problem bank make file physics121 solns

Wayne Hacker’s Problem Bank SolutionsCollege Physics

Version 1.0 (Draft)

Wayne HackerCopyright ©Wayne Hacker 2011. All rights reserved.

July 11, 2011

Physics Problem Bank Solutions Copyright ©Wayne Hacker 2009. All rights reserved. 1

Contents

0 About the mathematical prerequisites material 11

0.1 What is a College Physics Course? . . . . . . . . . . . . . . . . . . . . . 11

0.2 What are the math-prerequise topics for college physics? . . . . . . . . . 11

0.3 What is a University Physics course? . . . . . . . . . . . . . . . . . . . . 11

0.4 What is a college physics course in the state of Arizona? . . . . . . . . . 12

I Mathematical Prerequisites 13

1 Geometry 14

1.1 Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.1.1 Letter-based problems . . . . . . . . . . . . . . . . . . . . . . . . 14

1.1.2 Non-calculator-based problems . . . . . . . . . . . . . . . . . . . . 16

1.1.3 Calculator-based problems . . . . . . . . . . . . . . . . . . . . . . 19

1.2 Rectangles and rectangular solids . . . . . . . . . . . . . . . . . . . . . . 21




1.3 Right triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26




1.4 Right triangle word problems . . . . . . . . . . . . . . . . . . . . . . . . 33




2 Algebra 43

2.1 Evaluating functions with numerical arguments . . . . . . . . . . . . . . 43

2.1.1 Non-calculator based problems . . . . . . . . . . . . . . . . . . . . 43


2.2 Evaluating functions with variable-expression arguments . . . . . . . . . 49

2.3 Evaluating functions of multiple variables . . . . . . . . . . . . . . . . . . 52

2.3.1 Functions of two variables . . . . . . . . . . . . . . . . . . . . . . 52

2.3.2 Functions of three variables . . . . . . . . . . . . . . . . . . . . . 57

2.4 Solving linear equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

2.4.1 Linear equations with numerical solutions . . . . . . . . . . . . . 62

2.4.2 Linear equations with variable-expression solutions . . . . . . . . 64

2.5 Solving systems of linear equations . . . . . . . . . . . . . . . . . . . . . 67

2.6 Solving quadratic equations . . . . . . . . . . . . . . . . . . . . . . . . . 78

2.6.1 Factoring quadratic equations . . . . . . . . . . . . . . . . . . . . 78

2.6.2 Quadratic formula . . . . . . . . . . . . . . . . . . . . . . . . . . 82

2.7 Algebra word problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

3 Graphs 88

3.1 Single points on graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

3.2 Matching graphs and equations . . . . . . . . . . . . . . . . . . . . . . . 96

II Mathematical Preliminaries 107


4 Basic Trigonometry 108

4.1 Arc-Length Problems using s = θ r . . . . . . . . . . . . . . . . . . . . . 108

4.1.1 Degrees to radians: formula . . . . . . . . . . . . . . . . . . . . . 108

4.1.2 Radians to degrees: formula . . . . . . . . . . . . . . . . . . . . . 110

4.1.3 Degrees to radians: calculator . . . . . . . . . . . . . . . . . . . . 112

4.1.4 Radians to degrees: calculator . . . . . . . . . . . . . . . . . . . . 114

4.1.5 Arc length to radians: pictures . . . . . . . . . . . . . . . . . . . 116

4.1.6 Arc length to radians: descriptions . . . . . . . . . . . . . . . . . 117

4.1.7 Radians to arc length: pictures . . . . . . . . . . . . . . . . . . . 119

4.1.8 Radians to arc length: descriptions . . . . . . . . . . . . . . . . . 120

4.1.9 Word problems: arc length and radians . . . . . . . . . . . . . . . 122

4.2 Basic Right-Triangle Trigonometry . . . . . . . . . . . . . . . . . . . . . 124

4.2.1 Basic trig functions: finding . . . . . . . . . . . . . . . . . . . . . 124

4.2.2 Basic trig functions: identifying . . . . . . . . . . . . . . . . . . . 127

4.2.3 Basic trig functions: calculator; radians . . . . . . . . . . . . . . . 133

4.2.4 Basic trig functions: calculator; degrees . . . . . . . . . . . . . . . 135

4.2.5 Using basic trig functions: formulas . . . . . . . . . . . . . . . . . 137

4.2.6 Using basic trig functions: calculator . . . . . . . . . . . . . . . . 139

4.2.7 Reference angles . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

4.2.8 Arc functions: definition . . . . . . . . . . . . . . . . . . . . . . . 148

4.2.9 Arc functions: definition; diagrams . . . . . . . . . . . . . . . . . 150

4.2.10 Arc functions: calculator; radians . . . . . . . . . . . . . . . . . . 151

4.2.11 Arc function: calculator; degrees . . . . . . . . . . . . . . . . . . . 153

4.2.12 Arc functions: calculator; diagram; radians . . . . . . . . . . . . . 154

4.2.13 Arc functions: calculator; diagram; degrees . . . . . . . . . . . . . 157

4.2.14 Word problems: basic trig functions . . . . . . . . . . . . . . . . . 160

4.2.15 Word problems: arc functions . . . . . . . . . . . . . . . . . . . . 163


5 Introduction to measurement: dimensions, units, scientific notation,and significant figures 166

5.1 Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

5.1.1 Dimension or unit? . . . . . . . . . . . . . . . . . . . . . . . . . . 166

5.1.2 Dimensional consistency . . . . . . . . . . . . . . . . . . . . . . . 167

5.1.3 Practical questions involving dimensions . . . . . . . . . . . . . . 173

5.2 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

5.2.1 Converting between different sets of units . . . . . . . . . . . . . . 176

5.3 Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

5.4 Significant figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

5.5 Determining derived units from equations . . . . . . . . . . . . . . . . . . 193

6 Introduction to Vectors 196

6.1 Identifying Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

6.1.1 Given vector, identify on figure . . . . . . . . . . . . . . . . . . . 196

6.1.2 Given vector on figure, identify formula . . . . . . . . . . . . . . . 201

6.1.3 Given direction, identify on figure . . . . . . . . . . . . . . . . . . 207

6.1.4 Given vectors on figure, name quadrant of sum/difference . . . . . 212

6.1.5 Mixing it up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

6.2 Geometric Vector Addition and Subtraction . . . . . . . . . . . . . . . . 220

6.3 Position vs. Displacement Vector Problems . . . . . . . . . . . . . . . . . 225

6.4 Finding Components of Vectors . . . . . . . . . . . . . . . . . . . . . . . 229

6.5 Algebraic Vector Addition and Subtraction . . . . . . . . . . . . . . . . . 231

6.6 Concept vector and scalar questions . . . . . . . . . . . . . . . . . . . . . 235

6.7 Applications of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240

6.7.1 Breaking vectors into components . . . . . . . . . . . . . . . . . . 240

6.7.2 Introduction to force vectors . . . . . . . . . . . . . . . . . . . . . 248


III Kinematics 254

7 One-dimensional linear kinematics 255

7.1 Concept questions: kinematics . . . . . . . . . . . . . . . . . . . . . . . . 255

7.2 Qualitative kinematics: descriptions . . . . . . . . . . . . . . . . . . . . . 258

7.3 Qualitative kinematics: from graph . . . . . . . . . . . . . . . . . . . . . 266

7.4 Introducing the fundamental one-dimensional kinematic equations . . . . 271

7.4.1 Dimensional consistency of the fundamental equations . . . . . . 272

7.5 Quantitative kinematics: horizontal . . . . . . . . . . . . . . . . . . . . . 273

7.6 Quantitative kinematics: vertical . . . . . . . . . . . . . . . . . . . . . . 287

7.6.1 The difference between average speed and average velocity in one-dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292

7.7 Similarity problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294

7.8 Lab-application problems: One-Dimensional Kinematics . . . . . . . . . . 300

7.8.1 Ball-drop apparatus . . . . . . . . . . . . . . . . . . . . . . . . . 300

7.8.2 Spring-cannon apparatus . . . . . . . . . . . . . . . . . . . . . . . 300

7.9 Mixing It Up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307

8 One-dimensional rotational kinematics 310

8.1 Angular speed, period, and frequency . . . . . . . . . . . . . . . . . . . . 310

8.2 Fundamental equation 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 313




9 Two-dimensional kinematics 328

9.1 The basics of velocity and acceleration in two dimensions . . . . . . . . . 328

9.1.1 The difference between average speed and average velocity in a plane328

9.1.2 Average acceleration in a plane . . . . . . . . . . . . . . . . . . . 329

9.1.3 Relative velocity in a plane . . . . . . . . . . . . . . . . . . . . . 329

9.1.4 Graphical interpretation of velocity and acceleration . . . . . . . . 330

9.2 Uniform circular motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 332

9.2.1 Similarity problems . . . . . . . . . . . . . . . . . . . . . . . . . . 336

9.3 Projectile motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338

9.3.1 General equations for projectile motion . . . . . . . . . . . . . . . 338

9.3.2 Velocity and acceleration of a projectile . . . . . . . . . . . . . . . 339

9.3.3 Special Case I: The Half-Parabola . . . . . . . . . . . . . . . . . . 341

9.3.4 Special Case II: The Full Parabola . . . . . . . . . . . . . . . . . 348

9.4 Lab application problems: Two-Dimensional Kinematics . . . . . . . . . 359

9.4.1 Spring-cannon apparatus . . . . . . . . . . . . . . . . . . . . . . . 359

9.5 Advanced-Level Problems (Kinematics) . . . . . . . . . . . . . . . . . . . 366

IV Newton’s Laws 372

10 Introduction to Newton’s Laws 373

10.1 Concept questions involving Newton’s three laws . . . . . . . . . . . . . . 373

10.2 Free-body diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379

10.3 Pushing and pulling objects on horizontal frictionless surfaces . . . . . . 380


11 Applications of Newton’s Laws to Linear Motion 381

11.1 Newton’s laws in one-dimension . . . . . . . . . . . . . . . . . . . . . . . 381

11.1.1 One-dimensional kinematics and Newton’s second law . . . . . . . 381

11.1.2 Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388

11.1.3 Apparent weight problems (The Elevator Equation) . . . . . . . . 395

11.1.4 Spring Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 402

11.2 Newton’s laws in two-dimensions . . . . . . . . . . . . . . . . . . . . . . 408

11.2.1 Static equilibrium problems . . . . . . . . . . . . . . . . . . . . . 408

11.2.2 Block and pulley frictionless systems . . . . . . . . . . . . . . . . 420

11.2.3 Accelerometers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434

11.2.4 Static and kinetic friction problems . . . . . . . . . . . . . . . . . 437

11.2.5 Stopping distance problems . . . . . . . . . . . . . . . . . . . . . 446

12 Applications of Newton’s Laws to Uniform Circular Motion 449

12.1 Centripetal force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 449

12.1.1 Conceptual questions: centripetal force . . . . . . . . . . . . . . . 449

12.1.2 Horizontal Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 453

12.1.3 Vertical Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461

12.2 Gravitation and circular orbits . . . . . . . . . . . . . . . . . . . . . . . . 473

12.2.1 Gravation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473

12.2.2 Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475

13 Applications of Newton’s Laws to Rotating Bodies 484

13.1 Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484

13.2 Angular acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496

13.3 Moment of inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503

V Work, Energy, and Momentum 509


14 Work and mechanical energy 510

14.1 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510

14.2 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513

14.3 Kinetic energy (K.E.) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514

14.4 Potential energy (P.E.) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 517

14.5 Work-Mechanical-Energy Theorem . . . . . . . . . . . . . . . . . . . . . 519

14.5.1 Work-Kinetic-Energy Theorem . . . . . . . . . . . . . . . . . . . 519

14.5.2 Work-Potential-Energy Theorem . . . . . . . . . . . . . . . . . . 522

14.6 Conservation of mechanical energy (M.E.) . . . . . . . . . . . . . . . . . 529

15 Conservation of Linear Momentum 543

15.1 Computing Linear Momentum p = mv . . . . . . . . . . . . . . . . . . . 543

15.2 Applying conservation of momentum to isolated systems . . . . . . . . . 544

15.3 Impulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548

15.4 Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553

16 Rotational kinetic energy and angular momentum 568

VI Applications of Mechanics 578

17 Applications of Mechanics 579

17.1 Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579

17.2 Mechanical waves and sound . . . . . . . . . . . . . . . . . . . . . . . . . 581

17.3 Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584

17.4 Static Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587

17.4.1 Pascal’s Law and Archimedes’s principle . . . . . . . . . . . . . . 587

17.4.2 Density, specific volume, specific weight . . . . . . . . . . . . . . . 587

17.4.3 Static Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589


18 Introduction to Thermodynamics 593

18.1 Introductory Concepts and Definitions . . . . . . . . . . . . . . . . . . . 593

18.1.1 What is the study of thermodynamics? . . . . . . . . . . . . . . . 593

18.1.2 Defining Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 593

18.1.3 Closed systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594

18.1.4 Control volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594

18.1.5 Property, state, and process . . . . . . . . . . . . . . . . . . . . . 595

18.1.6 Extensive and intensive properties . . . . . . . . . . . . . . . . . . 595

18.1.7 Equilibrium, quasi-equilibrium, and processes . . . . . . . . . . . 595

18.1.8 Base SI units for mass, length, time, Force, energy, and pressure . 596

18.2 Temperature Scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 598

Physics Problem Bank Solutions Copyright ©Wayne Hacker 2009. All rights reserved.10

List of Figures

1 A car is being driven erratically along a straight stretch of highway. Thegraph shows its position x as a function of the time t. . . . . . . . . . . 266

2 A car is being driven erratically along a straight stretch of highway. Thegraph shows its position x as a function of the time t. . . . . . . . . . . 268

3 Spring cannon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367

4 Spring cannon and bucket . . . . . . . . . . . . . . . . . . . . . . . . . . 368

5 Spring cannon and bucket . . . . . . . . . . . . . . . . . . . . . . . . . . 369

6 Side view of the spring cannon firing ball into a bucket mounted on top ofa glider . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370

7 Earth-mass system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 528

8 Ball in the bucket. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 540

9 The loop-the-loop. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 541

10 Block sliding off a hemispherical piece of ice. . . . . . . . . . . . . . . . . 542

11 Bullet-wooden-block collision. . . . . . . . . . . . . . . . . . . . . . . . . 559

12 Ballistic Pendulum found in lab . . . . . . . . . . . . . . . . . . . . . . . 566

13 Ballistic Pendulum: Stage 1 . . . . . . . . . . . . . . . . . . . . . . . . . 566



16 Vertical glass tube attached to horizontal water pipe. . . . . . . . . . . . 592


0 About the mathematical prerequisites material

0.1 What is a College Physics Course?

Typically, in the United States a college physics course is a physics course that is taughtwithout the use of calculus. The prerequisites are high school algebra (a.k.a. AlgebraII from high school, or equivalently intermediate algebra from a community college) andtrigonometry. It is also assumed that the student has had an introduction to dimensions,units, and significant figures.

0.2 What are the math-prerequise topics for college physics?

College Physics Prerequisite Topics:

• Geometry section 1 and all subsections therein

• Algebra section 2 and all subsections therein

• Graphing section 3 and all subsections therein

• Basic Trigonometry section 4 and all subsections therein (except identities)

• An introduction to vectors section 6 and all subsections therein (except the dot andcross-product)

• Introduction to measurement and all subsections therein

0.3 What is a University Physics course?

Typically, in the United States a university physics course is a calculus-based physicscourse taken by engineering, physics, and math majors. The prerequisites are a high-school physics course and a minimum mathematical background of Calculus I from highschool, or equivalently from a college, or university. Remember, this is an absoluteminimum! A background of calculus II and a strong grasp of trigonometry will make thecourse more bearable.

University Physics Prerequisite Topics:

• Geometry section 1 and all subsections therein

• Algebra section 2 and all subsections therein

• Graphing section 3 and all subsections therein


• Basic Trigonometry section 4 and all subsections therein

• Calculus section ?? and all subsections therein

• An introduction to vectors section 6 and all subsections therein

• Introduction to measurement and all subsections therein

0.4 What is a college physics course in the state of Arizona?

Unfortunately, the University of Arizona and Pima community college only require high-school algebra as a prerequisite for college physics. As such, the physics 121 course taughtat Pima presupposes no trigonometry background what-so-ever. However, a backgroundin basic trigonometry will make the first few weeks of the course more bearable. Collegealgebra is also a required co-requisite for this course.

13

Part I

Mathematical Prerequisites

14

1 Geometry

1.1 Circles

1.1.1 Letter-based problems

Problem 1. If the radius of a circle is R, what is its diameter?

(a) R/2 *(b) 2R(c) πR/2 (d) 2πR(e) None of these

Solution: You should know this by memory.

Problem 2. If the radius of a circle is R, what is its circumference?

(a) R/2 (b) 2R(c) πR/2 *(d) 2πR(e) None of these


Problem 3. If the radius of a circle is R, what is its area?

(a) πR (b) 2πR*(c) πR2 (d) 2πR2

(e) None of these


Problem 4. If the diameter of a circle is D, what is its radius?

(a) πD2/2 (b) 2πD2

*(c) D/2 (d) 2D(e) None of these

Solution: D = 2R ⇒ R = D/2

Problem 5. If the diameter of a circle is D, what is its circumference?

*(a) πD (b) 2πD(c) πD/2 (d) πD/4(e) None of these

Solution: C = 2πR and D = 2R; so C = 2π(D/2) = πD

Problem 6. If the diameter of a circle is D, what is its area?

(a) πD (b) 2πD(c) πD2/2 *(d) πD2/4(e) None of these

Solution: A = πR2 and D = 2R; so A = π(D/2)2 = πD2/4.

15

Problem 7. If the area of a circle is A, what is its radius?

(a)

√A

2π(b)

√2A

π

(c)

√A

π*(d)

√A

π

(e) None of these

Solution: A = πR2 ⇒ R =

√A

π

Problem 8. If the area of a circle is A, what is its diameter?

*(a) 2

√A

π(b)

√2A

π

(c)2√A

π(d)

√2A

π

(e) None of these

Solution: A = πR2 and D = 2R; so R =

√A

π⇒ D = 2

√A

π

Problem 9. If the area of a circle is A, what is its circumference?

(a) 2π√A *(b) 2

√πA

(c)

√A

π(d)

√2A

π

(e) None of these

Solution: A = πR2 and C = 2πR; so R =

√A

π⇒ C = 2π

√A

π= 2√πA

Problem 10. If the circumference of a circle is C, what is its radius?

(a)

√2C

π(b)

2C

π

(c)

√C

2π*(d)

C

2π

(e) None of these

Solution: C = 2πR ⇒ R =C

2π

16

Problem 11. If the circumference of a circle is C, what is its diameter?

(a)

√C

2π(b)

√2C

π

(c)C

2π*(d)

C

π

(e) None of these

Solution: C = 2πR and D = 2R; so R =C

2π⇒ D =

C

π

Problem 12. If the circumference of a circle is C, what is its area?

(a)

√2C

π*(b)

C2

4π

(c)

√2C

π(d)

πC2

4

(e) None of these

Solution: C = 2πR and A = πR2; so R =C

2π⇒ A = π

(C

2π

)2

=C2

4π

1.1.2 Non-calculator-based problems

Problem 13. What is the circumference of a circle with radius 6 cm?

(a) 6π cm (b) 9π cm*(c) 12π cm (d) 36π cm(e) None of these

Solution: C = 2πR = 2π(6 cm) = 12π cm

Problem 14. What is the circumference of a circle with diameter 3 m?

*(a) 3π m (b) 6π m(c) 9π m (d) 36π m(e) None of these

Solution: C = 2πR and D = 2R; so C = πD = 3π m

Problem 15. What is the area of a circle with diameter 5 m?

(a)5π

2m2 *(b)

25π

4m2

(c) 5π2 m2 (d) 10π2 m2

(e) None of these

Solution: A = πR2 and D = 2R; so A = π

(D

2

)2

= π

(5 m

2

)2

=25π

4m2

17

Problem 16. What is the area of a circle with radius 3 cm?

(a)3π

4cm2 (b)

9π

4cm2

(c)9π

2cm2 *(d) 9π cm2

(e) None of these

Solution: A = πR2 = π(3 cm)2 = 9 cm2

Problem 17. What is the circumference of a circle with diameter 10 cm?

(a) 5π cm *(b) 10π cm(c) 25π cm (d) 100π cm(e) None of these

Solution: C = 2πR and D = 2R; so C = πD = 10π m

Problem 18. What is the area of a circle with diameter 10 m?

(a) 5π m2 (b) 10π m2

(c)25π

4cm2 *(d) 25π cm2

(e) None of these

Solution: A = πR2 and D = 2R; so A = π(D/2)2 = π(5 cm)2 = 25π cm2

Problem 19. What is the circumference of a circle with radius 4 cm?

(a) 2π cm (b) 4π cm*(c) 8π cm (d) 16π cm(e) None of these

Solution: C = 2πR = 2π(4 cm) = 8π cm

Problem 20. What is the area of a circle with radius 6 cm?

(a) 3π cm2 (b) 6π cm2

(c) 9π cm2 *(d) 36π cm2

(e) None of these

Solution: A = πR2 = π(6 cm)2 = 36π cm2

18

Problem 21. If the circumference of a circle is 5 cm, what is its area?

*(a)25

4πcm2 (b) 50π cm2

(c)25

2πcm2 (d)

25π

2cm2

(e) None of these

Solution: C = 2πR and A = πR2; so

R =C

2π⇒ A = π

(C

2π

)2

=C2

4π=

25

4π

Problem 22. If the area of a circle is 9 cm2, what is its radius?

(a) 3π cm2 (b) 3√π cm

(c)3

πcm *(d)

3√π

cm

(e) None of these

Solution: A = πR2 ⇒ R =

√A

π=

√9

π=

3√π

Problem 23. If the circumference of a circle is 4 cm, what is its diameter?

*(a)4

πcm (b) 8 cm

(c)8

πcm (d)

2√π

cm2

(e) None of these

Solution: C = 2πR and D = 2R; so R =C

2π⇒ D =

C

π=

4

π

Problem 24. If the area of a circle is 4 m2, what is its diameter?

(a)2

πm *(b)

4√π

m

(c) 16π m (d)16

πm

(e) None of these

Solution: A = πR2 and D = 2R; so

R =

√A

π⇒ D = 2

√A

π= 2

√4

π=

4√π

19

Problem 25. If the circumference of a circle is 10 m, what is its radius?

(a)10

πm (b)

√10

πm

*(c)5

πm (d)

√5

πm

(e) None of these

Solution: C = 2πR ⇒ R =C

2π=

10

2π=

5

π

Problem 26. If the area of a circle is 3 m2, what is its circumference?

(a)2√

3√π

m *(b) 2√

3π m

(c)

√3

2πm (d)

√6

πm

(e) None of these

Solution: A = πR2 and C = 2πR; so

R =

√A

π⇒ C = 2π

√A

π= 2√πA = 2

√3π

1.1.3 Calculator-based problems

Problem 27. What is the circumference of a circle with radius 11.9 cm? Round youranswer to the nearest 0.1 cm.

*(a) 74.8 cm (b) 82.2 cm(c) 90.5 cm (d) 99.5 cm(e) None of these

Solution: C = 2πr = 2π(11.9 cm) = 74.8 cm

Problem 28. What is the circumference of a circle with radius 8.3 cm? Round youranswer to the nearest centimeter.

(a) 38 cm (b) 42 cm(c) 47 cm *(d) 52 cm(e) None of these

Solution: C = 2πr = 2π(8.3 cm) = 52 cm

20

Problem 29. What is the circumference of a circle with diameter 5.6 cm? Round youranswer to the nearest centimeter.

*(a) 18 cm (b) 19 cm(c) 21 cm (d) 23 cm(e) None of these

Solution: C = 2πr and d = 2r; so C = πd = π(5.6 cm) = 18 cm

Problem 30. What is the circumference of a circle with diameter 6.9 cm? Round youranswer to the nearest centimeter.


Solution: C = 2πr and d = 2r; so C = πd = π(6.9 cm) = 22 cm

Problem 31. What is the area of a circle with radius 1.1 cm? Round your answer tothe nearest 0.1 cm2.

*(a) 3.8 cm2 (b) 4.2 cm2

(c) 4.6 cm2 (d) 5.1 cm2

(e) None of these

Solution: A = πr2 = π(1.1 cm)2 = 3.8 cm2

Problem 32. What is the area of a circle with radius 6.1 cm? Round your answer tothe nearest 10 cm2.

(a) 90 cm2 (b) 100 cm2

(c) 110 cm2 *(d) 120 cm2

(e) None of these

Solution: A = πr2 = π(6.1 cm)2 = 120 cm2

Problem 33. What is the area of a circle with diameter 4.4 cm? Round your answer tothe nearest cm2.

(a) 12 cm2 (b) 14 cm2

*(c) 15 cm2 (d) 17 cm2

(e) None of these

Solution: A = πr2 and d = 2r; so A = π

(d

2

)2

=πd2

4=π(4.4 cm)2

4= 15 cm2

21

Problem 34. What is the area of a circle with diameter 7.8 cm? Round your answer tothe nearest cm2.

(a) 39 cm2 (b) 43 cm2

*(c) 48 cm2 (d) 53 cm2

(e) None of these

Solution: A = πr2 and d = 2r; so A = π

(d

2

)2

=πd2

4=π(7.8 cm)2

4= 48 cm2

Problem 35. If the circumference of a circle is 4 cm, what is its area?

(a) 4π cm2 (b) 16π cm2

*(c)4

πcm2 (d)

16

πcm2

(e) None of these

Solution: C = 2πR and A = πR2; so

R =C

2π⇒ A = π

(C

2π

)2

=C2

4π=

42

4π=

4

π

1.2 Rectangles and rectangular solids


Problem 36. What is the area of a square with sides of length s?

(a) 4s (b) 2πs*(c) s2 (d) 4πs2

(e) None of these


Problem 37. What is the area of a rectangle with length l and width w?

(a) 2l + 2w *(b) lw

(c)√l2 + w2 (d) l2 + w2

(e) None of these


Problem 38. What is the volume of a cube with sides of length s?

(a) 8s (b) 6s2

*(c) s3 (d)√

3 s(e) None of these


22

Problem 39. What is the volume of a rectangular solid with length l, width w, andheight h?

(a) 2l + 2w + 2h *(b) lwh

(c)√l2 + w2 + h2 (d) l3 + w3 + h3

(e) None of these



Problem 40. What is the area of a square tabletop measuring 3 feet on a side?

*(a) 9 ft2 (b) 3π ft2

(c) 12 ft2 (d) 6π ft2

(e) None of these

Solution: A = s2 = (3 ft)2 = 9 ft2

Problem 41. What is the area of a rectangular sheet of paper measuring 10 inches longby 8 inches wide?

(a) 40 in2 *(b) 80 in2

(c) 40π in2 (d) 80π in2

(e) None of these

Solution: A = lw = (10 in)(8 in) = 80 in2

Problem 42. What is the volume of a cube measuring 2 inches on a side?

(a) 4 in3 (b) 4π in3

*(c) 8 in3 (d) 8π in3

(e) None of these

Solution: V = s3 = (2 in)3 = 8 in3

Problem 43. What is the volume of a rectangular box measuring 2 cm long by 3 cmwide by 5 cm high?

(a) 10 cm3 (b) 20 cm3

*(c) 30 cm3 (d) 60 cm3

(e) None of these

Solution: V = lwh = (2 cm)(3 cm)(5 cm) = 30 cm3

23

Problem 44. What is the area of a square room measuring 5 m on a side?

(a) 5π m2 (b) 25π m2

(c)25π

2m2 *(d) 25 m2

(e) None of these

Solution: A = s2 = (5 m)2 = 25 m2

Problem 45. What is the volume of a rectangular room measuring 4 m long by 5 mwide by 3 m high?

(a) 12 m3 (b) 24 m3

(c) 30 m3 *(d) 60 m3

(e) None of these

Solution: V = lwh = (4 m)(5 m)(3 m) = 60 m3

Problem 46. What is the volume of a cube measuring 10 cm on a side?

(a)1000

3cm3 (b) 500 cm3

*(c) 1000 cm3 (d) 2000 cm3

(e) None of these

Solution: V = s3 = (10 cm)3 = 1000 cm3

Problem 47. What is the area of a rectangle measuring 6 m long by 2 m wide?

*(a) 12 m2 (b) 24 m2

(c) 48 m2 (d) 6π m2

(e) None of these

Solution: A = lw = (6 m)(2 m) = 12 m2

Problem 48. What is the area of a square tabletop measuring 3 feet on a side?

*(a) 9 ft2 (b) 3π ft2

(c) 12 ft2 (d) 6π ft2

(e) None of these

Solution: A = s2 = (3 ft)2 = 9 ft2

24


Problem 49. What is the area of a square piece of ground whose sides are each 62 ftlong? Round your answer to the nearest 100 ft2.

(a) 3100 ft2 (b) 3500 ft2

*(c) 3800 ft2 (d) 4200 ft2

(e) None of these

Solution: A = s2 = (62 ft)2 = 3800 ft2

Problem 50. What is the area of a square tabletop whose sides are each 2.7 ft long?Round your answer to the nearest 0.1 ft2.

*(a) 7.3 ft2 (b) 8.0 ft2

(c) 8.8 ft2 (d) 9.7 ft2

(e) None of these

Solution: A = s2 = (2.7 ft)2 = 7.3 ft2

Problem 51. What is the area of a rectangular carpet measuring 2.4 m long by 1.7 mwide? Round your answer to the nearest 0.1 m2.

(a) 3.7 m2 *(b) 4.1 m2

(c) 4.5 m2 (d) 4.9 m2

(e) None of these

Solution: A = lw = (2.4 m)(1.7 m) = 4.1 m2

Problem 52. What is the area of a rectangular windowpane measuring 21 cm wide by27 cm high? Round your answer to the nearest 10 cm2.

