curvature. (See small inset.) Because ac = Δv / Δt , the acceleration is also toward the center; ac is called centripetal acceleration.
(Because Δθ is very small, the arc length Δs is equal to the chord length Δr for small time differences.)


The direction of centripetal acceleration is toward the center of curvature, but what is its magnitude? Note that the triangle
formed by the velocity vectors and the one formed by the radii r and Δs are similar. Both the triangles ABC and PQR are
isosceles triangles (two equal sides). The two equal sides of the velocity vector triangle are the speeds v1 = v2 = v . Using the
properties of two similar triangles, we obtain


(6.13)Δv
v =


Δs
r .


Acceleration is Δv / Δt , and so we first solve this expression for Δv :
(6.14)Δv = vrΔs.


Then we divide this by Δt , yielding
(6.15)Δv


Δt =
v

Δs
Δt .


Finally, noting that Δv / Δt = ac and that Δs / Δt = v , the linear or tangential speed, we see that the magnitude of the
centripetal acceleration is


(6.16)
ac = v


2
r ,


which is the acceleration of an object in a circle of radius r at a speed v . So, centripetal acceleration is greater at high speeds
and in sharp curves (smaller radius), as you have noticed when driving a car. But it is a bit surprising that ac is proportional to


speed squared, implying, for example, that it is four times as hard to take a curve at 100 km/h than at 50 km/h. A sharp corner
has a small radius, so that ac is greater for tighter turns, as you have probably noticed.


It is also useful to express ac in terms of angular velocity. Substituting v = rω into the above expression, we find


ac = (rω)2 / r = rω2 . We can express the magnitude of centripetal acceleration using either of two equations:


(6.17)
ac = v


2
r ; ac = rω


2.


Recall that the direction of ac is toward the center. You may use whichever expression is more convenient, as illustrated in


examples below.


A centrifuge (see Figure 6.9b) is a rotating device used to separate specimens of different densities. High centripetal
acceleration significantly decreases the time it takes for separation to occur, and makes separation possible with small samples.
Centrifuges are used in a variety of applications in science and medicine, including the separation of single cell suspensions
such as bacteria, viruses, and blood cells from a liquid medium and the separation of macromolecules, such as DNA and protein,


Chapter 6 | Uniform Circular Motion and Gravitation 205




from a solution. Centrifuges are often rated in terms of their centripetal acceleration relative to acceleration due to gravity (g) ;
maximum centripetal acceleration of several hundred thousand g is possible in a vacuum. Human centrifuges, extremely large


centrifuges, have been used to test the tolerance of astronauts to the effects of accelerations larger than that of Earth’s gravity.


Example 6.2 How Does the Centripetal Acceleration of a Car Around a Curve Compare with That
Due to Gravity?


What is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m/s
(about 90 km/h)? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed. See
Figure 6.9(a).


Strategy


Because v and r are given, the first expression in ac = v
2
r ; ac = rω


2 is the most convenient to use.


Solution


Entering the given values of v = 25.0 m/s and r = 500 m into the first expression for ac gives


(6.18)
ac = v


2
r =


(25.0 m/s)2
500 m = 1.25 m/s


2.


Discussion


To compare this with the acceleration due to gravity (g = 9.80 m/s2) , we take the ratio of


ac / g = ⎛⎝1.25 m/s2

⎠ /

⎝9.80 m/s2



⎠ = 0.128 . Thus, ac = 0.128 g and is noticeable especially if you were not wearing a


seat belt.


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Figure 6.9 (a) The car following a circular path at constant speed is accelerated perpendicular to its velocity, as shown. The magnitude of this
centripetal acceleration is found in Example 6.2. (b) A particle of mass in a centrifuge is rotating at constant angular velocity . It must be accelerated
perpendicular to its velocity or it would continue in a straight line. The magnitude of the necessary acceleration is found in Example 6.3.


Example 6.3 How Big Is the Centripetal Acceleration in an Ultracentrifuge?


Calculate the centripetal acceleration of a point 7.50 cm from the axis of an ultracentrifuge spinning at


7.5 × 104 rev/min. Determine the ratio of this acceleration to that due to gravity. See Figure 6.9(b).
Strategy


The term rev/min stands for revolutions per minute. By converting this to radians per second, we obtain the angular velocity


ω . Because r is given, we can use the second expression in the equation ac = v
2
r ; ac = rω


2 to calculate the


centripetal acceleration.


Solution


To convert 7.50×104 rev / min to radians per second, we use the facts that one revolution is 2πrad and one minute is
60.0 s. Thus,


(6.19)ω = 7.50×104 revmin×
2π rad
1 rev ×


1 min
60.0 s = 7854 rad/s.


Now the centripetal acceleration is given by the second expression in ac = v
2
r ; ac = rω


2 as


(6.20)ac = rω2.


Converting 7.50 cm to meters and substituting known values gives


(6.21)ac = (0.0750 m)(7854 rad/s)2 = 4.63×106 m/s2.


Chapter 6 | Uniform Circular Motion and Gravitation 207




Note that the unitless radians are discarded in order to get the correct units for centripetal acceleration. Taking the ratio of
ac to g yields


(6.22)ac
g =


4.63×106
9.80 = 4.72×10


5.


Discussion


This last result means that the centripetal acceleration is 472,000 times as strong as g . It is no wonder that such high ω


centrifuges are called ultracentrifuges. The extremely large accelerations involved greatly decrease the time needed to
cause the sedimentation of blood cells or other materials.


Of course, a net external force is needed to cause any acceleration, just as Newton proposed in his second law of motion. So a
net external force is needed to cause a centripetal acceleration. In Centripetal Force, we will consider the forces involved in
circular motion.


PhET Explorations: Ladybug Motion 2D


Learn about position, velocity and acceleration vectors. Move the ladybug by setting the position, velocity or acceleration,
and see how the vectors change. Choose linear, circular or elliptical motion, and record and playback the motion to analyze
the behavior.


Figure 6.10 Ladybug Motion 2D (http://cnx.org/content/m42084/1.8/ladybug-motion-2d_en.jar)


6.3 Centripetal Force
Any force or combination of forces can cause a centripetal or radial acceleration. 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 spinning centrifuge.


Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the
center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force
is mass times acceleration: net F = ma . For uniform circular motion, the acceleration is the centripetal acceleration— a = ac .
Thus, the magnitude of centripetal force Fc is


(6.23)Fc = mac.


By using the expressions for centripetal acceleration ac from ac = v
2
r ; ac = rω


2 , we get two expressions for the centripetal


force Fc in terms of mass, velocity, angular velocity, and radius of curvature:


(6.24)
Fc = mv


2
r ; Fc = mrω


2.


You may use whichever expression for centripetal force is more convenient. Centripetal force Fc is always perpendicular to the


path and pointing to the center of curvature, because ac is perpendicular to the velocity and pointing to the center of curvature.


Note that if you solve the first expression for r , you get


(6.25)
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 tight curve.


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Figure 6.11 The frictional force supplies the centripetal force and is numerically equal to it. Centripetal force is perpendicular to velocity and causes
uniform 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
a larger Fc produces a smaller r′ .


Example 6.4 What Coefficient of Friction Do Car Tires Need on a Flat Curve?


(a) Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 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 (see Figure 6.12).


Strategy and Solution for (a)


We know that Fc = mv
2


r . Thus,


(6.26)
Fc = mv


2
r =


(900 kg)(25.0 m/s)2
(500 m) = 1125 N.


Strategy for (b)


Figure 6.12 shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car
from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We
know that the maximum static friction (at which the tires roll but do 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 that N = mg . Thus the
centripetal force in this situation is


(6.27)Fc = f = µsN = µsmg.


Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for Fc from


the equation


(6.28)


Fc = mv
2
r


Fc = mrω2





⎬,


(6.29)
mv


2
r = µsmg.


We solve this for µs , noting that mass cancels, and obtain


Chapter 6 | Uniform Circular Motion and Gravitation 209




(6.30)
µs = v


2
rg .


Solution for (b)


Substituting the knowns,


(6.31)
µs =


(25.0 m/s)2


(500 m)(9.80 m/s2)
= 0.13.


(Because coefficients of friction are approximate, the answer is given to only two digits.)


Discussion


We could also solve part (a) using the first expression in Fc = m
v2
r


Fc = mrω2





⎬, because m, v, and r are given. The coefficient


of friction found in part (b) is much smaller than is typically found between tires and roads. The car will still negotiate the
curve if the coefficient is greater than 0.13, because static friction is a responsive force, being 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 than 25 m/s. Note that mass cancels, implying that in this
example, it does not matter how heavily loaded the car is to negotiate the turn. Mass cancels because friction is assumed
proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal
force would be less as will be discussed below.


210 Chapter 6 | Uniform Circular Motion and Gravitation


Figure 6.12 This car on level ground is moving away and turning to the left. The centripetal force causing the car to turn in a circular path is due to
friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.


Let us now consider banked curves, where the slope of the road helps you negotiate the curve. See Figure 6.13. The greater
the 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.


For ideal banking, the net external force equals the horizontal centripetal force in the absence of friction. The components of 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 perpendicular axes—in this case, the
vertical and horizontal directions.


Figure 6.13 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 will equal 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


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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, and so this must equal the centripetal force—that is,


(6.32)
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 vertical
components of the two external forces must be equal in magnitude and opposite in direction. From the figure, we see that the
vertical component of the normal force is N cos θ , and the only other vertical force is the car’s weight. These must be equal in
magnitude; thus,


(6.33)N cos θ = mg.


Now we can combine the last two equations to eliminate N and get an expression for θ , as desired. Solving the second
equation for N = mg / (cos θ) , and substituting this into the first yields


(6.34)
mg sin θcos θ =


mv2
r


(6.35)
mg tan(θ) = mv


2
r


tan θ = v
2


rg.


Taking the inverse tangent gives


(6.36)
θ = tan−1⎛⎝


v2
rg

⎠ (ideally banked curve, no friction).


This expression can be understood by considering how θ depends on v and r . A large θ will be 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 to take
the curve at greater or lower speed than if the curve is frictionless. Note that θ does not depend on the mass of the vehicle.


Figure 6.13 The car on this banked curve is moving away and turning to the left.


Example 6.5 What Is the Ideal Speed to Take a Steeply Banked Tight Curve?


Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply
banked. This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very
high speed. To illustrate, calculate the speed at which a 100 m radius curve banked at 65.0° should be driven if the road is
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


(6.37)
tan θ = v


2
rg


Chapter 6 | Uniform Circular Motion and Gravitation 211




we get


(6.38)v = (rg tan θ)1 / 2.


Noting that tan 65.0º = 2.14, we obtain


(6.39)
v = ⎡⎣(100 m)(9.80 m/s2)(2.14)





1 / 2


= 45.8 m/s.
Discussion


This is just about 165 km/h, consistent with a very steeply banked and rather sharp curve. Tire friction enables a vehicle to
take the curve at significantly higher speeds.


Calculations similar to those in the preceding examples can be performed for a host of interesting situations in which
centripetal force is involved—a number of these are presented in this chapter’s Problems and Exercises.


Take-Home Experiment


Ask a friend or relative to swing a golf club or a tennis racquet. Take appropriate measurements to estimate the centripetal
acceleration of the end of the club or racquet. You may choose to do this in slow motion.


PhET Explorations: Gravity and Orbits


Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the
sizes and distances between different heavenly bodies, and turn off gravity to see what would happen without it!


Figure 6.14 Gravity and Orbits (http://cnx.org/content/m42086/1.9/gravity-and-orbits_en.jar)


6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
What do taking off in a jet airplane, turning a corner in a car, riding a merry-go-round, and the circular motion of a tropical cyclone
have in common? Each exhibits fictitious forces—unreal forces that arise from motion and may seem real, because the
observer’s frame of reference is accelerating or rotating.


When taking off in a jet, most people would agree it feels as if you are being pushed back into the seat as the airplane
accelerates down the runway. Yet a physicist would say that you tend to remain stationary while the seat pushes forward on you,
and there is no real force backward on you. An even more common experience occurs when you make a tight curve in your
car—say, to the right. You feel as if you are thrown (that is, forced) toward the left relative to the car. Again, a physicist would say
that you are going in a straight line but the car moves to the right, and there is no real force on you to the left. Recall Newton’s
first law.


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Figure 6.15 (a) The car driver feels herself forced to the left relative to the car when she makes a right turn. This is a fictitious force arising from the
use of the car as a frame of reference. (b) In the Earth’s frame of reference, the driver moves in a straight line, obeying Newton’s first law, and the car
moves to the right. There is no real force to the left on the driver relative to Earth. There is a real force 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, while a physicist uses Earth. The physicist chooses Earth because
it is very nearly an inertial frame of reference—one in which all forces are real (that is, 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 Dynamics: Newton's Laws of
Motion The car is a non-inertial frame of reference because it is accelerated to the side. The force to the left sensed by car
passengers is a fictitious force having no physical origin. There is nothing real pushing them left—the car, as well as the driver,
is actually accelerating to the right.


Let us now take a mental ride on a merry-go-round—specifically, a rapidly rotating playground merry-go-round. You take the
merry-go-round to be your frame of reference because you rotate together. In that non-inertial frame, you feel a fictitious force,
named centrifugal force (not to be confused with centripetal force), trying to throw you off. You must hang on tightly to
counteract the centrifugal force. In Earth’s frame of reference, there is no force trying to throw you off. Rather you must hang on
to make yourself go in a circle because otherwise you would go in a straight line, right off the merry-go-round.


Figure 6.16 (a) A rider on a merry-go-round feels as if he is being thrown off. This fictitious force is called the centrifugal force—it explains the rider’s
motion in the rotating frame of reference. (b) In an inertial frame of reference and according to Newton’s laws, it is his inertia that carries him off and not
a real force (the unshaded rider has Fnet = 0 and heads in a straight line). A real 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, is put to
good use in centrifuges (see Figure 6.17). A centrifuge spins a sample very rapidly, as mentioned earlier in this chapter. Viewed
from the rotating frame of reference, the fictitious centrifugal force throws particles outward, hastening their sedimentation. The
greater the angular velocity, the greater the centrifugal force. But what really happens is that the inertia of the particles carries
them along a line tangent to the circle while the test tube is forced in a circular path by a centripetal force.


Chapter 6 | Uniform Circular Motion and Gravitation 213




Figure 6.17 Centrifuges use inertia to perform their task. Particles in the fluid sediment come out because their inertia carries them away from the
center of rotation. The large angular velocity of the centrifuge quickens the sedimentation. Ultimately, the particles will come into contact with the test
tube walls, which will then supply the centripetal force needed to make them move in a circle of constant radius.


Let us now consider what happens if something moves in a frame of reference that rotates. For example, what if you slide a ball
directly away from the center of the merry-go-round, as shown in Figure 6.18? The ball follows a straight path relative to Earth
(assuming negligible friction) and a path curved to the right on the merry-go-round’s surface. A person standing next to the
merry-go-round sees the ball moving straight and the merry-go-round rotating underneath it. In the merry-go-round’s frame of
reference, we explain the apparent curve to the right by using a fictitious force, called the Coriolis force, that causes the ball to
curve to the right. The fictitious Coriolis force can be used by anyone in that frame of reference to explain why objects follow
curved paths and allows us to apply Newton’s Laws in non-inertial frames of reference.


214 Chapter 6 | Uniform Circular Motion and Gravitation


Figure 6.18 Looking down on the counterclockwise rotation of a merry-go-round, we see that a ball slid straight toward the edge follows a path curved
to the right. The person slides the ball toward point B, starting at point A. Both points rotate to the shaded positions (A’ and B’) shown in the time that
the ball follows the curved path in the rotating frame and a straight path in Earth’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 its rotation.
Yet such effects do exist—in the rotation of weather systems, for example. Most consequences of Earth’s rotation can be
qualitatively understood by analogy with the merry-go-round. Viewed from above the North Pole, Earth rotates counterclockwise,
as does the merry-go-round in Figure 6.18. As on the merry-go-round, any motion in Earth’s northern hemisphere experiences a


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Coriolis force to the right. Just the opposite occurs in the southern hemisphere; there, the force is 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, while the tropical
cyclones (what hurricanes are called below the equator) in the southern hemisphere rotate in the clockwise direction. The terms
hurricane, typhoon, and tropical storm are regionally-specific names for tropical cyclones, storm systems characterized by low
pressure centers, strong winds, and heavy rains. Figure 6.19 helps show how these rotations take place. Air flows toward any
region of low pressure, and tropical cyclones contain particularly low pressures. Thus winds flow toward the center of a tropical
cyclone or 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. Low
pressure at the surface is associated with rising air, which also produces cooling and cloud formation, making low-pressure
patterns quite visible from space. Conversely, wind circulation around high-pressure zones is clockwise in the northern
hemisphere but is less visible because high pressure is associated with sinking air, producing clear skies.


The rotation of tropical cyclones and the path of a ball on a merry-go-round can just as well be explained by inertia and the
rotation of the system underneath. When non-inertial frames are used, fictitious forces, such as the Coriolis force, must be
invented to explain the curved path. There is no identifiable physical source for these fictitious 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 and truest, in the sense that all forces have real origins and explanations.


Figure 6.19 (a) The counterclockwise rotation of this northern hemisphere hurricane is a major consequence of the Coriolis force. (credit: NASA) (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 a high-pressure zone is also deflected to the right, producing a
clockwise rotation. (e) The opposite direction of rotation is produced by the Coriolis force in the southern hemisphere, leading to tropical cyclones.
(credit: NASA)


6.5 Newton’s Universal Law of Gravitation
What do aching feet, a falling apple, and the orbit of the Moon have in common? Each is caused by the gravitational force. Our
feet are strained by supporting our weight—the force of Earth’s gravity on us. An apple falls from a tree because of the same
force acting a few meters above Earth’s surface. And the Moon orbits Earth because gravity is able to supply the necessary
centripetal force at a distance of hundreds of millions of meters. In fact, the same force causes planets to orbit the Sun, stars to
orbit the center of the galaxy, and galaxies to cluster together. Gravity is another example of underlying simplicity in nature. It is
the weakest of the four basic forces found in nature, and in some ways the least understood. It is a force that acts at a distance,
without physical contact, and is expressed by a formula that is valid everywhere in the universe, for masses and distances that
vary from the tiny to the immense.


Sir Isaac Newton was the first scientist to precisely define the gravitational force, and to show that it could explain both falling
bodies and astronomical motions. See Figure 6.20. But Newton was not the first to suspect that the same force caused both our
weight and the motion of planets. His forerunner Galileo Galilei had contended that falling bodies and planetary motions had the
same cause. Some of Newton’s contemporaries, such as Robert Hooke, Christopher Wren, and Edmund Halley, had also made
some progress toward understanding gravitation. But Newton was the first to propose an exact mathematical form and to use
that form to show that the motion of heavenly bodies should be conic sections—circles, ellipses, parabolas, and hyperbolas. This
theoretical prediction was a major triumph—it had been known for some time that moons, planets, and comets follow such paths,
but no one had been able to propose a mechanism that caused them to follow these paths and not others.


Chapter 6 | Uniform Circular Motion and Gravitation 215




Figure 6.20 According to early accounts, Newton was inspired to make the connection between falling bodies and astronomical motions when he saw
an apple fall from a tree and realized that if the gravitational force could extend above the ground to a tree, it might also reach the Sun. The inspiration
of Newton’s apple is a part of worldwide folklore and may even be based in fact. Great importance is attached to it because Newton’s universal law of
gravitation and his laws of motion answered very old questions about nature and gave tremendous support to the notion of underlying simplicity and
unity in nature. Scientists still expect underlying simplicity to emerge from their ongoing inquiries into nature.


The gravitational force is relatively simple. It is always attractive, and it depends only on the masses involved and the distance
between them. Stated in modern language, Newton’s universal law of gravitation states that every particle in the universe
attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses
and inversely proportional to the square of the distance between them.


Figure 6.21 Gravitational attraction is along a line joining the centers of mass of these two bodies. The magnitude of the force is the same on each,
consistent with Newton’s third law.


Misconception Alert


The magnitude of the force on each object (one has larger mass than the other) is the same, consistent with Newton’s third
law.


216 Chapter 6 | Uniform Circular Motion and Gravitation


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The bodies we are dealing with tend to be large. To simplify the situation we assume that the body acts as if its entire mass is
concentrated at one specific point called the center of mass (CM), which will be further explored in Linear Momentum and
Collisions. For two bodies having masses m and M with a distance r between their centers of mass, the equation for
Newton’s universal law of gravitation is


(6.40)F = GmM
r2
,


where F is the magnitude of the gravitational force and G is a proportionality factor called the gravitational constant. G is a
universal gravitational constant—that is, it is thought to be the same everywhere in the universe. It has been measured
experimentally to be


(6.41)
G = 6.673×10−11N ⋅ m


2


kg2


in SI units. Note that the units of G are such that a force in newtons is obtained from F = GmM
r2


, when considering masses in


kilograms and distance in meters. For example, two 1.000 kg masses separated by 1.000 m will experience a gravitational


attraction of 6.673×10−11 N . This is an extraordinarily small force. The small magnitude of the gravitational force is consistent
with everyday experience. We are unaware that even large objects like mountains exert gravitational forces on us. In fact, our


body weight is the force of attraction of the entire Earth on us with a mass of 6×1024 kg .


Recall that the acceleration due to gravity g is about 9.80 m/s2 on Earth. We can now determine why this is so. The weight of
an object mg is the gravitational force between it and Earth. Substituting mg for F in Newton’s universal law of gravitation gives


(6.42)mg = GmM
r2
,


where m is the mass of the object, M is the mass of Earth, and r is the distance to the center of Earth (the distance between
the centers of mass of the object and Earth). See Figure 6.22. The mass m of the object cancels, leaving an equation for g :


(6.43)g = GM
r2
.


Substituting known values for Earth’s mass and radius (to three significant figures),


(6.44)
g =



⎜6.67×10−11N ⋅ m


2


kg2



⎟× 5.98×10


24 kg
(6.38×106 m)2


,


and we obtain a value for the acceleration of a falling body:


(6.45)g = 9.80 m/s2.


Figure 6.22 The distance between the centers of mass of Earth and an object on its surface is very nearly the same as the radius of Earth, because
Earth is so much larger than the object.


This is the expected value and is independent of the body’s mass. Newton’s law of gravitation takes Galileo’s observation that all
masses fall with the same acceleration a step further, explaining the observation in terms of a force that causes objects to fall—in
fact, in terms of a universally existing force of attraction between masses.


Take-Home Experiment


Take a marble, a ball, and a spoon and drop them from the same height. Do they hit the floor at the same time? If you drop a
piece of paper as well, does it behave like the other objects? Explain your observations.


Chapter 6 | Uniform Circular Motion and Gravitation 217




Making Connections


Attempts are still being made to understand the gravitational force. As we shall see in Particle Physics, modern physics is
exploring the connections of gravity to other forces, space, and time. General relativity alters our view of gravitation, leading
us to think of gravitation as bending space and time.


In the following example, we make a comparison similar to one made by Newton himself. He noted that if the gravitational force
caused the Moon to orbit Earth, then the acceleration due to gravity should equal the centripetal acceleration of the Moon in its
orbit. Newton found that the two accelerations agreed “pretty nearly.”


Example 6.6 Earth’s Gravitational Force Is the Centripetal Force Making the Moon Move in a
Curved Path


(a) Find the acceleration due to Earth’s gravity at the distance of the Moon.


(b) Calculate the centripetal acceleration needed to keep the Moon in its orbit (assuming a circular orbit about a fixed Earth),
and compare it with the value of the acceleration due to Earth’s gravity that you have just found.


Strategy for (a)


This calculation is the same as the one finding the acceleration due to gravity at Earth’s surface, except that r is the


distance from the center of Earth to the center of the Moon. The radius of the Moon’s nearly circular orbit is 3.84×108 m .
Solution for (a)


Substituting known values into the expression for g found above, remembering that M is the mass of Earth not the Moon,


yields


(6.46)
g = GM


r2
=



⎜6.67×10−11N ⋅ m


2


kg2



⎟× 5.98×10


24 kg
(3.84×108 m)2


= 2.70×10−3 m/s.2


Strategy for (b)


Centripetal acceleration can be calculated using either form of


(6.47)


ac = v
2
r


ac = rω2





⎬.


We choose to use the second form:


(6.48)ac = rω2,


where ω is the angular velocity of the Moon about Earth.