(a) 510 cm2 *(b) 570 cm2

(c) 620 cm2 (d) 690 cm2

(e) None of these

Solution: A = lw = (21 cm)(27 cm) = 570 m2

Problem 53. What is the volume of a cube measuring 13.4 cm on a side? Round youranswer to the nearest 10 cm3.

(a) 2170 cm3 *(b) 2410 cm3

(c) 2650 cm3 (d) 2910 cm3

(e) None of these

Solution: V = s3 = (13.4 cm)3 = 2410 cm3

25

Problem 54. What is the volume of a cube measuring 0.88 ft on a side? Round youranswer to the nearest 0.01 ft3.

(a) 0.50 ft3 (b) 0.55 ft3

(c) 0.61 ft3 *(d) 0.68 ft3

(e) None of these

Solution: V = s3 = (0.88 ft)3 = 0.68 ft3

Problem 55. A rectangular fish-tank is 53 cm long, 22 cm wide, and 24 cm high. Whatis its volume? Round your answer to the nearest 1000 cm3

(a) 20, 000 cm3 (b) 23, 000 cm3

(c) 25, 000 cm3 *(d) 28, 000 cm3

(e) None of these

Solution: V = lwh = (53 cm)(22 cm)(24 cm) = 28, 000 cm3

Problem 56. A room is 5.7 m long, 4.8 m wide, and 2.4 m high. What is its volume?Round your answer to the nearest cubic meter.

*(a) 66 m3 (b) 72 m3

(c) 79 m3 (d) 87 m3

(e) None of these

Solution: V = lwh = (5.7 m)(4.8 m)(2.4 m) = 66 m3

26

1.3 Right triangles


Problem 57. In the right triangle at right, which of thefollowing equations is true?

(a) r = x+ y (b) r = 12(x+ y)

*(c) r2 = x2 + y2 (d) r = 12(x− y)

(e) None of these


��

x

yr


(a) a2 = b2 + c2 (b) b2 = a2 + c2

*(c) c2 = a2 + b2 (d) a2 + b2 + c2 = 1

(e) None of these


��

a

bc


(a) x2 = r2 + y2 *(b) x2 = r2 − y2

(c) x2 = 12(r2 + y2) (d) x2 = 1

2(r2 − y2)

(e) None of these

Solution: r2 = x2 + y2 ⇒ x2 = r2 − y2

��

x

yr


(a) a2 = 12(c− b)2 (b) a2 = 1

2(c2 − b2)

(c) a2 = (c− b)2 *(d) a2 = c2 − b2

(e) None of these

Solution: c2 = a2 + b2 ⇒ a2 = c2 − b2

��

a

bc


*(a) y2 = r2 − x2 (b) y2 = x2 − r2

(c) y2 = 12(x2 + r2) (d) y2 = 1

2(x2 − r2)

(e) None of these

Solution: r2 = x2 + y2 ⇒ y2 = r2 − x2

��

x

yr

27


(a) b2 = 12(c− a)2 (b) b2 = 1

2(c+ a)2

*(c) b2 = c2 − a2 (d) b2 = c2 + a2

(e) None of these

Solution: c2 = a2 + b2 ⇒ b2 = c2 − a2

��

a

bc


*(a) r =√x2 + y2 (b) r = x2 + y2

(c) r =√

(x+ y)2 (d) r = 12(x+ y)

(e) None of these��

x

yr

Solution: Lengths like x, y, and r are not negative. Hence

r2 = x2 + y2√−−→ r =

√x2 + y2


(a) c = 12(a2 + b2) (b) c = 1

2

√a2 + b2

(c) c = a2 + b2 *(d) c =√a2 + b2


a

bc

Solution: Lengths like a, b, and c are not negative. Hence

c2 = a2 + b2√−−→ c =

√a2 + b2


(a) x =√

(r − y)2 (b) x =√

(r + y)2

*(c) x =√r2 − y2 (d) x =

√r2 + y2


x

yr

Solution: Lengths like x, y, and r are not negative. Hence

r2 = x2 + y2 −y2−−−→ x2 = r2 − y2√−−→ x =

√r2 − y2

28


(a) a =bc

2(b) a =

2c

b

(c) a =c2 − b2

2*(d) a =

√c2 − b2

(e) None of these

��

a

bc

Solution: Lengths like a, b, and c are not negative. Hence

c2 = a2 + b2 −b2−−−→ a2 = c2 − b2√−−→ a =

√c2 − b2


Problem 67. Find r in the right triangle at right.

(a)√

7 *(b) 5(c) 7 (d) 25(e) None of these �

��

4

3r

Solution: r is a length, so it must be positive. Hence

r2 = x2 + y2√−−→ r =

√x2 + y2 =

√(4)2 + (3)2 =

√16 + 9 =

√25 = 5


(a) 3 (b)√

9

(c) 9 *(d)√

41(e) None of these �

��

5

4r


r2 = x2 + y2√−−→ r =

√x2 + y2 =

√(5)2 + (4)2 =

√25 + 16 =

√41


(a)√

8 (b)√

15

(c) 4 *(d)√


��

5

3r


r2 = x2 + y2√−−→ r =

√x2 + y2 =

√(5)2 + (3)2 =

√25 + 9 =

√34

29

Problem 70. Find x in the right triangle at right.

(a) 1 (b)√

2

*(c)√

3 (d)√


��

x

12

Solution: x is a length, so it must be positive. Hence

r2 = x2 + y2

−y2−−−→ x2 = r2 − y2

√−−→ x =

√r2 − y2 =

√(2)2 − (1)2 =

√4− 1 =

√3

Problem 71. Find x in the right triangle at right.

(a) 2 (b)√

8

*(c) 4 (d)√


��

x

35


r2 = x2 + y2

−y2−−−→ x2 = r2 − y2

√−−→ x =

√r2 − y2 =

√(5)2 − (3)2 =

√25− 9 =

√16 = 4

Problem 72. Find y in the right triangle at right.

*(a) 3 (b) 9

(c)√

21 (d)√


��

4

y5

Solution: y is a length, so it must be positive. Hence

r2 = x2 + y2

−x2

−−−→ y2 = r2 − x2

√−−→ y =

√r2 − x2 =

√(5)2 − (4)2 =

√25− 16 =

√9 = 3

30


(a) 1 *(b)√

7(c) 5 (d) 7(e) None of these �

��

3

y4


r2 = x2 + y2

−x2

−−−→ y2 = r2 − x2

√−−→ y =

√r2 − x2 =

√(4)2 − (3)2 =

√16− 9 =

√7


(a) 9 (b)√

17

(c)√

30 *(d)√


��

7

y10


r2 = x2 + y2

−x2

−−−→ y2 = r2 − x2

√−−→ y =

√r2 − x2 =

√(10)2 − (7)2 =

√100− 49 =

√51


Problem 75. Find r in the right triangle at right. Roundyour answer to one decimal place.

(a) 6.0 (b) 6.6(c) 7.4 *(d) 8.2(e) None of these �

��

6.3

5.2r


r2 = x2 + y2√−−→ r =

√x2 + y2 =

√(6.3)2 + (5.2)2 = 8.2

31

Problem 76. Find r in the right triangle at right. Roundyour answer to the nearest integer.

(a) 29 *(b) 32(c) 35 (d) 39(e) None of these �

��

26

19r


r2 = x2 + y2√−−→ r =

√x2 + y2 =

√(26)2 + (19)2 = 32

Problem 77. Find r in the right triangle at right. Roundyour answer to two decimal places.

(a) 0.46 *(b) 0.51(c) 0.56 (d) 0.62(e) None of these �

��

0.43

0.28r


r2 = x2 + y2√−−→ r =

√x2 + y2 =

√(0.43)2 + (0.28)2 = 0.51

Problem 78. Find x in the right triangle at right. Roundyour answer to one decimal place.


��

x

1.83.5


r2 = x2 + y2

−y2−−−→ x2 = r2 − y2

√−−→ x =

√r2 − y2 =

√(3.5)2 − (1.8)2 = 3.0

Problem 79. Find x in the right triangle at right. Roundyour answer to the nearest integer.

(a) 33 (b) 36*(c) 40 (d) 44(e) None of these �

��

x

6274


r2 = x2 + y2

−y2−−−→ x2 = r2 − y2

√−−→ x =

√r2 − y2 =

√(74)2 − (62)2 = 40

32

Problem 80. Find x in the right triangle at right. Roundyour answer to two decimal places.


��

x

0.200.97


r2 = x2 + y2

−y2−−−→ x2 = r2 − y2

√−−→ x =

√r2 − y2 =

√(0.97)2 − (0.20)2 = 0.95

Problem 81. Find y in the right triangle at right. Roundyour answer to two decimal places.


��

0.31

y0.34


r2 = x2 + y2

−x2

−−−→ y2 = r2 − x2

√−−→ y =

√r2 − x2 =

√(0.34)2 − (0.41)2 = 0.14

Problem 82. Find y in the right triangle at right. Roundyour answer to one decimal place.

(a) 5.2 (b) 5.8*(c) 6.4 (d) 7.0(e) None of these �

��

4.1

y7.6


r2 = x2 + y2

−x2

−−−→ y2 = r2 − x2

√−−→ y =

√r2 − x2 =

√(7.6)2 − (4.1)2 = 6.4

33

Problem 83. Find y in the right triangle at right. Roundyour answer to the nearest integer.

(a) 25 (b) 28(c) 31 *(d) 34(e) None of these �

��

39

y52


r2 = x2 + y2

−x2

−−−→ y2 = r2 − x2

√−−→ y =

√r2 − x2 =

√(52)2 − (39)2 = 34

1.4 Right triangle word problems


Problem 84. Pirates have buried their treasure in the desert, and you have obtaineda copy of the directions to it. They tell you to begin at the entrance of the physicslab and go 2 km east, then to go north for another 5 km to the treasure. What is thestraight-line distance from the physics lab to the treasure site?

(a) 4 km (b) 7 km

(c)√

21 km *(d)√

29 km(e) None of these

Solution: Sketch a diagram, as at right. The route formsa right triangle, with the legs from the physics lab east to theturning point, and from the turning point north to the treasure.The straight line from the physics lab to the treasure is thehypotenuse. Use the theorem of Pythagoras:

r2 = x2 + y2

√−−→ r =

√x2 + y2 =

√(2 km)2 + (5 km)2 =

√29 km

rlab

r��

rtreasure

2 km

5 km

r

34

Problem 85. Pirates have buried their treasure in the desert, and you have obtaineda copy of the directions to it. They tell you to begin at the entrance of the physicslab and go 3 km south, then to go west for another 4 km to the treasure. What is thestraight-line distance from the physics lab to the treasure site?

*(a) 5 km (b)√

7 km

(c)√

12 km (d) 7 km(e) None of these

Solution: Sketch a diagram, as at right. The route formsa right triangle, with the legs from the physics lab east to theturning point, and from the turning point north to the treasure.The straight line from the physics lab to the treasure is thehypotenuse. Use the theorem of Pythagoras:

r2 = x2 + y2

√−−→ r =

√x2 + y2 =

√(4 km)2 + (3 km)2 = 5 km

rtreasure

r��

rlab

4 km

3 km

r

Problem 86. Your house is 5 km from the Pima Community College Missile Test Site.A missile is fired from the test site; it goes 3 km straight up, then explodes. How far isthe missile from your house when it explodes?

(a) 4 km (b)√

8 km

(c)√

15 km *(d)√


Solution: Sketch a diagram, as at right. Your house, thetest site, and the missile form a right triangle, with a horizontalleg from the house to the test site,and a vertical leg from thetest site to the missile. The straight line from the house to themissile is the hypotenuse. Use the theorem of Pythagoras:

r2 = x2 + y2

√−−→ r =

√x2 + y2 =

√(5 km)2 + (3 km)2 =

√34 km

rhouse

rsite

��

rmissile

5 km

3 km

r

35

Problem 87. Your house is 7 miles from the Pima Community College Missile TestSite. A missile is fired from the test site; it goes 1 mile straight up, then explodes. Howfar is the missile from your house when it explodes?

(a) 7.5 mi (b) 8 mi

*(c)√

50 mi (d) 10 mi(e) None of these

Solution: Sketch a diagram, as at right. Your house, thetest site, and the missile form a right triangle, with a horizontalleg from the house to the test site,and a vertical leg from thetest site to the missile. The straight line from the house to themissile is the hypotenuse. Use the theorem of Pythagoras:

r2 = x2 + y2

√−−→ r =

√x2 + y2 =

√(7 mi)2 + (1 mi)2 =

√50 mi

rhouse

rsite

��

rmissile

7 mi

1 mir

Problem 88. A vertical telephone pole is stabilized by adiagonal guy wire anchored in the ground, as shown at right.The wire is 10 m long; it is attached to the pole 7 m aboveground level. How far from the base of the pole does the wiremeet the ground?

(a) 9 m *(b)√

51 m

(c)√

93 m (d)√

149 m(e) None of these

@@@@@@

10 m 7 m

x

Solution: We have a sketch already; we can label the sides that we know and theside that we want to find. This is a right triangle where the hypotenuse is 34 ft and thevertical side is 21 ft; we want to know the horizontal side x. Use Pythagoras:

r2 = x2 + y2

−y2−−−→ x2 = r2 − y2

√−−→ x =

√r2 − y2 =

√(10 m)2 − (7 m)2 =

√100− 49 m =

√51 m

36

Problem 89. A physics instructor has locked himself out of his office, and tries to climbin through an upper window. He leans a 5-meter ladder against the side of the buildingso that the top of the ladder is 3 m above the ground. How far from the building is thebottom of the ladder?

*(a) 4 m (b) 8 m

(c)√

34 m (d)√

44 m(e) None of these

Solution: Sketch a diagram, as at right. The ground, thebuilding, and the ladder form a right triangle whose verticalside is 3 m and whose hypotenuse is 5 m. We want to know thehorizontal side x. Use the theorem of Pythagoras:

r2 = x2 + y2

−y2−−−→ x2 = r2 − y2

√−−→ x =

√r2 − y2 =

√(5 m)2 − (3 m)2 =

√16 m = 4 m

��

x

3 m5 m

Problem 90. A blimp is attached to a cable 3 kmlong. The other end of the cable is fastened to theground. The wind is strong enough to pull the cableinto a straight line. When you are standing directlybelow the blimp, you are 2 km from the place wherethe cable is anchored in the ground. How high aboveground level is the blimp?

(a) 1 km *(b)√

5 km

(c) 5 km (d)√


��

2 km

y3 km

Solution: We have a sketch already; we can label the sides that we know and theside that we want to find. The system is a right triangle whose hypotenuse is 3 km andwhose horizontal side is 2 km; we want to know the vertical side y. Use Pythagoras:

r2 = x2 + y2

−x2

−−−→ y2 = r2 − x2

√−−→ y =

√r2 − x2 =

√(3 km)2 − (2 km)2 =

√5 km

37


Problem 91. An airplane flies a route involving three cities. From Rio Cordaro, it flies210 miles straight east to Hackerville. From Hackerville, it flies 140 miles straight northto San Pitucco. How far does it fly from San Pitucco to Rio Cordaro? Round youranswer to the nearest 10 miles.

*(a) 250 miles (b) 280 miles(c) 310 miles (d) 340 miles(e) None of these

Solution: Sketch a diagram, as at right. The route forms aright triangle, with the legs from Rio Cordaro to Hackerville andfrom Hackerville to San Pitucco on either side of the right angle.The leg from San Pitucco to Rio Cordaro is the hypotenuse. Usethe theorem of Pythagoras:

r2 = x2 + y2

√−−→ r =

√x2 + y2 =

√(210 mi)2 + (140 mi)2 = 250 mi

rRM

rHV

��

rSP

210 mi

140 mir

Problem 92. An airplane flies a route involving three cities. From Rio Cordaro, it flies140 miles straight east to Hackerville. From Hackerville, it flies 320 miles straight northto San Pitucco. How far does it fly from San Pitucco to Rio Cordaro? Round youranswer to the nearest 10 miles.

(a) 330 miles *(b) 350 miles(c) 370 miles (d) 390 miles(e) None of these

Solution: Sketch a diagram, as at right. The route forms aright triangle, with the legs from Rio Cordaro to Hackerville andfrom Hackerville to San Pitucco on either side of the right angle.The leg from San Pitucco to Rio Cordaro is the hypotenuse. Usethe theorem of Pythagoras:

r2 = x2 + y2

√−−→ r =

√x2 + y2 =

√(140 mi)2 + (320 mi)2 = 350 mi

rRM

rHV

��

rSP

140 mi

320 mir

38

Problem 93. A ship sails 58 miles straight south. It then sails another 44 miles straightwest. At this point, what is its straight-line distance from its starting point? Roundyour answer to the nearest mile.

*(a) 73 miles (b) 80 miles(c) 88 miles (d) 97 miles(e) None of these

Solution: Sketch a diagram, as at right. The route forms aright triangle, with the southward and the westward legs on ei-ther side of the right angle. The final distance from the startingpoint is the hypotenuse. Use the theorem of Pythagoras:

r2 = x2 + y2

√−−→ r =

√x2 + y2 =

√(58 mi)2 + (44 mi)2 = 73 mi

rfinish��

rstart

44 mi

58 mir

Problem 94. A ship sails 79 miles straight south. It then sails another 34 miles straightwest. At this point, what is its straight-line distance from its starting point? Roundyour answer to the nearest mile.

(a) 80 miles (b) 83 miles*(c) 86 miles (d) 89 miles(e) None of these

Solution: Sketch a diagram, as at right. The route forms aright triangle, with the southward and the westward legs on ei-ther side of the right angle. The final distance from the startingpoint is the hypotenuse. Use the theorem of Pythagoras:

r2 = x2 + y2

√−−→ r =

√x2 + y2 =

√(79 mi)2 + (34 mi)2 = 86 mi

rfinish��

rstart

34 mi

79 mir

39

Problem 95. A vertical telephone pole is stabilized by adiagonal guy wire anchored in the ground, as shown at right.The wire is 34 feet long; it is attached to the pole 21 feet aboveground level. How far from the base of the pole does the wiremeet the ground? Round your answer to the nearest foot.

(a) 24 ft *(b) 27 ft(c) 29 ft (d) 32 ft(e) None of these

@@@@@@

21 ft 34 ft

x


r2 = x2 + y2

−y2−−−→ x2 = r2 − y2

√−−→ x =

√r2 − y2 =

√(34 ft)2 − (21 ft)2 = 27 ft

Problem 96. A vertical telephone pole is stabilized by adiagonal guy wire anchored in the ground, as shown at right.The wire is 47 feet long; it is attached to the pole 28 feet aboveground level. How far from the base of the pole does the wiremeet the ground? Round your answer to the nearest foot.

(a) 31 ft (b) 34 ft*(c) 38 ft (d) 42 ft(e) None of these

@@@@@@

28 ft 47 ft

x


r2 = x2 + y2

−y2−−−→ x2 = r2 − y2

√−−→ x =

√r2 − y2 =

√(47 ft)2 − (28 ft)2 = 38 ft

40

Problem 97. A physics instructor has locked himself out of his office, and tries to climbin through an upper window. He leans a 21-foot ladder against the side of the buildingso that the top of the ladder is 18 feet above the ground. How far from the building isthe bottom of the ladder? Round your answer to the nearest foot.

*(a) 11 ft (b) 12 ft(c) 13 ft (d) 14 ft(e) None of these

Solution: Sketch a diagram, as at right. The ground, thebuilding, and the ladder form a right triangle whose vertical sideis 18 ft and whose hypotenuse is 21 ft. We want to know thehorizontal side x. Use the theorem of Pythagoras:

r2 = x2 + y2

−y2−−−→ x2 = r2 − y2

√−−→ x =

√r2 − y2 =

√(21 ft)2 − (18 ft)2 = 11 ft

��

x

18 ft

21 ft

Problem 98. A physics instructor has locked himself out of his office, and tries to climbin through an upper window. He leans a 13-foot ladder against the side of the buildingso that the top of the ladder is 9 feet above the ground. How far from the building is thebottom of the ladder? Round your answer to the nearest foot.

(a) 6 ft (b) 7 ft(c) 8 ft *(d) 9 ft(e) None of these

Solution: Sketch a diagram, as at right. The ground, thebuilding, and the ladder form a right triangle whose verticalside is 9 ft and whose hypotenuse is 13 ft. We want to know thehorizontal side x. Use the theorem of Pythagoras:

r2 = x2 + y2

−y2−−−→ x2 = r2 − y2

√−−→ x =

√r2 − y2 =

√(13 ft)2 − (9 ft)2 = 9 ft

��

x

9 ft

13 ft

41

Problem 99. A blimp is attached to a cable 890meters long. The other end of the cable is fastenedto the ground. The wind is strong enough to pull thecable into a straight line. When you are standingdirectly below the blimp, you are 660 meters fromthe place where the cable is anchored in the ground.How high above ground level is the blimp? Roundyour answer to the nearest 10 m.

(a) 540 m *(b) 600 m(c) 660 m (d) 720 m(e) None of these

��

660 m

y890 m

Solution: We have a sketch already; we can label the sides that we know and theside that we want to find. The system is a right triangle whose hypotenuse is 890 m andwhose horizontal side is 660 m; we want to know the vertical side y. Use Pythagoras:

r2 = x2 + y2

−x2

−−−→ y2 = r2 − x2

√−−→ y =

√r2 − x2 =

√(890 m)2 − (660 m)2 = 600 m

Problem 100. A blimp is attached to a cable 820meters long. The other end of the cable is fastenedto the ground. The wind is strong enough to pull thecable into a straight line. When you are standingdirectly below the blimp, you are 640 meters fromthe place where the cable is anchored in the ground.How high above ground level is the blimp? Roundyour answer to the nearest 10 m.

(a) 420 m (b) 460 m*(c) 510 m (d) 560 m(e) None of these

��

640 m

y820 m


r2 = x2 + y2

−x2

−−−→ y2 = r2 − x2

√−−→ y =

√r2 − x2 =

√(820 m)2 − (640 m)2 = 510 m

42

Problem 101. A kite is attached to a string 310 ftlong. The wind is strong enough to stretch the stringinto a straight line. When you are standing directlyunder the kite, you are 270 ft from the personholding the string. How high above the ground isthe kite? Round your answer to the nearest 10 ft.


��

270 m

y310 m


r2 = x2 + y2

−x2

−−−→ y2 = r2 − x2

√−−→ y =

√r2 − x2 =

√(310 m)2 − (270 m)2 = 150 m

Problem 102. A kite is attached to a string 560 ftlong. The wind is strong enough to stretch the stringinto a straight line. When you are standing directlyunder the kite, you are 330 ft from the personholding the string. How high above the ground isthe kite? Round your answer to the nearest 10 ft.

(a) 370 ft (b) 410 ft*(c) 450 ft (d) 500 ft(e) None of these

��

330 m

y560 m


r2 = x2 + y2

−x2

−−−→ y2 = r2 − x2

√−−→ y =

√r2 − x2 =

√(560 m)2 − (330 m)2 = 450 m

43

2 Algebra

2.1 Evaluating functions with numerical arguments

2.1.1 Non-calculator based problems

Problem 103. If f(x) = x2 − 5, find f(5).

(a) −5 (b) 0(c) 9 *(d) 20(e) None of these

Solution: f(5) = 52 − 5 = 25− 5 = 20

Problem 104. If f(x) =x+ 3

7, find f(2).

*(a)5

7(b)

10

7

(c)17

7(d)

23

7

(e) None of these

Solution: f(2) =2 + 3

7=

5

7


3, find f(4).

(a)5

7(b)

2

3

*(c) 2 (d)10

3

(e) None of these


3=

6

3= 2


2, find f(2).

(a)3

4(b) 1

(c)5

4*(d)

3

2

(e) None of these


2=

3

2

44

Problem 107. If f(x) =x2 + 1

3, find f(2).

(a)5

9(b) 1

*(c)5

3(d) 3

(e) None of these


3=

4 + 1

3=

5

3

Problem 108. If f(x) =x2 − 1

3, find f(2).

(a)1

9(b)

4

9

(c)5

3*(d) 1

(e) None of these

Solution: f(2) =22 − 1

3=

4− 1

3=

3

3= 1


3, find f(3).

(a) 0 (b)1

9

(c)1

3*(d) 2

(e) None of these


3=

9− 3

3=

6

3= 2

Problem 110. If f(x) =√x2 − 5, find f(1).

(a) −2 (b) 2

(c) −√

6 *(d) No real value(e) None of these

Solution: f(1) =√

12 − 5 =√

1− 5 =√−4.

Since the quantity under the square root sign is negative, there is no real value.

45

Problem 111. If f(x) =√x2 + 2, find f(3).

*(a)√

11 (b) 5(c) 25 (d) No real value(e) None of these

Solution: f(3) =√

32 + 2 =√

9 + 2 =√

11

Problem 112. If f(x) =√x2 − 3, find f(2).

(a) −1 *(b) 1

(c)√

5 (d) No real value(e) None of these

Solution: f(2) =√

22 − 3 =√

4− 3 =√

1 = 1

Problem 113. If f(x) = 3x+ 2, find f(4).

(a) 12 *(b) 14(c) 16 (d) 18(e) None of these

Solution: f(4) = 3(4) + 2 = 12 + 2 = 14

Problem 114. If f(x) = 3x+ 2, find f(7).

(a) 15 (b) 21*(c) 23 (d) 27(e) None of these

Solution: f(7) = 3(7) + 2 = 21 + 2 = 23


(a) −5 (b) 0(c) 9 *(d) 20(e) None of these

Solution: f(5) = 52 − 5 = 25− 5 = 20


*(a) −1 (b) 1(c) 9 (d) 49(e) None of these

Solution: f(2) = 22 − 5 = 4− 5 = −1

46


Problem 117. If f(x) = 6.3x− 4.4, find f(3.9).

(a) 14.70 (b) 16.34(c) 18.15 *(d) 20.17(e) None of these

Solution: f(3.9) = (6.3)(3.9)− 4.4 = 20.17

Problem 118. If f(x) = 6.3x− 4.4, find f(2.1).

(a) 7.15 (b) 7.95*(c) 8.83 (d) 9.71(e) None of these

Solution: f(2.1) = (6.3)(2.1)− 4.4 = 8.83

Problem 119. If f(x) = −2.3x+ 7.5, find f(1.1).

*(a) 4.97 (b) 5.47(c) 6.01 (d) 6.62(e) None of these

Solution: f(1.1) = (−2.3)(1.1) + 7.5 = 4.97

Problem 120. If f(x) = −2.3x+ 7.5, find f(4.6).

(a) −2.77 *(b) −3.08(c) −3.39 (d) −3.73(e) None of these

Solution: f(4.6) = (−2.3)(4.6) + 7.5 = −3.08


7, find f(5). Round your answer to one decimal place.



7=

8

7= 1.1


7, find f(9). Round your answer to one decimal place.

(a) 0.9 *(b) 1.7(c) 4.3 (d) 9.4(e) None of these


7=

12

7= 1.7

47


4, find f(5).

(a) 0 (b) 1.25*(c) 5 (d) 23.75(e) None of these


4= 5


4, find f(9).

(a) 4 (b) 15.25*(c) 19 (d) 79.75(e) None of these


4= 19

Problem 125. If f(x) =(x− 5)2

4, find f(2).

(a) −5.75 (b) −4.25(c) −0.25 *(d) 2.25(e) None of these

Solution: f(2) =(2− 5)2

4= 2.25

Problem 126. If f(x) =(x− 5)2

4, find f(7).

(a) −4.5 (b) 0.75*(c) 1 (d) 11(e) None of these

Solution: f(7) =(7− 5)2

4= 1

Problem 127. If f(x) =√x2 − 5, find f(7). Round your answer to one decimal place.

(a) 4.0 (b) 4.8*(c) 6.6 (d) No real value(e) None of these

Solution: f(7) =√

72 − 5 = 6.6

48


(a) 4.0 (b) 6.8*(c) 8.7 (d) No real value(e) None of these

Solution: f(9) =√

92 − 5 = 8.7

Problem 129. If f(x) =√x2 − 5, find f(−4). Round your answer to one decimal place.

(a) −6.2 *(b) 3.3(c) 9.0 (d) No real value(e) None of these

Solution: f(−4) =√

(−4)2 − 5 = 3.3


(a) −0.2 (b) 3.0(c) 9.0 *(d) No real value(e) None of these

Solution: f(2) =√

22 − 5. Since the quantity under the square root sign is negative,there is no real value.

49

2.2 Evaluating functions with variable-expression arguments

Problem 131. If f(x) = 3x+ 2, find f(a+ 1).

(a) 3a+ 1 (b) 3a+ 2(c) 3a+ 3 *(d) 3a+ 5(e) None of these

Solution: f(a+ 1) = 3(a+ 1) + 2 = 3a+ 3 + 2 = 3a+ 5

Problem 132. If f(x) = 3x+ 2, find f(a− 5).

(a) 3a− 3 (b) 3a− 7*(c) 3a− 13 (d) 3a− 15(e) None of these

Solution: f(a− 5) = 3(a− 5) + 2 = 3a− 15 + 2 = 3a− 13

Problem 133. If f(x) = x2, find f(a+ 4).

(a) a2 + 4 (b) a2 + 4a+ 4(c) a2 + 16 *(d) a2 + 8a+ 16(e) None of these

Solution: f(a+ 4) = (a+ 4)2 = a2 + 8a+ 16

Problem 134. If f(x) = x2, find f(a− 3).

(a) a2 + 3 (b) a2 − 3(c) a2 + 9 *(d) a2 − 6a+ 9(e) None of these

Solution: f(a− 3) = (a− 3)2 = a2 − 6a+ 9

Problem 135. If f(x) = x2, find f(a− 2).

(a) a2 − 2 (b) a2 + 4*(c) a2 − 4a+ 4 (d) a2 − 2a− 4(e) None of these

Solution: f(a− 2) = (a− 2)2 = a2 − 4a+ 4

Problem 136. If f(x) = x2 − 2, find f(a+ 3).

*(a) a2 + 6a+ 7 (b) a2 + 7(c) a2 + 2a+ 1 (d) a2 − 10a+ 25(e) None of these

Solution: f(a+ 3) = (a+ 3)2 − 2 = a2 + 6a+ 9− 2 = a2 + 6a+ 7

50

Problem 137. If f(x) = x2 − 2, find f(a− 1).

*(a) a2 − 2a− 1 (b) a2 − 2a+ 1(c) a2 − 6a− 9 (d) a2 − 6a+ 9(e) None of these

Solution: f(a− 1) = (a− 1)2 − 2 = a2 − 2a+ 1− 2 = a2 − 2a− 1

Problem 138. If f(x) = x2 − 2, find f(a+ 2).

(a) a2 (b) a2 + 4*(c) a2 + 4a+ 2 (d) a2 − 2a+ 4(e) None of these

Solution: f(a+ 2) = (a+ 2)2 − 2 = a2 + 4a+ 4− 2 = a2 + 4a+ 2

Problem 139. If f(x) =x− 3

4, find f(a+ 3).

*(a)a

4(b)

4a+ 9

4

(c)a+ 9

4(d)

4a− 9

4

(e) None of these

Solution: f(a+ 3) =(a+ 3)− 3

4=a

4


4, find f(a− 1).

(a)a− 1

4*(b)

a− 4

4

(c)a+ 1

4(d)

a+ 4

4

(e) None of these

Solution: f(a− 1) =(a− 1)− 3

4=a− 4

4


4, find f(a− 5).

(a)a− 23

4*(b)

a− 8

4

(c)a+ 23

4(d)

a+ 8

4

(e) None of these

Solution: f(a− 5) =(a− 5)− 3

4=a− 8

4

51

Problem 142. If f(x) = (x− 2)2, find f(a+ 2).

*(a) a2 (b) a2 + 4(c) a2 + 4a+ 2 (d) a2 − 2a+ 4(e) None of these

Solution: f(a+ 2) = ((a+ 2)− 2)2 = (a)2 = a2

Problem 143. If f(x) = (x− 2)2, find f(a− 1).

(a) a2 − 2a− 1 *(b) a2 − 6a+ 9(c) a2 − 4a+ 3 (d) a2 + 4a+ 4(e) None of these

Solution: f(a− 1) = ((a− 1)− 2)2 = (a− 3)2 = a2 − 6a+ 9

Problem 144. If f(x) = (x− 2)2, find f(a− 2).

(a) a2 (b) a2 − 4a+ 2(c) a2 − 4a− 6 *(d) a2 − 8a+ 16(e) None of these

Solution: f(a− 2) = ((a− 2)− 2)2 = (a− 4)2 = a2 − 8a+ 16

52

2.3 Evaluating functions of multiple variables

2.3.1 Functions of two variables

Non-calculator based problems

Problem 145. The formula for the volume of a cylinder is: V = πr2l, where r is theradius of the cylinder and l is its length. If a cylinder has radius 3 cm and length 2 cm,what is its volume?

(a) 9π cm3 (b) 12π cm3

*(c) 18π cm3 (d) 36π cm3

(e) None of these

Solution: V = πr2l = π(3 cm)2(2 cm) = 18π cm3


(a) 9π cm3 *(b) 12π cm3

(c) 18π cm3 (d) 36π cm3

(e) None of these



(a) 9π cm3 (b) 12π cm3

(c) 18π cm3 *(d) 36π cm3

(e) None of these


Problem 148. The formula for the centripetal acceleration of an object moving in acircle is: a = v2/r, where v is the object’s speed and r is the radius of the circle. (Ifyou’ve never heard of “centripetal acceleration”, don’t worry; you’ll learn about it duringthis course.) If an object is moving in a circle with radius 6 at a speed of 9, what is itscentripetal acceleration?

(a)9

4(b) 4

*(c)27

2(d) 54

(e) None of these

Solution: a =v2

r=

92

6=

81

6=

27

2.

53


(a)4

9(b)

9

4

(c)8

3*(d) 9

(e) None of these

Solution: a =v2

r=

62

4=

36

4= 9.


(a)4

9(b)

9

4

*(c)8

3(d) 9

(e) None of these

Solution: a =v2

r=

42

6=

16

6=

8

3.

Problem 151. If an object is dropped from a height h, the speed with which it hits theground is: v =

√2gh, where g is the gravitational acceleration. (If you don’t know what

gravitational acceleration is, don’t worry; you’ll learn about it during this course.) If anobject is dropped from a height of 5, and the gravitational acceleration is 10, what is thespeed with which the object hits the ground? Round your answer to the nearest integer.

(a) 5 (b)√

50

*(c) 10 (d)√

200(e) None of these

Solution: v =√

2gh =√

2 · 10 · 5 =√

100 = 10

54




(a)√

12 (b) 3

*(c)√

24 (d) 9(e) None of these

Solution: v =√

2gh =√

2 · 2 · 6 =√

24




*(a)√

30 (b) 30

(c)√

60 (d) 60(e) None of these

Solution: v =√

2gh =√

2 · 5 · 3 =√

30

Calculator-based problems

Problem 154. The formula for the volume of a cylinder is: V = πr2l, where r is theradius of the cylinder and l is its length. If a cylinder has radius 5.22 and length 7.80,what is its volume? Round your answer to the nearest integer.

(a) 487 (b) 541(c) 601 *(d) 668(e) None of these

Solution: V = πr2l = π(5.22)2(7.80) = 668



Solution: V = πr2l = π(3.03)2(5.51) = 159

55


*(a) 373 (b) 410(c) 451 (d) 496(e) None of these

Solution: V = πr2l = π(6.99)2(2.43) = 373

Problem 157. The formula for the centripetal acceleration of an object moving in acircle is: a = v2/r, where v is the object’s speed and r is the radius of the circle. (Ifyou’ve never heard of “centripetal acceleration”, don’t worry; you’ll learn about it duringthis course.) If an object is moving in a circle with radius 4.8 at a speed of 13, what isits centripetal acceleration? Round your answer to the nearest integer.


Solution: a =v2

r=

(13)2

4.8= 35

Problem 158. The formula for the centripetal acceleration of an object moving in acircle is: a = v2/r, where v is the object’s speed and r is the radius of the circle. (Ifyou’ve never heard of “centripetal acceleration”, don’t worry; you’ll learn about it duringthis course.) If an object is moving in a circle with radius 4.4 at a speed of 19, what isits centripetal acceleration? Round your answer to the nearest integer.


Solution: a =v2

r=

(19)2

4.4= 82

Problem 159. The formula for the centripetal acceleration of an object moving in acircle is: a = v2/r, where v is the object’s speed and r is the radius of the circle. (If youdon’t know what centripetal acceleration is, don’t worry; you’ll learn about it during thiscourse.) If an object is moving in a circle with radius 8.2 at a speed of 27, what is itscentripetal acceleration? Round your answer to the nearest integer.


Solution: a =v2

r=

(27)2

8.2= 89

56



gravitational acceleration is, don’t worry; you’ll learn about it during this course.) Ifan object is dropped from a height of 23, and the gravitational acceleration is 32, whatis the speed with which the object hits the ground? Round your answer to the nearestinteger.


Solution: v =√

2gh =√

2(32)(23) = 38



gravitational acceleration is, don’t worry; you’ll learn about it during this course.) Ifan object is dropped from a height of 39, and the gravitational acceleration is 9.8, whatis the speed with which the object hits the ground? Round your answer to the nearestinteger.


Solution: v =√

2gh =√

2(9.8)(39) = 28



gravitational acceleration is, don’t worry; you’ll learn about it during this course.) If anobject is dropped from a height of 120, and the gravitational acceleration is 32, whatis the speed with which the object hits the ground? Round your answer to the nearestinteger.


Solution: v =√

2gh =√

2(32)(120) = 88

57



gravitational acceleration is, don’t worry; you’ll learn about it during this course.) Ifan object is dropped from a height of 56, and the gravitational acceleration is 9.8, whatis the speed with which the object hits the ground? Round your answer to the nearestinteger.


Solution: v =√

2gh =√

2(9.8)(56) = 33

2.3.2 Functions of three variables

Non-calculator based problems

Problem 164. The speed of an object is given by the formula: v = v0 + at, where v0 isthe initial speed, a is the acceleration, and t is the time. What is the speed of an objectif v0 = 3, a = 2, and t = 5?


Solution: v = v0 + at = 3 + (2)(5) = 3 + 10 = 13



Solution: v = v0 + at = 2 + (3)(5) = 2 + 15 = 17



Solution: v = v0 + at = 5 + (2)(3) = 5 + 6 = 11

58

Problem 167. The force exerted by a stretched spring is given by the formula: F =k(x− x0), where k is the spring constant, x is the stretched length of the spring, and x0

is its unstretched length. (You will learn about forces and spring constants during thiscourse.) What is the force F if k = 10, x = 5, and x0 = 4?


Solution: F = k(x− x0) = 10(5− 4) = 10 · 1 = 10




Solution: F = k(x− x0) = 5(10− 4) = 5 · 6 = 30




Solution: F = k(x− x0) = 4(10− 5) = 4 · 5 = 20

Problem 170. The mass of a cylindrical weight is given by the formula: m = πr2lρ,where r is the weight’s radius, l is its length, and ρ is its density. (You will learn aboutmass and density in this course. You will also learn a number of Greek letters, includingρ, which is “rho”.) What is the mass of such a weight if its radius is 2, its length is 3,and its density is 5?

(a) 10π (b) 30π*(c) 60π (d) 90π(e) None of these

Solution: m = πr2lρ = π(22)(3)(5) = π(4)(3)(5) = 60π

59


(a) 10π (b) 30π(c) 60π *(d) 90π(e) None of these



(a) 10π (b) 30π(c) 60π *(d) 90π(e) None of these


Calculator-based problems

Problem 173. The speed of an object is given by the formula: v = v0 + at, where v0 isthe initial speed, a is the acceleration, and t is the time. What is the speed of an objectif v0 = 12, a = 1.8, and t = 14? Round your answer to the nearest integer.


Solution: v = v0 + at = 12 + (1.8)(14) = 37

Problem 174. The speed of an object is given by the formula: v = v0 + at, where v0 isthe initial speed, a is the acceleration, and t is the time. What is the speed of an objectif v0 = 45, a = 2.9, and t = 8.3? Round your answer to the nearest integer.


Solution: v = v0 + at = 45 + (2.9)(8.3) = 69

60

Problem 175. The speed of an object is given by the formula: v = v0 + at, where v0 isthe initial speed, a is the acceleration, and t is the time. What is the speed of an objectif v0 = 28, a = −2.1, and t = 5.6? Round your answer to the nearest integer.


Solution: v = v0 + at = 28 + (−2.1)(5.6) = 16


is its unstretched length. (You will learn about forces and spring constants during thiscourse.) What is the force F if k = 28, x = 19, and x0 = 12? Round your answer to thenearest integer.


Solution: F = k(x− x0) = (28)(19− 12) = 196


is its unstretched length. (You will learn about forces and spring constants during thiscourse.) What is the force F if k = 1700, x = 0.23, and x0 = 0.19? Round your answerto the nearest integer.


Solution: F = k(x− x0) = (1700)(0.23− 0.19) = 68


is its unstretched length. (You will learn about forces and spring constants during thiscourse.) What is the force F if k = 2700, x = 0.41, and x0 = 0.34? Round your answerto the nearest integer.


Solution: F = k(x− x0) = (2700)(0.41− 0.34) = 189

61

Problem 179. The mass of a cylindrical weight is given by the formula: m = πr2lρ,where r is the weight’s radius, l is its length, and ρ is its density. (You will learn aboutmass and density in this course. You will also learn a number of Greek letters, includingρ, which is “rho”.) What is the mass of such a weight if its radius is 1.7, its length is5.1, and its density is 7.9? Round your answer to the nearest integer.


Solution: m = πr2lρ = π(1.7)2(5.1)(7.9) = 366

Problem 180. The mass of a cylindrical weight is given by the formula: m = πr2lρ,where r is the weight’s radius, l is its length, and ρ is its density. (You will learn aboutmass and density in this course. You will also learn a number of Greek letters, includingρ, which is “rho”.) What is the mass of such a weight if its radius is 0.32, its length is1.1, and its density is 8.8? Round your answer to one decimal place.


Solution: m = πr2lρ = π(0.32)2(1.1)(8.8) = 3.1

Problem 181. The mass of a cylindrical weight is given by the formula: m = πr2lρ,where r is the weight’s radius, l is its length, and ρ is its density. (You will learn aboutmass and density in this course. You will also learn a number of Greek letters, includingρ, which is “rho”.) What is the mass of such a weight if its radius is 0.55, its length is1.8, and its density is 19.3? Round your answer to the nearest integer.


Solution: m = πr2lρ = π(0.55)2(1.8)(19.3) = 33

62

2.4 Solving linear equations

2.4.1 Linear equations with numerical solutions

Problem 182. If 7x+ 11 = 8, find x. Which of the following statements is true?

(a) x < −1 *(b) −1 ≤ x < 0

(c) 0 ≤ x < 1 (d) x ≥ 1

(e) None of these

Solution: 7x+ 11 = 8−11−−−→ 7x = −3

÷7−−→ x = −3

7Hence −1 ≤ x < 0

Problem 183. If 9x− 11 = 7, find x. Which of the following statements is true?

(a) x < −1 (b) −1 ≤ x < 0

(c) 0 ≤ x < 1 *(d) x ≥ 1

(e) None of these

Solution: 9x− 11 = 7+11−−−→ 9x = 18

÷9−−→ x =18

9= 2

Hence x ≥ 1

Problem 184. If 5x+ 3 = 10, find x. Which of the following statements is true?

(a) x < −1 (b) −1 ≤ x < 0

(c) 0 ≤ x < 1 *(d) x ≥ 1

(e) None of these

Solution: 5x+ 3 = 10−3−−→ 5x = 7

÷5−−→ x =7

5Hence x ≥ 1

Problem 185. If 3x− 2 = −9, find x. Which of the following statements is true?

*(a) x < −1 (b) −1 ≤ x < 0

(c) 0 ≤ x < 1 (d) x ≥ 1

(e) None of these

Solution: 3x− 2 = −9+2−−→ 3x = −7

÷3−−→ x = −7

3Hence x < −1

63

Problem 186. If −4x+ 5 = 8, find x. Which of the following statements is true?

(a) x < −1 *(b) −1 ≤ x < 0

(c) 0 ≤ x < 1 (d) x ≥ 1

(e) None of these

Solution: −4x+ 5 = 8−5−−→ −4x = 3

÷(−4)−−−−→ x = −3

4Hence −1 ≤ x < 0

Problem 187. If 5x− 7 = 3x+ 8, find x. Which of the following statements is true?

(a) x < −1 (b) −1 ≤ x < 0

(c) 0 ≤ x < 1 *(d) x ≥ 1

(e) None of these

Solution: We need to collect all the terms with x on one side of the equation, andall of the terms with no x on the other.

5x− 7 = 3x+ 8−3x−−−→ 2x− 7 = 8

+7−−→ 2x = 15÷2−−→ x =

15

2

Hence x ≥ 1

Problem 188. If 2x− 4 = 5x+ 7, find x. Which of the following statements is true?

*(a) x < −1 (b) −1 ≤ x < 0

(c) 0 ≤ x < 1 (d) x ≥ 1

(e) None of these


2x− 4 = 5x+ 7−5x−−−→ −3x− 4 = 7

+4−−→ −3x = 11÷(−3)−−−−→ x = −11

3

Hence x < −1

Problem 189. If −3x− 8 = x− 11, find x. Which of the following statements is true?

(a) x < −1 (b) −1 ≤ x < 0

*(c) 0 ≤ x < 1 (d) x ≥ 1

(e) None of these


−3x− 8 = x− 11−x−−→ −4x− 8 = −11

+8−−→ −4x = −3÷(−4)−−−−→ x =

3

4

Hence 0 ≤ x < 1

64

Problem 190. If −3x− 4 = 2x+ 7, find x. Which of the following statements is true?

*(a) x < −1 (b) −1 ≤ x < 0

(c) 0 ≤ x < 1 (d) x ≥ 1

(e) None of these


−3x− 4 = 2x+ 7−2x−−−→ −5x− 4 = 7

+4−−→ −5x = 11÷(−5)−−−−→ x = −11

5

Hence x < −1

Problem 191. If x+ 7 = −4x+ 9, find x. Which of the following statements is true?

(a) x < −1 (b) −1 ≤ x < 0

*(c) 0 ≤ x < 1 (d) x ≥ 1

(e) None of these


x+ 7 = −4x+ 9+4x−−−→ 5x+ 7 = 9

−7−−→ 5x = 2÷5−−→ x =

2

5

Hence 0 ≤ x < 1

2.4.2 Linear equations with variable-expression solutions

Problem 192. If PV = nRT , find P .

(a) P =V

nRT*(b) P =

nRT

V

(c) P = V − nRT (d) P = nRT − V

(e) None of these

Solution: PV = nRT÷V−−→ P =

nRT

V

Problem 193. If PV = nRT , find T .

*(a) T =PV

nR(b) T =

nR

PV

(c) T = PV − nR (d) T = nR− PV

(e) None of these

Solution: PV = nRTswitch sides−−−−−−→ nRT = PV

÷nR−−−→ T =PV

nR

65

Problem 194. If m = πr2lρ, find ρ.

(a) ρ = m− πr2l (b) ρ = πr2l −m

*(c) ρ =m

πr2l(d) ρ =

πr2l

m

(e) None of these

Solution: m = πr2lρswitch sides−−−−−−→ πr2lρ = m

÷πr2l−−−−→ ρ =m

πr2l

Problem 195. If m = πr2lρ, find l.

(a) l = m− πr2ρ (b) l = πr2ρ−m

*(c) l =m

πr2ρ(d) l =

πr2ρ

m

(e) None of these

Solution: m = πr2lρswitch sides−−−−−−→ πr2lρ = m

÷πr2ρ−−−−→ l =m

πr2ρ

Problem 196. If v = v0 + at, find v0.

(a) v0 =v

at(b) v0 = − v

at

(c) v0 = v + at *(d) v0 = v − at

(e) None of these

Solution: v = v0 + atswitch sides−−−−−−→ v0 + at = v

−at−−−→ v0 = v − atProblem 197. If v = v0 + at, find a.

*(a) a =v − v0

t(b) a =

v

t− v0

(c) a =vt− v0

t(d) a = v − v0

t

(e) None of these

Solution: v0 + at = v−v0−−−→ at = v − v0

÷t−−→ a =v − v0

t

Problem 198. If v = v0 + at, find t.

(a) t =v

a− v0 (b) t =

va− v0

a

(c) t = v − v0

a*(d) t =

v − v0

a

(e) None of these

Solution: v0 + at = v−v0−−−→ at = v − v0

÷a−−→ t =v − v0

a

66

Problem 199. If F = k(x− x0), find k.

(a) k =x0 − xF

(b) k =x− x0

F

(c) k =F

x0 − x*(d) k =

F

x− x0

(e) None of these

Solution: k(x− x0) = F÷(x−x0)−−−−−→ k =

F

x− x0

Problem 200. If F = k(x− x0), find x.

(a) x =k − Fx0

*(b) x =F + kx0

k

(c) x = F − kx0 (d) x = F + kx0

(e) None of these

Solution: Here, it’s a good idea to start by expanding the expression to get rid ofthe parentheses.

F = k(x− x0) = kx− kx0+kx0−−−→ kx = F + kx0

÷k−→ x =F + kx0

k

Problem 201. If F = k(x− x0), find x0.

*(a) x0 =kx− F

k(b) x0 =

F + kx

k

(c) x0 =F − xk

(d) x0 =F + x

k

(e) None of these

Solution: Here, it’s a good idea to start by expanding the expression to get rid ofthe parentheses.

F = k(x− x0) = kx− kx0−kx−−→ −kx0 = F − kx ÷(−k)−−−→ x0 =

F − kx−k

=kx− F

k

67

2.5 Solving systems of linear equations

Problem 202. For the following pair of equations, find x and y:

5x− 2y = 1

−2x+ y = 2

What is the product xy?

(a) xy = −48 (b) xy = −60(c) xy = 48 *(d) xy = 60(e) None of these

Solution: There are two different approaches that we can use to solve this. We’ll gothrough both.

The first approach is to solve one of the equations for one of the variables, then tosubstitute that expression into the other equation. In this case, since y has a coefficientof 1 in the second equation, it’s easy to solve it for y:

−2x+ y = 2+2x−−→ y = 2x+ 2

Now substitute this into the first equation:

5x− 2y = 1y=2x+2−−−−→ 5x− 2(2x+ 2) = 1

simplify−−−−→ x− 4 = 1+4−→ x = 5

Now that we know x, we can substitute into the expression for y:

y = 2x+ 2 = 2(5) + 2 = 12

It’s always a good idea to check your solutions in the original equations:

5x− 2y = 5(5)− 2(12) = 25− 24 = 1 − 2x+ y = −2(5) + 12 = −10 + 12 = 2

Then xy = (5)(12) = 60.

68

The second approach is to multiply one or both of the equations by constants so that thecoefficient of one of the variables is the same in both equations (up to sign); then addor subtract the two equations to eliminate that variable. In this case, if we multiply thesecond equation by 2, we can eliminate y.

5x− 2y = 1 −→ 5x− 2y = 1

−2x+ y = 2×2−−→ −4x+ 2y = 4

add−−−→ x+ 0y = 5 ⇒ x = 5

Now we can substitute this value into either of the original two equations and solve fory. Since the coefficient of y in the second equation is 1, it’s easiest to use that one:

−2x+ y = 2+2x−−→ y = 2x+ 2 = 2(5) + 2 = 12


2x+ y = 4

3x+ 2y = 12


*(a) xy = −48 (b) xy = −60(c) xy = 48 (d) xy = 60(e) None of these

69


The first approach is to solve one of the equations for one of the variables, then tosubstitute that expression into the other equation. In this case, since y has a coefficientof 1 in the first equation, it’s easy to solve it for y:

2x+ y = 4−2x−−→ y = −2x+ 4

Now substitute this into the second equation:

3x+ 2y = 12y=−2x+4−−−−−→ 3x+ 2(−2x+ 4) = 12

simplify−−−−→ −x+ 8 = 12

−8−→ −x = 4×(−1)−−−→ x = −4


y = −2x+ 4 = −2(−4) + 4 = 12


2x+ y = 2(−4) + 12 = 4 3x+ 2y = 3(−4) + 2(12) = 12

Then xy = (−4)(12) = −48.

The second approach is to multiply one or both of the equations by constants so that thecoefficient of one of the variables is the same in both equations (up to sign); then addor subtract the two equations to eliminate that variable. In this case, if we multiply thefirst equation by 2, we can eliminate y.

2x+ y = 4×2−→ 4x+ 2y = 8

3x+ 2y = 12 −→ 3x+ 2y = 12

subtract−−−−→ x+ 0y = −4 ⇒ x = −4

Now we can substitute this value into either of the original two equations and solve fory. Since the coefficient of y in the first equation is 1, it’s easiest to use that one:

2x+ y = 4−2x−−→ y = −2x+ 4 = −2(−4) + 4 = 12

70


6x+ y = −2

x− y = 23


(a) xy = −48 *(b) xy = −60(c) xy = 48 (d) xy = 60(e) None of these


The first approach is to solve one of the equations for one of the variables, then tosubstitute that expression into the other equation. In this case, it’s easy to solve thesecond equation for x:

x− y = 23+y−→ x = y + 23


6x+ y = −2x=y+23−−−−→ 6(y + 23) + y = −2

simplify−−−−→ 7y + 138 = −2−138−−−→ 7y = −140

÷7−→ y = −20

Now that we know y, we can substitute into the expression for x:

x = y + 23 = −20 + 23 = 3


6x+ y = 6(3) + (−20) = 18− 20 = −2 x− y = 3− (−20) = 3 + 20 = 23

Then xy = (3)(−20) = −60.

The second approach is to multiply one or both of the equations by constants so that thecoefficient of one of the variables is the same in both equations (up to sign); then addor subtract the two equations to eliminate that variable. In this case, we don’t have tomultiply by constants: we can add the two equations and eliminate y.

6x+ y = −2

x− y = 23

add−−→ 7x+ 0y = 21 ⇒ 7x = 21÷7−→ x = 3


6x+ y = −2−6x−−→ y = −6x− 2 = −6(3)− 2 = −18− 2 = −20

71


−x+ y = 2

3x− 2y = 2


(a) xy = −48 (b) xy = −60*(c) xy = 48 (d) xy = 60(e) None of these



−x+ y = 2+x−→ y = x+ 2


3x− 2y = 2y=x+2−−−−→ 3x− 2(x+ 2) = 2

simplify−−−−→ x− 4 = 2+4−→ x = 6


y = x+ 2 = 6 + 2 = 8


−x+ y = −6 + 8 = 2 3x− 2y = 3(6)− 2(8) = 18− 16 = 2

Then xy = (6)(8) = 48.

The second approach is to multiply one or both of the equations by constants so that thecoefficient of one of the variables is the same in both equations (up to sign); then addor subtract the two equations to eliminate that variable. In this case, if we multiply thefirst equation by 2, we can eliminate y.

−x+ y = 2×2−→ −2x+ 2y = 4

3x− 2y = 2 −→ 3x− 2y = 2

add−−−→ x+ 0y = 6 ⇒ x = 6


−x+ y = 2+x−→ y = x+ 2 = 6 + 2 = 8

72


−x+ y = 15

2x+ 3y = 0


*(a) xy = −54 (b) xy = −56(c) xy = 54 (d) xy = 56(e) None of these



−x+ y = 15+x−→ y = x+ 15


2x+ 3y = 0y=x+15−−−−→ 2x+ 3(x+ 15) = 0

simplify−−−−→ 5x+ 45 = 0−45−−→ 5x = −45

÷5−→ x = −9


y = x+ 15 = −9 + 15 = 6


−x+ y = −(−9) + 6 = 9 + 6 = 15 2x+ 3y = 2(−9) + 3(6) = −18 + 18 = 0

Then xy = (−9)(6) = −54.

73

The second approach is to multiply one or both of the equations by constants so that thecoefficient of one of the variables is the same in both equations (up to sign); then addor subtract the two equations to eliminate that variable. In this case, if we multiply thefirst equation by 2, we can eliminate x.

−x+ y = 15×2−→ −2x+ 2y = 30

2x+ 3y = 0 −→ 2x+ 3y = 0

add−−−→ 0x+ 5y = 30 ⇒ 5y = 30÷5−→ y = 6

Now we can substitute this value into either of the original two equations and solve fory. We’ll choose the second equation, because of the zero on the right:

2x+ 3y = 0−3y−−→ 2x = −3y

÷2−→ x = −3y

2= −3(6)

2= −9


x+ 3y = 2

2x+ 5y = 8


(a) xy = −54 *(b) xy = −56(c) xy = 54 (d) xy = 56(e) None of these

74


The first approach is to solve one of the equations for one of the variables, then tosubstitute that expression into the other equation. In this case, since x has a coefficientof 1 in the first equation, it’s easy to solve it for x:

x+ 3y = 2−3y−−→ x = −3y + 2


2x+ 5y = 8x=−3y+2−−−−−→ 2(−3y + 2) + 5y = 8

simplify−−−−→ −y + 4 = 8

−4−→ −y = 4×(−1)−−−→ y = −4


x = −3y + 2 = −3(−4) + 2 = 12 + 2 = 14


x+ 3y = 14 + 3(−4) = 14− 12 = 2 2x+ 5y = 2(14) + 5(−4) = 28− 20 = 8

Then xy = (14)(−4) = −56.

The second approach is to multiply one or both of the equations by constants so that thecoefficient of one of the variables is the same in both equations (up to sign); then addor subtract the two equations to eliminate that variable. In this case, if we multiply thefirst equation by 2, we can eliminate x.

x+ 3y = 2×2−→ 2x+ 6y = 4

2x+ 5y = 8 −→ 2x+ 5y = 8

subtract−−−−→ 0x+ y = −4 ⇒ y = −4

Now we can substitute this value into either of the original two equations and solve fory. Since the coefficient of x in the first equation is 1, it’s easiest to use that one:

x+ 3y = 2−3y−−→ x = −3y + 2 = −3(−4) + 2 = 12 + 2 = 14

75


5x− y = −3

−3x+ y = 9


(a) xy = −54 (b) xy = −56*(c) xy = 54 (d) xy = 56(e) None of these


The first approach is to solve one of the equations for one of the variables, then tosubstitute that expression into the other equation. In this case, since y has a coefficientof 1 in the second equation, it’s easy to solve it for y:

−3x+ y = 9+3x−−→ y = 3x+ 9


5x− y = 3y=3x+9−−−−→ 5x− (3x+ 9) = −3

simplify−−−−→ 2x− 9 = −3+9−→ 2x = 6

÷2−→ x = 3


y = 3x+ 9 = 3(3) + 9 = 18


5x− y = 5(3)− 18 = 15− 18 = −3 − 3x+ y = −3(3) + 18 = −9 + 18 = 9

Then xy = (3)(18) = 54.

76


5x− y = −3

−3x+ y = 9

add−−→ 2x+ 0y = 6 ⇒ 2x = 6÷2−→ x = 3

Now we can substitute this value into either of the original two equations and solve fory. Since the coefficient of y in the second equation is 1, it’s easiest to use that one:

−3x+ y = 9+3x−−→ y = 3x+ 9 = 3(3) + 9 = 9 + 9 = 18


x+ y = 15

x− y = 1


(a) xy = −54 (b) xy = −56(c) xy = 54 *(d) xy = 56(e) None of these


The first approach is to solve one of the equations for one of the variables, then tosubstitute that expression into the other equation. We’ll solve the second equation for x:

x− y = 1+y−→ x = y + 1


x+ y = 15x=y+1−−−−→ (y + 1) + y = 15

simplify−−−−→ 2y + 1 = 15−1−→ 2y = 14

÷2−→ y = 7


x = y + 1 = 7 + 1 = 8


x+ y = 8 + 7 = 15 x− y = 8− 7 = 1

Then xy = (8)(7) = 56.