Solution for (b)


Given that the period (the time it takes to make one complete rotation) of the Moon’s orbit is 27.3 days, (d) and using


(6.49)1 d×24hrd ×60
min
hr ×60


s
min = 86,400 s


we see that


(6.50)ω = ΔθΔt =
2π rad


(27.3 d)(86,400 s/d) = 2.66×10
−6rad


s .


The centripetal acceleration is


(6.51)ac = rω2 = (3.84×108 m)(2.66×10−6 rad/s)2


= 2.72×10−3 m/s.2


The direction of the acceleration is toward the center of the Earth.


Discussion


The centripetal acceleration of the Moon found in (b) differs by less than 1% from the acceleration due to Earth’s gravity
found in (a). This agreement is approximate because the Moon’s orbit is slightly elliptical, and Earth is not stationary (rather
the Earth-Moon system rotates about its center of mass, which is located some 1700 km below Earth’s surface). The clear
implication is that Earth’s gravitational force causes the Moon to orbit Earth.


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Why does Earth not remain stationary as the Moon orbits it? This is because, as expected from Newton’s third law, if Earth exerts
a force on the Moon, then the Moon should exert an equal and opposite force on Earth (see Figure 6.23). We do not sense the
Moon’s effect on Earth’s motion, because the Moon’s gravity moves our bodies right along with Earth but there are other signs on
Earth that clearly show the effect of the Moon’s gravitational force as discussed in Satellites and Kepler's Laws: An Argument
for Simplicity.


Figure 6.23 (a) Earth and the Moon rotate approximately once a month around their common center of mass. (b) Their center of mass orbits the Sun in
an elliptical orbit, but Earth’s path around the Sun has “wiggles” in it. Similar wiggles in the paths of stars have been observed and are considered
direct evidence of planets orbiting those stars. This is important because the planets’ reflected light is often too dim to be observed.


Tides
Ocean tides are one very observable result of the Moon’s gravity acting on Earth. Figure 6.24 is a simplified drawing of the
Moon’s position relative to the tides. Because water easily flows on Earth’s surface, a high tide is created on the side of Earth
nearest to the Moon, where the Moon’s gravitational pull is strongest. Why is there also a high tide on the opposite side of Earth?
The answer is that Earth is pulled toward the Moon more than the water on the far side, because Earth is closer to the Moon. So
the water on the side of Earth closest to the Moon is pulled away from Earth, and Earth is pulled away from water on the far side.
As Earth rotates, the tidal bulge (an effect of the tidal forces between an orbiting natural satellite and the primary planet that it
orbits) keeps its orientation with the Moon. Thus there are two tides per day (the actual tidal period is about 12 hours and 25.2
minutes), because the Moon moves in its orbit each day as well).


Figure 6.24 The Moon causes ocean tides by attracting the water on the near side more than Earth, and by attracting Earth more than the water on the
far side. The distances and sizes are not to scale. For this simplified representation of the Earth-Moon system, there are two high and two low tides per
day at any location, because Earth rotates under the tidal bulge.


The Sun also affects tides, although it has about half the effect of the Moon. However, the largest tides, called spring tides, occur
when Earth, the Moon, and the Sun are aligned. The smallest tides, called neap tides, occur when the Sun is at a 90º angle to
the Earth-Moon alignment.


Chapter 6 | Uniform Circular Motion and Gravitation 219




Figure 6.25 (a, b) Spring tides: The highest tides occur when Earth, the Moon, and the Sun are aligned. (c) Neap tide: The lowest tides occur when the
Sun lies at 90º to the Earth-Moon alignment. Note that this figure is not drawn to scale.


Tides are not unique to Earth but occur in many astronomical systems. The most extreme tides occur where the gravitational
force is the strongest and varies most rapidly, such as near black holes (see Figure 6.26). A few likely candidates for black holes
have been observed in our galaxy. These have masses greater than the Sun but have diameters only a few kilometers across.
The tidal forces near them are so great that they can actually tear matter from a companion star.


220 Chapter 6 | Uniform Circular Motion and Gravitation


Figure 6.26 A black hole is an object with such strong gravity that not even light can escape it. This black hole was created by the supernova of one
star in a two-star system. The tidal forces created by the black hole are so great that it tears matter from the companion star. This matter is
compressed and heated as it is sucked into the black hole, creating light and X-rays observable from Earth.


”Weightlessness” and Microgravity
In contrast to the tremendous gravitational force near black holes is the apparent gravitational field experienced by astronauts
orbiting Earth. What is the effect of “weightlessness” upon an astronaut who is in orbit for months? Or what about the effect of
weightlessness upon plant growth? Weightlessness doesn’t mean that an astronaut is not being acted upon by the gravitational
force. There is no “zero gravity” in an astronaut’s orbit. The term just means that the astronaut is in free-fall, accelerating with the
acceleration due to gravity. If an elevator cable breaks, the passengers inside will be in free fall and will experience
weightlessness. You can experience short periods of weightlessness in some rides in amusement parks.


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Figure 6.27 Astronauts experiencing weightlessness on board the International Space Station. (credit: NASA)


Microgravity refers to an environment in which the apparent net acceleration of a body is small compared with that produced by
Earth at its surface. Many interesting biology and physics topics have been studied over the past three decades in the presence
of microgravity. Of immediate concern is the effect on astronauts of extended times in outer space, such as at the International
Space Station. Researchers have observed that muscles will atrophy (waste away) in this environment. There is also a
corresponding loss of bone mass. Study continues on cardiovascular adaptation to space flight. On Earth, blood pressure is
usually higher in the feet than in the head, because the higher column of blood exerts a downward force on it, due to gravity.
When standing, 70% of your blood is below the level of the heart, while in a horizontal position, just the opposite occurs. What
difference does the absence of this pressure differential have upon the heart?


Some findings in human physiology in space can be clinically important to the management of diseases back on Earth. On a
somewhat negative note, spaceflight is known to affect the human immune system, possibly making the crew members more
vulnerable to infectious diseases. Experiments flown in space also have shown that some bacteria grow faster in microgravity
than they do on Earth. However, on a positive note, studies indicate that microbial antibiotic production can increase by a factor
of two in space-grown cultures. One hopes to be able to understand these mechanisms so that similar successes can be
achieved on the ground. In another area of physics space research, inorganic crystals and protein crystals have been grown in
outer space that have much higher quality than any grown on Earth, so crystallography studies on their structure can yield much
better results.


Plants have evolved with the stimulus of gravity and with gravity sensors. Roots grow downward and shoots grow upward. Plants
might be able to provide a life support system for long duration space missions by regenerating the atmosphere, purifying water,
and producing food. Some studies have indicated that plant growth and development are not affected by gravity, but there is still
uncertainty about structural changes in plants grown in a microgravity environment.


The Cavendish Experiment: Then and Now


As previously noted, the universal gravitational constant G is determined experimentally. This definition was first done
accurately by Henry Cavendish (1731–1810), an English scientist, in 1798, more than 100 years after Newton published his
universal law of gravitation. The measurement of G is very basic and important because it determines the strength of one of the
four forces in nature. Cavendish’s experiment was very difficult because he measured the tiny gravitational attraction between
two ordinary-sized masses (tens of kilograms at most), using apparatus like that in Figure 6.28. Remarkably, his value for G
differs by less than 1% from the best modern value.


One important consequence of knowing G was that an accurate value for Earth’s mass could finally be obtained. This was done
by measuring the acceleration due to gravity as accurately as possible and then calculating the mass of Earth M from the
relationship Newton’s universal law of gravitation gives


(6.52)mg = GmM
r2
,


where m is the mass of the object, M is the mass of Earth, and r is the distance to the center of Earth (the distance between
the centers of mass of the object and Earth). See Figure 6.21. The mass m of the object cancels, leaving an equation for g :


(6.53)g = GM
r2
.


Rearranging to solve for M yields


(6.54)
M = gr


2


G
.


Chapter 6 | Uniform Circular Motion and Gravitation 221




So M can be calculated because all quantities on the right, including the radius of Earth r , are known from direct
measurements. We shall see in Satellites and Kepler's Laws: An Argument for Simplicity that knowing G also allows for the
determination of astronomical masses. Interestingly, of all the fundamental constants in physics, G is by far the least well
determined.


The Cavendish experiment is also used to explore other aspects of gravity. One of the most interesting questions is whether the
gravitational force depends on substance as well as mass—for example, whether one kilogram of lead exerts the same
gravitational pull as one kilogram of water. A Hungarian scientist named Roland von Eötvös pioneered this inquiry early in the
20th century. He found, with an accuracy of five parts per billion, that the gravitational force does not depend on the substance.
Such experiments continue today, and have improved upon Eötvös’ measurements. Cavendish-type experiments such as those
of Eric Adelberger and others at the University of Washington, have also put severe limits on the possibility of a fifth force and
have verified a major prediction of general relativity—that gravitational energy contributes to rest mass. Ongoing measurements
there use a torsion balance and a parallel plate (not spheres, as Cavendish used) to examine how Newton’s law of gravitation
works over sub-millimeter distances. On this small-scale, do gravitational effects depart from the inverse square law? So far, no
deviation has been observed.


Figure 6.28 Cavendish used an apparatus like this to measure the gravitational attraction between the two suspended spheres (m ) and the two on
the stand (M ) by observing the amount of torsion (twisting) created in the fiber. Distance between the masses can be varied to check the
dependence of the force on distance. Modern experiments of this type continue to explore gravity.


6.6 Satellites and Kepler’s Laws: An Argument for Simplicity
Examples of gravitational orbits abound. Hundreds of artificial satellites orbit Earth together with thousands of pieces of debris.
The Moon’s orbit about Earth has intrigued humans from time immemorial. The orbits of planets, asteroids, meteors, and comets
about the Sun are no less interesting. If we look further, we see almost unimaginable numbers of stars, galaxies, and other
celestial objects orbiting one another and interacting through gravity.


All these motions are governed by gravitational force, and it is possible to describe them to various degrees of precision. Precise
descriptions of complex systems must be made with large computers. However, we can describe an important class of orbits
without the use of computers, and we shall find it instructive to study them. These orbits have the following characteristics:


1. A small mass m orbits a much larger mass M . This allows us to view the motion as if M were stationary—in fact, as if
from an inertial frame of reference placed on M —without significant error. Mass m is the satellite of M , if the orbit is
gravitationally bound.


2. The system is isolated from other masses. This allows us to neglect any small effects due to outside masses.


222 Chapter 6 | Uniform Circular Motion and Gravitation


The conditions are satisfied, to good approximation, by Earth’s satellites (including the Moon), by objects orbiting the Sun, and by
the satellites of other planets. Historically, planets were studied first, and there is a classical set of three laws, called Kepler’s
laws of planetary motion, that describe the orbits of all bodies satisfying the two previous conditions (not just planets in our solar
system). These descriptive laws are named for the German astronomer Johannes Kepler (1571–1630), who devised them after
careful study (over some 20 years) of a large amount of meticulously recorded observations of planetary motion done by Tycho
Brahe (1546–1601). Such careful collection and detailed recording of methods and data are hallmarks of good science. Data
constitute the evidence from which new interpretations and meanings can be constructed.


Kepler’s Laws of Planetary Motion
Kepler’s First Law


The orbit of each planet about the Sun is an ellipse with the Sun at one focus.


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Figure 6.29 (a) An ellipse is a closed curve such that the sum of the distances from a point on the curve to the two foci ( f1 and f2 ) is a constant.
You can draw an ellipse as shown by putting a pin at each focus, and then placing a string around a pencil and the pins and tracing a line on paper. A
circle is a special case of an ellipse in which the two foci coincide (thus any point on the circle is the same distance from the center). (b) For any closed
gravitational orbit, m follows an elliptical path with M at one focus. Kepler’s first law states this fact for planets orbiting the Sun.


Kepler’s Second Law


Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times (see
Figure 6.30).


Kepler’s Third Law


The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average
distances from the Sun. In equation form, this is


(6.55)T1
2


T2
2 =


r1
3


r2
3,


where T is the period (time for one orbit) and r is the average radius. This equation is valid only for comparing two small
masses orbiting the same large one. Most importantly, this is a descriptive equation only, giving no information as to the cause of
the equality.


Chapter 6 | Uniform Circular Motion and Gravitation 223




Figure 6.30 The shaded regions have equal areas. It takes equal times for m to go from A to B, from C to D, and from E to F. The mass m moves
fastest when it is closest to M . Kepler’s second law was originally devised for planets orbiting the Sun, but it has broader validity.


Note again that while, for historical reasons, Kepler’s laws are stated for planets orbiting the Sun, they are actually valid for all
bodies satisfying the two previously stated conditions.


Example 6.7 Find the Time for One Orbit of an Earth Satellite


Given that the Moon orbits Earth each 27.3 d and that it is an average distance of 3.84×108 m from the center of Earth,
calculate the period of an artificial satellite orbiting at an average altitude of 1500 km above Earth’s surface.


Strategy


The period, or time for one orbit, is related to the radius of the orbit by Kepler’s third law, given in mathematical form in


T1
2


T2
2 =


r1
3


r2
3 . Let us use the subscript 1 for the Moon and the subscript 2 for the satellite. We are asked to find T2 . The


given information tells us that the orbital radius of the Moon is r1 = 3.84×10
8 m , and that the period of the Moon is


T1 = 27.3 d . The height of the artificial satellite above Earth’s surface is given, and so we must add the radius of Earth
(6380 km) to get r2 = (1500 + 6380) km = 7880 km . Now all quantities are known, and so T2 can be found.


Solution


Kepler’s third law is


(6.56)T1
2


T2
2 =


r1
3


r2
3.


To solve for T2 , we cross-multiply and take the square root, yielding


(6.57)
T2
2 = T1


2 ⎛

r2
r1



3


(6.58)
T2 = T1




r2
r1



3 / 2
.


Substituting known values yields


(6.59)


T2 = 27.3 d×24.0 hd ×



7880 km
3.84×105 km





3 / 2


= 1.93 h.


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Discussion This is a reasonable period for a satellite in a fairly low orbit. It is interesting that any satellite at this altitude will
orbit in the same amount of time. This fact is related to the condition that the satellite’s mass is small compared with that of
Earth.


People immediately search for deeper meaning when broadly applicable laws, like Kepler’s, are discovered. It was Newton who
took the next giant step when he proposed the law of universal gravitation. While Kepler was able to discover what was
happening, Newton discovered that gravitational force was the cause.


Derivation of Kepler’s Third Law for Circular Orbits
We shall derive Kepler’s third law, starting with Newton’s laws of motion and his universal law of gravitation. The point is to
demonstrate that the force of gravity is the cause for Kepler’s laws (although we will only derive the third one).


Let us consider a circular orbit of a small mass m around a large mass M , satisfying the two conditions stated at the beginning
of this section. Gravity supplies the centripetal force to mass m . Starting with Newton’s second law applied to circular motion,


(6.60)
Fnet = mac = mv


2
r .


The net external force on mass m is gravity, and so we substitute the force of gravity for Fnet :


(6.61)
GmM


r2
= mv


2
r .


The mass m cancels, yielding


(6.62)GMr = v
2.


The fact that m cancels out is another aspect of the oft-noted fact that at a given location all masses fall with the same
acceleration. Here we see that at a given orbital radius r , all masses orbit at the same speed. (This was implied by the result of
the preceding worked example.) Now, to get at Kepler’s third law, we must get the period T into the equation. By definition,
period T is the time for one complete orbit. Now the average speed v is the circumference divided by the period—that is,


(6.63)v = 2πr
T
.


Substituting this into the previous equation gives


(6.64)
GMr =


4π2 r2
T 2


.


Solving for T 2 yields


(6.65)
T 2 = 4π


2


GM
r3.


Using subscripts 1 and 2 to denote two different satellites, and taking the ratio of the last equation for satellite 1 to satellite 2
yields


(6.66)T1
2


T2
2 =


r1
3


r2
3.


This is Kepler’s third law. Note that Kepler’s third law is valid only for comparing satellites of the same parent body, because only
then does the mass of the parent body M cancel.


Now consider what we get if we solve T 2 = 4π
2


GM
r3 for the ratio r3 / T 2 . We obtain a relationship that can be used to


determine the mass M of a parent body from the orbits of its satellites:


(6.67)r3


T 2
= G
4π2


M.


If r and T are known for a satellite, then the mass M of the parent can be calculated. This principle has been used


extensively to find the masses of heavenly bodies that have satellites. Furthermore, the ratio r3 / T 2 should be a constant for all
satellites of the same parent body (because r3 / T 2 = GM / 4π2 ). (See Table 6.2).


Chapter 6 | Uniform Circular Motion and Gravitation 225




It is clear from Table 6.2 that the ratio of r3 / T 2 is constant, at least to the third digit, for all listed satellites of the Sun, and for
those of Jupiter. Small variations in that ratio have two causes—uncertainties in the r and T data, and perturbations of the
orbits due to other bodies. Interestingly, those perturbations can be—and have been—used to predict the location of new planets
and moons. This is another verification of Newton’s universal law of gravitation.


Making Connections


Newton’s universal law of gravitation is modified by Einstein’s general theory of relativity, as we shall see in Particle
Physics. Newton’s gravity is not seriously in error—it was and still is an extremely good approximation for most situations.
Einstein’s modification is most noticeable in extremely large gravitational fields, such as near black holes. However, general
relativity also explains such phenomena as small but long-known deviations of the orbit of the planet Mercury from classical
predictions.


The Case for Simplicity
The development of the universal law of gravitation by Newton played a pivotal role in the history of ideas. While it is beyond the
scope of this text to cover that history in any detail, we note some important points. The definition of planet set in 2006 by the
International Astronomical Union (IAU) states that in the solar system, a planet is a celestial body that:


1. is in orbit around the Sun,


2. has sufficient mass to assume hydrostatic equilibrium and


3. has cleared the neighborhood around its orbit.


A non-satellite body fulfilling only the first two of the above criteria is classified as “dwarf planet.”


In 2006, Pluto was demoted to a ‘dwarf planet’ after scientists revised their definition of what constitutes a “true” planet.


Table 6.2 Orbital Data and Kepler’s Third Law


Parent Satellite Average orbital radius r(km) Period T(y) r3 / T2 (km3 / y2)


Earth Moon 3.84×105 0.07481 1.01×1018


Sun Mercury 5.79×107 0.2409 3.34×1024


Venus 1.082×108 0.6150 3.35×1024


Earth 1.496×108 1.000 3.35×1024


Mars 2.279×108 1.881 3.35×1024


Jupiter 7.783×108 11.86 3.35×1024


Saturn 1.427×109 29.46 3.35×1024


Neptune 4.497×109 164.8 3.35×1024


Pluto 5.90×109 248.3 3.33×1024


Jupiter Io 4.22×105 0.00485 (1.77 d) 3.19×1021


Europa 6.71×105 0.00972 (3.55 d) 3.20×1021


Ganymede 1.07×106 0.0196 (7.16 d) 3.19×1021


Callisto 1.88×106 0.0457 (16.19 d) 3.20×1021


226 Chapter 6 | Uniform Circular Motion and Gravitation


The universal law of gravitation is a good example of a physical principle that is very broadly applicable. That single equation for
the gravitational force describes all situations in which gravity acts. It gives a cause for a vast number of effects, such as the
orbits of the planets and moons in the solar system. It epitomizes the underlying unity and simplicity of physics.


Before the discoveries of Kepler, Copernicus, Galileo, Newton, and others, the solar system was thought to revolve around Earth
as shown in Figure 6.31(a). This is called the Ptolemaic view, for the Greek philosopher who lived in the second century AD.
This model is characterized by a list of facts for the motions of planets with no cause and effect explanation. There tended to be
a different rule for each heavenly body and a general lack of simplicity.


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angular velocity:


arc length:


banked curve:


center of mass:


centrifugal force:


centripetal acceleration:


centripetal force:


Coriolis force:


fictitious force:


gravitational constant, G:


ideal angle:


ideal banking:


ideal speed:


microgravity:


Newton’s universal law of gravitation:


non-inertial frame of reference:


pit:


Figure 6.31(b) represents the modern or Copernican model. In this model, a small set of rules and a single underlying force
explain not only all motions in the solar system, but all other situations involving gravity. The breadth and simplicity of the laws of
physics are compelling. As our knowledge of nature has grown, the basic simplicity of its laws has become ever more evident.


Figure 6.31 (a) The Ptolemaic model of the universe has Earth at the center with the Moon, the planets, the Sun, and the stars revolving about it in
complex superpositions of circular paths. This geocentric model, which can be made progressively more accurate by adding more circles, is purely
descriptive, containing no hints as to what are the causes of these motions. (b) The Copernican model has the Sun at the center of the solar system. It
is fully explained by a small number of laws of physics, including Newton’s universal law of gravitation.


Glossary
ω , the rate of change of the angle with which an object moves on a circular path


Δs , the distance traveled by an object along a circular path


the curve in a road that is sloping in a manner that helps a vehicle negotiate the curve


the point where the entire mass of an object can be thought to be concentrated


a fictitious force that tends to throw an object off when the object is rotating in a non-inertial frame of
reference


the acceleration of an object moving in a circle, directed toward the center


any net force causing uniform circular motion


the fictitious force causing the apparent deflection of moving objects when viewed in a rotating frame of
reference


a force having no physical origin


a proportionality factor used in the equation for Newton’s universal law of gravitation; it is a
universal constant—that is, it is thought to be the same everywhere in the universe


the angle at which a car can turn safely on a steep curve, which is in proportion to the ideal speed


the sloping of a curve in a road, where the angle of the slope allows the vehicle to negotiate the curve at a
certain speed without the aid of friction between the tires and the road; the net external force on the vehicle equals the
horizontal centripetal force in the absence of friction


the maximum safe speed at which a vehicle can turn on a curve without the aid of friction between the tire and
the road


an environment in which the apparent net acceleration of a body is small compared with that produced by Earth
at its surface


every particle in the universe attracts every other particle with a force along a line
joining them; the force is directly proportional to the product of their masses and inversely proportional to the square of
the distance between them


an accelerated frame of reference


a tiny indentation on the spiral track moulded into the top of the polycarbonate layer of CD


Chapter 6 | Uniform Circular Motion and Gravitation 227




radians:


radius of curvature:


rotation angle:


ultracentrifuge:


uniform circular motion:


a unit of angle measurement


radius of a circular path


the ratio of the arc length to the radius of curvature on a circular path:


Δθ = Δsr


a centrifuge optimized for spinning a rotor at very high speeds


the motion of an object in a circular path at constant speed


Section Summary


6.1 Rotation Angle and Angular Velocity
• Uniform circular motion is motion in a circle at constant speed. The rotation angle Δθ is defined as the ratio of the arc


length to the radius of curvature:


Δθ = Δsr ,
where arc length Δs is distance traveled along a circular path and r is the radius of curvature of the circular path. The
quantity Δθ is measured in units of radians (rad), for which


2π rad = 360º= 1 revolution.
• The conversion between radians and degrees is 1 rad = 57.3º .
• Angular velocity ω is the rate of change of an angle,


ω = ΔθΔt ,


where a rotation Δθ takes place in a time Δt . The units of angular velocity are radians per second (rad/s). Linear velocity
v and angular velocity ω are related by


v = rω or ω = vr .


6.2 Centripetal Acceleration
• Centripetal acceleration ac is the acceleration experienced while in uniform circular motion. It always points toward the


center of rotation. It is perpendicular to the linear velocity v and has the magnitude


ac = v
2
r ; ac = rω


2.
• The unit of centripetal acceleration is m / s2 .


6.3 Centripetal Force
• Centripetal force Fc is any force causing uniform circular motion. It is a “center-seeking” force that always points toward


the center of rotation. It is perpendicular to linear velocity v and has magnitude
Fc = mac,


which can also be expressed as


Fc = mv
2
r


or
Fc = mrω2


,











6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
• Rotating and accelerated frames of reference are non-inertial.
• Fictitious forces, such as the Coriolis force, are needed to explain motion in such frames.


6.5 Newton’s Universal Law of Gravitation
• Newton’s universal law of gravitation: Every particle in the universe attracts every other particle with a force along a line


joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the
distance between them. In equation form, this is


F = GmM
r2
,


228 Chapter 6 | Uniform Circular Motion and Gravitation


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where F is the magnitude of the gravitational force. G is the gravitational constant, given by


G = 6.673×10–11 N ⋅ m2/kg2 .
• Newton’s law of gravitation applies universally.