77


x+ y = 15

x− y = 1

add−−→ 2x+ 0y = 16 ⇒ 2x = 16÷2−→ x = 8

Now we can substitute this value into either of the original two equations and solve fory. We’ll use the first equation:

x+ y = 15−x−→ y = −x+ 15 = −8 + 15 = 7

78

2.6 Solving quadratic equations

2.6.1 Factoring quadratic equations

Problem 210. Solve the equation: x2 − 2bx + b2 = 0. There are two solutions, x1 andx2, with x1 ≥ x2. (It is possible that x1 = x2.) What is the difference x1 − x2?

(a) x1 − x2 = 0 (b) x1 − x2 = 2b(c) x1 − x2 = −2b (d) x1 − x2 = 1(e) None of these

Solution: We can solve this by factoring, then by setting each factor equal to zero.

x2 − 2bx+ b2 = 0

Factor: (x− b)(x− b) = (x− b)2 = 0 (a perfect square)

Set factors equal to zero: x− b = 0 (a double root)

Solve: x1 = b and x2 = b

Answer: x1 − x2 = b− b = 0

Problem 211. Solve the equation: x2 − 11x+ 28 = 0. There are two solutions, x1 andx2, with x1 ≥ x2. (It is possible that x1 = x2.) What is the difference x1 − x2?

(a) x1 − x2 = 0 *(b) x1 − x2 = 3(c) x1 − x2 = 8 (d) x1 − x2 = 11(e) None of these


x2 − 11x+ 28 = 0

Factor: (x− 7)(x− 4) = 0

Set factors equal to zero: x− 7 = 0 or x− 4 = 0

Solve: x1 = 7 and x2 = 4

Answer: x1 − x2 = 7− 4 = 3

79

Problem 212. Solve the equation: x2 + 4x − 21 = 0. There are two solutions, x1 andx2, with x1 ≥ x2. (It is possible that x1 = x2.) What is the difference x1 − x2?

(a) x1 − x2 = 0 (b) x1 − x2 = 4*(c) x1 − x2 = 10 (d) x1 − x2 = 17(e) None of these


x2 + 4x− 21 = 0

Factor: (x− 3)(x+ 7) = 0

Set factors equal to zero: x− 3 = 0 or x+ 7 = 0

Solve: x1 = 3 and x2 = −7

Answer: x1 − x2 = 3− (−7) = 10

Problem 213. Solve the equation: x2 − 3x − 18 = 0. There are two solutions, x1 andx2, with x1 ≥ x2. (It is possible that x1 = x2.) What is the difference x1 − x2?

(a) x1 − x2 = 0 (b) x1 − x2 = 3(c) x1 − x2 = 6 *(d) x1 − x2 = 9(e) None of these


x2 − 3x− 18 = 0

Factor: (x− 6)(x+ 3) = 0



Answer: x1 − x2 = 6− (−3) = 9

Problem 214. Solve the equation: x2 + 8x + 15 = 0. There are two solutions, x1 andx2, with x1 ≥ x2. (It is possible that x1 = x2.) What is the difference x1 − x2?



x2 + 8x+ 15 = 0

Factor: (x+ 3)(x+ 5) = 0

Set factors equal to zero: x+ 3 = 0 or x+ 5 = 0

Solve: x1 = −3 and x2 = −5

Answer: x1 − x2 = −3− (−5) = 2

80

Problem 215. Solve the equation: 2x2 − 5x + 3 = 0. There are two solutions, x1 andx2, with x1 ≥ x2. (It is possible that x1 = x2.) What is the difference x1 − x2?

(a) x1 − x2 = 0 *(b) x1 − x2 = 12

(c) x1 − x2 = 32

(d) x1 − x2 = 52

(e) None of these


2x2 − 5x+ 3 = 0

Factor: (2x− 3)(x− 1) = 0

Set factors equal to zero: 2x− 3 = 0 or x− 1 = 0

Solve: x1 =3

2and x2 = 1

Answer: x1 − x2 =3

2− 1 =

3

2− 2

2=

1

2

Problem 216. Solve the equation: 2x2 + 11x+ 5 = 0. There are two solutions, x1 andx2, with x1 ≥ x2. (It is possible that x1 = x2.) What is the difference x1 − x2?

(a) x1 − x2 = 0 (b) x1 − x2 = 52

(c) x1 − x2 = 3 *(d) x1 − x2 = 92

(e) None of these


2x2 + 11x+ 5 = 0

Factor: (2x+ 1)(x+ 5) = 0

Set factors equal to zero: 2x+ 1 = 0 or x+ 5 = 0

Solve: x1 = −1

2and x2 = −5

Answer: x1 − x2 = −1

2− (−5) = −1

2+

10

2=

9

2

81

Problem 217. Solve the equation: x2 − 4x = 5. There are two solutions, x1 and x2,with x1 ≥ x2. (It is possible that x1 = x2.) What is the difference x1 − x2?

(a) x1 − x2 = 0 (b) x1 − x2 = 1(c) x1 − x2 = 4 *(d) x1 − x2 = 6(e) None of these

Solution: First, we need an equation with zero on one side. We can then solve byfactoring, then by setting each factor equal to zero.

x2 − 4x = 5−5−→ x2 − 4x− 5 = 0

Factor: (x− 5)(x+ 1) = 0



Answer: x1 − x2 = 5− (−1) = 6

Problem 218. Solve the equation: x2 + 7x = −10. There are two solutions, x1 and x2,with x1 ≥ x2. (It is possible that x1 = x2.) What is the difference x1 − x2?


Solution: First, we need an equation with zero on one side. We can then solve byfactoring, then by setting each factor equal to zero.

x2 + 7x = −10+10−−→ x2 + 7x+ 10 = 0

Factor: (x+ 2)(x+ 5) = 0

Set factors equal to zero: x+ 2 = 0 or x+ 5 = 0

Solve: x1 = −2 and x2 = −5

Answer: x1 − x2 = −2− (−5) = −2 + 5 = 3

82

2.6.2 Quadratic formula

Problem 219. Use a calculator or equivalent to solve the equation:

1.1x2 + 3.4x− 5.5 = 0

There are two solutions, x1 and x2, with x1 ≥ x2. (It is possible that x1 = x2.) Whichof the following statements is true of the larger solution x1?

(a) x1 < 1 *(b) 1 ≤ x1 < 1.5

(c) 1.5 ≤ x1 < 2 (d) 2 ≤ x1

(e) No real solution

Solution: We will need the quadratic formula to solve this equation:

x =−b±

√b2 − 4ac

2a

Since√b2 − 4ac ≥ 0, if it’s not imaginary, and a > 0, the larger solution is

x1 =−b+

√b2 − 4ac

2a=−3.4 +

√(3.4)2 − 4(1.1)(−5.5)

2(1.1)≈ 1.2 ⇒ 1 ≤ x1 < 1.5


2.4x2 + 3.7x− 0.61 = 0


(a) x1 < −1 (b) −1 ≤ x1 < 0

*(c) 0 ≤ x1 < 1 (d) 1 ≤ x1



x =−b±

√b2 − 4ac

2a


x1 =−b+

√b2 − 4ac

2a=−3.7 +

√(3.7)2 − 4(2.4)(−0.61)

2(2.4)≈ 0.15 ⇒ 0 ≤ x1 < 1

83


0.82x2 − 2.9x+ 0.32 = 0


(a) x1 < −1 (b) −1 ≤ x1 < −0.5

(c) −0.5 ≤ x1 < 0 *(d) 0 ≤ x1



x =−b±

√b2 − 4ac

2a


x1 =−b+

√b2 − 4ac

2a=

2.9 +√

(−2.9)2 − 4(0.82)(0.32)

2(0.82)≈ 3.4 ⇒ 0 ≤ x1


1.7x2 + 8.9x− 0.77 = 0


(a) x1 < −2 (b) −2 ≤ x1 < 0

*(c) 0 ≤ x1 < 2 (d) 2 ≤ x1



x =−b±

√b2 − 4ac

2a

Since√b2 − 4ac ≥ 0, if it’s not imaginary, the larger solution is

x1 =−b+

√b2 − 4ac

2a=−8.9 +

√(8.9)2 − 4(1.7)(−0.77)

2(1.7)≈ 0.085 ⇒ 0 ≤ x1 < 2

84

2.7 Algebra word problems

Problem 223. The length of a carpet is 3 feet greater than its width. The area of thecarpet is 80 square feet. Which of the following equations describes the carpet’s width?

*(a) w2 + 3w − 80 = 0 (b) w2 − 3w + 80 = 0

(c) w2 − 20 = 0 (d) w2 + 20 = 0

(e) None of these

Solution: If w is the carpet’s width, then the length is l = w + 3. Then

A = lw = w(w + 3) = w2 + 3w = 80 ⇒ w2 + 3w − 80 = 0

Problem 224. A window is twice as long as it is wide. The area of the window is 40square feet. Which of the following equations describes the window’s length?

(a) 3l2 − 40 = 0 (b) l2 − 20 = 0

(c) l2 − 120 = 0 *(d) l2 − 80 = 0

(e) None of these

Solution: Let l be the window’s length and w its width. Since l = 2w, we know thatw = l/2. Hence

A = lw = l · l2

=l2

2= 40 ⇒ l2 = 80 ⇒ l2 − 80 = 0

Problem 225. In the physics lab, you find two circular pieces of sheet metal. The radiusof one of the circles is 3 centimeters greater than the radius of the other. The area ofthe larger circle is twice the area of the smaller one. Which of the following equationsdescribes the radius of the smaller circle?

*(a) r2 − 6r − 9 = 0 (b) r2 − 3r − 9 = 0

(c) r2 + 6r + 9 = 0 (d) r2 + 3r + 9 = 0

(e) None of these

Solution: Let r be the radius of the smaller circle. Then r + 3 is the radius of thelarger one. The area of the small circle is πr2; the area of the large circle is π(r + 3)2.Since the area of the large circle is twice the area of the small one,

π(r + 3)2 = 2πr2

⇒ (r + 3)2 = 2r2

⇒ r2 + 6r + 9 = 2r2

⇒ r2 − 6r − 9 = 0

85

Problem 226. A rectangular window is twice as long as it is wide. If its length wereincreased by 3 feet and its width were decreased by 1 foot, it would have the same area.Which of the following equations describes the window’s width?

(a) w2 − w + 3 = 0 *(b) w − 3 = 0

(c) w2 + w − 3 = 0 (d) w2 − w − 3 = 0

(e) None of these

Solution: Let w be the window’s width. Then its length is l = 2w. Its area islw = (2w)w = 2w2. Decreasing the width by 1 and increasing the length by 3 gives anarea of (w − 1)(2w + 3). Hence

2w2 = (w − 1)(2w + 3) = 2w2 + w − 3 ⇒ w − 3 = 0

Problem 227. The distance from Tucson to Phoenix is 120 miles. You want to drivethere and back at an average speed of 60 miles per hour. Because of traffic congestion,your average speed from Tucson to Phoenix is 40 mph. How fast do you have to driveon the return trip?

(a) 60 miles per hour (b) 80 miles per hour

(c) 90 miles per hour *(d) 120 miles per hour

(e) None of these

Solution: The round-trip distance is 240 miles. If you drive that distance at anaverage speed of 60 miles per hour, then the time that it will take is 240/60 = 4 hours.If your average speed from Tucson to Phoenix is 40 mph, then the time for that leg ofthe trip is 120/40 = 3 hours. Hence you have 4 − 3 = 1 hour for the return trip. Thismeans you will have to cover 120 miles in 1 hour, so your speed must be 120 miles perhour.

Problem 228. You drive 40 miles at 60 miles per hour, then bicycle an additional 10miles at 12 miles per hour. What is your average speed for the entire trip?

*(a) 3313

miles per hour (b) 36 miles per hour

(c) 50 miles per hour (d) 50.4 miles per hour

(e) None of these

Solution: Your average speed for the trip is the total distance divided by the totaltime. The total distance is 40 + 10 = 50 miles. The total time is

40 miles

60 miles/hr+

10 miles

12 miles/hr=

2

3+

5

6=

9

6=

3

2hours

Hence the average speed is

50 miles32

hours= 50 · 2

3=

100

3= 33

1

3miles per hour

86

Problem 229. A boat moves at 5 miles per hour in still water. It is launched in a riverthat flows at 3 miles per hour. From its launch point, it goes downstream for 4 miles,then turns around and comes back upstream to the launch point. How long does theround trip take?

(a)4

5hours (b)

8

5hours

(c) 2 hours *(d)5

2hours

(e) None of these

Solution: The time for the round trip is the time that it takes to go downstream at5 + 3 = 8 miles per hour, plus the time that it takes to come back upstream at 5− 3 = 2miles per hour. Thus

t =4 miles

8 miles/hour+

4 miles

2 miles/hour=

1

2+ 2 =

5

2hours

Problem 230. An airplane flies at a speed of 80 miles per hour in still air. On a daywhen the wind is blowing from the north at 20 miles per hour, the airplane flies 200 milesstraight north, then turns around and returns to its starting point. What is its averagespeed on the round trip?

(a) 64 miles per hour (b) 6623

miles per hour

*(c) 75 miles per hour (d) 80 miles per hour

(e) None of these

Solution: The average speed for the round trip is the total distance divided by thetotal time. The total distance is 2× 200 = 400 miles. The total time is

200

80− 20+

200

80 + 20=

200

60+

200

100=

10

3+ 2 =

16

3hours

Hence the average speed is

400 miles163

hours= 400 · 3

16= 75 miles/hour

87

Problem 231. A runner and a bicyclist start from the same point at the same time,with the runner going straight north and the bicyclist going straight south. The bicyclistis 7 miles per hour faster than the runner. At the end of two hours, the two are 60 milesapart. What is the bicyclist’s speed?

(a) 1112

miles per hour (b) 14 miles per hour

*(c) 1812

miles per hour (d) 23 miles per hour

(e) None of these

Solution: Let b be the bicyclist’s speed. Then the runner’s speed is b− 7. Since thetwo are going in opposite directions, after two hours the distance between them is

2b+ 2(b− 7) = 4b− 14 = 60 miles ⇒ b =60 + 14

4=

37

2= 18

1

2miles/hour

Problem 232. You drive from Smithtown to Jonesville at a speed of v, making thetrip in time t. On the return trip, you are able to drive 10 miles per hour faster, whichshortens your travel time by one hour. Which of the following equations is true?

(a) (v − 10)t = v(t+ 1) (b) (v + 10)t = v(t− 1)

(c) (v − 10)(t+ 1) = vt *(d) (v + 10)(t− 1) = vt

(e) None of these

Solution: On the first leg of the trip, your speed is v and your time is t. On thereturn leg, your speed is v + 10 and your time is t− 1. The distance is the same in eachdirection; so

vt = (v + 10)(t− 1)

88

3 Graphs

3.1 Single points on graphs

Problem 233. The graph at right shows four pointslabelled with letters. The points are

(3, 3), (2,−5), (−6, 5), and (−4,−2).

Which of the four points is (3, 3)?

(a) A *(b) B(c) C (d) D

6y

-x

rBrA

rC r

D

Solution: Each of the four points is in a different quadrant, so we can use that toidentify them. Point A is in the second quadrant, with a negative x-coordinate and apositive y-coordinate. It must be (−6, 5). Point B is in the first quadrant; its x- andy-coordinates are both positive. It must be (3, 3). Point C is in the third quadrant; itsx- and y-coordinates are both negative. It must be (−4,−2). Point D is in the fourthquadrant; it has a positive x-coordinate and a negative y-coordinate. It must be (2,−5).


(3, 3), (2,−5), (−6, 5), and (−4,−2).

Which of the four points is (2,−5)?

(a) A (b) B(c) C *(d) D

6y

-x

rBrA

rC r

D


89


(3, 3), (2,−5), (−6, 5), and (−4,−2).

Which of the four points is (−6, 5)?

*(a) A (b) B(c) C (d) D

6y

-x

rBrA

rC r

D



(3, 3), (2,−5), (−6, 5), and (−4,−2).

Which of the four points is (−4,−2)?

(a) A (b) B*(c) C (d) D

6y

-x

rBrA

rC r

D


90


(3, 3), (2,−5), (−6, 5), and (−4,−2).

Which of the four points is A?

(a) (3, 3) (b) (2,−5)*(c) (−6, 5) (d) (−4,−2)

6y

-x

rBrA

rC r

D



(3, 3), (2,−5), (−6, 5), and (−4,−2).

Which of the four points is B?

*(a) (3, 3) (b) (2,−5)(c) (−6, 5) (d) (−4,−2)

6y

-x

rBrA

rC r

D


91


(3, 3), (2,−5), (−6, 5), and (−4,−2).

Which of the four points is C?

(a) (3, 3) (b) (2,−5)(c) (−6, 5) *(d) (−4,−2)

6y

-x

rBrA

rC r

D



(3, 3), (2,−5), (−6, 5), and (−4,−2).

Which of the four points is D?

(a) (3, 3) *(b) (2,−5)(c) (−6, 5) (d) (−4,−2)

6y

-x

rBrA

rC r

D


92


(2, 2), (3, 9), (7, 2), and (8, 8).

Which of the four points is A?

(a) (2, 2) *(b) (3, 9)(c) (7, 2) (d) (8, 8)

6y

-x

rC

rA rB

rD

Solution: The point (3, 9) is the only one where the x-coordinate is small and the y-coordinate is large. That must correspond to A. The point (7, 2) has a large x-coordinateand a small y-coordinate, so it must correspond to D. For the point (2, 2) the x- andy-coordinates are both small; that point must be C. For the point (8, 8), the x- andy-coordinates are both large; that point must be B.


(2, 2), (3, 9), (7, 2), and (8, 8).

Which of the four points is B?

(a) (2, 2) (b) (3, 9)(c) (7, 2) *(d) (8, 8)

6y

-x

rC

rA rB

rD


93


(2, 2), (3, 9), (7, 2), and (8, 8).

Which of the four points is C?

*(a) (2, 2) (b) (3, 9)(c) (7, 2) (d) (8, 8)

6y

-x

rC

rA rB

rD



(2, 2), (3, 9), (7, 2), and (8, 8).

Which of the four points is D?

(a) (2, 2) (b) (3, 9)*(c) (7, 2) (d) (8, 8)

6y

-x

rC

rA rB

rD


94


(2, 2), (3, 9), (7, 2), and (8, 8).


(a) A (b) B*(c) C (d) D

6y

-x

rC

rA rB

rD



(2, 2), (3, 9), (7, 2), and (8, 8).


*(a) A (b) B(c) C (d) D

6y

-x

rC

rA rB

rD


95


(2, 2), (3, 9), (7, 2), and (8, 8).


(a) A (b) B(c) C *(d) D

6y

-x

rC

rA rB

rD



(2, 2), (3, 9), (7, 2), and (8, 8).


(a) A *(b) B(c) C (d) D

6y

-x

rC

rA rB

rD


96

3.2 Matching graphs and equations

Problem 249. Which equation is shown on thegraph at right?

(a) y = 2x

*(b) y = 2x+ 1

(c) y = −2x

(d) y = −2x− 1

Solution: If the equation of a line is written inthe form: y = mx + b, then m is the slope and b isthe y-intercept.

6y

-x

��

The line in the figure has a positive slope (as x increases, y increases) and a positivey-intercept (when x = 0, y > 0). Of the answers given, only y = 2x + 1 has m > 0 andb > 0.


*(a) y = 2x

(b) y = 2x+ 1

(c) y = −2x

(d) y = −2x− 1


6y

-x

��

The line in the figure has a positive slope (as x increases, y increases) and a y-interceptof 0 (when x = 0, y = 0). Of the answers given, only y = 2x has m > 0 and b = 0.


(a) y = 2x

(b) y = 2x+ 1

*(c) y = −2x

(d) y = −2x− 1


6y

-x

AAAAAAAAAAAAAAA

The line in the figure has a negative slope (as x increases, y decreases) and a y-interceptof 0 (when x = 0, y = 0). Of the answers given, only y = −2x has m < 0 and b = 0.

97


(a) y = 2x

(b) y = 2x+ 1

(c) y = −2x

*(d) y = −2x− 1


6y

-x

AAAAAAAAAAAAAAA

The line in the figure has a negative slope (as x increases, y increases) and a negativey-intercept (when x = 0, y < 0). Of the answers given, only y = −2x− 1 has m < 0 andb < 0.

Problem 253. Which equation is shown on thegraph at right? (The scale is the same for the x- andy-axes.)

(a) y = 3x

(b) y = −3x

*(c) y =x

3

(d) y = −x3

6y

-x

��

��

��

��

Solution: If the equation of a line is written in the form: y = mx+ b, then m is theslope and b is the y-intercept. In this case, all of the possible answers have b = 0, so wemust focus on the slope m.

The line in the figure has a positive slope (as x increases, y increases), so it must bem = 3 or m = 1/3. Since the value of y increases more slowly than the value of x, m < 1.Hence m = 1/3; so y = x/3.

98


*(a) y = 3x

(b) y = −3x

(c) y =x

3

(d) y = −x3

6y

-x

��


The line in the figure has a positive slope (as x increases, y increases), so it must bem = 3 or m = 1/3. Since the value of y increases more rapidly than the value of x,m > 1. Hence m = 3; so y = 3x.


(a) y = 3x

(b) y = −3x

(c) y =x

3

*(d) y = −x3

6y

-x

PPPPPPPPPPPPPPP


The line in the figure has a negative slope (as x increases, y decreases), so it must bem = −3 or m = −1/3. Since the value of y decreases more slowly than the value of xincreases, m > −1. Hence m = −1/3; so y = −x/3.

99


(a) y = 3x

*(b) y = −3x

(c) y =x

3

(d) y = −x3

6y

-x

BBBBBBBBBBBBBBB


The line in the figure has a negative slope (as x increases, y decreases), so it must bem = −3 or m = −1/3. Since the value of y decreases more rapidly than the value of xincreases, m < −1. Hence m = −3; so y = −3x.

Problem 257. The four graphs (a), (b), (c), and (d) below represent four differentequations:

y = x+ 1 y = x− 1 y = −x+ 1 y = −x− 1

Which of the four graphs represents y = x+ 1?

(a) 6y

-x

@@@@@@@@

(b) 6y

-x

@@@@@@@@

(c) 6y

-x

��

*(d) 6y

-x

��

Solution: The equation y = x+1 is written in the form: y = mx+b. Here m = 1 > 0and b = 1 > 0. Since m > 0, the graph should have a positive slope (as x increases, yincreases). Since b > 0, the graph should have a positive y-intercept (when x = 0, y > 0).Only graph (d) slopes upward and crosses the y-axis above the x-axis.

100


y = x+ 1 y = x− 1 y = −x+ 1 y = −x− 1

Which of the four graphs represents y = x− 1?

(a) 6y

-x

@@@@@@@@

(b) 6y

-x

@@@@@@@@

*(c) 6y

-x

��

(d) 6y

-x

��

Solution: The equation y = x−1 is written in the form: y = mx+b. Here m = 1 > 0and b = −1 < 0. Since m > 0, the graph should have a positive slope (as x increases,y increases). Since b < 0, the graph should have a negative y-intercept (when x = 0,y < 0). Only graph (c) slopes upward and crosses the y-axis below the x-axis.


y = x+ 1 y = x− 1 y = −x+ 1 y = −x− 1

Which of the four graphs represents y = −x+ 1?

*(a) 6y

-x

@@@@@@@@

(b) 6y

-x

@@@@@@@@

(c) 6y

-x

��

(d) 6y

-x

��

Solution: The equation y = −x + 1 is written in the form: y = mx + b. Herem = −1 < 0 and b = 1 > 0. Since m < 0, the graph should have a negative slope (as xincreases, y decreases). Since b > 0, the graph should have a positive y-intercept (whenx = 0, y > 0). Only graph (a) slopes downward and crosses the y-axis above the x-axis.

101


y = x+ 1 y = x− 1 y = −x+ 1 y = −x− 1

Which of the four graphs represents y = −x− 1?

(a) 6y

-x

@@@@@@@@

*(b) 6y

-x

@@@@@@@@

(c) 6y

-x

��

(d) 6y

-x

��

Solution: The equation y = −x − 1 is written in the form: y = mx + b. Herem = −1 < 0 and b = −1 < 0. Since m < 0, the graph should have a negative slope (as xincreases, y decreases). Since b < 0, the graph should have a negative y-intercept (whenx = 0, y < 0). Only graph (b) slopes downward and crosses the y-axis below the x-axis.


(a) y = x2 + 1

*(b) y = x2 − 1

(c) y = −x2 + 1

(d) y = −x2 − 1

6y

-x

Solution: The graph has a negative y-intercept: it crosses the y-axis below the x-axis.This means that when x = 0, y < 0. That allows us to rule out equations (a) and (c).In the graph, when x has large absolute values, y > 0. This is consistent with (b), wherethe coefficient of x2 is positive; but not with (d), where the coefficient of x2 is negative.

102


(a) y = x2 + 1

(b) y = x2 − 1

*(c) y = −x2 + 1

(d) y = −x2 − 1

6y

-x

Solution: The graph has a positive y-intercept: it crosses the y-axis above the x-axis.This means that when x = 0, y > 0. That allows us to rule out equations (b) and (d).In the graph, when x has large absolute values, y < 0. This is consistent with (c), wherethe coefficient of x2 is negative, but not with (a), where the coefficient of x2 is positive.


*(a) y = x2 + 1

(b) y = x2 − 1

(c) y = −x2 + 1

(d) y = −x2 − 1

6y

-x

Solution: The graph has a positive y-intercept: it crosses the y-axis above the x-axis.This means that when x = 0, y > 0. That allows us to rule out equations (b) and (d).In the graph, when x has large absolute values, y > 0. This is consistent with (a), wherethe coefficient of x2 is positive, but not with (c), where the coefficient of x2 is negative.


(a) y = x2 + 1

(b) y = x2 − 1

(c) y = −x2 + 1

*(d) y = −x2 − 1

6y

-x

Solution: The graph has a negative y-intercept: it crosses the y-axis below the x-axis.This means that when x = 0, y < 0. That allows us to rule out equations (a) and (c).In the graph, when x has large absolute values, y < 0. This is consistent with (d), wherethe coefficient of x2 is negative, but not with (b), where the coefficient of x2 is positive.

103

Problem 265. The four graphs (a), (b), (c), and (d) below are all drawn on the samescale. They represent four different equations:

y = 2x2 y =x2

2y = −2x2 y = −x

2

2

Which of the four graphs represents y = 2x2?

(a) 6y

-x

(b) 6y

-x

(c) 6y

-x

*(d) 6y

-x

Solution: All four equations have y-intercepts of zero, so that won’t help us. Ingraphs (a) and (b), y ≤ 0 for all values of x. They must represent the two equationswhere the coefficient of x2 is negative. That leaves us with graphs (c) and (d), whichmust represent the two equations with positive coefficients of x2. Of these two equations,the curve of y = 2x2 will rise faster than the curve of y = x2/2: for example, the first ofthese includes the point (1, 2), whereas the second includes the point (1, 1

2). Hence the

graph of y = 2x2 is (d).

104


y = 2x2 y =x2

2y = −2x2 y = −x

2

2

Which of the four graphs represents y =x2

2?

(a) 6y

-x

(b) 6y

-x

*(c) 6y

-x

(d) 6y

-x

Solution: All four equations have y-intercepts of zero, so that won’t help us. Ingraphs (a) and (b), y ≤ 0 for all values of x. They must represent the two equationswhere the coefficient of x2 is negative. That leaves us with graphs (c) and (d), whichmust represent the two equations with positive coefficients of x2. Of these two equations,the curve of y = 2x2 will rise faster than the curve of y = x2/2: for example, the first ofthese includes the point (1, 2), whereas the second includes the point (1, 1

2). Hence the

graph of y = x2/2 is (c).

105


y = 2x2 y =x2

2y = −2x2 y = −x

2

2

Which of the four graphs represents y = −2x2?

(a) 6y

-x

*(b) 6y

-x

(c) 6y

-x

(d) 6y

-x

Solution: All four equations have y-intercepts of zero, so that won’t help us. Ingraphs (c) and (d), y ≥ 0 for all values of x. They must represent the two equationswhere the coefficient of x2 is positive. That leaves us with graphs (a) and (b), which mustrepresent the two equations with negative coefficients of x2. Of these two equations, thecurve of y = −2x2 will fall faster than the curve of y = −x2/2: for example, the first ofthese includes the point (1,−2), whereas the second includes the point (1,−1

2). Hence

the graph of y = −2x2 is (b).

106


y = 2x2 y =x2

2y = −2x2 y = −x

2

2

Which of the four graphs represents y = −x2

2?

*(a) 6y

-x

(b) 6y

-x

(c) 6y

-x

(d) 6y

-x

Solution: All four equations have y-intercepts of zero, so that won’t help us. Ingraphs (c) and (d), y ≥ 0 for all values of x. They must represent the two equationswhere the coefficient of x2 is positive. That leaves us with graphs (a) and (b), which mustrepresent the two equations with negative coefficients of x2. Of these two equations, thecurve of y = −2x2 will fall faster than the curve of y = −x2/2: for example, the first ofthese includes the point (1,−2), whereas the second includes the point (1,−1

2). Hence

the graph of y = −x2/2 is (a).

107

Part II

Mathematical Preliminaries

108

4 Basic Trigonometry

4.1 Arc-Length Problems using s = θ r

4.1.1 Degrees to radians: formula

Problem 269. An angle measures 29◦. What is its measurement in radians?

(a)29π

360rad *(b)

29π

180rad

(c)29 · 360

πrad (d)

29 · 180

πrad

(e) None of these


*(a)47π

180rad (b)

47

360πrad

(c)47

180πrad (d)

47

2πrad

(e) None of these


*(a)61π

180rad (b)

61π

90rad

(c)61

πrad (d)

61 · 360

πrad

(e) None of these


(a)83

πrad *(b)

83π

180rad

(c)83 · 360

πrad (d)

83

2πrad

(e) None of these

109


(a)19

πrad *(b)

19π

180rad

(c)19 · 360

πrad (d)

19π

90rad

(e) None of these


(a)53 · 360

πrad (b)

53

2πrad

*(c)53π

180rad (d)

53π

360rad

(e) None of these


(a)7 · 90

πrad (b)

7 · 180

πrad

*(c)7π

180rad (d)

7

2πrad

(e) None of these


(a)73 · 180

πrad *(b)

73π

180rad

(c)73 · 360

πrad (d)

73π

360rad

(e) None of these


*(a)43π

180rad (b)

43 · 90

πrad

(c)43π

360rad (d)

43

πrad

(e) None of these

110


(a)59 · 180

πrad (b)

59

πrad

*(c)59π

180rad (d)

59π

360rad

(e) None of these

4.1.2 Radians to degrees: formula

Problem 279. An angle measures 0.40 radians. What is its measurement in degrees?

(a)0.40π

360degrees (b)

0.40π

180degrees

(c)(0.40)(360)

πdegrees *(d)

(0.40)(180)

πdegrees

(e) None of these


(a)1.06π

180degrees (b)

(1.06)(90)

πdegrees

(c)1.06

2πdegrees *(d)

(1.06)(180)

πdegrees

(e) None of these


*(a)(0.88)(180)

πdegrees (b)

(0.88)(360)

πdegrees

(c)(0.88)(90)

πdegrees (d)

(0.88)

2πdegrees

(e) None of these


(a)(0.04)(360)

πdegrees (b)

0.04π

360degrees

(c)0.04π

90degrees *(d)

(0.04)(180)

πdegrees

(e) None of these

111


(a)0.15

πdegrees (b)

0.15π

90degrees

(c)0.15

2πdegrees *(d)

(0.15)(180)

πdegrees

(e) None of these


*(a)(1.34)(180)

πdegrees (b)

1.34

πdegrees

(c)1.34π

180degrees (d)

(1.34)(90)

πdegrees

(e) None of these


(a)0.77

2πdegrees (b)

0.77π

360degrees

(c)0.77

πdegrees *(d)

(0.77)(180)

πdegrees

(e) None of these


*(a)(0.12)(180)

πdegrees (b)

0.12

2πdegrees

(c)(0.12)(360)

πdegrees (d)

0.12π

360degrees

(e) None of these


(a)(0.85)(360)

πdegrees (b)

0.85

πdegrees

*(c)(0.85)(180)

πdegrees (d)

0.85π

90degrees

(e) None of these


(a)0.34π

360degrees *(b)

(0.34)(180)

πdegrees

(c)(0.34)(360)

πdegrees (d)

0.34π

180degrees

(e) None of these

112

4.1.3 Degrees to radians: calculator

Problem 289. An angle measures 38◦. Use a calculator or equivalent to find its mea-surement in radians. Round your answer to the nearest 0.01 rad.

(a) 1.51 rad *(b) 0.66 rad(c) 2.65 rad (d) 0.33 rad(e) None of these

Solution:38π

180= 0.66 rad


(a) 3.04 rad (b) 0.38 rad*(c) 1.52 rad (d) 6.07 rad(e) None of these

Solution:87π

180= 1.52 rad



Solution:7π

180= 0.12 rad



Solution:72π

180= 1.26 rad



Solution:12π

180= 0.21 rad

113



Solution:84π

180= 1.47 rad


(a) 2.73 rad (b) 0.18 rad(c) 1.47 rad *(d) 0.37 rad(e) None of these

Solution:21π

180= 0.37 rad


*(a) 1.31 rad (b) 2.63 rad(c) 0.65 rad (d) 0.76 rad(e) None of these

Solution:75π

180= 1.31 rad



Solution:63π

180= 1.10 rad



Solution:42π

180= 0.73 rad

114

4.1.4 Radians to degrees: calculator

Problem 299. An angle measures 0.52 rad. Use a calculator or equivalent to find itsmeasurement in degrees. Round your answer to the nearest degree.

*(a) 30◦ (b) 7◦

(c) 119◦ (d) 15◦

(e) None of these

Solution:(0.52)(180)

π= 30◦


(a) 131◦ (b) 16◦

(c) 7◦ *(d) 65◦

(e) None of these


π= 65◦


(a) 57◦ *(b) 6◦

(c) 1◦ (d) 13◦

(e) None of these


π= 6◦


(a) 348◦ (b) 4◦

*(c) 87◦ (d) 10◦

(e) None of these


π= 87◦


(a) 5◦ *(b) 21◦

(c) 42◦ (d) 11◦

(e) None of these


π= 21◦

115


*(a) 51◦ (b) 7◦

(c) 102◦ (d) 13◦

(e) None of these


π= 51◦


(a) 286◦ (b) 143◦

*(c) 72◦ (d) 5◦

(e) None of these


π= 72◦


(a) 151◦ (b) 4◦

*(c) 38◦ (d) 19◦

(e) None of these


π= 38◦


(a) 4◦ *(b) 99◦

(c) 50◦ (d) 36◦

(e) None of these


π= 99◦


(a) 13◦ (b) 6◦

(c) 14◦ *(d) 25◦

(e) None of these


π= 25◦

116

4.1.5 Arc length to radians: pictures

Problem 309. In the figure at right, what is themeasure of the angle θ?

(a)√

41 rad (b) 3π rad*(c) 5/4 rad (d) 4π/5 rad(e) None of these

��

θ4

5


(a) 10π/9 rad (b)√

19 π rad(c) 9π/10 rad *(d) 10/9 rad(e) None of these

��

θ9

10


(a) 5/3 rad *(b) 3/5 rad(c) 4 rad (d) 4π rad(e) None of these

��

θ5

3


*(a) 1 rad (b)√

2 π rad(c) 3 rad (d) π rad(e) None of these

θ9

9


(a)√

14 rad (b) 5π/7 rad(c) 5/7 rad *(d) 7/5 rad(e) None of these

��

θ5

7

117


*(a) 1/3 rad (b)√

10 π rad(c) 3 rad (d) π/3 rad(e) None of these

��θ

@R

31


*(a) 4/9 rad (b) 9/4 rad

(c)√

65 π rad (d)√

97 rad(e) None of these

��

θ9

4


(a) 4/7 rad *(b) 7/4 rad

(c) 4π/7 rad (d)√

33 π rad(e) None of these

BBBBB

θ4

7

4.1.6 Arc length to radians: descriptions

Problem 317. A circle has a radius of 5 cm. A central angle θ intercepts an arc whoselength is 4 cm. What is the measure of θ?

(a) 3 rad (b) 4π/5 rad

(c)√

41 π rad *(d) 4/5 rad(e) None of these


*(a) 5/13 rad (b) 5π/13 rad

(c)√

194 π rad (d) 13π/5 rad(e) None of these

118


(a)√

34 π rad *(b) 5/3 rad

(c) 4π rad (d)√



*(a) 7/6 rad (b)√

85 rad

(c)√

85 π rad (d)√



*(a) 2/3 rad (b) 3/2 rad

(c) 2π/3 rad (d) 2√

13 π rad(e) None of these

Problem 322. A circle has a radius of 8 cm. A central angle θ intercepts an arc whoselength is 4π cm. What is the measure of θ?

(a) 1/2 rad *(b) π/2 rad(c) 2 rad (d) 2π rad(e) None of these


(a) 3π rad *(b) π/3 rad(c) 3 rad (d) 1/3 rad(e) None of these

Problem 324. A circle has a radius of 15π cm. A central angle θ intercepts an arcwhose length is 3π cm. What is the measure of θ?

(a) π/5 rad (b) 5π rad*(c) 1/5 rad (d) 5 rad(e) None of these


(a) 2/3 rad (b) 2π/3 rad(c) 3/2 rad *(d) 3π/2 rad(e) None of these

119

Problem 326. A circle has a radius of 6π cm. A central angle θ intercepts an arc whoselength is 2 cm. What is the measure of θ?

(a) 2π/3 rad (b) 3π rad*(c) 1/3π rad (d) 3/2π rad(e) None of these

4.1.7 Radians to arc length: pictures

Problem 327. In the figure at right, if θ = 1.4 rad,what is the arc length s?

*(a) 9.8 cm (b) 5π cm(c) π/5 cm (d) 10π cm(e) None of these

Solution: (7 cm)(1.4) = 9.8 cm

��

θ7 cm

s

Problem 328. In the figure at right, if θ = 1.2 rad,what is the arc length s?

(a) 5 cm *(b) 7.2 cm(c) π/5 cm (d) 14.4π cm(e) None of these

Solution: (6 cm)(1.2) = 7.2 cm

��

θ6 cm

s

Problem 329. In the figure at right, if θ = 4/9 rad,what is the arc length s?

(a) 27π/2 cm *(b) 8/3 cm(c) 27π cm (d) 2/27 cm(e) None of these

Solution: (6 cm)(4/9) = 8/3 cm

��

��

θ6 cm

s


(a) 15π cm *(b) 10/3 cm(c) 15π/2 cm (d) 2/15 cm(e) None of these

Solution: (5 cm)(2/3) = 10/3 cm

��

θ5 cm

s

120


(a) 9π/10 cm (b) 18π/5 cm*(c) 18/5 cm (d) 20π/9 cm(e) None of these

Solution: (2 cm)(9/5) = 18/5 cm

CCCCC

θ2 cm

s


(a) 5π/8 cm (b) 15π/4 cm(c) 12π/5 cm *(d) 15/4 cm(e) None of these

Solution: (3 cm)(5/4) = 15/4 cm

��

θ3 cm

s

4.1.8 Radians to arc length: descriptions

Problem 333. A circle has a radius of 10 cm. A central angle has measure θ = 3/5 rad.What is the length of the arc intercepted by the angle?

(a) 6π/50 cm *(b) 6 cm(c) 12π cm (d) 6π cm(e) None of these

Solution: (10 cm)(3/5) = 6 cm


(a) π/4 cm *(b) 2 cm(c) 2π cm (d) 4π cm(e) None of these


Problem 335. A circle has a radius of 5 cm. A central angle has measure θ = π/3 rad.What is the length of the arc intercepted by the angle?

(a) 2/15 cm *(b) 5π/3 cm(c) 5/3 cm (d) 10π/3 cm(e) None of these

Solution: (5 cm)(π/3) = 5π/3 cm

121


*(a) 3π/2 cm (b) 3/2 cm(c) π/24 cm (d) π/12 cm(e) None of these

Solution: (6 cm)(π/4) = 3π/2 cm


(a) 6π cm (b) 3 cm(c) 2π/3 cm *(d) 16/3 cm(e) None of these

Solution: (4 cm)(4/3) = 16/3 cm


*(a) 2π cm (b) π2/25 cm(c) 50π cm (d) 2 cm(e) None of these

Solution: (10 cm)(π/5) = 2π cm


(a) π/8 cm *(b) 18 cm(c) 8π cm (d) 1/8 cm(e) None of these


Problem 340. A circle has a radius of 10 cm. A central angle has measure θ = 2π/5rad. What is the length of the arc intercepted by the angle?

(a) 2π/50 cm (b) 2/25 cm(c) 25/π cm *(d) 4π cm(e) None of these

Solution: (10 cm)(2π/5) = 4π cm

122

4.1.9 Word problems: arc length and radians

Problem 341. A wheel has a radius of 28 inches. It turns through an angle of 1.2 rad.How far does a point on the rim travel during the turn? Round your answer to thenearest inch.

(a) 11 in *(b) 34 in(c) 53 in (d) 106 in(e) None of these

Solution: s = rθ = (28 in)(1.2) = 34 in

Problem 342. A wheel has a radius of 71 cm. It turns through an angle of 0.6 rad. Howfar does a point on the rim travel during the turn? Round your answer to the nearestcm.


Solution: s = rθ = (71 cm)(0.6) = 43 cm

Problem 343. A pizza has a radius of 8.2 inches. A slice cut from the pizza has a centralangle of 0.39 rad. What is the length of the crust on the slice? Round your answer tothe nearest 0.1 in.

(a) 0.5 in (b) 1.0 in(c) 2.0 in *(d) 3.2 in(e) None of these

Solution: s = rθ = (8.2 in)(0.39) = 3.2 in

Problem 344. A pie has a radius of 13 cm. A slice cut from the pie has a central angleof 0.29 rad. What is the length of the crust on the slice? Round your answer to thenearest 0.1 cm.

(a) 1.2 cm (b) 2.7 cm*(c) 3.8 cm (d) 4.1 cm(e) None of these

Solution: s = rθ = (13 cm)(0.29) = 3.8 cm

Problem 345. A door measures 34 in from the hinge to the outer edge. It swingsthrough an angle of 1.4 rad. How far does the outer edge travel? Round your answer tothe nearest inch.

(a) 45 in *(b) 48 in(c) 51 in (d) 54 in(e) None of these

Solution: s = rθ = (34 in)(1.4) = 48 in

123

Problem 346. A wheel has a radius of 63 cm. Two adjacent spokes meet at the centerat an angle of 0.51 rad. How long is the rim between the two spokes? Round your answerto the nearest cm.

*(a) 32 cm (b) 34 cm(c) 36 cm (d) 38 cm(e) None of these

Solution: s = rθ = (63 cm)(0.51) = 32 cm

Problem 347. A bicycle wheel has a radius of 27 inches. It turns so that a point on therim travels 11 inches. What angle has the wheel turned through? Round your answer tothe nearest 0.01 rad.

(a) 0.31 rad (b) 0.35 rad(c) 0.38 rad *(d) 0.41 rad(e) None of these

Solution: θ = s/r = (11 in)/(27 in) = 0.41 rad

Problem 348. A pizza has a radius of 13 cm. A slice cut from the pizza has 10 cm ofcrust on the outside. What is the angle of the slice? Round your answer to the nearest0.01 rad.


Solution: θ = s/r = (10 cm)/(13 cm) = 0.77 rad

124

4.2 Basic Right-Triangle Trigonometry

4.2.1 Basic trig functions: finding

Problem 349. In the figure at right, sin θ = ?

(a) x/y (b) y/x*(c) y/r (d) x/r(e) None of these �

��

x

yr

θ

φ

Problem 350. In the figure at right, cos θ = ?

(a) r/y *(b) x/r(c) r/x (d) y/x(e) None of these �

��

x

yr

θ

φ

Problem 351. In the figure at right, tan θ = ?

*(a) y/x (b) y/r(c) r/x (d) r/y(e) None of these �

��

x

yr

θ

φ

Problem 352. In the figure at right, sinφ = ?

*(a) x/r (b) r/x(c) y/r (d) y/x(e) None of these �

��

x

yr

θ

φ

Problem 353. In the figure at right, cosφ = ?

*(a) y/r (b) r/x(c) x/r (d) x/y(e) None of these �

��

x

yr

θ

φ

Problem 354. In the figure at right, tanφ = ?

(a) y/r (b) r/x*(c) x/y (d) y/x(e) None of these �

��

x

yr

θ

φ

125

Problem 355. In the figure at right, tan θ = ?

(a) 4/3 (b) 3/5(c) 4/5 *(d) 3/4(e) None of these �

��

4

35

θ

φ

Problem 356. In the figure at right, sinφ = ?


��

4

35

θ

φ

Problem 357. In the figure at right, cos θ = ?

*(a) 4/5 (b) 5/3(c) 3/4 (d) 4/3(e) None of these �

��

4

35

θ

φ

Problem 358. In the figure at right, sin θ = ?


��

4

35

θ

φ

Problem 359. In the figure at right, tanφ = ?

(a) 5/3 (b) 3/5*(c) 4/3 (d) 5/4(e) None of these �

��

4

35

θ

φ

Problem 360. In the figure at right, cosφ = ?


��

4

35

θ

φ

Problem 361. In the figure at right, tanB = ?

(a) c/b *(b) b/c(c) b/a (d) a/c(e) None of these �

��

b

ca

C

B

126

Problem 362. In the figure at right, sinB = ?

(a) c/b *(b) b/a(c) b/c (d) a/c(e) None of these �

��

b

ca

C

B

Problem 363. In the figure at right, cosB = ?

(a) b/c *(b) c/a(c) a/b (d) b/a(e) None of these �

��

b

ca

C

B

Problem 364. In the figure at right, sinC = ?

(a) a/b (b) c/b*(c) c/a (d) a/c(e) None of these �

��

b

ca

C

B

Problem 365. In the figure at right, cosC = ?

(a) a/c (b) a/b*(c) b/a (d) c/b(e) None of these �

��

b

ca

C

B

Problem 366. In the figure at right, tanC = ?

(a) a/b (b) b/c(c) c/a *(d) c/b(e) None of these �

��

b

ca

C

B

Problem 367. In the figure at right, cos β = ?

(a) u/v (b) w/u*(c) w/v (d) v/u(e) None of these �

��

u

wv

α

β

Problem 368. In the figure at right, tan β = ?

(a) v/w *(b) u/w(c) u/v (d) w/v(e) None of these �

��

u

wv

α

β

127

Problem 369. In the figure at right, sinα = ?

(a) u/w (b) v/w*(c) w/v (d) w/u(e) None of these �

��

u

wv

α

β

Problem 370. In the figure at right, sin β = ?

(a) w/u (b) w/v*(c) u/v (d) v/w(e) None of these �

��

u

wv

α

β

Problem 371. In the figure at right, cosα = ?

(a) v/w *(b) u/v(c) v/u (d) w/v(e) None of these �

��

u

wv

α

β

Problem 372. In the figure at right, tanα = ?

*(a) w/u (b) v/w(c) u/v (d) v/u(e) None of these �

��

u

wv

α

β

4.2.2 Basic trig functions: identifying

Problem 373. In the figure at right, y/x = ?

(a) cos θ (b) 1/ cos θ(c) sin θ *(d) tan θ(e) None of these �

��

x

yr

θ

Problem 374. In the figure at right, y/r = ?

(a) 1/ sin θ *(b) sin θ(c) 1/ tan θ (d) cos θ(e) None of these �

��

x

yr

θ

128

Problem 375. In the figure at right, x/r = ?

(a) 1/ tan θ (b) 1/ cos θ(c) sin θ *(d) cos θ(e) None of these �

��

x

yr

θ

Problem 376. In the figure at right, r/x = ?

*(a) 1/ cos θ (b) 1/ sin θ(c) sin θ (d) 1/ tan θ(e) None of these �

��

x

yr

θ

Problem 377. In the figure at right, r/y = ?

(a) sin θ *(b) 1/ sin θ(c) tan θ (d) 1/ tan θ(e) None of these �

��

x

yr

θ

Problem 378. In the figure at right, x/y = ?

*(a) 1/ tan θ (b) 1/ sin θ(c) 1/ cos θ (d) sin θ(e) None of these �

��

x

yr

θ

Problem 379. In the figure at right, y/r = ?

(a) 1/ sinφ (b) 1/ cosφ(c) tanφ *(d) cosφ(e) None of these �

��

x

yrφ

Problem 380. In the figure at right, r/x = ?

(a) 1/ cosφ *(b) 1/ sinφ(c) tanφ (d) sinφ(e) None of these �

��

x

yrφ

Problem 381. In the figure at right, x/y = ?

(a) 1/ tanφ (b) 1/ sinφ*(c) tanφ (d) sinφ(e) None of these �

��

x

yrφ

129

Problem 382. In the figure at right, x/r = ?

(a) 1/ tanφ *(b) sinφ(c) cosφ (d) 1/ cosφ(e) None of these �

��

x

yrφ

Problem 383. In the figure at right, r/y = ?

(a) tanφ (b) sinφ(c) 1/ tanφ *(d) 1/ cosφ(e) None of these �

��

x

yrφ

Problem 384. In the figure at right, y/x = ?

(a) 1/ cosφ (b) cosφ(c) 1/ sinφ *(d) 1/ tanφ(e) None of these �

��

x

yrφ

Problem 385. In the figure at right, 3/4 = ?

(a) 1/ cos θ *(b) tan θ(c) 1/ sin θ (d) sin θ(e) None of these �

��

4

35

θ

φ


(a) sin θ *(b) 1/ tan θ(c) 1/ cos θ (d) cos θ(e) None of these �

��

4

35

θ

φ


*(a) 1/ cos θ (b) cos θ(c) 1/ tan θ (d) sin θ(e) None of these �

��

4

35

θ

φ


(a) tan θ (b) cos θ(c) 1/ sin θ *(d) sin θ(e) None of these �

��

4

35

θ

φ

130


(a) sin θ (b) 1/ tan θ(c) 1/ cos θ *(d) 1/ sin θ(e) None of these �

��

4

35

θ

φ


*(a) cos θ (b) 1/ sin θ(c) 1/ tan θ (d) 1/ cos θ(e) None of these �

��

4

35

θ

φ


(a) 1/ cosφ *(b) tanφ(c) cosφ (d) sinφ(e) None of these �

��

4

35

θ

φ


(a) sinφ (b) 1/ cosφ*(c) cosφ (d) tanφ(e) None of these �

��

4

35

θ

φ


(a) cosφ (b) tanφ(c) 1/ tanφ *(d) 1/ cosφ(e) None of these �

��

4

35

θ

φ


(a) 1/ tanφ (b) 1/ cosφ*(c) 1/ sinφ (d) tanφ(e) None of these �

��

4

35

θ

φ


(a) tanφ (b) 1/ sinφ(c) cosφ *(d) sinφ(e) None of these �

��

4

35

θ

φ

131


*(a) 1/ tanφ (b) tanφ(c) 1/ cosφ (d) 1/ sinφ(e) None of these �

��

4

35

θ

φ

Problem 397. In the figure at right, a/c = ?

*(a) sinA (b) tanA(c) cosA (d) 1/ cosA(e) None of these �

��

b

ac

A

B

Problem 398. In the figure at right, c/b = ?

*(a) 1/ sinB (b) 1/ tanB(c) cosB (d) sinB(e) None of these �

��

b

ac

A

B

Problem 399. In the figure at right, a/b = ?

*(a) tanA (b) cosA(c) 1/ cosA (d) 1/ sinA(e) None of these �

��

b

ac

A

B

Problem 400. In the figure at right, c/a = ?

(a) 1/ tanB (b) sinB(c) 1/ sinB *(d) 1/ cosB(e) None of these �

��

b

ac

A

B

Problem 401. In the figure at right, b/c = ?

(a) tanB *(b) sinB(c) 1/ cosB (d) 1/ sinB(e) None of these �

��

b

ac

A

B

Problem 402. In the figure at right, b/a = ?

(a) 1/ tanB *(b) tanB(c) 1/ sinB (d) cosB(e) None of these �

��

b

ac

A

B

132

Problem 403. In the figure at right, u/v = ?

(a) 1/ tanα (b) sinα(c) 1/ cosα *(d) 1/ sinα(e) None of these �

��

w

vu

α

β

Problem 404. In the figure at right, w/v = ?

(a) cosα *(b) 1/ tanα(c) 1/ cosα (d) 1/ sinα(e) None of these �

��

w

vu

α

β

Problem 405. In the figure at right, v/w = ?

(a) sin β *(b) 1/ tan β(c) 1/ sin β (d) cos β(e) None of these �

��

w

vu

α

β

Problem 406. In the figure at right, w/u = ?

(a) cos β *(b) sin β(c) tan β (d) 1/ cos β(e) None of these �

��

w

vu

α

β

Problem 407. In the figure at right, u/w = ?

*(a) 1/ sin β (b) 1/ tan β(c) cos β (d) tan β(e) None of these �

��

w

vu

α

β

Problem 408. In the figure at right, v/u = ?

(a) 1/ sinα (b) tanα(c) 1/ tanα *(d) sinα(e) None of these �

��

w

vu

α

β

133

4.2.3 Basic trig functions: calculator; radians

Problem 409. Use a calculator or equivalent to find sin(0.35 rad). Round your answerto three decimal places.


Problem 410. Use a calculator or equivalent to find cos(1.08 rad). Round your answerto three decimal places.


Problem 411. Use a calculator or equivalent to find tan(0.89 rad). Round your answerto three decimal places.








134













135

4.2.4 Basic trig functions: calculator; degrees

Problem 421. Use a calculator or equivalent to find sin 37◦. Round your answer tothree decimal places.


Problem 422. Use a calculator or equivalent to find cos 81◦. Round your answer tothree decimal places.


Problem 423. Use a calculator or equivalent to find tan 52◦. Round your answer tothree decimal places.








136













137

4.2.5 Using basic trig functions: formulas

Problem 433. In the figure at right, y = ?

(a) r cos θ *(b) r sin θ(c) r/ sin θ (d) r/ tan θ(e) None of these

Solution: y/r = sin θ; so y = r sin θ.��

yr

θ

Problem 434. In the figure at right, x = ?

(a) r sin θ *(b) r cos θ(c) r/ cos θ (d) r tan θ(e) None of these

Solution: x/r = cos θ; so x = r cos θ.��

x

r

θ

Problem 435. In the figure at right, x = ?

(a) y cos θ (b) y tan θ(c) y sin θ *(d) y/ tan θ(e) None of these

Solution: y/x = tan θ; so x = y/ tan θ.��

x

y

θ

Problem 436. In the figure at right, r = ?

(a) x tan θ (b) x/ sin θ(c) x sin θ *(d) x/ cos θ(e) None of these

Solution: x/r = cos θ; so r = x/ cos θ��

x

r

θ

Problem 437. In the figure at right, r = ?

*(a) y/ sin θ (b) y/ cos θ(c) y tan θ (d) y/ tan θ(e) None of these

Solution: y/r = sin θ; so r = y/ sin θ.��

yr

θ

Problem 438. In the figure at right, y = ?

(a) x cos θ *(b) x tan θ(c) x sin θ (d) x/ sin θ(e) None of these

Solution: y/x = tan θ; so y = x tan θ.��

x

y

θ

138

Problem 439. In the figure at right, s = ?

(a) 5/ cos θ (b) 5/ tan θ*(c) 5 sin θ (d) 5/ sin θ(e) None of these

Solution: s/5 = sin θ; so s = 5 sin θ.��

s5

θ


*(a) 3/ sin θ (b) 3/ tan θ(c) 3 tan θ (d) 3 cos θ(e) None of these

Solution: 3/s = sin θ; so s = 3/ sin θ.��

3s

θ

Problem 441. In the figure at right, t = ?

(a) 13 tanα (b) 13/ tanα(c) (cosα)/13 *(d) 13/ cosα(e) None of these

Solution: 13/t = cosα; so t = 13/ cosα.��

13

t

α


(a) 8 sinα (b) 8/ cosα(c) 8/ tanα *(d) 8 cosα(e) None of these

Solution: t/8 = cosα; so t = 8 cosα.��

t

8

α

Problem 443. In the figure at right, u = ?

(a) (cosφ)/9 *(b) 9/ tanφ(c) 9 tanφ (d) (tanφ)/9(e) None of these

Solution: 9/u = tanφ; so u = 9/ tanφ.��

u

9

φ


(a) (cosφ)/5 (b) 5/ sinφ*(c) 5 tanφ (d) 5/ cosφ(e) None of these

Solution: u/5 = tanφ; so u = 5 tanφ.��

5

u

φ


(a) 5 tanφ (b) (sinφ)/5(c) 5/ tanφ *(d) 5/ cosφ(e) None of these

Solution: 5/u = cosφ; so u = 5/ cosφ.��

5u

φ

139


*(a) 9 cosφ (b) 9/ cosφ(c) (sinφ)/9 (d) 9/ sinφ(e) None of these

Solution: s/9 = cosφ; so s = 9 cosφ.��

s9φ


(a) (sin θ)/3 *(b) 3/ tan θ(c) 3 cos θ (d) (tan θ)/3(e) None of these

Solution: 3/t = tan θ; so t = 3/ tan θ.��

3

t

θ


*(a) 8 sinα (b) 8/ cosα(c) 8/ sinα (d) (tanα)/8(e) None of these

Solution: u/8 = sinα; so u = 8 sinα.��

u

8α


(a) (cos θ)/3 *(b) 3 tan θ(c) (sin θ)/3 (d) 3 cos θ(e) None of these

Solution: s/3 = tan θ; so s = 3 tan θ.��

s

3

φ

θ


(a) (cos θ)/13 (b) (tan θ)/13*(c) 13/ sin θ (d) 13/ cos θ(e) None of these

Solution: 13/u = sin θ; so u = 13/ sin θ.��

13

uθ

4.2.6 Using basic trig functions: calculator

Problem 451. In the figure at right, use a calculatoror equivalent to find s. Round your answer to twodecimal places. (The figure is not necessarily drawnto scale.)


Solution: s/6 = cos 57◦; so s = 6 cos 57◦ = 3.27

��

s6 57◦

140

Problem 452. In the figure at right, use a calculatoror equivalent to find t. Round your answer to twodecimal places. (The figure is not necessarily drawnto scale.)


Solution: 5/t = tan 23◦; so t = 5/ tan 23◦ =11.78.

��

t

5

23◦

Problem 453. In the figure at right, use a calculatoror equivalent to find w. Round your answer to twodecimal places. (The figure is not necessarily drawnto scale.)


Solution: w/8 = cos 34◦; so w = 8 cos 34◦ = 6.63.

��

w

8

34◦

Problem 454. In the figure at right, use a calculatoror equivalent to find u. Round your answer to twodecimal places. (The figure is not necessarily drawnto scale.)


Solution: 7/u = sin 64◦; so u = 7/ sin 64◦ = 7.79

��

7

u 64◦



Solution: w/8 = tan 52◦; so w = 8 tan 52◦ =10.24

��

w

852◦

141



Solution: 3/s = cos 34◦; so s = 3/ cos 34◦ = 3.62

��

3

s

34◦



Solution: s/8 = sin 65◦; so s = 8 sin 65◦ = 7.25.

��

s

8 65◦

Problem 458. In the figure at right, use a calculatoror equivalent to find a. Round your answer to twodecimal places. (The figure is not necessarily drawnto scale.)


Solution: a/13 = sin 33◦; so a = 13 sin 33◦ = 7.08.

��

a13

33◦



Solution: 4/a = sin 40◦; so a = 4/ sin 40◦ = 6.22.