6.6 Satellites and Kepler’s Laws: An Argument for Simplicity
• Kepler’s laws are stated for a small mass m orbiting a larger mass M in near-isolation. Kepler’s laws of planetary motion


are then as follows:
Kepler’s first law


The orbit of each planet about the Sun is an ellipse with the Sun at one focus.


Kepler’s second law


Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times.


Kepler’s third law


The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average
distances from the Sun:


T1
2


T2
2 =


r1
3


r2
3,


where T is the period (time for one orbit) and r is the average radius of the orbit.
• The period and radius of a satellite’s orbit about a larger body M are related by


T 2 = 4π
2


GM
r3


or


r3


T 2
= G
4π2


M.


Conceptual Questions


6.1 Rotation Angle and Angular Velocity
1. There is an analogy between rotational and linear physical quantities. What rotational quantities are analogous to distance and
velocity?


6.2 Centripetal Acceleration
2. Can centripetal acceleration change the speed of circular motion? Explain.


6.3 Centripetal Force
3. If you wish to reduce the stress (which is related to centripetal force) on high-speed tires, would you use large- or small-
diameter tires? Explain.


4. Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and so on) be a centripetal
force? Can any combination of forces be a centripetal force?


5. If centripetal force is directed toward the center, why do you feel that you are ‘thrown’ away from the center as a car goes
around a curve? Explain.


6. Race car drivers routinely cut corners as shown in Figure 6.32. Explain how this allows the curve to be taken at the greatest
speed.


Chapter 6 | Uniform Circular Motion and Gravitation 229




Figure 6.32 Two paths around a race track curve are shown. Race car drivers will take the inside path (called cutting the corner) whenever possible
because it allows them to take the curve at the highest speed.


7. A number of amusement parks have rides that make vertical loops like the one shown in Figure 6.33. For safety, the cars are
attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will
supply the centripetal force. What other force acts 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?


Figure 6.33 Amusement rides with a vertical loop are an example of a form of curved motion.


8.What is the direction of the force exerted by the car on the passenger as the car goes over the top of the amusement ride
pictured in Figure 6.33 under the following circumstances:


(a) The car goes over the top at such a speed that the gravitational force is the only force acting?


(b) The car goes over the top faster than this speed?


(c) The car goes over the top slower than this speed?


9. As a skater forms a circle, what force is responsible for making her turn? Use a free body diagram in your answer.


10. Suppose a child is riding on a merry-go-round at a distance about halfway between its center and edge. She has a lunch box
resting on wax paper, so that there is very little friction between it and the merry-go-round. Which path shown in Figure 6.34 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 to the right? Explain your answer.


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Figure 6.34 A child riding on a merry-go-round releases her lunch box at point P. This is a view from above the clockwise rotation. Assuming it slides
with negligible friction, will it follow path A, B, or C, as viewed from Earth’s frame of reference? What will be the shape of the path it leaves in the dust
on the merry-go-round?


11. Do you feel yourself thrown to either side when you negotiate a curve that is ideally banked for your car’s speed? What is the
direction of the force exerted on you by the car seat?


12. Suppose a mass is moving in a circular path on a frictionless table as shown in figure. In the Earth’s frame of reference, there
is no centrifugal force pulling the mass away from the centre of rotation, yet there is a very real force stretching the string
attaching the mass to the nail. Using concepts related to centripetal force and Newton’s third law, explain what force stretches
the string, identifying its physical origin.


Figure 6.35 A mass attached to a nail on a frictionless table moves in a circular path. The force stretching the string is real and not fictional. What is
the physical origin of the force on the string?


6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
13.When a toilet is flushed or a sink is drained, the water (and other material) begins to rotate about the drain on the way down.
Assuming no initial rotation and a flow initially directly straight toward the drain, explain what causes the rotation and which
direction it has in the northern hemisphere. (Note that this is a small effect and in most toilets the rotation is caused by directional
water jets.) Would the direction of rotation reverse if water were forced up the drain?


14. Is there a real force that throws water from clothes during the spin cycle of a washing machine? Explain how the water is
removed.


15. In one amusement park ride, riders enter a large vertical barrel and stand against the wall on its horizontal floor. The barrel is
spun up and the floor drops away. Riders feel as if they are pinned to the wall by a force something like the gravitational force.
This is a fictitious force sensed and used by the riders to explain events in the rotating frame of reference of the barrel. Explain in
an inertial frame of reference (Earth is nearly one) what pins the riders to the wall, and identify all of the real forces acting on
them.


Chapter 6 | Uniform Circular Motion and Gravitation 231




16. Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the
ultimate determinant of the truth in physics, and why was this action ultimately accepted?


17. Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward


Earth. Tom says a satellite in orbit is not in freefall because the acceleration due to gravity is not 9.80 m/s2 . Who do you agree
with and why?


18. A non-rotating frame of reference placed at the center of the Sun is very nearly an inertial one. Why is it not exactly an inertial
frame?


6.5 Newton’s Universal Law of Gravitation
19. Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the
ultimate determinant of the truth in physics, and why was this action ultimately accepted?


20. Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward


Earth. Tom says a satellite in orbit is not in freefall because the acceleration due to gravity is not 9.80 m/s2 . Who do you agree
with and why?


21. Draw a free body diagram for a satellite in an elliptical orbit showing why its speed increases as it approaches its parent body
and decreases as it moves away.


22. Newton’s laws of motion and gravity were among the first to convincingly demonstrate the underlying simplicity and unity in
nature. Many other examples have since been discovered, and we now expect to find such underlying order in complex
situations. Is there proof that such order will always be found in new explorations?


6.6 Satellites and Kepler’s Laws: An Argument for Simplicity
23. In what frame(s) of reference are Kepler’s laws valid? Are Kepler’s laws purely descriptive, or do they contain causal
information?


232 Chapter 6 | Uniform Circular Motion and Gravitation


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Problems & Exercises


6.1 Rotation Angle and Angular Velocity
1. Semi-trailer trucks have an odometer on one hub of a
trailer wheel. The hub is weighted so that it does not rotate,
but it contains gears to count the number of wheel
revolutions—it then calculates the distance traveled. If the
wheel has a 1.15 m diameter and goes through 200,000
rotations, how many kilometers should the odometer read?


2. Microwave ovens rotate at a rate of about 6 rev/min. What
is this in revolutions per second? What is the angular velocity
in radians per second?


3. An automobile with 0.260 m radius tires travels 80,000 km
before wearing them out. How many revolutions do the tires
make, neglecting any backing up and any change in radius
due to wear?


4. (a) What is the period of rotation of Earth in seconds? (b)
What is the angular velocity of Earth? (c) Given that Earth has


a radius of 6.4×106 m at its equator, what is the linear
velocity at Earth’s surface?


5. A baseball pitcher brings his arm forward during a pitch,
rotating the forearm about the elbow. If the velocity of the ball
in the pitcher’s hand is 35.0 m/s and the ball is 0.300 m from
the elbow joint, what is the angular velocity of the forearm?


6. In lacrosse, a ball is thrown from a net on the end of a stick
by rotating the stick and forearm about the elbow. If the
angular velocity of the ball about the elbow joint is 30.0 rad/s
and the ball is 1.30 m from the elbow joint, what is the velocity
of the ball?


7. A truck with 0.420-m-radius tires travels at 32.0 m/s. What
is the angular velocity of the rotating tires in radians per
second? What is this in rev/min?


8. Integrated ConceptsWhen kicking a football, the kicker
rotates his leg about the hip joint.


(a) If the velocity of the tip of the kicker’s shoe is 35.0 m/s and
the hip joint is 1.05 m from the tip of the shoe, what is the
shoe tip’s angular velocity?


(b) The shoe is in contact with the initially stationary 0.500 kg
football for 20.0 ms. What average force is exerted on the
football to give it a velocity of 20.0 m/s?


(c) Find the maximum range of the football, neglecting air
resistance.


9. Construct Your Own Problem


Consider an amusement park ride in which participants are
rotated about a vertical axis in a cylinder with vertical walls.
Once the angular velocity reaches its full value, the floor
drops away and friction between the walls and the riders
prevents them from sliding down. Construct a problem in
which you calculate the necessary angular velocity that
assures the riders will not slide down the wall. Include a free
body diagram of a single rider. Among the variables to
consider are the radius of the cylinder and the coefficients of
friction between the riders’ clothing and the wall.


6.2 Centripetal Acceleration
10. A fairground ride spins its occupants inside a flying
saucer-shaped container. If the horizontal circular path the
riders follow has an 8.00 m radius, at how many revolutions
per minute will the riders be subjected to a centripetal
acceleration whose magnitude is 1.50 times that due to
gravity?


11. A runner taking part in the 200 m dash must run around
the end of a track that has a circular arc with a radius of
curvature of 30 m. If he completes the 200 m dash in 23.2 s
and runs at constant speed throughout the race, what is the
magnitude of his centripetal acceleration as he runs the
curved portion of the track?


12. Taking the age of Earth to be about 4×109 years and
assuming its orbital radius of 1.5 ×1011 has not changed
and is circular, calculate the approximate total distance Earth
has traveled since its birth (in a frame of reference stationary
with respect to the Sun).


13. The propeller of a World War II fighter plane is 2.30 m in
diameter.


(a) What is its angular velocity in radians per second if it spins
at 1200 rev/min?


(b) What is the linear speed of its tip at this angular velocity if
the plane is stationary on the tarmac?


(c) What is the centripetal acceleration of the propeller tip
under these conditions? Calculate it in meters per second
squared and convert to multiples of g .


14. An ordinary workshop grindstone has a radius of 7.50 cm
and rotates at 6500 rev/min.


(a) Calculate the magnitude of the centripetal acceleration at
its edge in meters per second squared and convert it to
multiples of g .


(b) What is the linear speed of a point on its edge?


15. Helicopter blades withstand tremendous stresses. In
addition to supporting the weight of a helicopter, they are
spun at rapid rates and experience large centripetal
accelerations, especially at the tip.


(a) Calculate the magnitude of the centripetal acceleration at
the tip of a 4.00 m long helicopter blade that rotates at 300
rev/min.


(b) Compare the linear speed of the tip with the speed of
sound (taken to be 340 m/s).


16. Olympic ice skaters are able to spin at about 5 rev/s.


(a) What is their angular velocity in radians per second?


(b) What is the centripetal acceleration of the skater’s nose if
it is 0.120 m from the axis of rotation?


(c) An exceptional skater named Dick Button was able to spin
much faster in the 1950s than anyone since—at about 9 rev/
s. What was the centripetal acceleration of the tip of his nose,
assuming it is at 0.120 m radius?


(d) Comment on the magnitudes of the accelerations found. It
is reputed that Button ruptured small blood vessels during his
spins.


17.What percentage of the acceleration at Earth’s surface is
the acceleration due to gravity at the position of a satellite
located 300 km above Earth?


18. Verify that the linear speed of an ultracentrifuge is about
0.50 km/s, and Earth in its orbit is about 30 km/s by
calculating:


(a) The linear speed of a point on an ultracentrifuge 0.100 m
from its center, rotating at 50,000 rev/min.


(b) The linear speed of Earth in its orbit about the Sun (use
data from the text on the radius of Earth’s orbit and
approximate it as being circular).


Chapter 6 | Uniform Circular Motion and Gravitation 233




19. A rotating space station is said to create “artificial
gravity”—a loosely-defined term used for an acceleration that
would be crudely similar to gravity. The outer wall of the
rotating space station would become a floor for the
astronauts, and centripetal acceleration supplied by the floor
would allow astronauts to exercise and maintain muscle and
bone strength more naturally than in non-rotating space
environments. If the space station is 200 m in diameter, what
angular velocity would produce an “artificial gravity” of


9.80 m/s2 at the rim?
20. At takeoff, a commercial jet has a 60.0 m/s speed. Its tires
have a diameter of 0.850 m.


(a) At how many rev/min are the tires rotating?


(b) What is the centripetal acceleration at the edge of the tire?


(c) With what force must a determined 1.00×10−15 kg
bacterium cling to the rim?


(d) Take the ratio of this force to the bacterium’s weight.


21. Integrated Concepts


Riders in an amusement park ride shaped like a Viking ship
hung from a large pivot are rotated back and forth like a rigid
pendulum. Sometime near the middle of the ride, the ship is
momentarily motionless at the top of its circular arc. The ship
then swings down under the influence of gravity.


(a) Assuming negligible friction, find the speed of the riders at
the bottom of its arc, given the system's center of mass
travels in an arc having a radius of 14.0 m and the riders are
near the center of mass.


(b) What is the centripetal acceleration at the bottom of the
arc?


(c) Draw a free body diagram of the forces acting on a rider at
the bottom of the arc.


(d) Find the force exerted by the ride on a 60.0 kg rider and
compare it to her weight.


(e) Discuss whether the answer seems reasonable.


22. Unreasonable Results


A mother pushes her child on a swing so that his speed is
9.00 m/s at the lowest point of his path. The swing is
suspended 2.00 m above the child’s center of mass.


(a) What is the magnitude of the centripetal acceleration of
the child at the low point?


(b) What is the magnitude of the force the child exerts on the
seat if his mass is 18.0 kg?


(c) What is unreasonable about these results?


(d) Which premises are unreasonable or inconsistent?


6.3 Centripetal Force
23. (a) A 22.0 kg child is riding a playground merry-go-round
that is rotating at 40.0 rev/min. What centripetal force must
she exert to stay on if she is 1.25 m from its center?


(b) What centripetal force does she need to stay on an
amusement park merry-go-round that rotates at 3.00 rev/min
if she is 8.00 m from its center?


(c) Compare each force with her weight.


24. 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.


25.What is the ideal banking angle for a gentle turn of 1.20
km radius on a highway with a 105 km/h speed limit (about 65
mi/h), assuming everyone travels at the limit?


26.What is the ideal speed to take a 100 m radius curve
banked at a 20.0° angle?


27. (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 this acceleration seem large to you?


28. Part of riding a bicycle involves leaning at the correct
angle when making a turn, as seen in Figure 6.36. To be
stable, the force exerted by the ground must be on a line
going through the center of gravity. The force on the bicycle
wheel can be resolved into two perpendicular
components—friction parallel to the road (this must supply the
centripetal force), and the vertical normal force (which must
equal the system’s weight).


(a) Show that θ (as defined in the figure) is related to the
speed v and radius of curvature r of the turn in 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).


Figure 6.36 A bicyclist negotiating a turn on level ground must lean at
the correct angle—the ability to do this becomes instinctive. The force of
the ground on the wheel needs to be on a line through the center of
gravity. The net external force on the system is the centripetal force. The
vertical component of the force on the wheel cancels the weight of the
system while its horizontal component must supply the centripetal force.


This process produces a relationship among the angle θ , the speed v
, and the radius of curvature r of the turn similar to that for the ideal
banking of roadways.


29. A large centrifuge, like the one shown in Figure 6.37(a),
is used to expose aspiring astronauts to accelerations similar
to those experienced in rocket launches and atmospheric
reentries.


(a) At what angular velocity is the centripetal acceleration
10 g if the rider is 15.0 m from the center of rotation?


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(b) The rider’s cage hangs on a pivot at the end of the arm,
allowing it to swing outward during rotation as shown in
Figure 6.37(b). At what angle θ below the horizontal will the
cage hang when the centripetal acceleration is 10 g ? (Hint:
The arm supplies centripetal force and supports the weight of
the cage. Draw a free body diagram of the forces to see what
the angle θ should be.)


Figure 6.37 (a) NASA centrifuge used to subject trainees to
accelerations similar to those experienced in rocket launches and
reentries. (credit: NASA) (b) Rider in cage showing how the cage pivots
outward during rotation. This allows the total force exerted on the rider
by the cage to be along its axis at all times.


30. Integrated Concepts


If a car takes a banked curve at less than the ideal speed,
friction is needed to keep it from sliding toward the inside of
the curve (a real problem on icy mountain roads). (a)
Calculate the ideal speed to take a 100 m radius curve
banked at 15.0º. (b) What is the minimum coefficient of
friction needed for a frightened driver to take the same curve
at 20.0 km/h?


31. Modern roller coasters have vertical loops like the one
shown in Figure 6.38. The radius of curvature is smaller at
the top than on the sides so that the downward centripetal
acceleration at the top will be greater than the acceleration
due to gravity, keeping the passengers pressed firmly into
their seats. What is the speed of the roller coaster at the top
of the loop if the radius of curvature there is 15.0 m and the
downward acceleration of the car is 1.50 g?


Figure 6.38 Teardrop-shaped loops are used in the latest roller coasters
so that the radius of curvature gradually decreases to a minimum at the
top. This means that the centripetal acceleration builds from zero to a
maximum at the top and gradually decreases again. A circular loop
would cause a jolting change in acceleration at entry, a disadvantage
discovered long ago in railroad curve design. With a small radius of
curvature at the top, the centripetal acceleration can more easily be kept
greater than g so that the passengers do not lose contact with their
seats nor do they need seat belts to keep them in place.


32. Unreasonable Results


(a) Calculate the minimum coefficient of friction needed for a
car to negotiate an unbanked 50.0 m radius curve at 30.0 m/
s.


(b) What is unreasonable about the result?


(c) Which premises are unreasonable or inconsistent?


6.5 Newton’s Universal Law of Gravitation
33. (a) Calculate Earth’s mass given the acceleration due to


gravity at the North Pole is 9.830 m/s2 and the radius of the
Earth is 6371 km from pole to pole.


(b) Compare this with the accepted value of


5.979×1024 kg .


34. (a) Calculate the magnitude of the acceleration due to
gravity on the surface of Earth due to the Moon.


(b) Calculate the magnitude of the acceleration due to gravity
at Earth due to the Sun.


(c) Take the ratio of the Moon’s acceleration to the Sun’s and
comment on why the tides are predominantly due to the Moon
in spite of this number.


35. (a) What is the acceleration due to gravity on the surface
of the Moon?


(b) On the surface of Mars? The mass of Mars is


6.418×1023 kg and its radius is 3.38×106 m .


36. (a) Calculate the acceleration due to gravity on the
surface of the Sun.


(b) By what factor would your weight increase if you could
stand on the Sun? (Never mind that you cannot.)


Chapter 6 | Uniform Circular Motion and Gravitation 235




37. The Moon and Earth rotate about their common center of
mass, which is located about 4700 km from the center of
Earth. (This is 1690 km below the surface.)


(a) Calculate the magnitude of the acceleration due to the
Moon’s gravity at that point.


(b) Calculate the magnitude of the centripetal acceleration of
the center of Earth as it rotates about that point once each
lunar month (about 27.3 d) and compare it with the
acceleration found in part (a). Comment on whether or not
they are equal and why they should or should not be.


38. Solve part (b) of Example 6.6 using ac = v2 / r .


39. Astrology, that unlikely and vague pseudoscience, makes
much of the position of the planets at the moment of one’s
birth. The only known force a planet exerts on Earth is
gravitational.


(a) Calculate the magnitude of the gravitational force exerted
on a 4.20 kg baby by a 100 kg father 0.200 m away at birth
(he is assisting, so he is close to the child).


(b) Calculate the magnitude of the force on the baby due to
Jupiter if it is at its closest distance to Earth, some


6.29×1011 m away. How does the force of Jupiter on the
baby compare to the force of the father on the baby? Other
objects in the room and the hospital building also exert similar
gravitational forces. (Of course, there could be an unknown
force acting, but scientists first need to be convinced that
there is even an effect, much less that an unknown force
causes it.)


40. The existence of the dwarf planet Pluto was proposed
based on irregularities in Neptune’s orbit. Pluto was
subsequently discovered near its predicted position. But it
now appears that the discovery was fortuitous, because Pluto
is small and the irregularities in Neptune’s orbit were not well
known. To illustrate that Pluto has a minor effect on the orbit
of Neptune compared with the closest planet to Neptune:


(a) Calculate the acceleration due to gravity at Neptune due


to Pluto when they are 4.50×1012 m apart, as they are at
present. The mass of Pluto is 1.4×1022 kg .


(b) Calculate the acceleration due to gravity at Neptune due


to Uranus, presently about 2.50×1012 m apart, and
compare it with that due to Pluto. The mass of Uranus is


8.62×1025 kg .


41. (a) The Sun orbits the Milky Way galaxy once each


2.60 x 108 y , with a roughly circular orbit averaging


3.00 x 104 light years in radius. (A light year is the distance
traveled by light in 1 y.) Calculate the centripetal acceleration
of the Sun in its galactic orbit. Does your result support the
contention that a nearly inertial frame of reference can be
located at the Sun?


(b) Calculate the average speed of the Sun in its galactic
orbit. Does the answer surprise you?


42. Unreasonable Result


A mountain 10.0 km from a person exerts a gravitational force
on him equal to 2.00% of his weight.


(a) Calculate the mass of the mountain.


(b) Compare the mountain’s mass with that of Earth.


(c) What is unreasonable about these results?


(d) Which premises are unreasonable or inconsistent? (Note
that accurate gravitational measurements can easily detect
the effect of nearby mountains and variations in local
geology.)


6.6 Satellites and Kepler’s Laws: An Argument
for Simplicity
43. A geosynchronous Earth satellite is one that has an
orbital period of precisely 1 day. Such orbits are useful for
communication and weather observation because the satellite
remains above the same point on Earth (provided it orbits in
the equatorial plane in the same direction as Earth’s rotation).
Calculate the radius of such an orbit based on the data for the
moon in Table 6.2.


44. Calculate the mass of the Sun based on data for Earth’s
orbit and compare the value obtained with the Sun’s actual
mass.


45. Find the mass of Jupiter based on data for the orbit of one
of its moons, and compare your result with its actual mass.


46. Find the ratio of the mass of Jupiter to that of Earth based
on data in Table 6.2.


47. Astronomical observations of our Milky Way galaxy


indicate that it has a mass of about 8.0×1011 solar masses.
A star orbiting on the galaxy’s periphery is about 6.0×104
light years from its center. (a) What should the orbital period


of that star be? (b) If its period is 6.0×107 instead, what is
the mass of the galaxy? Such calculations are used to imply
the existence of “dark matter” in the universe and have
indicated, for example, the existence of very massive black
holes at the centers of some galaxies.


48. Integrated Concepts


Space debris left from old satellites and their launchers is
becoming a hazard to other satellites. (a) Calculate the speed
of a satellite in an orbit 900 km above Earth’s surface. (b)
Suppose a loose rivet is in an orbit of the same radius that
intersects the satellite’s orbit at an angle of 90º relative to
Earth. What is the velocity of the rivet relative to the satellite
just before striking it? (c) Given the rivet is 3.00 mm in size,
how long will its collision with the satellite last? (d) If its mass
is 0.500 g, what is the average force it exerts on the satellite?
(e) How much energy in joules is generated by the collision?
(The satellite’s velocity does not change appreciably, because
its mass is much greater than the rivet’s.)


49. Unreasonable Results


(a) Based on Kepler’s laws and information on the orbital
characteristics of the Moon, calculate the orbital radius for an
Earth satellite having a period of 1.00 h. (b) What is
unreasonable about this result? (c) What is unreasonable or
inconsistent about the premise of a 1.00 h orbit?


50. Construct Your Own Problem


On February 14, 2000, the NEAR spacecraft was successfully
inserted into orbit around Eros, becoming the first artificial
satellite of an asteroid. Construct a problem in which you
determine the orbital speed for a satellite near Eros. You will
need to find the mass of the asteroid and consider such
things as a safe distance for the orbit. Although Eros is not
spherical, calculate the acceleration due to gravity on its
surface at a point an average distance from its center of
mass. Your instructor may also wish to have you calculate the
escape velocity from this point on Eros.


236 Chapter 6 | Uniform Circular Motion and Gravitation


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7 WORK, ENERGY, AND ENERGY
RESOURCES


Figure 7.1 How many forms of energy can you identify in this photograph of a wind farm in Iowa? (credit: Jürgen from Sandesneben, Germany,
Wikimedia Commons)


Chapter Outline
7.1. Work: The Scientific Definition


• Explain how an object must be displaced for a force on it to do work.
• Explain how relative directions of force and displacement determine whether the work done is positive, negative, or


zero.
7.2. Kinetic Energy and the Work-Energy Theorem


• Explain work as a transfer of energy and net work as the work done by the net force.
• Explain and apply the work-energy theorem.