��

4a

40◦

142



Solution: t/9 = tan 36◦; so t = 9 tan 36◦ = 6.54.

��

9

t

36◦



Solution: w/5 = cos 49◦; so w = 5 cos 49◦ = 3.28

��

w5 49◦



Solution: 13/a = tan 60◦; so a = 13/ tan 60◦ =7.51

��

13

a60◦



Solution: t/13 = tan 49◦; so t = 13 tan 49◦ =14.95

��

t

1349◦

143



Solution: 8/a = cos 55◦; so a = 8/ cos 55◦ = 13.95

��

8a 55◦



Solution: t/5 = tan 21◦; so t = 5 tan 21◦ = 1.92

��

5

t

21◦



Solution: s/6 = sin 73◦; so s = 6 sin 73◦ = 5.74

��

s

6 73◦



Solution: a/8 = sin 28◦; so a = 8 sin 28◦ = 3.76

��

a8

28◦

144



Solution: a/6 = cos 40◦; so a = 6 cos 40◦ = 4.60

��

a

6

40◦

4.2.7 Reference angles

Problem 469. If θ = 320◦, which quadrant is it in?

(a) Quadrant I (b) Quadrant II(c) Quadrant III *(d) Quadrant IV(e) None of these

Solution: 270◦ < θ < 360◦; so θ is in quadrant IV.


(a) Quadrant I *(b) Quadrant II(c) Quadrant III (d) Quadrant IV(e) None of these

Solution: 90◦ < θ < 180◦; so θ is in quadrant II.


(a) Quadrant I (b) Quadrant II*(c) Quadrant III (d) Quadrant IV(e) None of these

Solution: 180◦ < θ < 270◦; so θ is in quadrant III.


*(a) Quadrant I (b) Quadrant II(c) Quadrant III (d) Quadrant IV(e) None of these

Solution: 0◦ < θ < 90◦; so θ is in quadrant I.

Problem 473. If θ = 5.91 rad, which quadrant is it in?

(a) Quadrant I (b) Quadrant II(c) Quadrant III *(d) Quadrant IV(e) None of these

Solution: 3π/2 < θ < 2π; so θ is in quadrant IV.

145


*(a) Quadrant I (b) Quadrant II(c) Quadrant III (d) Quadrant IV(e) None of these

Solution: 0 < θ < π/2; so θ is in quadrant I.


(a) Quadrant I *(b) Quadrant II(c) Quadrant III (d) Quadrant IV(e) None of these

Solution: π/2 < θ < π; so θ is in quadrant II.


(a) Quadrant I (b) Quadrant II*(c) Quadrant III (d) Quadrant IV(e) None of these

Solution: π < θ < 3π/2; so θ is in quadrant III.

Problem 477. An angle with measure θ degrees is in quadrant II. What is the referenceangle θref?

(a) θ − 180◦ *(b) 180◦ − θ(c) θ − 360◦ (d) 360◦ − θ(e) None of these

Problem 478. An angle with measure θ degrees is in quadrant IV. What is the referenceangle θref?

(a) θ − 180◦ (b) 180◦ − θ(c) θ − 360◦ *(d) 360◦ − θ(e) None of these

Problem 479. An angle with measure θ degrees is in quadrant III. What is thereference angle θref?

*(a) θ − 180◦ (b) 180◦ − θ(c) θ − 360◦ (d) 360◦ − θ(e) None of these

Problem 480. An angle with measure θ radians is in quadrant III. What is thereference angle θref?

(a) 2π − θ (b) θ − 2π(c) π − θ *(d) θ − π(e) None of these

146

Problem 481. An angle with measure θ radians is in quadrant II. What is the referenceangle θref?

(a) 2π − θ (b) θ − 2π*(c) π − θ (d) θ − π(e) None of these

Problem 482. An angle with measure θ radians is in quadrant IV. What is the referenceangle θref?

*(a) 2π − θ (b) θ − 2π(c) π − θ (d) θ − π(e) None of these

Problem 483. If θ = 311◦, what is θref?

(a) 31◦ (b) 39◦

(c) 41◦ *(d) 49◦

(e) None of these

Solution: 270◦ < θ < 360◦; so θ is in quadrant IV; so θref = 360◦ − θ = 49◦.


(a) 32◦ (b) 42◦

*(c) 58◦ (d) 68◦

(e) None of these

Solution: 90◦ < θ < 180◦; so θ is in quadrant II; so θref = 180◦ − θ = 58◦.


*(a) 15◦ (b) 25◦

(c) 65◦ (d) 75◦

(e) None of these

Solution: 180◦ < θ < 270◦; so θ is in quadrant III; so θref = θ − 180◦ = 15◦.


(a) 2◦ (b) 28◦

(c) 72◦ *(d) 88◦

(e) None of these

Solution: 270◦ < θ < 360◦; so θ is in quadrant IV; so θref = 360◦ − θ = 88◦.


*(a) 17◦ (b) 27◦

(c) 63◦ (d) 73◦

(e) None of these

Solution: 90◦ < θ < 180◦; so θ is in quadrant II; so θref = 180◦ − θ = 17◦.

147


(a) 10◦ *(b) 30◦

(c) 40◦ (d) 60◦

(e) None of these

Solution: 180◦ < θ < 270◦; so θ is in quadrant III; so θref = θ − 180◦ = 30◦.

Problem 489. If θ is in the first quadrant, which of the following is true?

(a) sin θ ≤ 0 and cos θ ≤ 0 (b) sin θ ≤ 0 and cos θ ≥ 0(c) sin θ ≥ 0 and cos θ ≤ 0 *(d) sin θ ≥ 0 and cos θ ≥ 0(e) None of these

Problem 490. If θ is in the second quadrant, which of the following is true?

(a) sin θ ≤ 0 and cos θ ≤ 0 (b) sin θ ≤ 0 and cos θ ≥ 0*(c) sin θ ≥ 0 and cos θ ≤ 0 (d) sin θ ≥ 0 and cos θ ≥ 0(e) None of these

Problem 491. If θ is in the third quadrant, which of the following is true?

*(a) sin θ ≤ 0 and cos θ ≤ 0 (b) sin θ ≤ 0 and cos θ ≥ 0(c) sin θ ≥ 0 and cos θ ≤ 0 (d) sin θ ≥ 0 and cos θ ≥ 0(e) None of these

Problem 492. If θ is in the fourth quadrant, which of the following is true?

(a) sin θ ≤ 0 and cos θ ≤ 0 *(b) sin θ ≤ 0 and cos θ ≥ 0(c) sin θ ≥ 0 and cos θ ≤ 0 (d) sin θ ≥ 0 and cos θ ≥ 0(e) None of these

Problem 493. An angle with measure θ degrees is in the third quadrant. What is sin θ?

(a) sin(θ − 180◦) (b) sin(360◦ − θ)*(c) − sin(θ − 180◦) (d) − sin(θ − 360◦)(e) None of these

Solution: Since θ is in the third quadrant, the reference angle is θref = θ − 180◦; andsin θ ≤ 0. Hence sin θ = − sin θref = − sin(θ − 180◦).

Problem 494. An angle with measure θ degrees is in the fourth quadrant. What issin θ?

(a) sin(θ − 180◦) (b) sin(360◦ − θ)(c) sin(θ + 180◦) *(d) − sin(360◦ − θ)(e) None of these

Solution: Since θ is in the fourth quadrant, the reference angle is θref = 360◦− θ; andsin θ ≤ 0. Hence sin θ = − sin θref = − sin(360◦ − θ).

148

Problem 495. An angle with measure θ degrees is in the second quadrant. What issin θ?

*(a) sin(180◦ − θ) (b) sin(180◦ + θ)(c) sin(360◦ − θ) (d) − sin(360◦ + θ)(e) None of these

Solution: Since θ is in the second quadrant, the reference angle is θref = 180◦ − θ;and sin θ ≥ 0. Hence sin θ = sin θref = sin(180◦ − θ).

Problem 496. An angle with measure θ degrees is in the second quadrant. What iscos θ?

(a) cos(180◦ − θ) *(b) − cos(180◦ − θ)(c) cos(θ − 180◦) (d) − cos(360◦ + θ)(e) None of these

Solution: Since θ is in the second quadrant, the reference angle is θref = 180◦ − θ;and cos θ ≤ 0. Hence cos θ = − cos θref = − cos(180◦ − θ).

Problem 497. An angle with measure θ degrees is in the third quadrant. What is cos θ?

(a) cos(θ − 180◦) (b) cos(180◦ − θ)*(c) − cos(θ − 180◦) (d) cos(180◦ + θ)(e) None of these

Solution: Since θ is in the third quadrant, the reference angle is θref = θ − 180◦; andcos θ ≤ 0. Hence cos θ = − cos θref = − cos(θ − 180◦).

Problem 498. An angle with measure θ degrees is in the fourth quadrant. What iscos θ?

(a) cos(θ − 180◦) (b) cos(180◦ − θ)(c) cos(θ + 180◦) *(d) cos(360◦ − θ)(e) None of these

Solution: Since θ is in the fourth quadrant, the reference angle is θref = 360◦− θ; andcos θ ≥ 0. Hence cos θ = cos θref = cos(360◦ − θ).

4.2.8 Arc functions: definition

Problem 499. Complete the statement: If θ is an acute angle and sin θ = 0.3, then

(a) θ = sin 0.3 (b) θ = cos 0.3*(c) θ = arcsin 0.3 (d) θ = 0.3/ sin(e) None of these

Problem 500. Complete the statement: If θ is an acute angle and cos θ = 0.4, then

(a) θ = 1/ cos 0.4 (b) θ = cos 0.4*(c) θ = cos−1 0.4 (d) θ = sin 0.4(e) None of these

149

Problem 501. Complete the statement: If θ is an acute angle and tan θ = 3.1, then

*(a) θ = tan−1 3.1 (b) θ = 3.1/ tan(c) θ = 1/ tan 3.1 (d) θ = tan(1/3.1)(e) None of these

Problem 502. Complete the statement: If θ is an acute angle and sin θ = 0.9, then

*(a) θ = sin−1 0.9 (b) θ = sin(1/0.9)(c) θ = cos 0.9 (d) θ = 1/ sin 0.9(e) None of these

Problem 503. Complete the statement: If θ is an acute angle and cos θ = 0.2, then

*(a) θ = arccos 0.2 (b) θ = cos 0.2(c) θ = sin 0.2 (d) θ = cos(5)(e) None of these

Problem 504. Complete the statement: If θ is an acute angle and tan θ = 0.5, then

*(a) θ = arctan 0.5 (b) θ = tan 2(c) θ = 1/ tan 0.5 (d) θ = 0.5/ tan(e) None of these

Problem 505. Complete the statement: If θ = arcsin 0.5, then

(a) sin θ = 2 *(b) sin θ = 0.5(c) θ = sin 2 (d) cos θ = 2(e) None of these

Problem 506. Complete the statement: If θ = arccos 0.2, then

(a) cos θ = 5 (b) θ = cos 5*(c) cos θ = 0.2 (d) θ = cos 0.2(e) None of these

Problem 507. Complete the statement: If θ = arctan 2, then

(a) tan θ = 0.5 (b) θ = tan 0.5(c) θ = 2/ tan *(d) tan θ = 2(e) None of these

Problem 508. Complete the statement: If θ = sin−1 0.4, then

(a) sin θ = 2.5 (b) 1/ sin θ = 0.4*(c) sin θ = 0.4 (d) θ = 2.5(e) None of these

150

Problem 509. Complete the statement: If θ = cos−1(1/3), then

(a) cos θ = 3 (b) 3θ = cos*(c) cos θ = 1/3 (d) sin θ = 3(e) None of these

Problem 510. Complete the statement: If θ = tan−1(2/3), then

(a) tan θ = 3/2 (b) θ = 1/ tan(2/3)*(c) tan θ = 2/3 (d) θ = tan(3/2)(e) None of these

4.2.9 Arc functions: definition; diagrams

Problem 511. Which statement is true of the figureat right?

(a) θ = sin(5/3) (b) θ = sin(3/5)(c) θ = arcsin(5/3) *(d) θ = arcsin(3/5)(e) None of these �

��

4

35

θ

φ


(a) θ = cos(5/4) (b) θ = arccos(5/4)(c) θ = cos(4/5) *(d) θ = arccos(4/5)(e) None of these �

��

4

35

θ

φ


(a) θ = tan(3/4) (b) θ = tan(4/3)*(c) θ = arctan(3/4) (d) θ = arctan(4/3)(e) None of these �

��

4

35

θ

φ


*(a) φ = sin−1(4/5) (b) φ = sin(4/5)(c) φ = sin−1(5/4) (d) φ = sin(5/4)(e) None of these �

��

4

35

θ

φ

151


(a) φ = cos(4/5) (b) φ = cos−1(4/5)(c) φ = cos(3/5) *(d) φ = cos−1(3/5)(e) None of these �

��

4

35

θ

φ


(a) φ = tan−1(3/4) *(b) φ = tan−1(4/3)(c) φ = tan(3/4) (d) φ = tan(4/3)(e) None of these �

��

4

35

θ

φ

4.2.10 Arc functions: calculator; radians

Problem 517. Use a calculator or equivalent to determine the value of θ if θ is an acuteangle and cos θ = 0.41. Round your answer to the nearest 0.01 radian.

(a) 0.39 rad (b) 36.83 rad*(c) 1.15 rad (d) No such angle(e) None of these

Solution: θ = cos−1 0.41 = 1.15 rad

Problem 518. Use a calculator or equivalent to determine the value of θ if θ is an acuteangle and sin θ = 0.83. Round your answer to the nearest 0.01 radian.

(a) 1.48 rad (b) 0.59 rad*(c) 0.98 rad (d) No such angle(e) None of these

Solution: θ = sin−1 0.83 = 0.98 rad

Problem 519. Use a calculator or equivalent to determine the value of θ if θ is an acuteangle and tan θ = 1.03. Round your answer to the nearest 0.01 radian.

(a) 0.86 rad *(b) 0.80 rad(c) 0.51 rad (d) No such angle(e) None of these

Solution: θ = tan−1 1.03 = 0.80 rad



Solution: θ = cos−1 0.77 = 0.69 rad

152


(a) 0.39 rad (b) 36.83 rad(c) 1.15 rad *(d) No such angle(e) None of these

Solution: No such angle; sin θ > 1 is impossible.


*(a) 0.22 rad (b) 0.29 rad(c) 18.45 rad (d) No such angle(e) None of these



(a) 1.17 rad (b) 1.67 rad(c) 0.86 rad *(d) No such angle(e) None of these

Solution: No such angle; cos θ > 1 is impossible.



Solution: θ = sin−1 0.55 = 0.58 rad


*(a) 1.04 rad (b) 0.14 rad(c) 89.48 rad (d) No such angle(e) None of these


153

4.2.11 Arc function: calculator; degrees

Problem 526. Use a calculator or equivalent to determine the value of θ if θ is an acuteangle and sin θ = 0.63. Round your answer to the nearest degree.

(a) 46◦ *(b) 39◦

(c) 32◦ (d) No such angle(e) None of these

Solution: θ = sin−1 0.63 = 39◦

Problem 527. Use a calculator or equivalent to determine the value of θ if θ is an acuteangle and cos θ = 1.19. Round your answer to the nearest degree.

(a) 21◦ (b) 50◦

(c) 53◦ *(d) No such angle(e) None of these

Solution: No such angle; cos θ > 1 is impossible.

Problem 528. Use a calculator or equivalent to determine the value of θ if θ is an acuteangle and tan θ = 1.30. Round your answer to the nearest degree.

*(a) 52◦ (b) 34◦


Solution: θ = tan−1 1.30 = 52◦


*(a) 36◦ (b) 39◦


Solution: θ = cos−1 0.81 = 36◦


(a) 39◦ *(b) 25◦



154


(a) 3◦ *(b) 19◦




*(a) 42◦ (b) 48◦




(a) 1◦ (b) 2◦

*(c) 16◦ (d) No such angle(e) None of these

Solution: θ = cos−1 0.96 = 16◦


(a) 57◦ *(b) 1◦



4.2.12 Arc functions: calculator; diagram; radians

Problem 535. In the figure at right, use a calculatoror equivalent to find θ. Round your answer to 0.01rad. (The figure is not necessarily drawn to scale.)


Solution: cos θ = 5/7; so θ = cos−1(5/7) = 0.78rad

��

5

7

θ

155



Solution: sin θ = 4/10; so θ = sin−1(4/10) = 0.41rad

��

410

θ



Solution: tan θ = 8/11; so θ = tan−1(8/11) =0.63 rad

��

11

8

θ



Solution: tan θ = 5/8; so θ = tan−1(5/8) = 0.56rad

��

8

5

θ



Solution: sin θ = 12/19; so θ = sin−1(12/19) =0.68 rad

��

1219

θ

156

Problem 540. In the figure at right, use a calculatoror equivalent to find φ. Round your answer to 0.01rad. (The figure is not necessarily drawn to scale.)


Solution: sinφ = 7/11; so φ = sin−1(7/11) = 0.69rad

��

7

11φ



Solution: tanφ = 10/6; so φ = tan−1(10/6) =1.03 rad

��

10

6

φ



Solution: cosφ = 9/14; so φ = cos−1(9/14) =0.87 rad

��

914

φ



Solution: tanφ = 9/5; so φ = tan−1(9/5) = 1.06rad

��

9

5

φ

157



Solution: cosφ = 3/20; so φ = cos−1(3/20) =1.42 rad

��

320

φ

4.2.13 Arc functions: calculator; diagram; degrees

Problem 545. In the figure at right, use a calculatoror equivalent to find φ. Round your answer to thenearest degree. (The figure is not necessarily drawnto scale.)

(a) 59◦ *(b) 53◦

(c) 40◦ (d) 47◦

(e) None of these

Solution: sinφ = 4/5; so φ = sin−1(4/5) = 53◦

��

4

5φ


(a) 24◦ (b) 36◦

*(c) 49◦ (d) 57◦

(e) None of these

Solution: tanφ = 8/7; so φ = tan−1(8/7) = 49◦

��

8

7

φ


(a) 52◦ (b) 65◦

*(c) 54◦ (d) 67◦

(e) None of these

Solution: cosφ = 7/12; so φ = cos−1(7/12) = 54◦

��

712

φ

158


(a) 50◦ *(b) 52◦

(c) 54◦ (d) 56◦

(e) None of these

Solution: tanφ = 9/7; so φ = tan−1(9/7) = 52◦

��

9

7

φ


(a) 60◦ (b) 63◦

*(c) 66◦ (d) 69◦

(e) None of these

Solution: sinφ = 11/12; so φ = sin−1(11/12) =66◦

��

11

12φ


(a) 76◦ *(b) 78◦

(c) 80◦ (d) 82◦

(e) None of these

Solution: cosφ = 1/5; so φ = cos−1(1/5) = 78◦

��

15

φ

Problem 551. In the figure at right, use a calculatoror equivalent to find θ. Round your answer to thenearest degree. (The figure is not necessarily drawnto scale.)

(a) 47◦ *(b) 50◦

(c) 53◦ (d) 56◦

(e) None of these

Solution: sin θ = 10/13; so θ = sin−1(10/13) =50◦

��

1013

θ

159


(a) 21◦ *(b) 23◦

(c) 25◦ (d) 27◦

(e) None of these

Solution: cos θ = 12/13; so θ = cos−1(12/13) =23◦

��

12

13

θ


*(a) 40◦ (b) 42◦

(c) 44◦ (d) 46◦

(e) None of these

Solution: tan θ = 17/20; so θ = tan−1(17/20) =40◦

��

20

17

θ


(a) 23◦ (b) 25◦

*(c) 27◦ (d) 29◦

(e) None of these

Solution: cos θ = 8/9; so θ = cos−1(8/9) = 27◦

��

8

9

θ


(a) 6◦ *(b) 8◦

(c) 10◦ (d) 12◦

(e) None of these

Solution: tan θ = 1/7; so θ = tan−1(1/7) = 8◦

��

7

1

θ

160


(a) 15◦ *(b) 17◦

(c) 19◦ (d) 21◦

(e) None of these

Solution: sin θ = 2/7; so θ = sin−1(2/7) = 17◦

��

27

θ

4.2.14 Word problems: basic trig functions

Problem 557. A tree casts a shadow 53 ftlong. The sun is θ = 49◦ above the horizon.How tall is the tree? Round your answer to thenearest foot.


Solution: Let h be the height. Then h/53 =tan 49◦; so h = 53 tan 49◦ = 61 ft.

Problem 558. You want to know the width ofa river. You begin by standing directly acrossfrom a tree on the opposite bank. You thenwalk 400 ft straight downstream. From thisnew point, the tree is at an angle of θ = 67◦ tothe upstream direction. How wide is the river?Round your answer to the nearest foot.


Solution: Let w be the width of the river.Then w/400 = tan 67◦; so w = 400 tan 67◦ =942 ft.

161

Problem 559. A prisoner is trying to escapeby digging a tunnel under a wall 147 ft away.Unfortunately, his compass is inaccurate, andhis tunnel is off course by θ = 19◦. How longdoes the tunnel have to be to reach the wall?Round your answer to the nearest foot.


Solution: Let t be the length of the tunnel.Then 147/t = cos 19◦; so t = 147/ cos 19◦ = 155ft.

Problem 560. A highway runs directly eastand west. An airplane flies across the highwayin a direction θ = 34◦ north of east. Whatis the total distance that the airplane will flybefore it is 21 miles north of the highway?Round your answer to the nearest tenth of amile.

(a) 25.3 miles (b) 29.4 miles(c) 33.5 miles *(d) 37.6 miles(e) None of these

Solution: Let t be the total distance flownby the airplane. Then 21/t = sin 34◦; so t =21/ sin 34◦ = 37.6 miles.

Problem 561. A car is driving up a mountainroad with a slope of θ = 4.5◦. It goes 9000 ftalong the road by the odometer. At the endof this time, how high is it above its startingpoint? Round your answer to the nearest 10 ft.


Solution: Let h be the height above thestarting point. Then h/9000 = sin 4.5◦; so h =9000 sin 4.5◦ = 710 ft.

162

Problem 562. From where you stand, amountain peak has an elevation of θ = 12.5◦.On a map, the horizontal distance between youand the peak is 4.11 miles. How high aboveyou is the peak? Round your answer to thenearest 0.01 miles.

(a) 0.83 miles (b) 0.87 miles*(c) 0.91 miles (d) 0.95 miles(e) None of these

Solution: Let h be the height of thepeak. Then h/4.11 = tan 12.5◦; so h =4.11 tan 12.5◦ = 0.91 miles.

Problem 563. You and a friend are trying tofind out how high airplanes fly over his house.He phones you when a plane is directly over-head. At that moment, the airplane is at anelevation of θ = 36.2◦, as seen from your house.If your house is 4100 ft from your friend’s house,how high is the airplane? Round your answerto the nearest 100 ft.

*(a) 3000 ft (b) 3300 ft(c) 3600 ft (d) 3900 ft(e) None of these

Solution: Let h be the height of theairplane. Then h/4100 = tan 36.2◦; so h =4100 tan 36.2◦ = 3000 ft.

Problem 564. A pole is supported by a di-agonal guy wire. The wire is anchored in theground 13.8 m from the base of the pole, andmeets the ground at an angle of θ = 59◦. Howlong is the wire? Round your answer to thenearest 0.1 m.

(a) 25.8 m (b) 26.3 m*(c) 26.8 m (d) 27.3 m(e) None of these

Solution: Let w be the length of the wire.Then 13.8/w = cos 59◦; so w = 13.8/ cos 59◦ =26.8 m.

163

Problem 565. You are flying a kite in awind strong enough to stretch the string intoa straight line. The string is 71 m long, andmakes an angle of θ = 51◦ with the horizontal.How high above the ground is the kite? Roundyour answer to the nearest meter.

*(a) 55 m (b) 57 m(c) 59 m (d) 61 m(e) None of these

Solution: Let h be the height of the kite.Then h/71 = sin 51◦; so h = 71 sin 51◦ = 55 m.

4.2.15 Word problems: arc functions

Problem 566. A tree is 73 ft tall. It casts ashadow 93 ft long. What is the sun’s angle ofelevation θ? Round your answer to the nearestdegree.

*(a) 38◦ (b) 40◦

(c) 42◦ (d) 44◦

(e) None of these

Solution: 73/93 = tan θ;so θ = tan−1(73/93) = 38◦.

Problem 567. A prisoner is trying to escapeby digging a tunnel under a wall 113 ft away.Unfortunately, his compass is inaccurate, andthe tunnel is off course by an angle of θ. Hedoes not reach the wall until the tunnel is 128ft long. What is the value of θ? Round youranswer to the nearest degree.

(a) 26◦ *(b) 28◦

(c) 30◦ (d) 32◦

(e) None of these

Solution: 113/128 = cos θ;so θ = cos−1(113/128) = 28◦.

164

Problem 568. A highway runs directly eastand west. An airplane flies across the highwayin a direction θ north of east. When the air-plane has flown a total distance of 19 miles, itis 5 miles north of the highway. What is θ?Round your answer to the nearest degree.

(a) 13◦ *(b) 15◦

(c) 17◦ (d) 19◦

(e) None of these

Solution: 5/19 = sin θ;so θ = sin−1(5/19) = 15◦.

Problem 569. A car is driving up a moun-tain road with a slope of θ. It goes 12,000 ftalong the road by the odometer. At the end ofthis time, it is 1800 ft above its starting point.What is θ? Round your answer to the nearest0.1◦.

(a) 8.0◦ (b) 8.3◦

*(c) 8.6◦ (d) 8.9◦

(e) None of these

Solution: 1800/12, 000 = sin θ;so θ = sin−1(1800/12, 000) = 8.6◦.

Problem 570. A mountain peak is 7200 fthigher than your house. On the map, there is ahorizontal distance of 53,000 ft from the houseto the peak. What is the peak’s elevation θ asseen from the house? Round your answer tothe nearest 0.1◦.

(a) 7.1◦ (b) 7.4◦

*(c) 7.7◦ (d) 8.0◦

(e) None of these

Solution: 7200/53, 000 = tan θ;so θ = tan−1(7200/53, 000) = 7.7◦.

165

Problem 571. Your house is 3800 ft fromyour friend’s house. An airplane flies over thefriend’s house at a height of 960 ft. Whenthe plane is directly above your friend’s house,what is its elevation θ as seen from your house?Round your answer to the nearest degree.

*(a) 14◦ (b) 15◦

(c) 16◦ (d) 17◦

(e) None of these

Solution: 960/3800 = tan θ;so θ = tan−1(960/3800) = 14◦.

Problem 572. A pole is supported by a diag-onal guy wire. The wire is 19.9 m long, and isanchored in the ground 15.2 m from the base ofthe pole. At what angle θ does the wire meetthe ground? Round your answer to the nearestdegree.

(a) 36◦ (b) 38◦

*(c) 40◦ (d) 42◦

(e) None of these

Solution: 15.2/19.9 = cos θ;so θ = cos−1(15.2/19.9) = 40◦.

Problem 573. You are flying a kite in awind strong enough to stretch the string intoa straight line. The string is 94 m long, andthe kite is 75 m above the ground. What angleθ does the string make with the ground? Roundyour answer to the nearest degree.

(a) 49◦ (b) 51◦

*(c) 53◦ (d) 55◦

(e) None of these

Solution: 75/94 = sin θ;so θ = sin−1(75/94) = 53◦.

166

5 Introduction to measurement: dimensions, units,

scientific notation, and significant figures

5.1 Dimensions

5.1.1 Dimension or unit?

Problem 574. What are the three fundamental dimensions?

Solution: length L, mass M , and time T .

Problem 575. What’s the difference between dimensions of the fundamental physicalquantities:length, mass, time and the base units: meter, kilogram, second?

Solution: Any physical quantity can be characterized by dimensions. You canthink of a dimension as a measurable extent of some kind, such as length. The arbitrarymagnitudes that we assign to assign to dimensions are called units. For example, in orderto give a precise quantitative measurement of the length (a dimension) of a particularobject, we need a standard reference length from which to compare our object’s length.

Problem 576. For the given physical quantities, identify the three fundamental physicalquantities:(a) mass (c) time (e) force(b) volume (d) density (f) length

Solution: (a), (c) and (f). The others are known as secondary dimensions or derivedquantities.

167

5.1.2 Dimensional consistency

In the following problems, determine which of the formulas could not be correct becauseit violates the consistency-of-dimensions principle discussed in class. Do not worry aboutthe origin or application of the formulas: just focus on the issue of consistency of units.That is, based solely on consistency of units in an equation, could the given formula becorrect?

Definition: The Dimensionator [ ] is the “take the fundamental dimensions of” op-erator.

Warning: Do not let the subscripts confuse you. The subscripts are only used to labelvariables, they are dimensionless numbers and letters and do not affect the outcome ofdimensionator one bit.

The dimensions of the fundamental quantities: {Length L, Mass M, Time T} are

• d = distance; [d] = L

• x = distance; [x] = L

• R = radius; [R] = L

• h = height; [h] = L

• m = mass; [m] = M

• t = time; [t] = T

The dimensions of the derived quantities (i.e., the quantities derived from the fundamen-tal quantities) are

• r = rate (speed); [r] = L/T

• v = velocity; [v] = L/T

• a = acceleration; [a] = L/T 2

• A = area; [A] = L2

• V = volume; [V ] = L3

Problem 577. d = rt

168

Solution: Look at the units. d has units of (length); r has units of (length/time);and t has units of (time). In units, the formula looks like:

(length) =

(length

time

)· (time)

Since the units of time on the right-hand side cancel, the two sides of the equation match,and the formula is dimensionally correct. In fact, this is the ubiquitous formula:distance = rate × time.

Problem 578. v = x+ t

Solution: Look at the units. v has units of length/time ([v] = LT

); x has units oflength ([x] = L); and t has units of time ([t] = T ). In units, the formula looks like:

L

T= L+ T .

Since the units on the two sides of the equation don’t match, the formula can’t be correct.In fact, every term is different!