7.3. Gravitational Potential Energy
• Explain gravitational potential energy in terms of work done against gravity.
• Show that the gravitational potential energy of an object of mass m at height h on Earth is given by PEg = mgh .
• Show how knowledge of the potential energy as a function of position can be used to simplify calculations and


explain physical phenomena.
7.4. Conservative Forces and Potential Energy


• Define conservative force, potential energy, and mechanical energy.
• Explain the potential energy of a spring in terms of its compression when Hooke’s law applies.
• Use the work-energy theorem to show how having only conservative forces implies conservation of mechanical


energy.
7.5. Nonconservative Forces


• Define nonconservative forces and explain how they affect mechanical energy.
• Show how the principle of conservation of energy can be applied by treating the conservative forces in terms of their


potential energies and any nonconservative forces in terms of the work they do.
7.6. Conservation of Energy


• Explain the law of the conservation of energy.
• Describe some of the many forms of energy.
• Define efficiency of an energy conversion process as the fraction left as useful energy or work, rather than being


transformed, for example, into thermal energy.
7.7. Power


• Calculate power by calculating changes in energy over time.
• Examine power consumption and calculations of the cost of energy consumed.


7.8. Work, Energy, and Power in Humans
• Explain the human body’s consumption of energy when at rest vs. when engaged in activities that do useful work.
• Calculate the conversion of chemical energy in food into useful work.


7.9. World Energy Use
• Describe the distinction between renewable and nonrenewable energy sources.
• Explain why the inevitable conversion of energy to less useful forms makes it necessary to conserve energy


resources.


Chapter 7 | Work, Energy, and Energy Resources 237




Introduction to Work, Energy, and Energy Resources
Energy plays an essential role both in everyday events and in scientific phenomena. You can no doubt name many forms of
energy, from that provided by our foods, to the energy we use to run our cars, to the sunlight that warms us on the beach. You
can also cite examples of what people call energy that may not be scientific, such as someone having an energetic personality.
Not only does energy have many interesting forms, it is involved in almost all phenomena, and is one of the most important
concepts of physics. What makes it even more important is that the total amount of energy in the universe is constant. Energy
can change forms, but it cannot appear from nothing or disappear without a trace. Energy is thus one of a handful of physical
quantities that we say is conserved.


Conservation of energy (as physicists like to call the principle that energy can neither be created nor destroyed) is based on
experiment. Even as scientists discovered new forms of energy, conservation of energy has always been found to apply. Perhaps
the most dramatic example of this was supplied by Einstein when he suggested that mass is equivalent to energy (his famous


equation E = mc 2 ).
From a societal viewpoint, energy is one of the major building blocks of modern civilization. Energy resources are key limiting
factors to economic growth. The world use of energy resources, especially oil, continues to grow, with ominous consequences
economically, socially, politically, and environmentally. We will briefly examine the world’s energy use patterns at the end of this
chapter.


There is no simple, yet accurate, scientific definition for energy. Energy is characterized by its many forms and the fact that it is
conserved. We can loosely define energy as the ability to do work, admitting that in some circumstances not all energy is
available to do work. Because of the association of energy with work, we begin the chapter with a discussion of work. Work is
intimately related to energy and how energy moves from one system to another or changes form.


7.1 Work: The Scientific Definition


What It Means to Do Work
The scientific definition of work differs in some ways from its everyday meaning. Certain things we think of as hard work, such as
writing an exam or carrying a heavy load on level ground, are not work as defined by a scientist. The scientific definition of work
reveals its relationship to energy—whenever work is done, energy is transferred.


For work, in the scientific sense, to be done, a force must be exerted and there must be motion or displacement in the direction
of the force.


Formally, the work done on a system by a constant force is defined to be the product of the component of the force in the
direction of motion times the distance through which the force acts. For one-way motion in one dimension, this is expressed in
equation form as


(7.1)W = ∣ F ∣ (cos θ) ∣ d ∣ ,


where W is work, d is the displacement of the system, and θ is the angle between the force vector F and the displacement
vector d , as in Figure 7.2. We can also write this as


(7.2)W = Fd cos θ.
To find the work done on a system that undergoes motion that is not one-way or that is in two or three dimensions, we divide the
motion into one-way one-dimensional segments and add up the work done over each segment.


What is Work?


The work done on a system by a constant force is the product of the component of the force in the direction of motion times
the distance through which the force acts. For one-way motion in one dimension, this is expressed in equation form as


(7.3)W = Fd cos θ,


where W is work, F is the magnitude of the force on the system, d is the magnitude of the displacement of the system,
and θ is the angle between the force vector F and the displacement vector d .


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Figure 7.2 Examples of work. (a) The work done by the force F on this lawn mower is Fd cos θ . Note that F cos θ is the component of the
force in the direction of motion. (b) A person holding a briefcase does no work on it, because there is no motion. No energy is transferred to or from the
briefcase. (c) The person moving the briefcase horizontally at a constant speed does no work on it, and transfers no energy to it. (d) Work is done on
the briefcase by carrying it up stairs at constant speed, because there is necessarily a component of force F in the direction of the motion. Energy is
transferred to the briefcase and could in turn be used to do work. (e) When the briefcase is lowered, energy is transferred out of the briefcase and into
an electric generator. Here the work done on the briefcase by the generator is negative, removing energy from the briefcase, because F and d are
in opposite directions.


To examine what the definition of work means, let us consider the other situations shown in Figure 7.2. The person holding the
briefcase in Figure 7.2(b) does no work, for example. Here d = 0 , so W = 0 . Why is it you get tired just holding a load? The
answer is that your muscles are doing work against one another, but they are doing no work on the system of interest (the
“briefcase-Earth system”—see Gravitational Potential Energy for more details). There must be motion for work to be done, and
there must be a component of the force in the direction of the motion. For example, the person carrying the briefcase on level


Chapter 7 | Work, Energy, and Energy Resources 239




ground in Figure 7.2(c) does no work on it, because the force is perpendicular to the motion. That is, cos 90º = 0 , and so
W = 0 .
In contrast, when a force exerted on the system has a component in the direction of motion, such as in Figure 7.2(d), work is
done—energy is transferred to the briefcase. Finally, in Figure 7.2(e), energy is transferred from the briefcase to a generator.
There are two good ways to interpret this energy transfer. One interpretation is that the briefcase’s weight does work on the
generator, giving it energy. The other interpretation is that the generator does negative work on the briefcase, thus removing
energy from it. The drawing shows the latter, with the force from the generator upward on the briefcase, and the displacement
downward. This makes θ = 180º , and cos 180º = –1 ; therefore, W is negative.


Calculating Work
Work and energy have the same units. From the definition of work, we see that those units are force times distance. Thus, in SI
units, work and energy are measured in newton-meters. A newton-meter is given the special name joule (J), and


1 J = 1 N ⋅ m = 1 kg ⋅ m2/s2 . One joule is not a large amount of energy; it would lift a small 100-gram apple a distance of
about 1 meter.


Example 7.1 Calculating the Work You Do to Push a Lawn Mower Across a Large Lawn


How much work is done on the lawn mower by the person in Figure 7.2(a) if he exerts a constant force of 75.0 N at an
angle 35º below the horizontal and pushes the mower 25.0 m on level ground? Convert the amount of work from joules to
kilocalories and compare it with this person’s average daily intake of 10,000 kJ (about 2400 kcal ) of food energy. One
calorie (1 cal) of heat is the amount required to warm 1 g of water by 1ºC , and is equivalent to 4.184 J , while one food
calorie (1 kcal) is equivalent to 4184 J .
Strategy


We can solve this problem by substituting the given values into the definition of work done on a system, 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


(7.4)W = Fd cos θ.
Substituting the known values gives


(7.5)W = (75.0 N)(25.0 m) cos (35.0º)
= 1536 J = 1.54×103 J.


Converting the work in joules to kilocalories yields W = (1536 J)(1 kcal / 4184 J) = 0.367 kcal . The ratio of the work
done to the daily consumption is


(7.6)W
2400 kcal = 1.53×10


−4.


Discussion


This ratio is a tiny fraction of what the person consumes, but it is typical. Very little of the energy released in the consumption
of food is used to do work. Even when we “work” all day long, less than 10% of our food energy intake is used to do work
and more than 90% is converted to thermal energy or stored as chemical energy in fat.


240 Chapter 7 | Work, Energy, and Energy Resources


7.2 Kinetic Energy and the Work-Energy Theorem


Work Transfers Energy
What happens to the work done on a system? Energy is transferred into the system, but in what form? Does it remain in the
system or move on? The answers depend on the situation. For example, if the lawn mower in Figure 7.2(a) is pushed just hard
enough to keep it going at a constant speed, then energy put into the mower by the person is removed continuously by friction,
and eventually leaves the system in the form of heat transfer. In contrast, work done on the briefcase by the person carrying it up
stairs in Figure 7.2(d) is stored in the briefcase-Earth system and can be recovered at any time, as shown in Figure 7.2(e). In
fact, the building of the pyramids in ancient Egypt is an example of storing energy in a system by doing work on the system.
Some of the energy imparted to the stone blocks in lifting them during construction of the pyramids remains in the stone-Earth
system and has the potential to do work.


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In this section we begin the study of various types of work and forms of energy. We will find that some types of work leave the
energy of a system constant, for example, whereas others change the system in some way, such as making it move. We will also
develop definitions of important forms of energy, such as the energy of motion.


Net Work and the Work-Energy Theorem
We know from the study of Newton’s laws in Dynamics: Force and Newton's Laws of Motion that net force causes
acceleration. We will see in this section that work done by the net force gives a system energy of motion, and in the process we
will also find an expression for the energy of motion.


Let us start by considering the total, or net, work done on a system. Net work is defined to be the sum of work done by all
external forces—that is, net work is the work done by the net external force Fnet . In equation form, this is
Wnet = Fnetd cos θ where θ is the angle between the force vector and the displacement vector.


Figure 7.3(a) shows a graph of force versus displacement for the component of the force in the direction of the
displacement—that is, an F cos θ vs. d graph. In this case, F cos θ is constant. You can see that the area under the graph is
Fd cos θ , or the work done. Figure 7.3(b) shows a more general process where the force varies. The area under the curve is
divided into strips, each having an average force (F cos θ)


i(ave) . The work done is (F cos θ)i(ave)di for each strip, and the


total work done is the sum of the Wi . Thus the total work done is the total area under the curve, a useful property to which we


shall refer later.


Figure 7.3 (a) A graph of F cos θ vs. d , when F cos θ is constant. The area under the curve represents the work done by the force. (b) A
graph of F cos θ vs. d in which the force varies. The work done for each interval is the area of each strip; thus, the total area under the curve
equals the total work done.


Net work will be simpler to examine if we consider a one-dimensional situation where a force is used to accelerate an object in a
direction parallel to its initial velocity. Such a situation occurs for the package on the roller belt conveyor system shown in Figure
7.4.


Chapter 7 | Work, Energy, and Energy Resources 241




Figure 7.4 A package on a roller belt is pushed horizontally through a distance d .


The force of gravity and the normal force acting on the package are perpendicular to the displacement and do no work.
Moreover, they are also equal in magnitude and opposite in direction so they cancel in calculating the net force. The net force
arises solely from the horizontal applied force Fapp and the horizontal friction force f . Thus, as expected, the net force is
parallel to the displacement, so that θ = 0º and cos θ = 1 , and the net work is given by


(7.7)Wnet = Fnetd.


The effect of the net force Fnet is to accelerate the package from v0 to v . The kinetic energy of the package increases,
indicating that the net work done on the system is positive. (See Example 7.2.) By using Newton’s second law, and doing some
algebra, we can reach an interesting conclusion. Substituting Fnet = ma from Newton’s second law gives


(7.8)Wnet = mad.


To get a relationship between net work and the speed given to a system by the net force acting on it, we take d = x − x0 and
use the equation studied in Motion Equations for Constant Acceleration in One Dimension for the change in speed over a


distance d if the acceleration has the constant value a ; namely, v2 = v0
2 + 2ad (note that a appears in the expression for


the net work). Solving for acceleration gives a = v
2 − v0


2


2d . When a is substituted into the preceding expression for Wnet , we


obtain


(7.9)
Wnet = m






v2 − v0


2


2d



⎟d.


The d cancels, and we rearrange this to obtain


(7.10)W = 12mv
2 − 12mv0


2 .


This expression is called the work-energy theorem, and it actually applies in general (even for forces that vary in direction and
magnitude), although we have derived it for the special case of a constant force parallel to the displacement. The theorem


implies that the net work on a system equals the change in the quantity 12mv
2 . This quantity is our first example of a form of


energy.


The Work-Energy Theorem


The net work on a system equals the change in the quantity 12mv
2 .


(7.11)W net = 12mv
2 − 12mv0


2


The quantity 12mv
2 in the work-energy theorem is defined to be the translational kinetic energy (KE) of a mass m moving at a


speed v . (Translational kinetic energy is distinct from rotational kinetic energy, which is considered later.) In equation form, the
translational kinetic energy,


(7.12)KE = 12mv
2,


242 Chapter 7 | Work, Energy, and Energy Resources


is the energy associated with translational motion. Kinetic energy is a form of energy associated with the motion of a particle,
single body, or system of objects moving together.


We are aware that it takes energy to get an object, like a car or the package in Figure 7.4, up to speed, but it may be a bit
surprising that kinetic energy is proportional to speed squared. This proportionality means, for example, that a car traveling at


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100 km/h has four times the kinetic energy it has at 50 km/h, helping to explain why high-speed collisions are so devastating. We
will now consider a series of examples to illustrate various aspects of work and energy.


Example 7.2 Calculating the Kinetic Energy of a Package


Suppose a 30.0-kg package on the roller belt conveyor system in Figure 7.4 is moving at 0.500 m/s. What is its kinetic
energy?


Strategy


Because the mass m and speed v are given, the kinetic energy can be calculated from its definition as given in the


equation KE = 12mv
2 .


Solution


The kinetic energy is given by


(7.13)KE = 12mv
2.


Entering known values gives


(7.14)KE = 0.5(30.0 kg)(0.500 m/s)2,


which yields


(7.15)KE = 3.75 kg ⋅ m2/s2 = 3.75 J.


Discussion


Note that the unit of kinetic energy is the joule, the same as the unit of work, as mentioned when work was first defined. It is
also interesting that, although this is a fairly massive package, its kinetic energy is not large at this relatively low speed. This
fact is consistent with the observation that people can move packages like this without exhausting themselves.


Example 7.3 Determining the Work to Accelerate a Package


Suppose that you push on the 30.0-kg package in Figure 7.4 with a constant force of 120 N through a distance of 0.800 m,
and that the opposing friction force averages 5.00 N.


(a) Calculate the net work done on the package. (b) Solve the same problem as in part (a), this time by finding the work
done by each force that contributes to the net force.


Strategy and Concept for (a)


This is a motion in one dimension problem, because the downward force (from the weight of the package) and the normal
force have equal magnitude and opposite direction, so that they cancel in calculating the net force, while the applied force,
friction, and the displacement are all horizontal. (See Figure 7.4.) As expected, the net work is the net force times distance.


Solution for (a)


The net force is the push force minus friction, or Fnet= 120 N – 5.00 N = 115 N . Thus the net work is


(7.16)Wnet = Fnetd = (115 N)(0.800 m)
= 92.0 N ⋅ m = 92.0 J.


Discussion for (a)


This value is the net work done on the package. The person actually does more work than this, because friction opposes the
motion. Friction does negative work and removes some of the energy the person expends and converts it to thermal energy.
The net work equals the sum of the work done by each individual force.


Strategy and Concept for (b)


The forces acting on the package are gravity, the normal force, the force of friction, and the applied force. The normal force
and force of gravity are each perpendicular to the displacement, and therefore do no work.


Solution for (b)


The applied force does work.


(7.17)Wapp = Fappd cos(0º) = Fappd
= (120 N)(0.800 m)
= 96.0 J


Chapter 7 | Work, Energy, and Energy Resources 243




The friction force and displacement are in opposite directions, so that θ = 180º , and the work done by friction is
(7.18)Wfr = Ffrd cos(180º) = −Ffrd


= −(5.00 N)(0.800 m)
= −4.00 J.


So the amounts of work done by gravity, by the normal force, by the applied force, and by friction are, respectively,


(7.19)Wgr = 0,
WN = 0,
Wapp = 96.0 J,
Wfr = − 4.00 J.


The total work done as the sum of the work done by each force is then seen to be


(7.20)Wtotal = Wgr +WN +Wapp +Wfr = 92.0 J.


Discussion for (b)


The calculated total work Wtotal as the sum of the work by each force agrees, as expected, with the work Wnet done by


the net force. The work done by a collection of forces acting on an object can be calculated by either approach.


Example 7.4 Determining Speed from Work and Energy


Find the speed of the package in Figure 7.4 at the end of the push, using work and energy concepts.


Strategy


Here the work-energy theorem can be used, because we have just calculated the net work, Wnet , and the initial kinetic


energy, 12mv0
2 . These calculations allow us to find the final kinetic energy, 12mv


2 , and thus the final speed v .


Solution


The work-energy theorem in equation form is


(7.21)Wnet = 12mv
2 − 12mv0


2.


Solving for 12mv
2 gives


(7.22)1
2mv


2 = Wnet + 12mv0
2.


Thus,


(7.23)1
2mv


2 = 92.0 J+3.75 J = 95.75 J.


Solving for the final speed as requested and entering known values gives


(7.24)
v = 2(95.75 J)m =


191.5 kg ⋅ m2/s2
30.0 kg


= 2.53 m/s.
Discussion


Using work and energy, we not only arrive at an answer, we see that the final kinetic energy is the sum of the initial kinetic
energy and the net work done on the package. This means that the work indeed adds to the energy of the package.


Example 7.5 Work and Energy Can Reveal Distance, Too


How far does the package in Figure 7.4 coast after the push, assuming friction remains constant? Use work and energy
considerations.


Strategy


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We know that once the person stops pushing, friction will bring the package to rest. In terms of energy, friction does negative
work until it has removed all of the package’s kinetic energy. The work done by friction is the force of friction times the
distance traveled times the cosine of the angle between the friction force and displacement; hence, this gives us a way of
finding the distance traveled after the person stops pushing.


Solution


The normal force and force of gravity cancel in calculating the net force. The horizontal friction force is then the net force,
and it acts opposite to the displacement, so θ = 180º . To reduce the kinetic energy of the package to zero, the work Wfr
by friction must be minus the kinetic energy that the package started with plus what the package accumulated due to the
pushing. Thus Wfr = −95.75 J . Furthermore, Wfr = f d′ cos θ = – f d′ , where d′ is the distance it takes to stop. Thus,


(7.25)
d′ = −Wfr


f
= −−95.75 J5.00 N ,


and so


(7.26)d′ = 19.2 m.
Discussion


This is a reasonable distance for a package to coast on a relatively friction-free conveyor system. Note that the work done
by friction is negative (the force is in the opposite direction of motion), so it removes the kinetic energy.


Some of the examples in this section can be solved without considering energy, but at the expense of missing out on gaining
insights about what work and energy are doing in this situation. On the whole, solutions involving energy are generally shorter
and easier than those using kinematics and dynamics alone.


7.3 Gravitational Potential Energy


Work Done Against Gravity
Climbing stairs and lifting objects is work in both the scientific and everyday sense—it is work done against the gravitational
force. When there is work, there is a transformation of energy. The work done against the gravitational force goes into an
important form of stored energy that we will explore in this section.


Let us calculate the work done in lifting an object of mass m through a height h , such as in Figure 7.5. If the object is lifted
straight up at constant speed, then the force needed to lift it is equal to its weight mg . The work done on the mass is then


W = Fd = mgh . We define this to be the gravitational potential energy (PEg) put into (or gained by) the object-Earth
system. This energy is associated with the state of separation between two objects that attract each other by the gravitational
force. For convenience, we refer to this as the PEg gained by the object, recognizing that this is energy stored in the
gravitational field of Earth. Why do we use the word “system”? Potential energy is a property of a system rather than of a single
object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the
Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it
increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position,
we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but
this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to
the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a
ladder will be the same for the first two rungs as for the last two rungs.


Converting Between Potential Energy and Kinetic Energy
Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass,
gravitational force will do an amount of work equal to mgh on it, thereby increasing its kinetic energy by that same amount (by


the work-energy theorem). We will find it more useful to consider just the conversion of PEg to KE without explicitly
considering the intermediate step of work. (See Example 7.7.) This shortcut makes it is easier to solve problems using energy (if
possible) rather than explicitly using forces.


Chapter 7 | Work, Energy, and Energy Resources 245




Figure 7.5 (a) The work done to lift the weight is stored in the mass-Earth system as gravitational potential energy. (b) As the weight moves downward,
this gravitational potential energy is transferred to the cuckoo clock.


More precisely, we define the change in gravitational potential energy ΔPEg to be


(7.27)ΔPEg = mgh,


where, for simplicity, we denote the change in height by h rather than the usual Δh . Note that h is positive when the final
height is greater than the initial height, and vice versa. For example, if a 0.500-kg mass hung from a cuckoo clock is raised 1.00
m, then its change in gravitational potential energy is


(7.28)mgh = ⎛⎝0.500 kg⎞⎠⎛⎝9.80 m/s2

⎠(1.00 m)


246 Chapter 7 | Work, Energy, and Energy Resources


= 4.90 kg ⋅ m2/s2 = 4.90 J.
Note that the units of gravitational potential energy turn out to be joules, the same as for work and other forms of energy. As the
clock runs, the mass is lowered. We can think of the mass as gradually giving up its 4.90 J of gravitational potential energy,
without directly considering the force of gravity that does the work.


Using Potential Energy to Simplify Calculations


The equation ΔPEg = mgh applies for any path that has a change in height of h , not just when the mass is lifted straight up.
(See Figure 7.6.) It is much easier to calculate mgh (a simple multiplication) than it is to calculate the work done along a
complicated path. The idea of gravitational potential energy has the double advantage that it is very broadly applicable and it
makes calculations easier. From now on, we will consider that any change in vertical position h of a mass m is accompanied
by a change in gravitational potential energy mgh , and we will avoid the equivalent but more difficult task of calculating work


done by or against the gravitational force.


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Figure 7.6 The change in gravitational potential energy (ΔPEg) between points A and B is independent of the path. ΔPEg = mgh for any path
between the two points. Gravity is one of a small class of forces where the work done by or against the force depends only on the starting and ending
points, not on the path between them.


Example 7.6 The Force to Stop Falling


A 60.0-kg person jumps onto the floor from a height of 3.00 m. If he lands stiffly (with his knee joints compressing by 0.500
cm), calculate the force on the knee joints.


Strategy


This person’s energy is brought to zero in this situation by the work done on him by the floor as he stops. The initial PEg is
transformed into KE as he falls. The work done by the floor reduces this kinetic energy to zero.
Solution


The work done on the person by the floor as he stops is given by


(7.29)W = Fd cos θ = −Fd,


with a minus sign because the displacement while stopping and the force from floor are in opposite directions
(cos θ = cos 180º = − 1) . The floor removes energy from the system, so it does negative work.


The kinetic energy the person has upon reaching the floor is the amount of potential energy lost by falling through height h :


(7.30)KE = −ΔPEg = −mgh,


The distance d that the person’s knees bend is much smaller than the height h of the fall, so the additional change in
gravitational potential energy during the knee bend is ignored.


The work W done by the floor on the person stops the person and brings the person’s kinetic energy to zero:


(7.31)W = −KE = mgh.


Combining this equation with the expression for W gives


(7.32)−Fd = mgh.


Recalling that h is negative because the person fell down, the force on the knee joints is given by


Chapter 7 | Work, Energy, and Energy Resources 247




(7.33)
F = −mgh


d
= −



⎝60.0 kg⎞⎠⎛⎝9.80 m/s2



⎠(−3.00 m)


5.00×10−3 m
= 3.53×105 N.


Discussion


Such a large force (500 times more than the person’s weight) over the short impact time is enough to break bones. A much
better way to cushion the shock is by bending the legs or rolling on the ground, increasing the time over which the force
acts. A bending motion of 0.5 m this way yields a force 100 times smaller than in the example. A kangaroo's hopping shows
this method in action. The kangaroo is the only large animal to use hopping for locomotion, but the shock in hopping is
cushioned by the bending of its hind legs in each jump.(See Figure 7.7.)


Figure 7.7 The work done by the ground upon the kangaroo reduces its kinetic energy to zero as it lands. However, by applying the force of the ground
on the hind legs over a longer distance, the impact on the bones is reduced. (credit: Chris Samuel, Flickr)


Example 7.7 Finding the Speed of a Roller Coaster from its Height


(a) What is the final speed of the roller coaster shown in Figure 7.8 if it starts from rest at the top of the 20.0 m hill and work
done by frictional forces is negligible? (b) What is its final speed (again assuming negligible friction) if its initial speed is 5.00
m/s?


Figure 7.8 The speed of a roller coaster increases as gravity pulls it downhill and is greatest at its lowest point. Viewed in terms of energy, the
roller-coaster-Earth system’s gravitational potential energy is converted to kinetic energy. If work done by friction is negligible, all ΔPEg is
converted to KE .


Strategy


The roller coaster loses potential energy as it goes downhill. We neglect friction, so that the remaining force exerted by the
track is the normal force, which is perpendicular to the direction of motion and does no work. The net work on the roller


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coaster is then done by gravity alone. The loss of gravitational potential energy from moving downward through a distance
h equals the gain in kinetic energy. This can be written in equation form as −ΔPEg = ΔKE . Using the equations for
PEg and KE , we can solve for the final speed v , which is the desired quantity.


Solution for (a)


Here the initial kinetic energy is zero, so that ΔKE = 12mv
2 . The equation for change in potential energy states that


ΔPEg = mgh . Since h is negative in this case, we will rewrite this as ΔPEg = −mg ∣ h ∣ to show the minus sign
clearly. Thus,


(7.34)−ΔPEg = ΔKE


becomes


(7.35)mg ∣ h ∣ = 12mv
2.


Solving for v , we find that mass cancels and that


(7.36)v = 2g ∣ h ∣ .


Substituting known values,


(7.37)v = 2⎛⎝9.80 m/s2

⎠(20.0 m)


= 19.8 m/s.
Solution for (b)


Again − ΔPEg = ΔKE . In this case there is initial kinetic energy, so ΔKE = 12mv
2 − 12mv0


2 . Thus,


(7.38)mg ∣ h ∣ = 12mv
2 − 12mv0


2.


Rearranging gives


(7.39)1
2mv


2 = mg ∣ h ∣ + 12mv0
2.


This means that the final kinetic energy is the sum of the initial kinetic energy and the gravitational potential energy. Mass
again cancels, and


(7.40)v = 2g ∣ h ∣ + v0
2.


This equation is very similar to the kinematics equation v = v0
2 + 2ad , but it is more general—the kinematics equation is


valid only for constant acceleration, whereas our equation above is valid for any path regardless of whether the object
moves with a constant acceleration. Now, substituting known values gives


(7.41)v = 2(9.80 m/s2)(20.0 m) + (5.00 m/s)2


= 20.4 m/s.
Discussion and Implications


First, note that mass cancels. This is quite consistent with observations made in Falling Objects that all objects fall at the
same rate if friction is negligible. Second, only the speed of the roller coaster is considered; there is no information about its
direction at any point. This reveals another general truth. When friction is negligible, the speed of a falling body depends
only on its initial speed and height, and not on its mass or the path taken. For example, the roller coaster will have the same
final speed whether it falls 20.0 m straight down or takes a more complicated path like the one in the figure. Third, and
perhaps unexpectedly, the final speed in part (b) is greater than in part (a), but by far less than 5.00 m/s. Finally, note that
speed can be found at any height along the way by simply using the appropriate value of h at the point of interest.


We have seen that work done by or against the gravitational force depends only on the starting and ending points, and not on the
path between, allowing us to define the simplifying concept of gravitational potential energy. We can do the same thing for a few
other forces, and we will see that this leads to a formal definition of the law of conservation of energy.


Chapter 7 | Work, Energy, and Energy Resources 249




Making Connections: Take-Home Investigation—Converting Potential to Kinetic Energy


One can study the conversion of gravitational potential energy into kinetic energy in this experiment. On a smooth, level
surface, use a ruler of the kind that has a groove running along its length and a book to make an incline (see Figure 7.9).
Place a marble at the 10-cm position on the ruler and let it roll down the ruler. When it hits the level surface, measure the
time it takes to roll one meter. Now place the marble at the 20-cm and the 30-cm positions and again measure the times it
takes to roll 1 m on the level surface. Find the velocity of the marble on the level surface for all three positions. Plot velocity
squared versus the distance traveled by the marble. What is the shape of each plot? If the shape is a straight line, the plot
shows that the marble’s kinetic energy at the bottom is proportional to its potential energy at the release point.


Figure 7.9 A marble rolls down a ruler, and its speed on the level surface is measured.


7.4 Conservative Forces and Potential Energy


Potential Energy and Conservative Forces
Work is done by a force, and some forces, such as weight, have special characteristics. A conservative force is one, like the
gravitational force, for which work done by or against it depends only on the starting and ending points of a motion and not on the
path taken. We can define a potential energy (PE) for any conservative force, just as we did for the gravitational force. For
example, when you wind up a toy, an egg timer, or an old-fashioned watch, you do work against its spring and store energy in it.
(We treat these springs as ideal, in that we assume there is no friction and no production of thermal energy.) This stored energy
is recoverable as work, and it is useful to think of it as potential energy contained in the spring. Indeed, the reason that the spring
has this characteristic is that its force is conservative. That is, a conservative force results in stored or potential energy.
Gravitational potential energy is one example, as is the energy stored in a spring. We will also see how conservative forces are
related to the conservation of energy.


Potential Energy and Conservative Forces


Potential energy is the energy a system has due to position, shape, or configuration. It is stored energy that is completely
recoverable.


A conservative force is one for which work done by or against it depends only on the starting and ending points of a motion
and not on the path taken.


We can define a potential energy (PE) for any conservative force. The work done against a conservative force to reach a
final configuration depends on the configuration, not the path followed, and is the potential energy added.


Potential Energy of a Spring


First, let us obtain an expression for the potential energy stored in a spring ( PEs ). We calculate the work done to stretch or
compress a spring that obeys Hooke’s law. (Hooke’s law was examined in Elasticity: Stress and Strain, and states that the
magnitude of force F on the spring and the resulting deformation ΔL are proportional, F = kΔL .) (See Figure 7.10.) For our
spring, we will replace ΔL (the amount of deformation produced by a force F ) by the distance x that the spring is stretched or
compressed along its length. So the force needed to stretch the spring has magnitude F = kx , where k is the spring’s force
constant. The force increases linearly from 0 at the start to kx in the fully stretched position. The average force is kx / 2 . Thus


the work done in stretching or compressing the spring is Ws = Fd =


kx
2

⎠x =


1
2kx


2 . Alternatively, we noted in Kinetic Energy


and the Work-Energy Theorem that the area under a graph of F vs. x is the work done by the force. In Figure 7.10(c) we


see that this area is also 12kx
2 . We therefore define the potential energy of a spring, PEs , to be


(7.42)PEs = 12kx
2,


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where k is the spring’s force constant and x is the displacement from its undeformed position. The potential energy represents
the work done on the spring and the energy stored in it as a result of stretching or compressing it a distance x . The potential
energy of the spring PEs does not depend on the path taken; it depends only on the stretch or squeeze x in the final
configuration.


Figure 7.10 (a) An undeformed spring has no PEs stored in it. (b) The force needed to stretch (or compress) the spring a distance x has a


magnitude F = kx , and the work done to stretch (or compress) it is 12kx
2 . Because the force is conservative, this work is stored as potential


energy (PEs) in the spring, and it can be fully recovered. (c) A graph of F vs. x has a slope of k , and the area under the graph is 12kx
2 . Thus


the work done or potential energy stored is 12kx
2 .


The equation PEs = 12kx
2 has general validity beyond the special case for which it was derived. Potential energy can be stored


in any elastic medium by deforming it. Indeed, the general definition of potential energy is energy due to position, shape, or


configuration. For shape or position deformations, stored energy is PEs = 12kx
2 , where k is the force constant of the particular


system and x is its deformation. Another example is seen in Figure 7.11 for a guitar string.


Figure 7.11Work is done to deform the guitar string, giving it potential energy. When released, the potential energy is converted to kinetic energy and
back to potential as the string oscillates back and forth. A very small fraction is dissipated as sound energy, slowly removing energy from the string.


Conservation of Mechanical Energy
Let us now consider what form the work-energy theorem takes when only conservative forces are involved. This will lead us to
the conservation of energy principle. The work-energy theorem states that the net work done by all forces acting on a system
equals its change in kinetic energy. In equation form, this is


Chapter 7 | Work, Energy, and Energy Resources 251




(7.43)Wnet = 12mv
2 − 12mv0


2 = ΔKE.


If only conservative forces act, then


(7.44)Wnet = Wc,


where Wc is the total work done by all conservative forces. Thus,


(7.45)Wc = ΔKE.


Now, if the conservative force, such as the gravitational force or a spring force, does work, the system loses potential energy.
That is, Wc = −ΔPE . Therefore,


(7.46)−ΔPE = ΔKE
or


(7.47)ΔKE + ΔPE = 0.
This equation means that the total kinetic and potential energy is constant for any process involving only conservative forces.
That is,


(7.48)KE + PE = constant
or
KEi + PEi = KEf + PEf





⎬(conservative forces only),


where i and f denote initial and final values. This equation is a form of the work-energy theorem for conservative forces; it is
known as the conservation of mechanical energy principle. Remember that this applies to the extent that all the forces are
conservative, so that friction is negligible. The total kinetic plus potential energy of a system is defined to be its mechanical
energy, (KE + PE) . In a system that experiences only conservative forces, there is a potential energy associated with each
force, and the energy only changes form between KE and the various types of PE , with the total energy remaining constant.


Example 7.8 Using Conservation of Mechanical Energy to Calculate the Speed of a Toy Car


A 0.100-kg toy car is propelled by a compressed spring, as shown in Figure 7.12. The car follows a track that rises 0.180 m
above the starting point. The spring is compressed 4.00 cm and has a force constant of 250.0 N/m. Assuming work done by
friction to be negligible, find (a) how fast the car is going before it starts up the slope and (b) how fast it is going at the top of
the slope.


Figure 7.12 A toy car is pushed by a compressed spring and coasts up a slope. Assuming negligible friction, the potential energy in the spring is
first completely converted to kinetic energy, and then to a combination of kinetic and gravitational potential energy as the car rises. The details of
the path are unimportant because all forces are conservative—the car would have the same final speed if it took the alternate path shown.


Strategy


The spring force and the gravitational force are conservative forces, so conservation of mechanical energy can be used.
Thus,


(7.49)KEi+PEi = KEf + PEf
or


(7.50)1
2mv i


2 + mghi + 12kxi
2 = 12mvf


2 + mghf + 12kxf
2,


where h is the height (vertical position) and x is the compression of the spring. This general statement looks complex but
becomes much simpler when we start considering specific situations. First, we must identify the initial and final conditions in
a problem; then, we enter them into the last equation to solve for an unknown.


Solution for (a)


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This part of the problem is limited to conditions just before the car is released and just after it leaves the spring. Take the
initial height to be zero, so that both hi and hf are zero. Furthermore, the initial speed vi is zero and the final


compression of the spring xf is zero, and so several terms in the conservation of mechanical energy equation are zero and


it simplifies to


(7.51)1
2kxi


2 = 12mvf
2.


In other words, the initial potential energy in the spring is converted completely to kinetic energy in the absence of friction.
Solving for the final speed and entering known values yields


(7.52)
vf = kmxi


= 250.0 N/m0.100 kg (0.0400 m)


= 2.00 m/s.
Solution for (b)


One method of finding the speed at the top of the slope is to consider conditions just before the car is released and just after
it reaches the top of the slope, completely ignoring everything in between. Doing the same type of analysis to find which
terms are zero, the conservation of mechanical energy becomes


(7.53)1
2kx i


2 = 1 2mvf
2 + mghf.


This form of the equation means that the spring’s initial potential energy is converted partly to gravitational potential energy
and partly to kinetic energy. The final speed at the top of the slope will be less than at the bottom. Solving for vf and


substituting known values gives


(7.54)
vf =


kxi
2


m − 2ghf


= ⎛⎝
250.0 N/m
0.100 kg



⎠(0.0400 m)


2 − 2(9.80 m/s2)(0.180 m)


= 0.687 m/s.
Discussion


Another way to solve this problem is to realize that the car’s kinetic energy before it goes up the slope is converted partly to
potential energy—that is, to take the final conditions in part (a) to be the initial conditions in part (b).


Note that, for conservative forces, we do not directly calculate the work they do; rather, we consider their effects through their
corresponding potential energies, just as we did in Example 7.8. Note also that we do not consider details of the path
taken—only the starting and ending points are important (as long as the path is not impossible). This assumption is usually a
tremendous simplification, because the path may be complicated and forces may vary along the way.


PhET Explorations: Energy Skate Park


Learn about conservation of energy with a skater dude! Build tracks, ramps and jumps for the skater and view the kinetic
energy, potential energy and friction as he moves. You can also take the skater to different planets or even space!


Figure 7.13 Energy Skate Park (http://cnx.org/content/m42149/1.4/energy-skate-park_en.jar)


7.5 Nonconservative Forces


Nonconservative Forces and Friction
Forces are either conservative or nonconservative. Conservative forces were discussed in Conservative Forces and Potential
Energy. A nonconservative force is one for which work depends on the path taken. Friction is a good example of a
nonconservative force. As illustrated in Figure 7.14, work done against friction depends on the length of the path between the


Chapter 7 | Work, Energy, and Energy Resources 253




starting and ending points. Because of this dependence on path, there is no potential energy associated with nonconservative
forces. An important characteristic is that the work done by a nonconservative force adds or removes mechanical energy from a
system. Friction, for example, creates thermal energy that dissipates, removing energy from the system. Furthermore, even if
the thermal energy is retained or captured, it cannot be fully converted back to work, so it is lost or not recoverable in that sense
as well.


Figure 7.14 The amount of the happy face erased depends on the path taken by the eraser between points A and B, as does the work done against
friction. Less work is done and less of the face is erased for the path in (a) than for the path in (b). The force here is friction, and most of the work goes
into thermal energy that subsequently leaves the system (the happy face plus the eraser). The energy expended cannot be fully recovered.


How Nonconservative Forces Affect Mechanical Energy
Mechanical energy may not be conserved when nonconservative forces act. For example, when a car is brought to a stop by
friction on level ground, it loses kinetic energy, which is dissipated as thermal energy, reducing its mechanical energy. Figure
7.15 compares the effects of conservative and nonconservative forces. We often choose to understand simpler systems such as
that described in Figure 7.15(a) first before studying more complicated systems as in Figure 7.15(b).


Figure 7.15 Comparison of the effects of conservative and nonconservative forces on the mechanical energy of a system. (a) A system with only
conservative forces. When a rock is dropped onto a spring, its mechanical energy remains constant (neglecting air resistance) because the force in the
spring is conservative. The spring can propel the rock back to its original height, where it once again has only potential energy due to gravity. (b) A
system with nonconservative forces. When the same rock is dropped onto the ground, it is stopped by nonconservative forces that dissipate its
mechanical energy as thermal energy, sound, and surface distortion. The rock has lost mechanical energy.


How the Work-Energy Theorem Applies
Now let us consider what form the work-energy theorem takes when both conservative and nonconservative forces act. We will
see that the work done by nonconservative forces equals the change in the mechanical energy of a system. As noted in Kinetic
Energy and the Work-Energy Theorem, the work-energy theorem states that the net work on a system equals the change in its
kinetic energy, or Wnet = ΔKE . The net work is the sum of the work by nonconservative forces plus the work by conservative
forces. That is,


(7.55)


(7.56)


254 Chapter 7 | Work, Energy, and Energy Resources


Wnet = Wnc + Wc,


so that


Wnc + Wc = ΔKE,


where Wnc is the total work done by all nonconservative forces and Wc is the total work done by all conservative forces.


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Figure 7.16 A person pushes a crate up a ramp, doing work on the crate. Friction and gravitational force (not shown) also do work on the crate; both
forces oppose the person’s push. As the crate is pushed up the ramp, it gains mechanical energy, implying that the work done by the person is greater
than the work done by friction.


Consider Figure 7.16, in which a person pushes a crate up a ramp and is opposed by friction. As in the previous section, we
note that work done by a conservative force comes from a loss of gravitational potential energy, so that Wc = −ΔPE .
Substituting this equation into the previous one and solving for Wnc gives


(7.57)Wnc = ΔKE + ΔPE.


This equation means that the total mechanical energy (KE + PE) changes by exactly the amount of work done by
nonconservative forces. In Figure 7.16, this is the work done by the person minus the work done by friction. So even if energy is
not conserved for the system of interest (such as the crate), we know that an equal amount of work was done to cause the
change in total mechanical energy.


We rearrange Wnc = ΔKE + ΔPE to obtain


(7.58)KEi+PEi +Wnc = KEf + PEf .


This means that the amount of work done by nonconservative forces adds to the mechanical energy of a system. If Wnc is


positive, then mechanical energy is increased, such as when the person pushes the crate up the ramp in Figure 7.16. If Wnc is


negative, then mechanical energy is decreased, such as when the rock hits the ground in Figure 7.15(b). If Wnc is zero, then


mechanical energy is conserved, and nonconservative forces are balanced. For example, when you push a lawn mower at
constant speed on level ground, your work done is removed by the work of friction, and the mower has a constant energy.


Applying Energy Conservation with Nonconservative Forces


When no change in potential energy occurs, applying KEi+PEi +Wnc = KEf + PEf amounts to applying the work-energy
theorem by setting the change in kinetic energy to be equal to the net work done on the system, which in the most general case
includes both conservative and nonconservative forces. But when seeking instead to find a change in total mechanical energy in
situations that involve changes in both potential and kinetic energy, the previous equation KEi + PEi +Wnc = KEf + PEf
says that you can start by finding the change in mechanical energy that would have resulted from just the conservative forces,
including the potential energy changes, and add to it the work done, with the proper sign, by any nonconservative forces
involved.


Example 7.9 Calculating Distance Traveled: How Far a Baseball Player Slides


Consider the situation shown in Figure 7.17, where a baseball player slides to a stop on level ground. Using energy
considerations, calculate the distance the 65.0-kg baseball player slides, given that his initial speed is 6.00 m/s and the force
of friction against him is a constant 450 N.


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Figure 7.17 The baseball player slides to a stop in a distance d . In the process, friction removes the player’s kinetic energy by doing an amount
of work fd equal to the initial kinetic energy.


Strategy


Friction stops the player by converting his kinetic energy into other forms, including thermal energy. In terms of the work-
energy theorem, the work done by friction, which is negative, is added to the initial kinetic energy to reduce it to zero. The
work done by friction is negative, because f is in the opposite direction of the motion (that is, θ = 180º , and so
cos θ = −1 ). Thus Wnc = − fd . The equation simplifies to


(7.59)1
2mv i


2 − fd = 0


or


(7.60)fd = 12mv i
2.


This equation can now be solved for the distance d .


Solution


Solving the previous equation for d and substituting known values yields


(7.61)
d =


mv i
2


2 f


= (65.0 kg)(6.00 m/s)
2


(2)(450 N)
= 2.60 m.


Discussion


The most important point of this example is that the amount of nonconservative work equals the change in mechanical
energy. For example, you must work harder to stop a truck, with its large mechanical energy, than to stop a mosquito.


Example 7.10 Calculating Distance Traveled: Sliding Up an Incline


Suppose that the player from Example 7.9 is running up a hill having a 5.00º incline upward with a surface similar to that in
the baseball stadium. The player slides with the same initial speed. Determine how far he slides.


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Figure 7.18 The same baseball player slides to a stop on a 5.00º slope.


Strategy


In this case, the work done by the nonconservative friction force on the player reduces the mechanical energy he has from
his kinetic energy at zero height, to the final mechanical energy he has by moving through distance d to reach height h
along the hill, with h = d sin 5.00º . This is expressed by the equation


(7.62)KE + PEi +Wnc = KEf + PEf .


Solution


The work done by friction is again Wnc = − fd ; initially the potential energy is PEi = mg ⋅ 0 = 0 and the kinetic energy


is KEi = 12mv i
2 ; the final energy contributions are KEf = 0 for the kinetic energy and PEf = mgh = mgd sin θ for


the potential energy.


Substituting these values gives


(7.63)1
2mv i


2 + 0 + ⎛⎝− fd

⎠ = 0 + mgd sin θ.


Solve this for d to obtain


(7.64)
d =




1
2

⎠mv i


2


f + mg sin θ


= (0.5)(65.0 kg)(6.00 m/s)
2


450 N+(65.0 kg)(9.80 m/s2) sin (5.00º)
= 2.31 m.


Discussion


As might have been expected, the player slides a shorter distance by sliding uphill. Note that the problem could also have
been solved in terms of the forces directly and the work energy theorem, instead of using the potential energy. This method
would have required combining the normal force and force of gravity vectors, which no longer cancel each other because
they point in different directions, and friction, to find the net force. You could then use the net force and the net work to find
the distance d that reduces the kinetic energy to zero. By applying conservation of energy and using the potential energy
instead, we need only consider the gravitational potential energy mgh , without combining and resolving force vectors. This


simplifies the solution considerably.


Making Connections: Take-Home Investigation—Determining Friction from the Stopping Distance


This experiment involves the conversion of gravitational potential energy into thermal energy. Use the ruler, book, and
marble from Take-Home Investigation—Converting Potential to Kinetic Energy. In addition, you will need a foam cup
with a small hole in the side, as shown in Figure 7.19. From the 10-cm position on the ruler, let the marble roll into the cup
positioned at the bottom of the ruler. Measure the distance d the cup moves before stopping. What forces caused it to
stop? What happened to the kinetic energy of the marble at the bottom of the ruler? Next, place the marble at the 20-cm and
the 30-cm positions and again measure the distance the cup moves after the marble enters it. Plot the distance the cup
moves versus the initial marble position on the ruler. Is this relationship linear?


With some simple assumptions, you can use these data to find the coefficient of kinetic friction µk of the cup on the table.


The force of friction f on the cup is µkN , where the normal force N is just the weight of the cup plus the marble. The


normal force and force of gravity do no work because they are perpendicular to the displacement of the cup, which moves


Chapter 7 | Work, Energy, and Energy Resources 257




horizontally. The work done by friction is fd . You will need the mass of the marble as well to calculate its initial kinetic


energy.


It is interesting to do the above experiment also with a steel marble (or ball bearing). Releasing it from the same positions on
the ruler as you did with the glass marble, is the velocity of this steel marble the same as the velocity of the marble at the
bottom of the ruler? Is the distance the cup moves proportional to the mass of the steel and glass marbles?


Figure 7.19 Rolling a marble down a ruler into a foam cup.


PhET Explorations: The Ramp


Explore forces, energy and work as you push household objects up and down a ramp. Lower and raise the ramp to see how
the angle of inclination affects the parallel forces acting on the file cabinet. Graphs show forces, energy and work.


Figure 7.20 The Ramp (http://cnx.org/content/m42150/1.6/the-ramp_en.jar)


7.6 Conservation of Energy


Law of Conservation of Energy
Energy, as we have noted, is conserved, making it one of the most important physical quantities in nature. The law of
conservation of energy can be stated as follows:


Total energy is constant in any process. It may change in form or be transferred from one system to another, but the total
remains the same.


We have explored some forms of energy and some ways it can be transferred from one system to another. This exploration led
to the definition of two major types of energy—mechanical energy (KE + PE) and energy transferred via work done by
nonconservative forces (Wnc) . But energy takes many other forms, manifesting itself in many different ways, and we need to be
able to deal with all of these before we can write an equation for the above general statement of the conservation of energy.


Other Forms of Energy than Mechanical Energy


At this point, we deal with all other forms of energy by lumping them into a single group called other energy (OE ). Then we can
state the conservation of energy in equation form as


(7.65)KEi + PEi +Wnc + OEi = KEf + PEf + OEf.


All types of energy and work can be included in this very general statement of conservation of energy. Kinetic energy is KE ,
work done by a conservative force is represented by PE , work done by nonconservative forces is Wnc , and all other energies
are included as OE . This equation applies to all previous examples; in those situations OE was constant, and so it subtracted
out and was not directly considered.


Making Connections: Usefulness of the Energy Conservation Principle


The fact that energy is conserved and has many forms makes it very important. You will find that energy is discussed in
many contexts, because it is involved in all processes. It will also become apparent that many situations are best understood
in terms of energy and that problems are often most easily conceptualized and solved by considering energy.


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When does OE play a role? One example occurs when a person eats. Food is oxidized with the release of carbon dioxide,
water, and energy. Some of this chemical energy is converted to kinetic energy when the person moves, to potential energy
when the person changes altitude, and to thermal energy (another form of OE ).