Problem 579. V = 43πR2

Solution: Here V has units of (length)3; R has units of (length); and 43

and π arepure numbers, which don’t have units. In units, the formula looks like:

(length)3 = (length)2

Since the units on the two sides of the equation don’t match, the formula can’t be correct.

Problem 580. V = πR2h

Solution: Here V has units of (length)3; R has units of (length); h has units of(length); and π is a pure number, which doesn’t have units. In units, the formula lookslike:

(length)3 = (length)2(length)

Since the units of time on the right-hand side cancel, the two sides of the equation match,and the formula is dimensionally correct. In fact, this is the formula for the volume of acylinder.

Problem 581. V = Ad

Solution: Here V has units of (length)3; A has units of (length)2; and d has units of(length). In units, the formula looks like:

(length)3 = (length)2 · (length)

We get units of (length)3 on both sides of the equation. From the standpoint of consis-tency of units, this formula could be correct.

169

Problem 582. V = Ah

Solution: Here V has units of (length)3; A has units of (length)2; and h has units of(length). In units, the formula looks like:


We get units of (length)3 on both sides of the equation. From the standpoint of consis-tency of units, this formula could be correct.

Problem 583. V = Ax2

Solution: Here V has units of (length)3; A has units of (length)2; and x has units of(length). In units, the formula looks like:

(length)3 = (length)2 · (length)2 = (length)4

Since the units on the two sides of the equation don’t match, the formula can’t be correct.

Problem 584. v =x

t

Solution: Here v has units of lengthtime

; x has units of (length); and t has units of (time).In units, the formula looks like:

length

time=

length

time

From the standpoint of consistency of units, this formula could be correct.

Problem 585. v22 = v2

1 + x2

Solution: Here v2 and v1 have units of lengthtime

; x has units of (length). In units, theformula looks like:

(length)2

(time)2=

(length)2

(time)2+ (length)2

We can’t even say what the units are on the right-hand side of the equation, since we’retrying to add two quantities measured in different kinds of units. This formula doesn’tmake any sense.

Problem 586. v22 = v2

1 +x2

t2

170

Solution: Here v2 and v1 have units of lengthtime

; x has units of (length); and t has unitsof (time). In units, the formula looks like:

(length)2

(time)2=

(length)2

(time)2+

(length)2

(time)2

We can deal with the right-hand side, since we’re adding two quantities measured in the

same units. Both sides of the formula have units of (length)2

(time)2. From the standpoint of

consistency of units, this formula could be correct.

Problem 587. Determine whether equations (i) and (ii) below are dimensionally con-sistent.

(i) v =x− x0

t(ii) x = a(t+ v)2 ,

where x = distance, v = velocity, t = time, and a = acceleration.

*(a) (i) is dimensionally consistent; (ii) is dimensionally inconsistent

(b) (i) is dimensionally inconsistent; (ii) is dimensionally consistent

(c) (i) and (ii) are both dimensionally consistent

(d) (i) and (ii) are both dimensionally inconsistent

Solution: Can’t add velocity and time since they are dimensionally inconsistent.

Problem 588. Consider the formulas (i) and (ii). Are the formulas dimensionally con-sistent? Here x = distance, v = velocity, V = volume, and A = area.

(i) V = Ax (ii) v22 = v2

1 + x2





Solution: In (i), V has units of (length)3; A has units of (length)2; and x has unitsof (length). In units, the formula looks like:


We get units of (length)3 on both sides of the equation. Hence (i) is dimensionallyconsistent.

In (ii), v2 and v1 have units of lengthtime

; x has units of (length). In units, the formula lookslike:

(length)2

(time)2=

(length)2

(time)2+ (length)2

Since the two terms on the right-hand side have different units, (ii) is not dimensionallyconsistent.

171

Problem 589. Consider the formulas (i) and (ii). Are the formulas dimensionally con-sistent? Here V = volume, r = distance, x = distance, and t = time.

(i) V = πxr2 (ii) t =x− t0

2





Solution: In (i), V has units of (length)3; π is dimensionless; and x and r both havedimensions of (length). In units, the formula looks like

(length)3 = (length)(length)2 = (length)3

Hence (i) is dimensionally consistent.

In (ii), t and t0 both have units of (time); x has units of (length); and 2 is dimensionless.In units, the formula looks like

(time) = (length)− (time)

The two terms on the right side are not dimensionally consistent; so the formula as awhole is not dimensionally consistent.


(i) a =x− x0

t2(ii) x2 = at2 + (x+ vt)2 ,

where x = distance, v = velocity, t = time, and a = acceleration.*(a) (i) is dimensionally consistent; (ii) is dimensionally inconsistent




Solution: Equation (i) is dimensional consistent. Equation (ii) is dimensionallyinconsistent. To see this apply the dimensionator.

[x]2 = [a][t]2 + ([x] + [vt])2 ⇒ L2 =L

T 2T 2 +

(L+

L

TT

)2

= L+ L2

The first term on the right-hand side of the equation is different from the other two.

172


(i)x = at2 + (1 + (vt)/x)2 (ii) v2f = v2

i + 2ax2 ,

where x = distance, v = velocity, t = time, a = acceleration, r = distance, g = magnitudeof gravity and µs is the coefficient of static friction.

(a) (i) is dimensionally consistent; (ii) is dimensionally inconsistent



*(d) (i) and (ii) are both dimensionally inconsistent

Solution: (i) Dimensionally inconsistent. [x] = [a][t2] + (1 + [(vt)/x])2 becomes

L = LT 2T

2 + (1 + ([vt]=L)L

)2 = L + 1, where we’ve used the fact that [vt] = distance =L. Since L and 1 are dimensional inconsistent, the RHS of the equation is dimensionalinconsistant.

(ii) Dimensionally inconsistent. [v2f ] = [v2

i ] = L2/T 2, but [2ax2] = (L/T 2)(L2) = L3/T 2

Problem 592. (Using dimensional analysis to get statistical-partial credit)Consider the following test question and multiple-choice answers. Without knowing howto solve the problem, can you identify the answer that cannot be a solution based solelyon dimensional analysis?

You are towing a crate of physics books down the hallway, using a rope that makes anangle of θ to the horizontal. The crate has a mass of m; the coefficient of kinetic frictionbetween the crate and the floor is µk. As you go down the hallway, you are giving thecrate a forward acceleration of a. What is the tension T in the rope, expressed in termsof the mass m, the acceleration a, the angle θ (dimensionless), µk (dimensionless), andthe gravitational constant of acceration g? Note: Tension has units of force.

*(a) (mgµk + a) cos θ (b)m(gµk + a)

cos θ + µk sin θ

(c)m(gµk + a)

cos θ(d) (mgµk + a)(cos θ + µk sin θ)

Solution: The answer is (a). Gravity g has units of acceleration, so each term in thefactor (gµk + a) has units of acceleration and is therefore dimensionally consistent withacceleration. The factor (cos θ + µk sin θ) is composed of non-dimensional terms, so it’sdimensionless. Thus, the equation in (a) has units of acceleration. The equation in (b)is mass times acceleration (a.k.a force), which is consistent. Similarly, the equations in(c) and (d) have units of force.

173

5.1.3 Practical questions involving dimensions

Problem 593. In the construction business, concrete is often sold by the yard. What isthe precise meaning of a yard of concrete?

Solution: It wouldn’t make sense to sell something like concrete, which has volume,by units of length. A “yard” of concrete is actually a cubic yard.

Problem 594. In the carpet business, carpet is sold by the yard. What do they reallymean by a yard of carpet?

Solution: It wouldn’t make sense to sell something like carpet, which has area, byunits of length. A “yard” of carpet is actually a square yard.

Problem 595. Name some physical phenomena that could be used as crude timingdevices.

Solution: There are lots of examples. The movement of shadows; the oscillations ofa pendulum (a weight suspended from a string would do); the filling of a glass left undera dripping tap; the burning down of a candle... You’ve probably thought of differentexamples, and they’re probably just as good as these.

Problem 596. Suppose your lab partner tells you that the volume of a cylinder isV = πr3h. Could this formula be correct?

Solution: In a cylinder, r is the radius, which has units of length; h is the height,which also has units of length; and π is a pure number, which has no units. On the rightside of your lab partner’s formula, the units are

(length)3 · (length) = (length)4

However, volume has units of (length)3. The two sides of the formula don’t match.Clearly, your lab partner is trying to use you to bring down the curve so that he’ll geta better grade in the course. This, he hopes, will allow him to get into medical school,become a wealthy plastic surgeon, retire early, and pass the time relaxing in a gold-platedhot tub full of Scotch while you are still spending eight hours a day asking people if theywant fries with that. You should consider asking for a new lab partner.

174

5.2 Units

Problem 597. What are the units of the number π?

Solution: Pure numbers are dimensionless. One way you can see this for π inparticular is this: π is defined as the ratio of a circle’s circumference to its diameter,and both of these are lengths. Suppose the circumference and the diameter are bothmeasured in cm. Then

π =circumference

diameter=

cm

cm

The cm in the numerator and denominator cancel, leaving no units.

Problem 598. What are the units of the number e?

Solution: e is a pure number, and has no units.

Problem 599. What are the units of the number −5?

Solution: −5 is a pure number, and has no units.

Problem 600. Give an example from everyday life of something that’s about 1 m long.

Solution: Find a tape measure and go looking for examples. Try the height of a lowbookshelf, or the width of a wide door, or the length of your great-great grandfather’scavalry saber, or...

Problem 601. Give an example from everyday life of something that has a mass ofabout 1 kg.

Solution: Check stuff in the kitchen. Your instructor’s local grocery store yielded aquart jar of mayo that weighed 946 g, which is pretty close.

Problem 602. Give an example from everyday life of something that weighs about 1 N.

Solution: You won’t find weights expressed in newtons on packages. Look forsomething that weighs about 100 g, or about 31

2oz. The easiest way to remember what

1 newton feels like is to imagine holding a small apple, and then to remember the storyof Newton and the apple falling on his head.

Problem 603. Give an example from everyday life of something with a volume of about1 l.

Solution: This is easier, since soft drinks are often sold in 2 l bottles. Get one ofthose and pour out half. Some convenience stores sell 1 liter bottles.

175

Problem 604. Give an example from everyday life of a distance that’s about 1 km.

Solution: The mile markers on Interstate 19 are actually kilometer markers. That is,the distance between markers is kilometers. If your neighborhood highways are markedin miles, you’ll have to find something else. A kilometer is about 0.6 miles.

Problem 605. How much do you weigh in pounds? Round your answer to the nearestpound. Guess if you don’t have a scale handy. If that’s your weight in pounds, what isyour mass in kg? Round to the nearest kg.

Solution: This will depend on your actual weight. Suppose you weigh 150 lbs. Usethe conversion factor 0.454 kg/lb. Then your weight is

(150 lb)

(0.454

kg

lb

)= 68 kg

Notice how the units work. We’re multiplying (lb) by (kg/lb). The pounds in thenumerator and denominator cancel, leaving only kg.

Problem 606. What is your height in inches? Round your answer to the nearest inch.Guess if you don’t have a tape measure handy. If that’s your height in inches, what isyour height in cm? Round to the nearest cm.

Solution: This will depend on your actual height. Suppose your height is 65 inches(5 ft 5 in). Use the conversion factor 2.54 cm/in. Then your height is

(65 in)(

2.54cm

in

)= 165 cm

If we ignore the numbers and write this in terms of units alone, it’s

(in)cm

in= cm

The inches in the numerator and the denominator cancel, leaving only cm in the numer-ator.

Problem 607. How far is it in miles from where you live to your physics classroom?Round your answer to the nearest 0.1 mile. Guess if you haven’t measured it. If that’sthe distance in miles, what is the distance in km? Round to the nearest 0.1 km.

Solution: Your answer will depend on the distance from you to your classroom.Suppose it’s 2.3 miles. Use the conversion factor 1.61 km/mi. Then the distance inkilometers is

(2.3 mi)

(1.61

km

mi

)= 3.7 km

Again, look at the units:

mi · km

mi= km

The miles in the numerator and the denominator cancel, leaving only km.

176

5.2.1 Converting between different sets of units

Problem 608. A box is 13 inches tall. What is its height in centimeters? Round to thenearest cm.

Solution: Use the conversion factor 2.54 cm/in. The height of the box is

(13 in)(

2.54cm

in

)= 33 cm

Notice that inches in the numerator and denominator cancel, leaving only cm.

Problem 609. A bookshelf is 2.3 m tall. How tall is it in feet? Round to the nearest0.1 ft.

Solution: Use the conversion factor 3.28 ft/m. The height of the bookshelf is

(2.3 m)

(3.28

ft

m

)= 7.5 ft

Notice that meters in the numerator and denominator cancel, leaving only ft.

Problem 610. Two towns are 37 miles apart. How far apart are they in kilometers?Round your answer to the nearest km.

Solution: Use the conversion factor 1.61 km/mi. The distance is

(37 mi)

(1.61

km

mi

)= 60 km

Notice that miles in the numerator and denominator cancel, leaving only km.

Problem 611. Two other towns are 113 km apart. How far apart are they in miles?Round your answer to the nearest mile.

Solution: Again, use the conversion factor 1.61 km/mi. To make the units comeout, we have to divide by the conversion factor. Equivalently, we have to multiply by thereciprocal:

km÷ km

mi= km ·

(km

mi

)−1

= km · mi

km= mi

The distance is

(113 km) ·(

1

1.61

mi

km

)= 70 mi

Notice that we manipulated the given conversion factor in such a way that the unwantedunits (km) cancelled out from the problem.

177

Problem 612. A house sits on a lot of 7260 square feet. What is the size of the lot insquare meters? Round your answer to the nearest m2.

Solution: We’ll use the conversion factor 3.28 ft/m. Let’s look at the units to seewhat we need to do with the conversion factor. We have a measurement in ft2. We wantto convert it to m2. Our conversion factor is in ft/m. If we divide by the square of theconversion factor, we get what we want:

ft2 ÷(

ft

m

)2

= ft2 ·(

ft

m

)−2

= ft2 · m2

ft2 = m2

Hence the size of the lot is

7260 ft2 ·(

1

3.28

m

ft

)2

= 675 m2

Notice that we are using the one-factor here:

1 m = 3.28 ft ⇒ 1 =

(1

3.28

m

ft

)=

(3.28

1

ft

m

).

Problem 613. A pond has a surface area of 3900 m2. What is its surface area in squarefeet? Round to the nearest 100 ft2.

Solution: Use the conversion factor 3.28 ft/m. We need to multiply by the square ofthe conversion factor to make the units turn out right:

m2 · ft2

m2= ft2

The surface area of the pond is

3900 m2 ·(

3.28ft

m

)2

= 42, 000 ft2

Problem 614. You are given a ticket for driving at 38 miles per hour in a school zone.What is your speed in meters per second? Round your answer to the nearest m/s.

(a) 15 m/s *(b) 17 m/s(c) 19 m/s (d) 20 m/s(e) None of these

178

Solution: Use the conversion factors: 1610 m/mi and 3600 s/hr. To make the unitswork out, we need to multiply by the first and the reciprocal of the second:

mi

hr· m

mi·( s

hr

)−1

=mi

hr· m

mi· hr

s=

m

s

The speed for which you got the ticket is

38mi

hr·(

1610m

mi

)( 1

3600

hr

s

)= 17

m

s

Problem 615. You are given a ticket for driving at 37 miles per hour in a school zone.What was your speed in m/s? Round your answer to the nearest m/s.

Solution: Use the conversion factors: 1610 m/mi and 3600 s/hr. To make the unitswork out, we need to multiply by the first and the reciprocal of the second:

mi

hr· m

mi·( s

hr

)−1

=mi

hr· m

mi· hr

s=

m

s

The speed for which you got the ticket is

37mi

hr·(

1610m

mi

)( 1

3600

hr

s

)= 16.5

m

s

Problem 616. A water-balloon dropped from a tenth-floor window hits the ground at24.5 m/s. What is its speed in miles per hour? Round your answer to the nearest 0.1mile/hour.

Solution: We’ll need to use two conversion factors: 1.61 km/mi = 1610 m/mi, and3600 s/hr. We’ll proceed in steps. In the first step we’ll convert m/s to m/hr. Then inthe second step we’ll convert m/hr to miles/hr. If we look at the units, we’ll see thatwe need to multiply by the reciprocal of the first conversion factor and multiply by thesecond:

Step 1:m

s· s

hr=

m

hr

Step 2:(m

s· s

hr

)(m

mi

)−1

=m

hr· mi

m=

mi

hr

The speed of the balloon at impact is

24.5m

s·(

3600s

hr

)( 1

1610

mi

m

)= 54.8

mi

hr

Problem 617. A water-balloon dropped from a 38th-floor window hits the ground at86 m/s. What is its speed in ft/s? Round your answer to the nearest ft/s.

(a) 26 ft/s (b) 29 ft/s(c) 257 ft/s *(d) 282 ft/s(e) None of these

179

Solution: Convert from SI system to British system: (86 m/s)(3.28 ft/m) = 282 ft/s.

Problem 618. An evil scientist grows a giant carrot with a mass of 60 kg. What is thecarrot’s weight in pounds? Round your answer to the nearest pound.

(a) 17 lb (b) 27 lb*(c) 132 lb (d) 207 lb(e) None of these

Solution: (60 kg)(2.2 lb/kg) = 132 lb

Problem 619. A can of beer has a volume of 355 cm3. What is its volume in cubicinches? Round your answer to the nearest 0.1 in3.

Solution: Use the conversion factor 2.54 cm/in. To get the units to work out, weneed to multiple by the cube of the reciprocal of the conversion factor:

cm3 ·(

in

cm

)3

= cm3 · in3

cm3= in3

The volume of the beer-can is

355 cm3 ·(

1 in

2.54 cm

)3

= 21.7 in3

Problem 620. A pond has a volume of 325,000 ft3. What is its volume in m3? Roundyour answer to the nearest 100 cubic meters.

Solution: Use the conversion factor 3.28 ft/m. To get the units to work, we have tomultiple by the cube of the reciprocal of the conversion factor:

ft3 ·(

ft

m

)−3

= ft3 · m3

ft3 = m3

The volume of the pond is

325, 000 ft3 ·(

1 m

3.28 ft

)3

= 9200 m3

Problem 621. A river is flowing at 46,500 cubic feet per second. What is the river’sflow rate in cubic meters per second? Round your answer to the nearest 10 m3/s.

180

Solution: Use the conversion factor 3.28 ft/m. To make the units work, we need todivide by the cube of the conversion factor:

ft3

s÷(

ft

m

)3

=ft3

s· m3

ft3 =m3

s

From the previous problem it is clear that dividing by the conversion factor yields thesame result as multiplying by the reciprocal. The flow rate of the river is

46, 500ft3

s·(

1 m

3.28 ft

)3

= 1320m3

s

Mass and weight. In the English system, the pound is a unit of force. In the metricsystem, the newton is a unit of force; the kilogram is a unit of mass. However, we oftenuse grams and kilograms as units of weight instead of mass. For example, a one-poundbox of crackers will inform you that it weighs 454 g, not 4.45 N. You should be ableto convert between pounds and kilograms for everyday use; and between pounds andnewtons for physics applications.

Problem 622. A piano is pushed with a force of 450 N. What is the equivalent force inpounds? Round your answer to the nearest 10 lb.

Solution: Use the conversion factor 4.45 N/lb. To make the units work, we need todivide by the conversion factor:

N÷ N

lb= N · lb

N= lb

The force on the piano is

450 N ·(

1 lb

4.45 N

)= 100 lb

Problem 623. A box weighs 46 lb. What is the equivalent weight in newtons? Roundyour answer to the nearest N.

Solution: Use the conversion factor 4.45 N/lb. To make the units work, we multiplyby the conversion factor:

lb · N

lb= N

The weight of the box is

46 lb ·(

4.45N

lb

)= 205 N

Problem 624. A sack of potatoes has a mass of 12 kg. What is the weight of thepotatoes in pounds? Round your answer to the nearest lb.

181

Solution: Use the conversion factor 2.2 lb/kg. To make the units work, multiply bythe conversion factor:

kg · lb

kg= lb

The weight of the potatoes is

12 kg ·(

2.2lb

kg

)= 26 lb

Problem 625. A sack of potatoes has a mass of 20 kg. What is the weight of thepotatoes in pounds? Round your answer to the nearest lb.

Solution: Fw = (20 kg)(2.2 lb/kg) = 44 lb.

Problem 626. A German shepherd weighs 72 lb. What is the dog’s mass in kg? Roundyour answer to the nearest kg.

Solution: Use the conversion factor 2.2 lb/kg. To make the units work, divide by theconversion factor:

lb÷ lb

kg= lb · kg

lb= kg

The mass of the dog is

72 lb ·(

1 N

2.2 lb

)= 33 kg

Problem 627. A cable weighs 5.2 lb/ft. What is the cable’s mass in kg/m? Round youranswer to the nearest 0.1 kg/m.

Solution: We need to convert lb to kg, and ft to m. Use the conversion factors2.2 lb/kg and 3.28 ft/m. To make the units work, we have to multiply by the secondconversion factor and to multiple by the reciprocal of the first:

lb

ft· ft

m·(

lb

kg

)−1

=lb

ft· ft

m· kg

lb=

kg

m

The linear density of the cable is

5.2lb

ft·(

3.28ft

m

)(1 kg

2.2 lb

)= 7.8

kg

m

Problem 628. The pressure at the bottom of a swimming pool is 18 lb/in2. What isthe pressure in N/cm2? Round your answer to the nearest N/cm2.

182

Solution: We need to convert pounds to newtons, and inches to centimeters. Use theconversion factors: 4.45 N/lb and 2.54 cm/in. To make the units work, multiply by thefirst conversion factor, and divide by the square of the second:

lb

in2 ·N

lb÷(cm

in

)2

=lb

in2 ·N

lb· in2

cm2=

N

cm2

The pressure is

18lb

in2 ·(

4.45N

lb

)(1 in

2.54 cm

)2

= 12N

cm2

Problem 629. An engine burns fuel at a rate of 11.2 g/s. What is the consumption ratein kg/hr? Round your answer to the nearest 0.1 kg/hr.

Solution: Use the conversion factors: 1000 g/kg and 3600 s/hr. To make the unitswork, multiply by the second factor and the reciprocal of the first:

g

s· s

hr·(

g

kg

)−1

=g

s· s

hr· kg

g=

kg

hr

The engine consumes fuel at a rate of

11.2g

s·(

3600s

hr

)( 1 kg

1000 g

)= 40.3

kg

hr

Problem 630. You have just discovered the new element explodium. It has a density of9.70 g/cm3. The USA Today reporter says that their readers don’t understand all thatcenti-whatsis stuff, and wants the density in pounds per cubic foot. Round your answerto the nearest lb/ft3.

Solution: We need to convert grams to pounds, and cubic centimeters to cubic feet.For the former, we’ll use the conversion factor 454 g/lb. For the latter, we’ll combinetwo conversion factors:

2.54cm

in· 12

in

ft= 30.48

cm

ft

To make the units work, we need to multiply by the cube of the second conversion factor,and divide by the first one:

g

cm3·(cm

ft

)3

÷ g

lb=

g

cm3· cm3

ft3 ·lb

g=

lb

ft3

The density of explodium is

9.70g

cm3·(

30.48cm

ft

)3

÷ 454g

lb= 605

lb

ft3 .

183

For pedagogical purposes, we write the conversion out in one long string out products:

9.70g

cm3= 9.70

g

cm3· 1 · 1 · 1

= 9.70g

cm3·(

2.54 cm

1 in

)3(12 in

1 ft

)3(1 lb

454 g

)=

9.70 · (2.54)3 · 123

454

lb

ft3

= 605lb

ft3

Notice that the second equality follows from the one-factor.

Problem 631. Your male friend is trying to impress a pretty German girl at your localpub. He says to her ”I can bench press 330lbs”. She then asks ”How much is that inkilograms?”. He is dumbfounded. Can you save him by converting 330 lbs to kgs, sothat she can be properly impressed?

Solution: We need to convert lb to kg. Use the conversion factor 2.2 lb/kg. Theweight in kilograms is

330 lb ·(

1 kg

2.2 lb

)= 150 kg .

Problem 632. You have a measuring stick that has centimeters and inches on it. Younotice that 1.00 in = 2.54 cm. Using only this information together with the relationship1 mile = 5,280 ft., find a relationship between miles and kilometers. That is, 1 mile =how many kilometers?

Solution: We will use a total of 5 conversion factors. That is, we will apply theone-factor 5 times (via multiplication).

1 mile = 1 mile · 1 · 1 · 1 · 1 · 1

= 1 mile ·(

5280ft

1mile

)(12in

1ft

)(2.54cm

1in

)(1m

100cm

)(1km

1000m

)=

5280 · 12 · 2.54

105km

= 1.61 km

where the second equality follows from the one-factor. Thus, 1 mile = 1.61 km. Crudely,1 mile ≈ 3

2km.

Problem 633. Use the relationship 1.00 in = 2.54 cm to find a relationship between ayard and a meter. Your answer should show that a yard is close in length to a meter.

184

Solution: Using the relationship 1.00 in = 2.54 cm and a number of conversion factorswe have

1 yd = 1 yd · 1 · 1 · 1 · 1 · 1

= 1 yd ·(

3ft

1yd

)(12in

1ft

)(2.54cm

1in

)(1m

100cm

)=

3 · 12 · 2.54

100m

= 0.914 m

Thus, 1m ≈ 1.09yd. So, 100 m is longer than a 1 football field.

5.3 Scientific Notation

For the following problems express the following numbers in scientific notation withoutusing a calculator. Assume 3 significant figures in each number with more than 3 trailingzeros.

Problem 634. 1 = 1× 100

Problem 635. 12 = 1.2× 101

Problem 636. 123 = 1.23× 102

Problem 637. 1234 = 1.234× 103

Problem 638. 12345 = 1.2345× 104

Problem 639. 0.1 = 1.× 10−1

Problem 640. 0.12 = 1.2× 10−1

Problem 641. 0.012 = 1.2× 10−2

Problem 642. 0.0012 = 1.2× 10−3

Problem 643. 0.00012 = 1.2× 10−4

Problem 644. 395 = 3.95× 102

Problem 645. 97, 000 = 9.70× 104

Problem 646. .0000345 = 3.45× 10−5

Problem 647. 10, 000, 000 = 1.00× 107

Problem 648. .00000000500 = 5.00× 10−9

185

For the following problems convert the expressions that are given in scientific notationinto standard form without using a calculator.

Problem 649. 1× 100 = 1

Problem 650. 1.2× 101 = 12

Problem 651. 1.23× 102 = 123

Problem 652. 1.234× 103 = 1234

Problem 653. 1.2345× 104 = 12345

Problem 654. 1× 10−1 = 0.1

Problem 655. 1.2× 10−1 = 0.12

Problem 656. 1.2× 10−2 = 0.012

Problem 657. 1.2× 10−3 = 0.0012

Problem 658. 1.2× 10−4 = 0.00012

Problem 659. 3.95× 102 = 395

Problem 660. 9.70× 104 = 97, 000

Problem 661. 3.45× 10−5 = .0000345

Problem 662. 1.00× 107 = 10, 000, 000

Problem 663. 5.00× 10−9 = .00000000500

Problem 664. Convert the number 6.2×10−3 from scientific notation to standard form.

(a) 0.00062 (b) −6200*(c) 0.0062 (d) −62, 000(e) None of these

Solution: Move the decimal 3 places to the left.

Problem 665. Convert the number 9.70× 10−4 to standard form.


Solution: Since 9.70 × 10−4 has three significant figures, the standard-form numbermust also have three. By rule 4, the final zero in (d) is significant.

186

Problem 666. Convert the number 97,000 to scientific notation. Your answer shouldhave three significant figures.

(a) 9.70× 103 *(b) 9.70× 104

(c) 9.7× 103 (d) 9.7× 104

(e) None of these

Solution: Answers (b) and (d) are both equal to 97,000, but (d) only has twosignificant figures.

For the following problems perform the following arithmetic without using a calculator.

Problem 667.(1.0× 103) · (9.0× 108)

3.0× 107= 3.0× 104

Problem 668.(4.67× 107) · (1.01× 102)

6.24× 103= 7.56× 105

Problem 669.(7.3× 102) · (2.23× 105)

2.45× 104= 6.6× 103

Problem 670. (3.4× 107)− (3.0× 106) = (3.4× 107)− (0.30× 107) = 3.1× 107.

Problem 671. (3.4× 103) + (3.0× 104) = (.34× 104) + (3.0× 104) = 3.3× 104.

187

5.4 Significant figures

Problem 672. A useful approximation for the number of seconds in a year is π × 107

s. Determine the number of significant figures in this approximation given that there are365.25 days in a year.

Solution: Let’s start by using conversion factors to calculate how many seconds thereare in a year.

1 year = 1 year · 1 · 1 · 1 · 1

= 1 year ·(

365.25 days

1 year

)(24 hours

1 day

)(60 minutes

1 hour

)(60 seconds

1 minute

)= 365.25 · 24 · 3600 s

= 31, 557, 600 s

Let’s compare this to π × 107 = 31415926.59 s. The approximation is only accurate to 2significant figures.

Problem 673. A common approximation for π = 3.141592659 . . . is 227

. That is, π ≈ 227

.To how many significant figures is this approximation accurate?

Solution: To 8 decimal places 227≈ 3.142857143. The two expressions agree to the

hundreds place, and if we round off they agree to three decimal places (the thousandsplace), hence they agree to 4 significant figures.

Problem 674. A common approximation for π = 3.141592659 . . . is 355113

. That is,π ≈ 355

113. To how many significant figures is this approximation accurate?

Solution: To 8 decimal places 355113≈ 3.14159292. The two expressions agree to the

6th decimal place. Hence they agree to 7 significant figures.

For the following problems preform the indicated operations. Assume that all of thenumbers come from real data with the correct number of significant figures. Be sure toexpress your answer in the correct number of significant figures. Please refer to rules 1-11in the lecture notes.

Problem 675. 566.3 + 32.50 + 2.197 = ?