Some of the Many Forms of Energy
What are some other forms of energy? You can probably name a number of forms of energy not yet discussed. Many of these
will be covered in later chapters, but let us detail a few here. Electrical energy is a common form that is converted to many other
forms and does work in a wide range of practical situations. Fuels, such as gasoline and food, carry chemical energy that can
be transferred to a system through oxidation. Chemical fuel can also produce electrical energy, such as in batteries. Batteries
can in turn produce light, which is a very pure form of energy. Most energy sources on Earth are in fact stored energy from the
energy we receive from the Sun. We sometimes refer to this as radiant energy, or electromagnetic radiation, which includes
visible light, infrared, and ultraviolet radiation. Nuclear energy comes from processes that convert measurable amounts of mass
into energy. Nuclear energy is transformed into the energy of sunlight, into electrical energy in power plants, and into the energy
of the heat transfer and blast in weapons. Atoms and molecules inside all objects are in random motion. This internal mechanical
energy from the random motions is called thermal energy, because it is related to the temperature of the object. These and all
other forms of energy can be converted into one another and can do work.


Table 7.1 gives the amount of energy stored, used, or released from various objects and in various phenomena. The range of
energies and the variety of types and situations is impressive.


Problem-Solving Strategies for Energy


You will find the following problem-solving strategies useful whenever you deal with energy. The strategies help in organizing
and reinforcing energy concepts. In fact, they are used in the examples presented in this chapter. The familiar general
problem-solving strategies presented earlier—involving identifying physical principles, knowns, and unknowns, checking
units, and so on—continue to be relevant here.


Step 1. Determine the system of interest and identify what information is given and what quantity is to be calculated. A
sketch will help.


Step 2. Examine all the forces involved and determine whether you know or are given the potential energy from the work
done by the forces. Then use step 3 or step 4.


Step 3. If you know the potential energies for the forces that enter into the problem, then forces are all conservative, and you
can apply conservation of mechanical energy simply in terms of potential and kinetic energy. The equation expressing
conservation of energy is


(7.66)KEi + PEi = KEf + PEf.


Step 4. If you know the potential energy for only some of the forces, possibly because some of them are nonconservative
and do not have a potential energy, or if there are other energies that are not easily treated in terms of force and work, then
the conservation of energy law in its most general form must be used.


(7.67)KEi + PEi +Wnc + OEi = KEf + PEf + OEf.


In most problems, one or more of the terms is zero, simplifying its solution. Do not calculate Wc , the work done by


conservative forces; it is already incorporated in the PE terms.
Step 5. You have already identified the types of work and energy involved (in step 2). Before solving for the unknown,
eliminate terms wherever possible to simplify the algebra. For example, choose h = 0 at either the initial or final point, so
that PEg is zero there. Then solve for the unknown in the customary manner.


Step 6. Check the answer to see if it is reasonable. Once you have solved a problem, reexamine the forms of work and
energy to see if you have set up the conservation of energy equation correctly. For example, work done against friction
should be negative, potential energy at the bottom of a hill should be less than that at the top, and so on. Also check to see
that the numerical value obtained is reasonable. For example, the final speed of a skateboarder who coasts down a 3-m-
high ramp could reasonably be 20 km/h, but not 80 km/h.


Transformation of Energy
The transformation of energy from one form into others is happening all the time. The chemical energy in food is converted into
thermal energy through metabolism; light energy is converted into chemical energy through photosynthesis. In a larger example,
the chemical energy contained in coal is converted into thermal energy as it burns to turn water into steam in a boiler. This
thermal energy in the steam in turn is converted to mechanical energy as it spins a turbine, which is connected to a generator to
produce electrical energy. (In all of these examples, not all of the initial energy is converted into the forms mentioned. This
important point is discussed later in this section.)


Another example of energy conversion occurs in a solar cell. Sunlight impinging on a solar cell (see Figure 7.21) produces
electricity, which in turn can be used to run an electric motor. Energy is converted from the primary source of solar energy into
electrical energy and then into mechanical energy.


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Figure 7.21 Solar energy is converted into electrical energy by solar cells, which is used to run a motor in this solar-power aircraft. (credit: NASA)


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Table 7.1 Energy of Various Objects and Phenomena


Object/phenomenon Energy in joules


Big Bang 1068


Energy released in a supernova 1044


Fusion of all the hydrogen in Earth’s oceans 1034


Annual world energy use 4×1020


Large fusion bomb (9 megaton) 3.8×1016


1 kg hydrogen (fusion to helium) 6.4×1014


1 kg uranium (nuclear fission) 8.0×1013


Hiroshima-size fission bomb (10 kiloton) 4.2×1013


90,000-ton aircraft carrier at 30 knots 1.1×1010


1 barrel crude oil 5.9×109


1 ton TNT 4.2×109


1 gallon of gasoline 1.2×108


Daily home electricity use (developed countries) 7×107


Daily adult food intake (recommended) 1.2×107


1000-kg car at 90 km/h 3.1×105


1 g fat (9.3 kcal) 3.9×104


ATP hydrolysis reaction 3.2×104


1 g carbohydrate (4.1 kcal) 1.7×104


1 g protein (4.1 kcal) 1.7×104


Tennis ball at 100 km/h 22


Mosquito ⎛⎝10–2 g at 0.5 m/s

⎠ 1.3×10−6


Single electron in a TV tube beam 4.0×10−15


Energy to break one DNA strand 10−19


Efficiency
Even though energy is conserved in an energy conversion process, the output of useful energy or work will be less than the
energy input. The efficiency Eff of an energy conversion process is defined as


(7.68)
Efficiency(Eff ) = useful energy or work outputtotal energy input =


Wout
Ein


.


Table 7.2 lists some efficiencies of mechanical devices and human activities. In a coal-fired power plant, for example, about 40%
of the chemical energy in the coal becomes useful electrical energy. The other 60% transforms into other (perhaps less useful)
energy forms, such as thermal energy, which is then released to the environment through combustion gases and cooling towers.


Chapter 7 | Work, Energy, and Energy Resources 261




Table 7.2 Efficiency of the Human Body and
Mechanical Devices


Activity/device Efficiency (%)[1]


Cycling and climbing 20


Swimming, surface 2


Swimming, submerged 4


Shoveling 3


Weightlifting 9


Steam engine 17


Gasoline engine 30


Diesel engine 35


Nuclear power plant 35


Coal power plant 42


Electric motor 98


Compact fluorescent light 20


Gas heater (residential) 90


Solar cell 10


PhET Explorations: Masses and Springs


A realistic mass and spring laboratory. Hang masses from springs and adjust the spring stiffness and damping. You can
even slow time. Transport the lab to different planets. A chart shows the kinetic, potential, and thermal energies for each
spring.


Figure 7.22 Masses and Springs (http://cnx.org/content/m42151/1.5/mass-spring-lab_en.jar)


7.7 Power


What is Power?
Power—the word conjures up many images: a professional football player muscling aside his opponent, a dragster roaring away
from the starting line, a volcano blowing its lava into the atmosphere, or a rocket blasting off, as in Figure 7.23.


Figure 7.23 This powerful rocket on the Space Shuttle Endeavor did work and consumed energy at a very high rate. (credit: NASA)


1. Representative values


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These images of power have in common the rapid performance of work, consistent with the scientific definition of power ( P ) as
the rate at which work is done.


Power


Power is the rate at which work is done.


(7.69)P = Wt


The SI unit for power is the watt (W ), where 1 watt equals 1 joule/second (1 W = 1 J/s).


Because work is energy transfer, power is also the rate at which energy is expended. A 60-W light bulb, for example, expends 60
J of energy per second. Great power means a large amount of work or energy developed in a short time. For example, when a
powerful car accelerates rapidly, it does a large amount of work and consumes a large amount of fuel in a short time.


Calculating Power from Energy


Example 7.11 Calculating the Power to Climb Stairs


What is the power output for a 60.0-kg woman who runs up a 3.00 m high flight of stairs in 3.50 s, starting from rest but
having a final speed of 2.00 m/s? (See Figure 7.24.)


Figure 7.24When this woman runs upstairs starting from rest, she converts the chemical energy originally from food into kinetic energy and
gravitational potential energy. Her power output depends on how fast she does this.


Strategy and Concept


The work going into mechanical energy is W= KE + PE . At the bottom of the stairs, we take both KE and PEg as


initially zero; thus, W = KEf + PEg = 12mvf
2 + mgh , where h is the vertical height of the stairs. Because all terms are


given, we can calculate W and then divide it by time to get power.


Solution


Substituting the expression for W into the definition of power given in the previous equation, P = W / t yields
(7.70)


P = Wt =
1
2mvf


2 + mgh
t .


Entering known values yields


Chapter 7 | Work, Energy, and Energy Resources 263




(7.71)
P =


0.5⎛⎝60.0 kg⎞⎠(2.00 m/s)2 + ⎛⎝60.0 kg⎞⎠⎛⎝9.80 m/s2

⎠(3.00 m)


3.50 s
= 120 J + 1764 J3.50 s
= 538 W.


Discussion


The woman does 1764 J of work to move up the stairs compared with only 120 J to increase her kinetic energy; thus, most
of her power output is required for climbing rather than accelerating.


It is impressive that this woman’s useful power output is slightly less than 1 horsepower (1 hp = 746 W) ! People can
generate more than a horsepower with their leg muscles for short periods of time by rapidly converting available blood sugar and
oxygen into work output. (A horse can put out 1 hp for hours on end.) Once oxygen is depleted, power output decreases and the
person begins to breathe rapidly to obtain oxygen to metabolize more food—this is known as the aerobic stage of exercise. If the
woman climbed the stairs slowly, then her power output would be much less, although the amount of work done would be the
same.


Making Connections: Take-Home Investigation—Measure Your Power Rating


Determine your own power rating by measuring the time it takes you to climb a flight of stairs. We will ignore the gain in
kinetic energy, as the above example showed that it was a small portion of the energy gain. Don’t expect that your output will
be more than about 0.5 hp.


Examples of Power
Examples of power are limited only by the imagination, because there are as many types as there are forms of work and energy.
(See Table 7.3 for some examples.) Sunlight reaching Earth’s surface carries a maximum power of about 1.3 kilowatts per


square meter (kW/m2). A tiny fraction of this is retained by Earth over the long term. Our consumption rate of fossil fuels is far
greater than the rate at which they are stored, so it is inevitable that they will be depleted. Power implies that energy is
transferred, perhaps changing form. It is never possible to change one form completely into another without losing some of it as
thermal energy. For example, a 60-W incandescent bulb converts only 5 W of electrical power to light, with 55 W dissipating into
thermal energy. Furthermore, the typical electric power plant converts only 35 to 40% of its fuel into electricity. The remainder
becomes a huge amount of thermal energy that must be dispersed as heat transfer, as rapidly as it is created. A coal-fired power


plant may produce 1000 megawatts; 1 megawatt (MW) is 106 W of electric power. But the power plant consumes chemical
energy at a rate of about 2500 MW, creating heat transfer to the surroundings at a rate of 1500 MW. (See Figure 7.25.)


Figure 7.25 Tremendous amounts of electric power are generated by coal-fired power plants such as this one in China, but an even larger amount of
power goes into heat transfer to the surroundings. The large cooling towers here are needed to transfer heat as rapidly as it is produced. The transfer
of heat is not unique to coal plants but is an unavoidable consequence of generating electric power from any fuel—nuclear, coal, oil, natural gas, or the
like. (credit: Kleinolive, Wikimedia Commons)


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Table 7.3 Power Output or Consumption


Object or Phenomenon Power in Watts


Supernova (at peak) 5×1037


Milky Way galaxy 1037


Crab Nebula pulsar 1028


The Sun 4×1026


Volcanic eruption (maximum) 4×1015


Lightning bolt 2×1012


Nuclear power plant (total electric and heat transfer) 3×109


Aircraft carrier (total useful and heat transfer) 108


Dragster (total useful and heat transfer) 2×106


Car (total useful and heat transfer) 8×104


Football player (total useful and heat transfer) 5×103


Clothes dryer 4×103


Person at rest (all heat transfer) 100


Typical incandescent light bulb (total useful and heat transfer) 60


Heart, person at rest (total useful and heat transfer) 8


Electric clock 3


Pocket calculator 10−3


Power and Energy Consumption
We usually have to pay for the energy we use. It is interesting and easy to estimate the cost of energy for an electrical appliance
if its power consumption rate and time used are known. The higher the power consumption rate and the longer the appliance is
used, the greater the cost of that appliance. The power consumption rate is P = W / t = E / t , where E is the energy supplied
by the electricity company. So the energy consumed over a time t is


(7.72)E = Pt.


Electricity bills state the energy used in units of kilowatt-hours (kW ⋅ h), which is the product of power in kilowatts and time in
hours. This unit is convenient because electrical power consumption at the kilowatt level for hours at a time is typical.


Example 7.12 Calculating Energy Costs


What is the cost of running a 0.200-kW computer 6.00 h per day for 30.0 d if the cost of electricity is $0.120 per kW ⋅ h ?
Strategy


Cost is based on energy consumed; thus, we must find E from E = Pt and then calculate the cost. Because electrical
energy is expressed in kW ⋅ h , at the start of a problem such as this it is convenient to convert the units into kW and
hours.


Solution


The energy consumed in kW ⋅ h is
(7.73)E = Pt = (0.200 kW)(6.00 h/d)(30.0 d)


= 36.0 kW ⋅ h,


and the cost is simply given by


Chapter 7 | Work, Energy, and Energy Resources 265




(7.74)cost = (36.0 kW ⋅ h)($0.120 per kW ⋅ h) = $4.32 per month.


Discussion


The cost of using the computer in this example is neither exorbitant nor negligible. It is clear that the cost is a combination of
power and time. When both are high, such as for an air conditioner in the summer, the cost is high.


The motivation to save energy has become more compelling with its ever-increasing price. Armed with the knowledge that
energy consumed is the product of power and time, you can estimate costs for yourself and make the necessary value
judgments about where to save energy. Either power or time must be reduced. It is most cost-effective to limit the use of high-
power devices that normally operate for long periods of time, such as water heaters and air conditioners. This would not include
relatively high power devices like toasters, because they are on only a few minutes per day. It would also not include electric
clocks, in spite of their 24-hour-per-day usage, because they are very low power devices. It is sometimes possible to use devices
that have greater efficiencies—that is, devices that consume less power to accomplish the same task. One example is the
compact fluorescent light bulb, which produces over four times more light per watt of power consumed than its incandescent
cousin.


Modern civilization depends on energy, but current levels of energy consumption and production are not sustainable. The
likelihood of a link between global warming and fossil fuel use (with its concomitant production of carbon dioxide), has made
reduction in energy use as well as a shift to non-fossil fuels of the utmost importance. Even though energy in an isolated system
is a conserved quantity, the final result of most energy transformations is waste heat transfer to the environment, which is no
longer useful for doing work. As we will discuss in more detail in Thermodynamics, the potential for energy to produce useful
work has been “degraded” in the energy transformation.


7.8 Work, Energy, and Power in Humans


Energy Conversion in Humans
Our own bodies, like all living organisms, are energy conversion machines. Conservation of energy implies that the chemical
energy stored in food is converted into work, thermal energy, and/or stored as chemical energy in fatty tissue. (See Figure 7.26.)
The fraction going into each form depends both on how much we eat and on our level of physical activity. If we eat more than is
needed to do work and stay warm, the remainder goes into body fat.


Figure 7.26 Energy consumed by humans is converted to work, thermal energy, and stored fat. By far the largest fraction goes to thermal energy,
although the fraction varies depending on the type of physical activity.


Power Consumed at Rest
The rate at which the body uses food energy to sustain life and to do different activities is called the metabolic rate. The total
energy conversion rate of a person at rest is called the basal metabolic rate (BMR) and is divided among various systems in the
body, as shown in Table 7.4. The largest fraction goes to the liver and spleen, with the brain coming next. Of course, during
vigorous exercise, the energy consumption of the skeletal muscles and heart increase markedly. About 75% of the calories
burned in a day go into these basic functions. The BMR is a function of age, gender, total body weight, and amount of muscle
mass (which burns more calories than body fat). Athletes have a greater BMR due to this last factor.


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Table 7.4 Basal Metabolic Rates (BMR)


Organ Power consumed at rest (W) Oxygen consumption (mL/min) Percent of BMR


Liver & spleen 23 67 27


Brain 16 47 19


Skeletal muscle 15 45 18


Kidney 9 26 10


Heart 6 17 7


Other 16 48 19


Totals 85 W 250 mL/min 100%


Energy consumption is directly proportional to oxygen consumption because the digestive process is basically one of oxidizing
food. We can measure the energy people use during various activities by measuring their oxygen use. (See Figure 7.27.)
Approximately 20 kJ of energy are produced for each liter of oxygen consumed, independent of the type of food. Table 7.5
shows energy and oxygen consumption rates (power expended) for a variety of activities.


Power of Doing Useful Work
Work done by a person is sometimes called useful work, which is work done on the outside world, such as lifting weights. Useful
work requires a force exerted through a distance on the outside world, and so it excludes internal work, such as that done by the
heart when pumping blood. Useful work does include that done in climbing stairs or accelerating to a full run, because these are
accomplished by exerting forces on the outside world. Forces exerted by the body are nonconservative, so that they can change
the mechanical energy (KE + PE ) of the system worked upon, and this is often the goal. A baseball player throwing a ball, for
example, increases both the ball’s kinetic and potential energy.


If a person needs more energy than they consume, such as when doing vigorous work, the body must draw upon the chemical
energy stored in fat. So exercise can be helpful in losing fat. However, the amount of exercise needed to produce a loss in fat, or
to burn off extra calories consumed that day, can be large, as Example 7.13 illustrates.


Example 7.13 Calculating Weight Loss from Exercising


If a person who normally requires an average of 12,000 kJ (3000 kcal) of food energy per day consumes 13,000 kJ per day,
he will steadily gain weight. How much bicycling per day is required to work off this extra 1000 kJ?


Solution


Table 7.5 states that 400 W are used when cycling at a moderate speed. The time required to work off 1000 kJ at this rate is
then


(7.75)Time = energy⎛

energy
time





= 1000 kJ400 W = 2500 s = 42 min.


Discussion


If this person uses more energy than he or she consumes, the person’s body will obtain the needed energy by metabolizing
body fat. If the person uses 13,000 kJ but consumes only 12,000 kJ, then the amount of fat loss will be


(7.76)
Fat loss = (1000 kJ)⎛⎝


1.0 g fat
39 kJ



⎠ = 26 g,


assuming the energy content of fat to be 39 kJ/g.


Chapter 7 | Work, Energy, and Energy Resources 267




Figure 7.27 A pulse oxymeter is an apparatus that measures the amount of oxygen in blood. Oxymeters can be used to determine a person’s
metabolic rate, which is the rate at which food energy is converted to another form. Such measurements can indicate the level of athletic conditioning
as well as certain medical problems. (credit: UusiAjaja, Wikimedia Commons)


Table 7.5 Energy and Oxygen Consumption Rates[2] (Power)


Activity Energy consumption in watts Oxygen consumption in liters O2/min


Sleeping 83 0.24


Sitting at rest 120 0.34


Standing relaxed 125 0.36


Sitting in class 210 0.60


Walking (5 km/h) 280 0.80


Cycling (13–18 km/h) 400 1.14


Shivering 425 1.21


Playing tennis 440 1.26


Swimming breaststroke 475 1.36


Ice skating (14.5 km/h) 545 1.56


Climbing stairs (116/min) 685 1.96


Cycling (21 km/h) 700 2.00


Running cross-country 740 2.12


Playing basketball 800 2.28


Cycling, professional racer 1855 5.30


Sprinting 2415 6.90


All bodily functions, from thinking to lifting weights, require energy. (See Figure 7.28.) The many small muscle actions
accompanying all quiet activity, from sleeping to head scratching, ultimately become thermal energy, as do less visible muscle
actions by the heart, lungs, and digestive tract. Shivering, in fact, is an involuntary response to low body temperature that pits
muscles against one another to produce thermal energy in the body (and do no work). The kidneys and liver consume a
surprising amount of energy, but the biggest surprise of all it that a full 25% of all energy consumed by the body is used to
maintain electrical potentials in all living cells. (Nerve cells use this electrical potential in nerve impulses.) This bioelectrical
energy ultimately becomes mostly thermal energy, but some is utilized to power chemical processes such as in the kidneys and
liver, and in fat production.


2. for an average 76-kg male


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Figure 7.28 This fMRI scan shows an increased level of energy consumption in the vision center of the brain. Here, the patient was being asked to
recognize faces. (credit: NIH via Wikimedia Commons)


7.9 World Energy Use
Energy is an important ingredient in all phases of society. We live in a very interdependent world, and access to adequate and
reliable energy resources is crucial for economic growth and for maintaining the quality of our lives. But current levels of energy
consumption and production are not sustainable. About 40% of the world’s energy comes from oil, and much of that goes to
transportation uses. Oil prices are dependent as much upon new (or foreseen) discoveries as they are upon political events and
situations around the world. The U.S., with 4.5% of the world’s population, consumes 24% of the world’s oil production per year;
66% of that oil is imported!


Renewable and Nonrenewable Energy Sources
The principal energy resources used in the world are shown in Figure 7.29. The fuel mix has changed over the years but now is
dominated by oil, although natural gas and solar contributions are increasing. Renewable forms of energy are those sources
that cannot be used up, such as water, wind, solar, and biomass. About 85% of our energy comes from nonrenewable fossil
fuels—oil, natural gas, coal. The likelihood of a link between global warming and fossil fuel use, with its production of carbon
dioxide through combustion, has made, in the eyes of many scientists, a shift to non-fossil fuels of utmost importance—but it will
not be easy.


Figure 7.29World energy consumption by source, in billions of kilowatt-hours: 2006. (credit: KVDP)


The World’s Growing Energy Needs
World energy consumption continues to rise, especially in the developing countries. (See Figure 7.30.) Global demand for
energy has tripled in the past 50 years and might triple again in the next 30 years. While much of this growth will come from the
rapidly booming economies of China and India, many of the developed countries, especially those in Europe, are hoping to meet
their energy needs by expanding the use of renewable sources. Although presently only a small percentage, renewable energy is
growing very fast, especially wind energy. For example, Germany plans to meet 20% of its electricity and 10% of its overall
energy needs with renewable resources by the year 2020. (See Figure 7.31.) Energy is a key constraint in the rapid economic
growth of China and India. In 2003, China surpassed Japan as the world’s second largest consumer of oil. However, over 1/3 of
this is imported. Unlike most Western countries, coal dominates the commercial energy resources of China, accounting for 2/3 of


Chapter 7 | Work, Energy, and Energy Resources 269




its energy consumption. In 2009 China surpassed the United States as the largest generator of CO2 . In India, the main energy
resources are biomass (wood and dung) and coal. Half of India’s oil is imported. About 70% of India’s electricity is generated by
highly polluting coal. Yet there are sizeable strides being made in renewable energy. India has a rapidly growing wind energy
base, and it has the largest solar cooking program in the world.


Figure 7.30 Past and projected world energy use (source: Based on data from U.S. Energy Information Administration, 2011)


Figure 7.31 Solar cell arrays at a power plant in Steindorf, Germany (credit: Michael Betke, Flickr)


Table 7.6 displays the 2006 commercial energy mix by country for some of the prime energy users in the world. While non-
renewable sources dominate, some countries get a sizeable percentage of their electricity from renewable resources. For
example, about 67% of New Zealand’s electricity demand is met by hydroelectric. Only 10% of the U.S. electricity is generated
by renewable resources, primarily hydroelectric. It is difficult to determine total contributions of renewable energy in some
countries with a large rural population, so these percentages in this table are left blank.


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Table 7.6 Energy Consumption—Selected Countries (2006)


Country
Consumption,
in EJ (1018 J)


Oil NaturalGas Coal Nuclear Hydro
Other
Renewables


Electricity
Use per
capita
(kWh/yr)


Energy
Use
per
capita
(GJ/yr)


Australia 5.4 34% 17% 44% 0% 3% 1% 10000 260


Brazil 9.6 48% 7% 5% 1% 35% 2% 2000 50


China 63 22% 3% 69% 1% 6% 1500 35


Egypt 2.4 50% 41% 1% 0% 6% 990 32


Germany 16 37% 24% 24% 11% 1% 3% 6400 173


India 15 34% 7% 52% 1% 5% 470 13


Indonesia 4.9 51% 26% 16% 0% 2% 3% 420 22


Japan 24 48% 14% 21% 12% 4% 1% 7100 176


New
Zealand 0.44 32% 26% 6% 0% 11% 19% 8500 102


Russia 31 19% 53% 16% 5% 6% 5700 202


U.S. 105 40% 23% 22% 8% 3% 1% 12500 340


World 432 39% 23% 24% 6% 6% 2% 2600 71


Energy and Economic Well-being
The last two columns in this table examine the energy and electricity use per capita. Economic well-being is dependent upon
energy use, and in most countries higher standards of living, as measured by GDP (gross domestic product) per capita, are
matched by higher levels of energy consumption per capita. This is borne out in Figure 7.32. Increased efficiency of energy use
will change this dependency. A global problem is balancing energy resource development against the harmful effects upon the
environment in its extraction and use.


Figure 7.32 Power consumption per capita versus GDP per capita for various countries. Note the increase in energy usage with increasing GDP.
(2007, credit: Frank van Mierlo, Wikimedia Commons)


Conserving Energy
As we finish this chapter on energy and work, it is relevant to draw some distinctions between two sometimes misunderstood
terms in the area of energy use. As has been mentioned elsewhere, the “law of the conservation of energy” is a very useful
principle in analyzing physical processes. It is a statement that cannot be proven from basic principles, but is a very good


Chapter 7 | Work, Energy, and Energy Resources 271




basal metabolic rate:


chemical energy:


conservation of mechanical energy:


conservative force:


efficiency:


electrical energy:


energy:


fossil fuels:


friction:


gravitational potential energy:


horsepower:


joule:


kilowatt-hour:


kinetic energy:


law of conservation of energy:


mechanical energy:


metabolic rate:


net work:


nonconservative force:


nuclear energy:


potential energy:


potential energy of a spring:


power:


bookkeeping device, and no exceptions have ever been found. It states that the total amount of energy in an isolated system will
always remain constant. Related to this principle, but remarkably different from it, is the important philosophy of energy
conservation. This concept has to do with seeking to decrease the amount of energy used by an individual or group through (1)
reduced activities (e.g., turning down thermostats, driving fewer kilometers) and/or (2) increasing conversion efficiencies in the
performance of a particular task—such as developing and using more efficient room heaters, cars that have greater miles-per-
gallon ratings, energy-efficient compact fluorescent lights, etc.


Since energy in an isolated system is not destroyed or created or generated, one might wonder why we need to be concerned
about our energy resources, since energy is a conserved quantity. The problem is that the final result of most energy
transformations is waste heat transfer to the environment and conversion to energy forms no longer useful for doing work. To
state it in another way, the potential for energy to produce useful work has been “degraded” in the energy transformation. (This
will be discussed in more detail in Thermodynamics.)


Glossary
the total energy conversion rate of a person at rest


the energy in a substance stored in the bonds between atoms and molecules that can be released in a
chemical reaction


the rule that the sum of the kinetic energies and potential energies remains constant if
only conservative forces act on and within a system


a force that does the same work for any given initial and final configuration, regardless of the path
followed


a measure of the effectiveness of the input of energy to do work; useful energy or work divided by the total input of
energy


the energy carried by a flow of charge


the ability to do work


oil, natural gas, and coal


the force between surfaces that opposes one sliding on the other; friction changes mechanical energy into thermal
energy


the energy an object has due to its position in a gravitational field


an older non-SI unit of power, with 1 hp = 746 W


SI unit of work and energy, equal to one newton-meter


(kW ⋅ h) unit used primarily for electrical energy provided by electric utility companies


the energy an object has by reason of its motion, equal to 12mv
2 for the translational (i.e., non-rotational)


motion of an object of mass m moving at speed v


the general law that total energy is constant in any process; energy may change in form or
be transferred from one system to another, but the total remains the same


the sum of kinetic energy and potential energy


the rate at which the body uses food energy to sustain life and to do different activities


work done by the net force, or vector sum of all the forces, acting on an object


a force whose work depends on the path followed between the given initial and final configurations


energy released by changes within atomic nuclei, such as the fusion of two light nuclei or the fission of a
heavy nucleus


energy due to position, shape, or configuration


the stored energy of a spring as a function of its displacement; when Hooke’s law applies, it is


given by the expression 12kx
2 where x is the distance the spring is compressed or extended and k is the spring


constant


the rate at which work is done


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radiant energy:


renewable forms of energy:


thermal energy:


useful work:


watt:


work:


work-energy theorem:


the energy carried by electromagnetic waves


those sources that cannot be used up, such as water, wind, solar, and biomass


the energy within an object due to the random motion of its atoms and molecules that accounts for the
object's temperature


work done on an external system


(W) SI unit of power, with 1 W = 1 J/s


the transfer of energy by a force that causes an object to be displaced; the product of the component of the force in the
direction of the displacement and the magnitude of the displacement


the result, based on Newton’s laws, that the net work done on an object is equal to its change in
kinetic energy


Section Summary


7.1 Work: The Scientific Definition
• Work is the transfer of energy by a force acting on an object as it is displaced.
• The work W that a force F does on an object is the product of the magnitude F of the force, times the magnitude d of


the displacement, times the cosine of the angle θ between them. In symbols,


W = Fd cos θ.
• The SI unit for work and energy is the joule (J), where 1 J = 1 N ⋅ m = 1 kg ⋅ m2/s2 .
• The work done by a force is zero if the displacement is either zero or perpendicular to the force.
• The work done is positive if the force and displacement have the same direction, and negative if they have opposite


direction.


7.2 Kinetic Energy and the Work-Energy Theorem
• The net work Wnet is the work done by the net force acting on an object.


• Work done on an object transfers energy to the object.


• The translational kinetic energy of an object of mass m moving at speed v is KE = 12mv
2 .


• The work-energy theorem states that the net work Wnet on a system changes its kinetic energy,


Wnet = 12mv
2 − 12mv0


2 .


7.3 Gravitational Potential Energy
• Work done against gravity in lifting an object becomes potential energy of the object-Earth system.
• The change in gravitational potential energy, ΔPEg , is ΔPEg = mgh , with h being the increase in height and g the


acceleration due to gravity.
• The gravitational potential energy of an object near Earth’s surface is due to its position in the mass-Earth system. Only


differences in gravitational potential energy, ΔPEg , have physical significance.
• As an object descends without friction, its gravitational potential energy changes into kinetic energy corresponding to


increasing speed, so that ΔKE= −ΔPEg .


7.4 Conservative Forces and Potential Energy
• A conservative force is one for which work depends only on the starting and ending points of a motion, not on the path


taken.
• We can define potential energy (PE) for any conservative force, just as we defined PEg for the gravitational force.


• The potential energy of a spring is PEs = 12kx
2 , where k is the spring’s force constant and x is the displacement from


its undeformed position.
• Mechanical energy is defined to be KE + PE for a conservative force.
• When only conservative forces act on and within a system, the total mechanical energy is constant. In equation form,


Chapter 7 | Work, Energy, and Energy Resources 273




KE + PE = constant
or
KEi + PEi = KEf + PEf







where i and f denote initial and final values. This is known as the conservation of mechanical energy.


7.5 Nonconservative Forces
• A nonconservative force is one for which work depends on the path.
• Friction is an example of a nonconservative force that changes mechanical energy into thermal energy.
• Work Wnc done by a nonconservative force changes the mechanical energy of a system. In equation form,


Wnc = ΔKE + ΔPE or, equivalently, KEi + PEi +Wnc = KEf + PEf .
• When both conservative and nonconservative forces act, energy conservation can be applied and used to calculate motion


in terms of the known potential energies of the conservative forces and the work done by nonconservative forces, instead
of finding the net work from the net force, or having to directly apply Newton’s laws.


7.6 Conservation of Energy
• The law of conservation of energy states that the total energy is constant in any process. Energy may change in form or be


transferred from one system to another, but the total remains the same.
• When all forms of energy are considered, conservation of energy is written in equation form as
KEi + PEi +Wnc + OEi = KEf + PEf + OEf , where OE is all other forms of energy besides mechanical energy.


• Commonly encountered forms of energy include electric energy, chemical energy, radiant energy, nuclear energy, and
thermal energy.


• Energy is often utilized to do work, but it is not possible to convert all the energy of a system to work.


• The efficiency Eff of a machine or human is defined to be Eff = Wout
Ein


, where Wout is useful work output and Ein is


the energy consumed.


7.7 Power
• Power is the rate at which work is done, or in equation form, for the average power P for work W done over a time t ,


P = W / t .
• The SI unit for power is the watt (W), where 1 W = 1 J/s .
• The power of many devices such as electric motors is also often expressed in horsepower (hp), where 1 hp = 746 W .


7.8 Work, Energy, and Power in Humans
• The human body converts energy stored in food into work, thermal energy, and/or chemical energy that is stored in fatty


tissue.
• The rate at which the body uses food energy to sustain life and to do different activities is called the metabolic rate, and the


corresponding rate when at rest is called the basal metabolic rate (BMR)
• The energy included in the basal metabolic rate is divided among various systems in the body, with the largest fraction


going to the liver and spleen, and the brain coming next.
• About 75% of food calories are used to sustain basic body functions included in the basal metabolic rate.
• The energy consumption of people during various activities can be determined by measuring their oxygen use, because the


digestive process is basically one of oxidizing food.


7.9 World Energy Use
• The relative use of different fuels to provide energy has changed over the years, but fuel use is currently dominated by oil,


although natural gas and solar contributions are increasing.
• Although non-renewable sources dominate, some countries meet a sizeable percentage of their electricity needs from


renewable resources.
• The United States obtains only about 10% of its energy from renewable sources, mostly hydroelectric power.
• Economic well-being is dependent upon energy use, and in most countries higher standards of living, as measured by GDP


(Gross Domestic Product) per capita, are matched by higher levels of energy consumption per capita.
• Even though, in accordance with the law of conservation of energy, energy can never be created or destroyed, energy that


can be used to do work is always partly converted to less useful forms, such as waste heat to the environment, in all of our
uses of energy for practical purposes.


Conceptual Questions


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7.1 Work: The Scientific Definition
1. Give an example of something we think of as work in everyday circumstances that is not work in the scientific sense. Is energy
transferred or changed in form in your example? If so, explain how this is accomplished without doing work.


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2. Give an example of a situation in which there is a force and a displacement, but the force does no work. Explain why it does
no work.


3. Describe a situation in which a force is exerted for a long time but does no work. Explain.


7.2 Kinetic Energy and the Work-Energy Theorem
4. The person in Figure 7.33 does work on the lawn mower. Under what conditions would the mower gain energy? Under what
conditions would it lose energy?


Figure 7.33


5.Work done on a system puts energy into it. Work done by a system removes energy from it. Give an example for each
statement.


6.When solving for speed in Example 7.4, we kept only the positive root. Why?


7.3 Gravitational Potential Energy
7. In Example 7.7, we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5
m/s downhill. Suppose the roller coaster had had an initial speed of 5 m/s uphill instead, and it coasted uphill, stopped, and then
rolled back down to a final point 20 m below the start. We would find in that case that it had the same final speed. Explain in
terms of conservation of energy.


8. Does the work you do on a book when you lift it onto a shelf depend on the path taken? On the time taken? On the height of
the shelf? On the mass of the book?


7.4 Conservative Forces and Potential Energy
9.What is a conservative force?


10. The force exerted by a diving board is conservative, provided the internal friction is negligible. Assuming friction is negligible,
describe changes in the potential energy of a diving board as a swimmer dives from it, starting just before the swimmer steps on
the board until just after his feet leave it.


11. Define mechanical energy. What is the relationship of mechanical energy to nonconservative forces? What happens to
mechanical energy if only conservative forces act?


12.What is the relationship of potential energy to conservative force?


7.6 Conservation of Energy
13. Consider the following scenario. A car for which friction is not negligible accelerates from rest down a hill, running out of
gasoline after a short distance. The driver lets the car coast farther down the hill, then up and over a small crest. He then coasts
down that hill into a gas station, where he brakes to a stop and fills the tank with gasoline. Identify the forms of energy the car
has, and how they are changed and transferred in this series of events. (See Figure 7.34.)


Chapter 7 | Work, Energy, and Energy Resources 275




Figure 7.34 A car experiencing non-negligible friction coasts down a hill, over a small crest, then downhill again, and comes to a stop at a gas station.


14. Describe the energy transfers and transformations for a javelin, starting from the point at which an athlete picks up the javelin
and ending when the javelin is stuck into the ground after being thrown.


15. Do devices with efficiencies of less than one violate the law of conservation of energy? Explain.


16. List four different forms or types of energy. Give one example of a conversion from each of these forms to another form.


17. List the energy conversions that occur when riding a bicycle.


7.7 Power
18. Most electrical appliances are rated in watts. Does this rating depend on how long the appliance is on? (When off, it is a zero-
watt device.) Explain in terms of the definition of power.


19. Explain, in terms of the definition of power, why energy consumption is sometimes listed in kilowatt-hours rather than joules.
What is the relationship between these two energy units?


20. A spark of static electricity, such as that you might receive from a doorknob on a cold dry day, may carry a few hundred watts
of power. Explain why you are not injured by such a spark.


7.8 Work, Energy, and Power in Humans
21. Explain why it is easier to climb a mountain on a zigzag path rather than one straight up the side. Is your increase in
gravitational potential energy the same in both cases? Is your energy consumption the same in both?


22. Do you do work on the outside world when you rub your hands together to warm them? What is the efficiency of this activity?


23. Shivering is an involuntary response to lowered body temperature. What is the efficiency of the body when shivering, and is
this a desirable value?


24. Discuss the relative effectiveness of dieting and exercise in losing weight, noting that most athletic activities consume food
energy at a rate of 400 to 500 W, while a single cup of yogurt can contain 1360 kJ (325 kcal). Specifically, is it likely that exercise
alone will be sufficient to lose weight? You may wish to consider that regular exercise may increase the metabolic rate, whereas
protracted dieting may reduce it.


7.9 World Energy Use
25.What is the difference between energy conservation and the law of conservation of energy? Give some examples of each.


26. If the efficiency of a coal-fired electrical generating plant is 35%, then what do we mean when we say that energy is a
conserved quantity?


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Problems & Exercises


7.1 Work: The Scientific Definition
1. How much work does a supermarket checkout attendant
do on a can of soup he pushes 0.600 m horizontally with a
force of 5.00 N? Express your answer in joules and
kilocalories.


2. A 75.0-kg person climbs stairs, gaining 2.50 meters in
height. Find the work done to accomplish this task.


3. (a) Calculate the work done on a 1500-kg elevator car by
its cable to lift it 40.0 m at constant speed, assuming friction
averages 100 N. (b) What is the work done on the lift by the
gravitational force in this process? (c) What is the total work
done on the lift?


4. 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 goes into
useful work by the force that keeps the car moving at
constant speed despite friction. (See Table 7.1 for the energy
content of gasoline.) (a) What is the magnitude of the force
exerted to keep the car moving at constant speed? (b) If the
required force is directly proportional to speed, how many
gallons will be used to drive 108 km at a speed of 28.0 m/s?


5. Calculate the work done by an 85.0-kg man who pushes a
crate 4.00 m up along a ramp that makes an angle of 20.0º
with the horizontal. (See Figure 7.35.) He exerts a force of
500 N on the crate parallel to the ramp and moves at a
constant speed. Be certain to include the work he does on the
crate and on his body to get up the ramp.


Figure 7.35 A man pushes a crate up a ramp.


6. How much work is done by the boy pulling his sister 30.0 m
in a wagon as shown in Figure 7.36? Assume no friction acts
on the wagon.


Figure 7.36 The boy does work on the system of the wagon and the
child when he pulls them as shown.


7. A shopper pushes a grocery cart 20.0 m at constant speed
on level ground, against a 35.0 N frictional force. He pushes
in a direction 25.0º below the horizontal. (a) What is the
work done on the cart by friction? (b) What is the work done
on the cart by the gravitational force? (c) What is the work
done on the cart by the shopper? (d) Find the force the


shopper exerts, using energy considerations. (e) What is the
total work done on the cart?


8. Suppose the ski patrol lowers a rescue sled and victim,
having a total mass of 90.0 kg, down a 60.0º slope at
constant speed, as shown in Figure 7.37. The coefficient of
friction between the sled and the snow is 0.100. (a) How
much work is done by friction as the sled moves 30.0 m along
the hill? (b) How much work is done by the rope on the sled in
this distance? (c) What is the work done by the gravitational
force on the sled? (d) What is the total work done?


Figure 7.37 A rescue sled and victim are lowered down a steep slope.


7.2 Kinetic Energy and the Work-Energy
Theorem
9. Compare the kinetic energy of a 20,000-kg truck moving at
110 km/h with that of an 80.0-kg astronaut in orbit moving at
27,500 km/h.


10. (a) How fast must a 3000-kg elephant move to have the
same kinetic energy as a 65.0-kg sprinter running at 10.0 m/
s? (b) Discuss how the larger energies needed for the
movement of larger animals would relate to metabolic rates.


11. Confirm the value given for the kinetic energy of an
aircraft carrier in Table 7.1. You will need to look up the
definition of a nautical mile (1 knot = 1 nautical mile/h).


12. (a) Calculate the force needed to bring a 950-kg car to
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) Suppose instead
the car hits a concrete abutment at full speed and is brought
to a stop in 2.00 m. Calculate the force exerted on the car and
compare it with the force found in part (a).


13. A car’s bumper is designed to withstand a 4.0-km/h
(1.1-m/s) collision with an immovable object without damage
to the body of the car. The bumper cushions the shock by
absorbing the force over a distance. Calculate the magnitude
of the average force on a bumper that collapses 0.200 m
while bringing a 900-kg car to rest from an initial speed of 1.1
m/s.


14. Boxing gloves are padded to lessen the force of a blow.
(a) Calculate the force exerted by a boxing glove on an
opponent’s face, if the glove and face compress 7.50 cm
during a blow in which the 7.00-kg arm and glove are brought
to rest from an initial speed of 10.0 m/s. (b) Calculate the
force exerted by an identical blow in the gory old days when
no gloves were used and the knuckles and face would
compress only 2.00 cm. (c) Discuss the magnitude of the


Chapter 7 | Work, Energy, and Energy Resources 277




force with glove on. Does it seem high enough to cause
damage even though it is lower than the force with no glove?


15. Using energy considerations, calculate the average force
a 60.0-kg sprinter exerts backward on the track to accelerate
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 of 30.0 N
against him.


7.3 Gravitational Potential Energy
16. A hydroelectric power facility (see Figure 7.38) converts
the gravitational potential energy of water behind a dam to
electric energy. (a) What is the gravitational potential energy


relative to the generators of a lake of volume 50.0 km3 (
mass = 5.00×1013 kg) , given that the lake has an
average height of 40.0 m above the generators? (b) Compare
this with the energy stored in a 9-megaton fusion bomb.


Figure 7.39 A toy car moves up a sloped track. (credit: Leszek
Leszczynski, Flickr)


21. In a downhill ski race, surprisingly, little advantage is
gained by getting a running start. (This is because the initial
kinetic energy is small compared with the gain in gravitational
potential energy on even small hills.) To demonstrate this, find
the final speed and the time taken for a skier who skies 70.0
m along a 30º slope neglecting friction: (a) Starting from
rest. (b) Starting with an initial speed of 2.50 m/s. (c) Does the
answer surprise you? Discuss why it is still advantageous to
get a running start in very competitive events.


7.4 Conservative Forces and Potential Energy


22. A 5.00×105-kg subway train is brought to a stop from a
speed of 0.500 m/s in 0.400 m by a large spring bumper at
the end of its track. What is the force constant k of the
spring?


23. A pogo stick has a spring with a force constant of


2.50×104 N/m , which can be compressed 12.0 cm. To
what maximum height can a child jump on the stick using only
the energy in the spring, if the child and stick have a total
mass of 40.0 kg? Explicitly show how you follow the steps in
the Problem-Solving Strategies for Energy.


7.5 Nonconservative Forces
24. A 60.0-kg skier with an initial speed of 12.0 m/s coasts up
a 2.50-m-high rise as shown in Figure 7.40. Find her final
speed at the top, given that the coefficient of friction between
her skis and the snow is 0.0800. (Hint: Find the distance
traveled up the incline assuming a straight-line path as shown
in the figure.)


Figure 7.40 The skier’s initial kinetic energy is partially used in coasting
to the top of a rise.


25. (a) How high a hill can a car coast up (engine
disengaged) if work done by friction is negligible and its initial
speed is 110 km/h? (b) If, in actuality, a 750-kg car with an
initial speed of 110 km/h is observed to coast up a hill to a
height 22.0 m above its starting point, how much thermal
energy was generated by friction? (c) What is the average
force of friction if the hill has a slope 2.5º above the
horizontal?


278 Chapter 7 | Work, Energy, and Energy Resources


Figure 7.38 Hydroelectric facility (credit: Denis Belevich, Wikimedia
Commons)


17. (a) How much gravitational potential energy (relative to
the ground on which it is built) is stored in the Great Pyramid


of Cheops, given that its mass is about 7 × 109 kg and its
center of mass is 36.5 m above the surrounding ground? (b)
How does this energy compare with the daily food intake of a
person?


18. Suppose a 350-g kookaburra (a large kingfisher bird)
picks up a 75-g snake and raises it 2.5 m from the ground to
a branch. (a) How much work did the bird do on the snake?
(b) How much work did it do to raise its own center of mass to
the branch?


19. In Example 7.7, we found that the speed of a roller
coaster that had descended 20.0 m was only slightly greater
when it had an initial speed of 5.00 m/s than when it started
from rest. This implies that ΔPE >> KEi . Confirm this
statement by taking the ratio of ΔPE to KEi . (Note that
mass cancels.)


20. A 100-g toy car is propelled by a compressed spring that
starts it moving. The car follows the curved track in Figure
7.39. Show that the final speed of the toy car is 0.687 m/s if
its initial speed is 2.00 m/s and it coasts up the frictionless
slope, gaining 0.180 m in altitude.


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7.6 Conservation of Energy
26. Using values from Table 7.1, how many DNA molecules
could be broken by the energy carried by a single electron in
the beam of an old-fashioned TV tube? (These electrons
were not dangerous in themselves, but they did create
dangerous x rays. Later model tube TVs had shielding that
absorbed x rays before they escaped and exposed viewers.)


27. Using energy considerations and assuming negligible air
resistance, show that a rock thrown from a bridge 20.0 m
above water with an initial speed of 15.0 m/s strikes the water
with a speed of 24.8 m/s independent of the direction thrown.


28. If the energy in fusion bombs were used to supply the
energy needs of the world, how many of the 9-megaton
variety would be needed for a year’s supply of energy (using
data from Table 7.1)? This is not as far-fetched as it may
sound—there are thousands of nuclear bombs, and their
energy can be trapped in underground explosions and
converted to electricity, as natural geothermal energy is.


29. (a) Use of hydrogen fusion to supply energy is a dream
that may be realized in the next century. Fusion would be a
relatively clean and almost limitless supply of energy, as can
be seen from Table 7.1. To illustrate this, calculate how many
years the present energy needs of the world could be
supplied by one millionth of the oceans’ hydrogen fusion
energy. (b) How does this time compare with historically
significant events, such as the duration of stable economic
systems?


7.7 Power
30. The Crab Nebula (see Figure 7.41) pulsar is the remnant
of a supernova that occurred in A.D. 1054. Using data from
Table 7.3, calculate the approximate factor by which the
power output of this astronomical object has declined since its
explosion.


Figure 7.41 Crab Nebula (credit: ESO, via Wikimedia Commons)


31. Suppose a star 1000 times brighter than our Sun (that is,
emitting 1000 times the power) suddenly goes supernova.
Using data from Table 7.3: (a) By what factor does its power
output increase? (b) How many times brighter than our entire
Milky Way galaxy is the supernova? (c) Based on your
answers, discuss whether it should be possible to observe
supernovas in distant galaxies. Note that there are on the


order of 1011 observable galaxies, the average brightness
of which is somewhat less than our own galaxy.


32. A person in good physical condition can put out 100 W of
useful power for several hours at a stretch, perhaps by
pedaling a mechanism that drives an electric generator.
Neglecting any problems of generator efficiency and practical
considerations such as resting time: (a) How many people
would it take to run a 4.00-kW electric clothes dryer? (b) How
many people would it take to replace a large electric power
plant that generates 800 MW?


33.What is the cost of operating a 3.00-W electric clock for a
year if the cost of electricity is $0.0900 per kW ⋅ h ?
34. A large household air conditioner may consume 15.0 kW
of power. What is the cost of operating this air conditioner
3.00 h per day for 30.0 d if the cost of electricity is $0.110 per
kW ⋅ h ?
35. (a) What is the average power consumption in watts of an
appliance that uses 5.00 kW ⋅ h of energy per day? (b)
How many joules of energy does this appliance consume in a
year?


36. (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 person to lift 2000 kg of
bricks 1.50 m to a platform? (Work done to lift his body can be
omitted because it is not considered useful output here.)


37. A 500-kg dragster accelerates from rest to a final speed of
110 m/s in 400 m (about a quarter of a mile) and encounters
an average frictional force of 1200 N. What is its average
power output in watts and horsepower if this takes 7.30 s?


38. (a) How long will it take an 850-kg car with a useful power
output of 40.0 hp (1 hp = 746 W) to reach a speed of 15.0 m/
s, neglecting friction? (b) How long will this acceleration take
if the car also climbs a 3.00-m-high hill in the process?


39. (a) Find the useful power output of an elevator motor that
lifts a 2500-kg load a height of 35.0 m in 12.0 s, if it also
increases the speed from rest to 4.00 m/s. Note that the total
mass of the counterbalanced system is 10,000 kg—so that
only 2500 kg is raised in height, but the full 10,000 kg is
accelerated. (b) What does it cost, if electricity is $0.0900 per
kW ⋅ h ?
40. (a) What is the available energy content, in joules, of a
battery that operates a 2.00-W electric clock for 18 months?


(b) How long can a battery that can supply 8.00×104 J run
a pocket calculator that consumes energy at the rate of


1.00×10−3 W ?


41. (a) How long would it take a 1.50×105 -kg airplane with
engines that produce 100 MW of power to reach a speed of
250 m/s and an altitude of 12.0 km if air resistance were
negligible? (b) If it actually takes 900 s, what is the power? (c)
Given this power, what is the average force of air resistance if
the airplane takes 1200 s? (Hint: You must find the distance
the plane travels in 1200 s assuming constant acceleration.)


42. Calculate the power output needed for a 950-kg car to
climb a 2.00º slope at a constant 30.0 m/s while
encountering wind resistance and friction totaling 600 N.
Explicitly show how you follow the steps in the Problem-
Solving Strategies for Energy.


43. (a) Calculate the power per square meter reaching Earth’s
upper atmosphere from the Sun. (Take the power output of


Chapter 7 | Work, Energy, and Energy Resources 279




the Sun to be 4.00×1026W.) (b) Part of this is absorbed
and reflected by the atmosphere, so that a maximum of


1.30 kW/m2 reaches Earth’s surface. Calculate the area in
km2 of solar energy collectors needed to replace an electric
power plant that generates 750 MW if the collectors convert
an average of 2.00% of the maximum power into electricity.
(This small conversion efficiency is due to the devices
themselves, and the fact that the sun is directly overhead only
briefly.) With the same assumptions, what area would be
needed to meet the United States’ energy needs


(1.05×1020 J)? Australia’s energy needs (5.4×1018 J)?


China’s energy needs (6.3×1019 J)? (These energy
consumption values are from 2006.)


7.8 Work, Energy, and Power in Humans
44. (a) How long can you rapidly climb stairs (116/min) on the
93.0 kcal of energy in a 10.0-g pat of butter? (b) How many
flights is this if each flight has 16 stairs?


45. (a) What is the power output in watts and horsepower of a
70.0-kg sprinter who accelerates from rest to 10.0 m/s in 3.00
s? (b) Considering the amount of power generated, do you
think a well-trained athlete could do this repetitively for long
periods of time?


46. Calculate the power output in watts and horsepower of a
shot-putter who takes 1.20 s to accelerate the 7.27-kg shot
from rest to 14.0 m/s, while raising it 0.800 m. (Do not include
the power produced to accelerate his body.)


Figure 7.42 Shot putter at the Dornoch Highland Gathering in 2007.
(credit: John Haslam, Flickr)


47. (a) What is the efficiency of an out-of-condition professor


who does 2.10×105 J of useful work while metabolizing
500 kcal of food energy? (b) How many food calories would a
well-conditioned athlete metabolize in doing the same work
with an efficiency of 20%?


48. Energy that is not utilized for work or heat transfer is
converted to the chemical energy of body fat containing about
39 kJ/g. How many grams of fat will you gain if you eat
10,000 kJ (about 2500 kcal) one day and do nothing but sit
relaxed for 16.0 h and sleep for the other 8.00 h? Use data
from Table 7.5 for the energy consumption rates of these
activities.


49. Using data from Table 7.5, calculate the daily energy
needs of a person who sleeps for 7.00 h, walks for 2.00 h,
attends classes for 4.00 h, cycles for 2.00 h, sits relaxed for


3.00 h, and studies for 6.00 h. (Studying consumes energy at
the same rate as sitting in class.)


50.What is the efficiency of a subject on a treadmill who puts
out work at the rate of 100 W while consuming oxygen at the
rate of 2.00 L/min? (Hint: See Table 7.5.)


51. Shoveling snow can be extremely taxing because the
arms have such a low efficiency in this activity. Suppose a
person shoveling a footpath metabolizes food at the rate of
800 W. (a) What is her useful power output? (b) How long will
it take her to lift 3000 kg of snow 1.20 m? (This could be the
amount of heavy snow on 20 m of footpath.) (c) How much
waste heat transfer in kilojoules will she generate in the
process?


52. Very large forces are produced in joints when a person
jumps from some height to the ground. (a) Calculate the
magnitude of the force produced if an 80.0-kg person jumps
from a 0.600–m-high ledge and lands stiffly, compressing joint
material 1.50 cm as a result. (Be certain to include the weight
of the person.) (b) In practice the knees bend almost
involuntarily to help extend the distance over which you stop.
Calculate the magnitude of the force produced if the stopping
distance is 0.300 m. (c) Compare both forces with the weight
of the person.


53. Jogging on hard surfaces with insufficiently padded shoes
produces large forces in the feet and legs. (a) Calculate the
magnitude of the force needed to stop the downward motion
of a jogger’s leg, if his leg has a mass of 13.0 kg, a speed of
6.00 m/s, and stops in a distance of 1.50 cm. (Be certain to
include the weight of the 75.0-kg jogger’s body.) (b) Compare
this force with the weight of the jogger.


54. (a) Calculate the energy in kJ used by a 55.0-kg woman
who does 50 deep knee bends in which her center of mass is
lowered and raised 0.400 m. (She does work in both
directions.) You may assume her efficiency is 20%. (b) What
is the average power consumption rate in watts if she does
this in 3.00 min?


55. Kanellos Kanellopoulos flew 119 km from Crete to
Santorini, Greece, on April 23, 1988, in the Daedalus 88, an
aircraft powered by a bicycle-type drive mechanism (see
Figure 7.43). His useful power output for the 234-min trip was
about 350 W. Using the efficiency for cycling from Table 7.2,
calculate the food energy in kilojoules he metabolized during
the flight.


Figure 7.43 The Daedalus 88 in flight. (credit: NASA photo by Beasley)


56. The swimmer shown in Figure 7.44 exerts an average
horizontal backward force of 80.0 N with his arm during each
1.80 m long stroke. (a) What is his work output in each


280 Chapter 7 | Work, Energy, and Energy Resources


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stroke? (b) Calculate the power output of his arms if he does
120 strokes per minute.


Figure 7.44


57. Mountain climbers carry bottled oxygen when at very high
altitudes. (a) Assuming that a mountain climber uses oxygen
at twice the rate for climbing 116 stairs per minute (because
of low air temperature and winds), calculate how many liters
of oxygen a climber would need for 10.0 h of climbing. (These
are liters at sea level.) Note that only 40% of the inhaled
oxygen is utilized; the rest is exhaled. (b) How much useful
work does the climber do if he and his equipment have a
mass of 90.0 kg and he gains 1000 m of altitude? (c) What is
his efficiency for the 10.0-h climb?


58. The awe-inspiring Great Pyramid of Cheops was built
more than 4500 years ago. Its square base, originally 230 m
on a side, covered 13.1 acres, and it was 146 m high, with a


mass of about 7×109 kg . (The pyramid’s dimensions are
slightly different today due to quarrying and some sagging.)
Historians estimate that 20,000 workers spent 20 years to
construct it, working 12-hour days, 330 days per year. (a)
Calculate the gravitational potential energy stored in the
pyramid, given its center of mass is at one-fourth its height.
(b) Only a fraction of the workers lifted blocks; most were
involved in support services such as building ramps (see
Figure 7.45), bringing food and water, and hauling blocks to
the site. Calculate the efficiency of the workers who did the
lifting, assuming there were 1000 of them and they consumed
food energy at the rate of 300 kcal/h. What does your answer
imply about how much of their work went into block-lifting,
versus how much work went into friction and lifting and
lowering their own bodies? (c) Calculate the mass of food that
had to be supplied each day, assuming that the average
worker required 3600 kcal per day and that their diet was 5%
protein, 60% carbohydrate, and 35% fat. (These proportions
neglect the mass of bulk and nondigestible materials
consumed.)


Figure 7.45 Ancient pyramids were probably constructed using ramps
as simple machines. (credit: Franck Monnier, Wikimedia Commons)


59. (a) How long can you play tennis on the 800 kJ (about
200 kcal) of energy in a candy bar? (b) Does this seem like a


long time? Discuss why exercise is necessary but may not be
sufficient to cause a person to lose weight.


7.9 World Energy Use
60. Integrated Concepts


(a) Calculate the force the woman in Figure 7.46 exerts to do
a push-up at constant speed, taking all data to be known to
three digits. (b) How much work does she do if her center of
mass rises 0.240 m? (c) What is her useful power output if
she does 25 push-ups in 1 min? (Should work done lowering
her body be included? See the discussion of useful work in
Work, Energy, and Power in Humans.


Figure 7.46 Forces involved in doing push-ups. The woman’s weight
acts as a force exerted downward on her center of gravity (CG).


61. Integrated Concepts


A 75.0-kg cross-country skier is climbing a 3.0º slope at a
constant speed of 2.00 m/s and encounters air resistance of
25.0 N. Find his power output for work done against the
gravitational force and air resistance. (b) What average force
does he exert backward on the snow to accomplish this? (c) If
he continues to exert this force and to experience the same
air resistance when he reaches a level area, how long will it
take him to reach a velocity of 10.0 m/s?


62. Integrated Concepts


The 70.0-kg swimmer in Figure 7.44 starts a race with an
initial velocity of 1.25 m/s and exerts an average force of 80.0
N backward with his arms during each 1.80 m long stroke. (a)
What is his initial acceleration if water resistance is 45.0 N?
(b) What is the subsequent average resistance force from the
water during the 5.00 s it takes him to reach his top velocity of
2.50 m/s? (c) Discuss whether water resistance seems to
increase linearly with velocity.


63. Integrated Concepts


A toy gun uses a spring with a force constant of 300 N/m to
propel a 10.0-g steel ball. If the spring is compressed 7.00 cm
and friction is negligible: (a) How much force is needed to
compress the spring? (b) To what maximum height can the
ball be shot? (c) At what angles above the horizontal may a
child aim to hit a target 3.00 m away at the same height as
the gun? (d) What is the gun’s maximum range on level
ground?


64. Integrated Concepts


(a) What force must be supplied by an elevator cable to


produce an acceleration of 0.800 m/s2 against a 200-N
frictional force, if the mass of the loaded elevator is 1500 kg?
(b) How much work is done by the cable in lifting the elevator
20.0 m? (c) What is the final speed of the elevator if it starts
from rest? (d) How much work went into thermal energy?


65. Unreasonable Results


A car advertisement claims that its 900-kg car accelerated
from rest to 30.0 m/s and drove 100 km, gaining 3.00 km in
altitude, on 1.0 gal of gasoline. The average force of friction


Chapter 7 | Work, Energy, and Energy Resources 281




282 Chapter 7 | Work, Energy, and Energy Resources


including air resistance was 700 N. Assume all values are
known to three significant figures. (a) Calculate the car’s
efficiency. (b) What is unreasonable about the result? (c)
Which premise is unreasonable, or which premises are
inconsistent?


66. Unreasonable Results


Body fat is metabolized, supplying 9.30 kcal/g, when dietary
intake is less than needed to fuel metabolism. The
manufacturers of an exercise bicycle claim that you can lose
0.500 kg of fat per day by vigorously exercising for 2.00 h per
day on their machine. (a) How many kcal are supplied by the
metabolization of 0.500 kg of fat? (b) Calculate the kcal/min
that you would have to utilize to metabolize fat at the rate of
0.500 kg in 2.00 h. (c) What is unreasonable about the
results? (d) Which premise is unreasonable, or which
premises are inconsistent?


67. Construct Your Own Problem


Consider a person climbing and descending stairs. Construct
a problem in which you calculate the long-term rate at which
stairs can be climbed considering the mass of the person, his
ability to generate power with his legs, and the height of a
single stair step. Also consider why the same person can
descend stairs at a faster rate for a nearly unlimited time in
spite of the fact that very similar forces are exerted going
down as going up. (This points to a fundamentally different
process for descending versus climbing stairs.)


68. Construct Your Own Problem


Consider humans generating electricity by pedaling a device
similar to a stationary bicycle. Construct a problem in which
you determine the number of people it would take to replace a
large electrical generation facility. Among the things to
consider are the power output that is reasonable using the
legs, rest time, and the need for electricity 24 hours per day.
Discuss the practical implications of your results.


69. Integrated Concepts


A 105-kg basketball player crouches down 0.400 m while
waiting to jump. After exerting a force on the floor through this
0.400 m, his feet leave the floor and his center of gravity rises
0.950 m above its normal standing erect position. (a) Using
energy considerations, calculate his velocity when he leaves
the floor. (b) What average force did he exert on the floor?
(Do not neglect the force to support his weight as well as that
to accelerate him.) (c) What was his power output during the
acceleration phase?


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8 LINEAR MOMENTUM AND COLLISIONS


Figure 8.1 Each rugby player has great momentum, which will affect the outcome of their collisions with each other and the ground. (credit: ozzzie,
Flickr)


Chapter Outline
8.1. Linear Momentum and Force


• Define linear momentum.
• Explain the relationship between momentum and force.
• State Newton’s second law of motion in terms of momentum.
• Calculate momentum given mass and velocity.


8.2. Impulse
• Define impulse.
• Describe effects of impulses in everyday life.
• Determine the average effective force using graphical representation.
• Calculate average force and impulse given mass, velocity, and time.


8.3. Conservation of Momentum
• Describe the principle of conservation of momentum.
• Derive an expression for the conservation of momentum.
• Explain conservation of momentum with examples.
• Explain the principle of conservation of momentum as it relates to atomic and subatomic particles.


8.4. Elastic Collisions in One Dimension
• Describe an elastic collision of two objects in one dimension.
• Define internal kinetic energy.
• Derive an expression for conservation of internal kinetic energy in a one dimensional collision.
• Determine the final velocities in an elastic collision given masses and initial velocities.


8.5. Inelastic Collisions in One Dimension
• Define inelastic collision.
• Explain perfectly inelastic collision.
• Apply an understanding of collisions to sports.
• Determine recoil velocity and loss in kinetic energy given mass and initial velocity.


8.6. Collisions of Point Masses in Two Dimensions
• Discuss two dimensional collisions as an extension of one dimensional analysis.
• Define point masses.
• Derive an expression for conservation of momentum along x-axis and y-axis.
• Describe elastic collisions of two objects with equal mass.
• Determine the magnitude and direction of the final velocity given initial velocity, and scattering angle.


8.7. Introduction to Rocket Propulsion
• State Newton’s third law of motion.
• Explain the principle involved in propulsion of rockets and jet engines.
• Derive an expression for the acceleration of the rocket and discuss the factors that affect the acceleration.
• Describe the function of a space shuttle.


Chapter 8 | Linear Momentum and Collisions 283




Introduction to Linear Momentum and Collisions
We use the term momentum in various ways in everyday language, and most of these ways are consistent with its precise
scientific definition. We speak of sports teams or politicians gaining and maintaining the momentum to win. We also recognize
that momentum has something to do with collisions. For example, looking at the rugby players in the photograph colliding and
falling to the ground, we expect their momenta to have great effects in the resulting collisions. Generally, momentum implies a
tendency to continue on course—to move in the same direction—and is associated with great mass and speed.


Momentum, like energy, is important because it is conserved. Only a few physical quantities are conserved in nature, and
studying them yields fundamental insight into how nature works, as we shall see in our study of momentum.


8.1 Linear Momentum and Force


Linear Momentum
The scientific definition of linear momentum is consistent with most people’s intuitive understanding of momentum: a large, fast-
moving object has greater momentum than a smaller, slower object. Linear momentum is defined as the product of a system’s
mass multiplied by its velocity. In symbols, linear momentum is expressed as


(8.1)p = mv.


Momentum is directly proportional to the object’s mass and also its velocity. Thus the greater an object’s mass or the greater its
velocity, the greater its momentum. Momentum p is a vector having the same direction as the velocity v . The SI unit for
momentum is kg ·m/s .


Linear Momentum


Linear momentum is defined as the product of a system’s mass multiplied by its velocity:


(8.2)p = mv.


Example 8.1 Calculating Momentum: A Football Player and a Football


(a) Calculate the momentum of a 110-kg football player running at 8.00 m/s. (b) Compare the player’s momentum with the
momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s.


Strategy


No information is given regarding direction, and so we can calculate only the magnitude of the momentum, p . (As usual, a


symbol that is in italics is a magnitude, whereas one that is italicized, boldfaced, and has an arrow is a vector.) In both parts
of this example, the magnitude of momentum can be calculated directly from the definition of momentum given in the
equation, which becomes


(8.3)p = mv


when only magnitudes are considered.


Solution for (a)


To determine the momentum of the player, substitute the known values for the player’s mass and speed into the equation.


(8.4)pplayer =

⎝110 kg⎞⎠(8.00 m/s) = 880 kg ·m/s


Solution for (b)


To determine the momentum of the ball, substitute the known values for the ball’s mass and speed into the equation.


(8.5)pball =

⎝0.410 kg⎞⎠(25.0 m/s) = 10.3 kg ·m/s


The ratio of the player’s momentum to that of the ball is


(8.6)pplayer
pball


= 88010.3 = 85.9.


Discussion


Although the ball has greater velocity, the player has a much greater mass. Thus the momentum of the player is much
greater than the momentum of the football, as you might guess. As a result, the player’s motion is only slightly affected if he
catches the ball. We shall quantify what happens in such collisions in terms of momentum in later sections.


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Momentum and Newton’s Second Law
The importance of momentum, unlike the importance of energy, was recognized early in the development of classical physics.
Momentum was deemed so important that it was called the “quantity of motion.” Newton actually stated his second law of
motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over
which it changes. Using symbols, this law is


(8.7)Fnet =
Δp
Δt ,


where Fnet is the net external force, Δp is the change in momentum, and Δt is the change in time.


Newton’s Second Law of Motion in Terms of Momentum


The net external force equals the change in momentum of a system divided by the time over which it changes.


(8.8)Fnet =
Δp
Δt


Making Connections: Force and Momentum


Force and momentum are intimately related. Force acting over time can change momentum, and Newton’s second law of
motion, can be stated in its most broadly applicable form in terms of momentum. Momentum continues to be a key concept
in the study of atomic and subatomic particles in quantum mechanics.


This statement of Newton’s second law of motion includes the more familiar Fnet=ma as a special case. We can derive this
form as follows. First, note that the change in momentum Δp is given by


(8.9)Δp = Δ⎛⎝mv⎞⎠.


If the mass of the system is constant, then


(8.10)Δ(mv) = mΔv.


So that for constant mass, Newton’s second law of motion becomes


(8.11)Fnet =
Δp
Δt =


mΔv
Δt .


Because ΔvΔt = a , we get the familiar equation


(8.12)Fnet=ma


when the mass of the system is constant.


Newton’s second law of motion stated in terms of momentum is more generally applicable because it can be applied to systems
where the mass is changing, such as rockets, as well as to systems of constant mass. We will consider systems with varying
mass in some detail; however, the relationship between momentum and force remains useful when mass is constant, such as in
the following example.


Example 8.2 Calculating Force: Venus Williams’ Racquet


During the 2007 French Open, Venus Williams hit the fastest recorded serve in a premier women’s match, reaching a speed
of 58 m/s (209 km/h). What is the average force exerted on the 0.057-kg tennis ball by Venus Williams’ racquet, assuming
that the ball’s speed just after impact is 58 m/s, that the initial horizontal component of the velocity before impact is
negligible, and that the ball remained in contact with the racquet for 5.0 ms (milliseconds)?


Strategy


This problem involves only one dimension because the ball starts from having no horizontal velocity component before
impact. Newton’s second law stated in terms of momentum is then written as


(8.13)Fnet =
Δp
Δt .


As noted above, when mass is constant, the change in momentum is given by


(8.14)Δp = mΔv = m(vf − vi).


Chapter 8 | Linear Momentum and Collisions 285




In this example, the velocity just after impact and the change in time are given; thus, once Δp is calculated, Fnet =
Δp
Δt


can be used to find the force.


Solution


To determine the change in momentum, substitute the values for the initial and final velocities into the equation above.


(8.15)Δp = m(vf – vi)
= ⎛⎝0.057 kg⎞⎠(58 m/s – 0 m/s)
= 3.306 kg ·m/s ≈ 3.3 kg ·m/s


Now the magnitude of the net external force can determined by using Fnet =
Δp
Δt :


(8.16)
Fnet =


Δp
Δt =


3.306 kg ⋅ m/s
5.0×10−3 s


= 661 N ≈ 660 N,


where we have retained only two significant figures in the final step.


Discussion


This quantity was the average force exerted by Venus Williams’ racquet on the tennis ball during its brief impact (note that
the ball also experienced the 0.56-N force of gravity, but that force was not due to the racquet). This problem could also be
solved by first finding the acceleration and then using Fnet =ma , but one additional step would be required compared with
the strategy used in this example.


8.2 Impulse
The effect of a force on an object depends on how long it acts, as well as how great the force is. In Example 8.1, a very large
force acting for a short time had a great effect on the momentum of the tennis ball. A small force could cause the same change
in momentum, but it would have to act for a much longer time. For example, if the ball were thrown upward, the gravitational
force (which is much smaller than the tennis racquet’s force) would eventually reverse the momentum of the ball. Quantitatively,
the effect we are talking about is the change in momentum Δp .


By rearranging the equation Fnet =
Δp
Δt to be


(8.17)Δp = FnetΔt,


we can see how the change in momentum equals the average net external force multiplied by the time this force acts. The
quantity FnetΔt is given the name impulse. Impulse is the same as the change in momentum.


Impulse: Change in Momentum


Change in momentum equals the average net external force multiplied by the time this force acts.


(8.18)Δp = FnetΔt


The quantity FnetΔt is given the name impulse.


There are many ways in which an understanding of impulse can save lives, or at least limbs. The dashboard padding in a
car, and certainly the airbags, allow the net force on the occupants in the car to act over a much longer time when there is a
sudden stop. The momentum change is the same for an occupant, whether an air bag is deployed or not, but the force (to
bring the occupant to a stop) will be much less if it acts over a larger time. Cars today have many plastic components. One
advantage of plastics is their lighter weight, which results in better gas mileage. Another advantage is that a car will crumple
in a collision, especially in the event of a head-on collision. A longer collision time means the force on the car will be less.
Deaths during car races decreased dramatically when the rigid frames of racing cars were replaced with parts that could
crumple or collapse in the event of an accident.


Bones in a body will fracture if the force on them is too large. If you jump onto the floor from a table, the force on your legs
can be immense if you land stiff-legged on a hard surface. Rolling on the ground after jumping from the table, or landing with
a parachute, extends the time over which the force (on you from the ground) acts.


286 Chapter 8 | Linear Momentum and Collisions


This content is available for free at http://cnx.org/content/col11406/1.9




Example 8.3 Calculating Magnitudes of Impulses: Two Billiard Balls Striking a Rigid Wall


Two identical billiard balls strike a rigid wall with the same speed, and are reflected without any change of speed. The first
ball strikes perpendicular to the wall. The second ball strikes the wall at an angle of 30º from the perpendicular, and
bounces off