Solution: 566.3 + 32.50 + 2.197 = 600.997round off to one decimal place−−−−−−−−−−−−−−−−−→ Answer =

601.0 . The last zero is significant. Notice that we used rule 10 for addition and/orsubtraction.

Problem 676. 261.72− 90.715 = ?

Solution: 261.72− 90.715 = 171.005round off: rule 9−−−−−−−−−→ Answer = 171.00

188

Problem 677. 523.498− 32.907527 + 0.52 = ?

Solution: 523.498 − 32.907527 + 0.52 = 491.110473round off: 2 dec.−−−−−−−−−→ Answer =

491.11

Problem 678. 225 + 1.087− 45.3975 = ?

Solution: There are two possible answers (even though I said all numbers come frommeasurements) If the number 225 is exact, then

225 + 1.087− 45.3975 = 180.6895round off: 3 dec.−−−−−−−−−→ Answer = 180.690

since the number 225 is exact so it does not affect the accuracy of the calculation. If 225is a measured number, then it has 3 significant figures and zero decimal digits. Thus thesum must be rounded to 3 significant figures: 225 + 1.087− 45.3975 = 181

Problem 679. 596.9/3.986 = ?

Solution: 596.9/3.986 = 149.7491...round off: 4 sig figs.−−−−−−−−−−−→ Answer = 149.7

Problem 680. 505× 0.0065

Solution: 505× 0.0065 = 3.2825round off: 2 sig figs.−−−−−−−−−−−→ Answer = 3.3

Since the second factor in the product only has 2 significant figures, namely 65, theanswer can only have two significant figures.

Problem 681. 2.00× 107 + 3.5× 332 = ?

Hint: First compute the exact numerical result, then express it the correct number ofsignificant figures.

Solution: 2.00× 107 + 3.5× 332 = 1376. The weakest link is 3.5, it is only know to2 significant figures. Thus the answer = 1400.

Problem 682. (3.7)3 = ?

Solution: (3.7)3 = 50.653 ⇒ Answer = 51. Since x3 = x · x · x, the expression(3.7)3 is really just a multiplication. From rule 11, we must treat this as a multiplication.

Problem 683. What is the average of 2.5212, 2.5214, 2.5213, 2.5212, and 2.5220?

Solution: Average =2.5212 + 2.5214 + 2.5213 + 2.5212 + 2.5220

5= 2.52142 Can

only keep 5 significant figures, so Answer = 2.5214.

189

Problem 684. Calculate 2.00× 1.07 + 3.5× 332. Your answer should have the correctnumber of significant figures.

*(a) 1200 (b) 1160(c) 1164 (d) 1164.14(e) None of these

Solution: Begin by calculating the products. 2.00 × 1.07 = 2.14; since both factorshave 3 significant figures, so does the product. 3.5× 332 = 1162; since 3.5 has only twosignificant figures, so does the product. That means that the 1162 is only accurate to thehundreds place. We add 2.14 (accurate to the one-hundredth place) to 1162 (accurateonly to the hundreds place) and get 2.14+1162 = 1164.14. Since the sum is only accurateto the decimal place of the least accurate term, it’s only accurate to the hundreds place;so we round it to 1200.

Problem 685. Evaluate the following expression. Round your answer to the correctnumber of significant figures: 1.24 + 24.038 + 6.925

(a) 32 (b) 32.2*(c) 32.20 (d) 32.203(e) None of these

Solution: Keep 2 decimal places (the weakest link).

Problem 686.(7.3× 102) · (2.23× 105)

2.45× 104

Solution: 6.6 × 103. The first factor in the numerator has 2 significant digits; theother factor in the numerator, and the denominator, have 3. By rule 11, our answershould be rounded to 2 significant digits.

Problem 687. How many significant digits are there in: m = 0.0507 kg?

Solution: There are three significant digits. Starting from the left, the first two zerosare place holders. The first significant digit is 5, after that all of the digits are significant.

Problem 688. How many significant digits are there in: m = 28.350 g?

Solution: The last zero is significant, since it is to the right of the decimal point, andalso to the right of a nonzero digit. Thus there are five significant digits.

Problem 689. How many significant digits are there in: v = 1.022 m/s?

190

Solution: The zero is significant, since it comes between two nonzero digits. Thusthere are four significant digits.

Problem 690. How many significant digits are there in: r = 0.028 cm?

Solution: Neither of the zeros is significant, since there are no nonzero digits to theleft of them. The zeros are there only to give place value to the 2 and the 8. Thus thereare two significant digits.

Problem 691. How many significant digits are there in: h = 0.08 cm?

Solution: one


Solution: 4


Solution: 5

Problem 694. How many significant digits are there in: V = 44, 000 ft3?

Problem 695. How many significant digits are there in: r = 0.0307 m?


Solution: The zeros to the left of the nonzero digits are placeholders, and are notsignificant digits.

Problem 696. How many significant digits are there in: v = 1.022 m/s?


Solution: The zero is significant because it’s between two nonzero digits.



Solution: The zeros to the left of the nonzero digits are not significant since they areplaceholders. There are 4 significant figures.

191



Solution: The zeros to the left of the first nonzero digit are not significant since theyare placeholders. All of the zeros after the nonzero digit are significant figures. Thereare 4 significant figures.

Problem 699. Use a calculator to determine the decimal value of 1/7. Round youranswer to 3 significant digits.


Solution: The calculator should give 1/7 = 0.142857 . . . The initial zero is notsignificant, so you need to round to three decimal places. Since the 2 is followed by an8, round it up to 3: 1/7 = 0.143.

Solution: The zeros are not significant, since they do not follow a decimal point orprecede a nonzero digit. They are there only to give place value to the two 4’s. Thusthere are two significant digits.


Solution: A calculator gives 3.142857... To get three significant digits, we round thisto 3.14. Recall that 22/7 is a rational number approximation to π.

Problem 701. Use a calculator to determine the decimal value of π. Round your answerto 3 significant digits. How does this answer compare to the previous answer for 22/7?

Solution: A calculator gives 3.1415926... To get three significant digits, we roundthis to 0.143. Notice that the zero before the decimal point is not significant.


Solution: A calculator gives 0.0042194... To get three significant digits, we roundthis to 0.00422. Notice that the three leading zeros are not significant.

Problem 703. You have calculated the volume of a pond as 83,512.338 ft3. Round thisanswer to 2 significant digits.

192

Solution: The first two digits are nonzero, so rounding to the thousands place is whatwe want: 84,000 ft3. Notice that we don’t have a decimal point with zeros after it, sincethis would indicate that all of the zeros were significant.

Problem 704. You have taken several measurements of the size of a very small particle.The mean of your measurements is 0.00013722 cm. Round this answer to 2 significantdigits.

Solution: The three zeros after the decimal point are not significant, since there’s nononzero digit before them. We round to two digits after the string of zeros: 0.0014 cm.

Problem 705. (Significant Figures and Conversion) If you were to convert the measuredlength 7.923 ft into yards by multiplying by the conversion factor (1 yd/3 ft), how manysignificant figures should the answer contain. Justify your answer.

Solution: The answer should contain 4 significant figures since the original measurednumber contains 4 significant figures and we are multiplying it by the conversion, whichis an exact number.

Problem 706. Evaluate the following expression. Round your answer to the correct

number of significant figures:(7.3× 102) · (2.23× 105)

2.45× 104

(a) 6.64× 103 *(b) 6.6× 103

(c) 6.6× 102 (d) 6.64× 10(e) None of these

Solution: 6.64449 × 103 → 6.6 × 103. The first factor in the numerator has 2significant digits; the other factor in the numerator, and the denominator, have 3. Byrule 11, our answer should be rounded to 2 significant digits.

Problem 707. Evaluate the following expression. Round your answer to the correctnumber of significant figures.

(3.021× 10−2)(8.446× 1011)

1.9× 103

Solution: 1.3 × 107. Doing this on a calculator yields 1.34291 × 107. Since thedenominator only has two significant figures, we must round the result likewise.

193

5.5 Determining derived units from equations

In the following problems use the given physical governing equation to determine thedimensions of the given derived quantity in terms of the fundamental dimensions oflength, time, and mass, or in terms of other derived quantities as specified in the problem.Use the SI units when expressing the dimensions.

Problem 708. Determine the expression for the unit of work (defined as a Joule J) interms of the fundamental units from the definition of work (in one spacial dimension bya constant force) as force times distance W = Fd, where W is the symbol for work (amethod of transferring energy), F is the force exerted on the object and d is the distancethat the force acts on the object.

Solution: Using our take-the-dimensions-of-the-quantity operator [ ] on the mathe-matical expression for work gives

[W ] = [Fd] = [F ][d] = N ·m ⇒ 1 Joule = 1N ·m = 1J

A Joule is pronounced a Newton-meter.

Problem 709. Determine the expression for the unit of kinetic energy (K.E.) in termsof the fundamental units from the definition of K.E. as K = 1

2mv2 where K is the symbol

for kinetic energy, m is the mass of the object, v is the speed of the object.

Solution: Using our take-the-dimensions-of-the-quantity operator [ ] on the mathe-matical expression for K.E. gives

[K] =

[1

2mv2

]= [m][v]2 = kg

m2

s2⇒ [K] = 1 kg

m2

s2= 1

(kg

m

s2

)·m = 1N ·m = 1J

Problem 710. Determine the expression for the unit of gravitational potential energy(P.E.) in terms of the fundamental units from the definition of P.E. as U = mgh, whereU is the symbol for potential energy, m is the mass of the object, g is the magnitude ofacceleration due to gravity, and h is the height of an object above some reference level.

Solution: Using our take-the-dimensions-of-the-quantity operator [ ] on the mathe-matical expression for gravitational P.E. gives

[U ] = [mgh] = [m][g][h] = kg(m

s2

)m ⇒ [U ] = 1N ·m = 1J

Problem 711. Determine the expression for the unit of elastic potential energy (P.E.)stored in a spring in terms of the fundamental units from the definition of P.E. as U =12kx2, where U is the symbol for potential energy, k is the spring constant, and and x is

the displacement from the equilibrium position.

194

Problem 712. Compare your results from the four previous expressions for work, kinetic,and potential energy. What do you conclude about the relation between the various units?

Solution: All four expressions have the same units, namely Joules. The Joule is theSI unit for energy. This result says that work, kinetic energy, and potential all have unitsof energy. As we’ll discover later in the course, by the principle of conservation of energy:energy can neither be created nor destroyed, but energy can transform from one form toanother via the mechanism of work.

Problem 713. Determine the fundamental dimensions for density = mass/volume, de-noted by ρ.

Solution: Start by writing ρ = density = massvolume

= mV

, where m is mass and V is

volume. Using the dimensionator: [ρ] = [m][V ]

= ML3 .

Problem 714. Express pressure, denoted by P , in terms of the fundamental dimensions.For now, just take pressure to be force/area. Note: In a later chapter we will use themathematical expression for Newton’s second law to show that the dimensions of force

are [F ] = ML

T 2. For now, just take this as a fact.

Solution: Start by writing P = pressure = forcearea

= FA

, where F is force and A is area.

Using the dimensionator: [P ] =[F ]

[A]=

[F ]

L2= M

L

T 2· 1

L2

Problem 715. Determine the expression for the unit of pressure (the Pascal Pa) interms of the SI units for pressure and area from the equation defining pressure P = F

A,

where F is force and A is area.

Solution: Using our take-the-dimensions-of-the-quantity operator [ ] on the mathe-matical expression for pressure gives

[P ] =

[F

A

]=

[F ]

[A]=

N

m2⇒ 1 Pascal =

N

m2

Problem 716. Determine the units of the universal gas constant R from the equationfor the ideal gas law: PV = nRT , where P is pressure, V is volume, n is moles (denotedmol), and T is temperature (with SI units of absolute temperature the kelvin K).

Solution: Using our take-the-dimensions-of-the-quantity operator [ ] on the equationfor the ideal gas law gives

[PV ] = [nRT ] = [n][R][T ]÷[n][T ]−−−−−−−→ [R] =

[P ][V ]

[n][T ]=

J

(mol) (K)

195

Problem 717. In the study of thermodynamics one of the common expressions that youencounter is pressure P times volume V . It turns out that this term has the dimensionsof work = force × distance. To see this we write volume = area × length, then using thedefinitions of pressure and work we have

pressure × volume = (force/area) × (area × length) = force × length = work.

Verify our word equation using the fundamental dimensions for the product of pressureand volume: pressure × volume = PV and work as work = Fd, where d is the distancethat the force acts over.

Solution: [PV ] = [P ][V ] = [F ]L2 · L3 = [F ] · L. Thus, PV has units of force times

distance. Later in the course we will learn that this is the units of work.

Using our take-the-dimensions-of-the-quantity operator [ ] on the mathematical expres-sion for PV gives

[PV ] = [P ][V ] =[F ]

[A][V ] =

N

m2m3 ⇒ [PV ] = N ·m = J

196

6 Introduction to Vectors

Unless stated otherwise, the direction of a vector in xy-space is the counterclockwiseangle θ that it makes with the x-axis; 0 ≤ θ < 360◦ or 0 ≤ θ < 2π rad., and angle mea-surements in your answers should be in the same units as in the statement of the problem.

6.1 Identifying Vectors

6.1.1 Given vector, identify on figure

Problem 718. Which of the vectors inthe graph corresponds to −10ı− 5?

(a) ~A (b) ~B

(c) ~C *(d) ~D(e) None of these

Solution: Only ~D has a negative x-component and a negative y-component.

-

6

x

y

��

~A

PPPP

PPi~C

��~D

AAAAAAAAK

~B

Problem 719. Which of the vectors inthe graph corresponds to −6ı + 2?

(a) ~A (b) ~B

*(c) ~C (d) ~D(e) None of these

Solution: Only ~B and ~C have anegative x-component and a positive y-component. In −6ı + 2, the magnitudeof the x-component (6) is greater than themagnitude of the y-component (2). This

is true of ~C but not of ~B.

-

6

x

y

��

~A

PPPP

PPi~C

��~D

AAAAAAAAK

~B

197

Problem 720. Which of the vectors inthe graph corresponds to 4ı + 4?

*(a) ~A (b) ~B

(c) ~C (d) ~D(e) None of these

Solution: Only ~A has a positive x-component and a positive y-component.

-

6

x

y

��

~A

PPPP

PPi~C

��~D

AAAAAAAAK

~B

Problem 721. Which of the vectors inthe graph corresponds to −ı + 2?

(a) ~A *(b) ~B


Solution: Only ~B and ~C have anegative x-component and a positive y-component. In −ı + 2, the magnitude ofthe x-component (1) is smaller than themagnitude of the y-component (2). This

is true of ~B but not of ~C.

-

6

x

y

��

~A

PPPP

PPi~C

��~D

AAAAAAAAK

~B

Problem 722. Which of the vectors inthe graph corresponds to −3ı + ?

(a) ~A (b) ~B


Solution: Only ~C has a negative x-component and a positive y-component.

-

6

x

y

��

~A

PPPP

PPi~C

SSSSSSSSw ~D

��~B

198

Problem 723. Which of the vectors inthe graph corresponds to 2ı + ?

*(a) ~A (b) ~B


Solution: Only ~A and ~B have apositive x-component and a positive y-component. In 2ı + , the magnitude ofthe x-component (2) is greater than themagnitude of the y-component (1). This

is true of ~A but not of ~B.

-

6

x

y

��

�* ~APP

PPPPi~C

SSSSSSSSw ~D

��~B

Problem 724. Which of the vectors inthe graph corresponds to ı + 2?

(a) ~A *(b) ~B


Solution: Only ~A and ~B have apositive x-component and a positive y-component. In ı + 2, the magnitude ofthe x-component (1) is smaller than themagnitude of the y-component (2). This

is true of ~B but not of ~A.

-

6

x

y

��

�* ~A@

@@@

@@I

~C

SSSSSSSSw ~D

��~B

Problem 725. Which of the vectors inthe graph corresponds to 3ı− 4?

(a) ~A (b) ~B


Solution: Only ~D has a positive x-component and a negative y-component.

-

6

x

y

��

~A

PPPP

PPi~C

SSSSSSSSw ~D

AAAAAAAAK

~B

199

Problem 726. Which of the vectors inthe graph corresponds to 〈−3, 1〉?(a) ~A (b) ~B


Solution: Only ~B and ~C have anegative x-component and a positive y-component. In 〈−3, 1〉, the magnitude ofthe x-component (3) is greater than themagnitude of the y-component (1). This

is true of ~C but not of ~B.

-

6

x

y

��

~A

AAAAAAAAK

~B

PPPP

PPi~C

��~D

Problem 727. Which of the vectors inthe graph corresponds to 〈2, 2〉?*(a) ~A (b) ~B



-

6

x

y

��

~A

AAAAAAAAK

~B

PPPP

PPi~C

��

��~D

Problem 728. Which of the vectors inthe graph corresponds to 〈−1, 2〉?(a) ~A *(b) ~B


Solution: Only ~B and ~C have anegative x-component and a positive y-component. In 〈−1, 2〉, the magnitude ofthe x-component (1) is smaller than themagnitude of the y-component (2). This

is true of ~B but not of ~C.

-

6

x

y

HHHHj ~A

AAAAAAAAK

~B

PPPP

PPi~C

��~D

200

Problem 729. Which of the vectors inthe graph corresponds to 〈−2,−1〉?(a) ~A (b) ~B


Solution: Only ~D has a negative x-component and a negative y-component.

-

6

x

y

AAAAAAAAU ~A

AAAAAAAAK

~B

PPPP

PPi~C

��~D

Problem 730. Which of the vectors inthe graph corresponds to 〈1, 3〉?*(a) ~A (b) ~B



-

6

x

y

��~A

AAAAAAAAK

~B

PPPP

PPi~C

HHHHj ~D

Problem 731. Which of the vectors inthe graph corresponds to 〈−4,−2〉?(a) ~A (b) ~B


Solution: Only ~C has a negative x-component and a negative y-component.

-

6

x

y

AAAAAAAAU ~D

AAAAAAAAK

~B

��

��~C

HHHHj ~A

201

Problem 732. Which of the vectors inthe graph corresponds to 〈10,−5〉?(a) ~A *(b) ~B


Solution: Only ~A and ~B have apositive x-component and a negative y-component. In 〈10,−5〉, the magnitudeof the x-component (10) is greater thanthe magnitude of the y-component (5).

This is true of ~B but not of ~A.

-

6

x

y

AAAAAAAAU ~A

HHHHHHj ~B

PPPP

PPi~C

��~D

Problem 733. Which of the vectors inthe graph corresponds to 〈3,−3〉?*(a) ~A (b) ~B


Solution: Only ~A has a positive x-component and a negative y-component.

-

6

x

y

@@@@R ~A

��

~B

@@

@@I

~C

��

��~D

6.1.2 Given vector on figure, identify formula

Problem 734. Which of the vectorsbelow corresponds to ~A on the graph atright?

(a) ı + (b) −ı + *(c) ı− (d) −ı− (e) None of these

Solution: ~A has a positive x-component and a negative y-component;so (c).

-

6

x

y

@@@@R ~A

��

~B

@@

@@I

~C

��

��~D

202

Problem 735. Which of the vectorsbelow corresponds to ~B on the graph atright?

*(a) ı + (b) −ı + (c) ı− (d) −ı− (e) None of these

Solution: ~B has a positive x-component and a positive y-component;so (a).

-

6

x

y

@@@@R ~A

��

~B

@@

@@I

~C

��

��~D

Problem 736. Which of the vectorsbelow corresponds to ~C on the graph atright?

(a) ı + *(b) −ı + (c) ı− (d) −ı− (e) None of these

Solution: ~C has a negative x-component and a positive y-component;so (b).

-

6

x

y

@@@@R ~A

��

~B

@@

@@I

~C

��

��~D

Problem 737. Which of the vectorsbelow corresponds to ~D on the graph atright?

(a) ı + (b) −ı + (c) ı− *(d) −ı− (e) None of these

Solution: ~D has a negative x-component and a negative y-component;so (d).

-

6

x

y

@@@@R ~A

��

~B

@@

@@I

~C

��

��~D

203


(a) 3ı + (b) −3ı + *(c) 2ı− (d) −ı− 3(e) None of these

Solution: ~A has a positive x-component and a negative y-component;so (c).

-

6

x

y

HHHHH

Hj ~A


*(a) 2ı + 6 (b) 3ı + (c) 2ı− (d) −ı− 3(e) None of these

Solution: ~A has a positive x-component and a positive y-component;so (a) or (b). ~A has a small x-componentand a large y-component; so (a).

-

6

x

y

��~A


(a) 2ı− (b) ı− 2(c) −2ı + *(d) −ı + 2(e) None of these

Solution: ~A has a negative x-component and a positive y-component;so (c) or (d). ~A has a small x-componentand a large y-component; so (d).

-

6

x

y

AAAAAAAAK

~A

204


(a) 6ı− 3 (b) −3ı + 6*(c) −6ı + 3 (d) 3ı− 6(e) None of these

Solution: ~A has a negative x-component and a positive y-component;so (b) or (c). ~A has a large x-componentand a small y-component; so (c).

-

6

x

y

HHH

HHHY

~A


(a) 〈−1, 2〉 (b) 〈−2, 1〉(c) 〈−1, 2〉 *(d) 〈−2,−1〉(e) None of these

Solution: ~A has a negative x-component and a negative y-component;so (d).

-

6

x

y

��

��~A


(a) 〈−1, 2〉 *(b) 〈1,−2〉(c) 〈−2, 1〉 (d) 〈2,−1〉(e) None of these

Solution: ~A has a positive x-component and a negative y-component;so (b) or (d). ~A has a small x-componentand a large y-component; so (b).

-

6

x

y

AAAAAAU ~A

205


(a) 〈−6,−3〉 *(b) 〈−3,−6〉(c) 〈6,−3〉 (d) 〈3,−6〉(e) None of these

Solution: ~A has a negative x-component and a negative y-component;so (a) or (b). ~A has a small x-componentand a large y-component; so (b).

-

6

x

y

��~A


(a) 〈−5,−10〉 (b) 〈5,−10〉*(c) 〈−10, 5〉 (d) 〈10,−5〉(e) None of these

Solution: ~A has a negative x-component and a positive y-component;so (c).

-

6

x

y

HHH

HHHY

~A


(a) 〈−2, 4〉 *(b) 〈2, 4〉(c) 〈−4, 2〉 (d) 〈4, 2〉(e) None of these

Solution: ~A has a positive x-component and a positive y-component;so (b) or (d). ~A has a small x-componentand a large y-component; so (b).

-

6

x

y

��~A

206


*(a) 〈−1, 2〉 (b) 〈−2, 1〉(c) 〈1,−2〉 (d) 〈2,−1〉(e) None of these

Solution: ~A has a negative x-component and a positive y-component;so (a) or (b). ~A has a small x-componentand a large y-component; so (a).

-

6

x

y

AAAAAAK

~A


*(a) 〈2, 1〉 (b) 〈1, 2〉(c) 〈−2, 1〉 (d) 〈−1, 2〉(e) None of these

Solution: ~A has a positive x-component and a positive y-component;so (a) or (b). ~A has a large x-componentand a small y-component; so (a).

-

6

x

y

��

��*

~A


(a) 〈−2,−1〉 (b) 〈−1,−2〉*(c) 〈2,−1〉 (d) 〈1,−2〉(e) None of these

Solution: ~A has a positive x-component and a negative y-component;so (c) or (d). ~A has a large x-componentand a small y-component; so (c).

-

6

x

y

HHHHH

Hj ~A

207

6.1.3 Given direction, identify on figure

Problem 750. Which of the vectors inthe graph corresponds to the direction20◦ east of north?

(a) ~A (b) ~B


Solution: Since we want “east ofnorth”, we take the north direction as thereference line. Going 20◦ east of northputs us in the first quadrant; so ~D.

-

6

E

N

�� ~D

AAAAAAK~A

��~B

AAAAAAU ~C

Problem 751. Which of the vectors inthe graph corresponds to the direction20◦ east of south?

(a) ~A (b) ~B


Solution: Since we want “east ofsouth”, we take the south direction as thereference line. Going 20◦ east of southputs us in the fourth quadrant; so ~C.

-

6

E

N

�� ~D

AAAAAAK~A

��~B

AAAAAAU ~C

Problem 752. Which of the vectors inthe graph corresponds to the direction20◦ west of north?

*(a) ~A (b) ~B


Solution: Since we want “west ofnorth”, we take the north direction as thereference line. Going 20◦ west of northputs us in the second quadrant; so ~A.

-

6

E

N

�� ~D

AAAAAAK~A

��~B

AAAAAAU ~C

208

Problem 753. Which of the vectors inthe graph corresponds to the direction20◦ west of south?

(a) ~A *(b) ~B


Solution: Since we want “west ofsouth”, we take the south direction as thereference line. Going 20◦ west of southputs us in the third quadrant; so ~B.

-

6

E

N

�� ~D

AAAAAAK~A

��~B

AAAAAAU ~C

Problem 754. Which of the vectors inthe graph corresponds to the direction30◦ south of west?

(a) ~A (b) ~B


Solution: Since we want “south ofwest”, we take the west direction as thereference line. Going 30◦ south of westputs us in the third quadrant; so ~C or~D. Since 30◦ < 45◦, our vector should becloser to the west direction than to thesouth direction; so ~C. ( ~D would be morelike 60◦ south of west.)

-

6

E

N

AAAAAAK

~A

HHH

HHHY

~B

��~C

��~D

Problem 755. Which of the vectors inthe graph corresponds to the direction30◦ north of west?

(a) ~A *(b) ~B


Solution: Since we want “north ofwest”, we take the west direction as thereference line. Going 30◦ north of westputs us in the second quadrant; so ~A or~B. Since 30◦ < 45◦, our vector should becloser to the west direction than to thenorth direction; so ~B. ( ~A would be morelike 60◦ north of west.)

-

6

E

N

AAAAAAK

~A

HHHHH

HY~B

��~C

��~D

209

Problem 756. Which of the vectors inthe graph corresponds to the direction60◦ south of west?

(a) ~A (b) ~B


Solution: Since we want “south ofwest”, we take the west direction as thereference line. Going 60◦ south of westputs us in the third quadrant; so ~C or~D. Since 60◦ > 45◦, our vector should becloser to the south direction than to thewest direction; so ~D. (~C would be morelike 30◦ south of west.)

-

6

E

N

AAAAAAK

~A

HHHHH

HY~B

��~C

��~D

Problem 757. Which of the vectors inthe graph corresponds to the direction60◦ north of west?

*(a) ~A (b) ~B


Solution: Since we want “north ofwest”, we take the west direction as thereference line. Going 60◦ north of westputs us in the second quadrant; so ~A or~B. Since 60◦ > 45◦, our vector should becloser to the north direction than to thewest direction; so ~A. ( ~B would be morelike 30◦ north of west.)

-

6

E

N

AAAAAAK

~A

HHHHH

HY~B

��~C

��~D

Problem 758. What is the direction ofvector ~A in the graph at right?

(a) 30◦ north of east*(b) 30◦ south of east(c) 30◦ north of west(d) 30◦ south of west(e) None of these

Solution: Of the solutions, only 30◦

south of east is in the fourth quadrant.

-

6

E

N

HHHHHHj ~A

210


*(a) 30◦ north of east(b) 30◦ south of east(c) 30◦ north of west(d) 30◦ south of west(e) None of these


north of east is in the first quadrant.

-

6

E

N

��

��*

~A


(a) 30◦ north of east(b) 30◦ south of east(c) 30◦ north of west*(d) 30◦ south of west(e) None of these


south of west is in the third quadrant.

-

6

E

N

��

��~A


(a) 30◦ north of east(b) 30◦ south of east*(c) 30◦ north of west(d) 30◦ south of west(e) None of these


north of west is in the second quadrant.

-

6

E

N

HHH

HHHY

~A

211


(a) 30◦ east of north(b) 60◦ east of north(c) 30◦ west of north*(d) 60◦ west of north(e) None of these

Solution: 30◦ west of north and 60◦

west of north are both in the second quad-rant. However, 30◦ west of north wouldbe closer to the north direction than tothe west direction. Hence: 60◦ west ofnorth.

-

6

E

N

HHHH

HHY~A


(a) 30◦ east of north(b) 60◦ east of north*(c) 30◦ west of north(d) 60◦ west of north(e) None of these

Solution: 30◦ west of north and 60◦

west of north are both in the second quad-rant. However, 60◦ west of north wouldbe closer to the west direction than tothe north direction. Hence: 30◦ west ofnorth.

-

6

E

N

AAAAAAK

~A


(a) 30◦ east of north*(b) 60◦ east of north(c) 30◦ west of north(d) 60◦ west of north(e) None of these

Solution: 30◦ east of north and 60◦

east of north are both in the first quad-rant. However, 30◦ east of north would becloser to the north direction than to theeast direction. Hence: 60◦ east of north.

-

6

E

N

��

��* ~A

212


*(a) 30◦ east of north(b) 60◦ east of north(c) 30◦ west of north(d) 60◦ west of north(e) None of these

Solution: 30◦ east of north and 60◦

east of north are both in the first quad-rant. However, 60◦ east of north wouldbe closer to the east direction than to thenorth direction. Hence: 30◦ east of north.

-

6

E

N

��~A

6.1.4 Given vectors on figure, name quadrant of sum/difference

Problem 766. The vectors ~A and ~B arelabelled on the figure at right. Sketch andlabel the vector ~A − ~B. What quadrantdoes ~A− ~B lie in?

(a) Quadrant I(b) Quadrant II*(c) Quadrant III(d) Quadrant IV(e) It lies on the x- or y-axis

Solution: Sketch a copy of − ~B, la-belled − ~B′, with its tail at the head of ~A.Then ~A − ~B has its tail at the tail of ~Aand its head at the head of − ~B′.

-

6

x

y

��9~A

��

~B

��

��

− ~B′~A− ~B

213

Problem 767. The vectors ~A and ~B arelabelled on the figure at right. Sketch andlabel the vector ~A + ~B. What quadrantdoes ~A+ ~B lie in?

(a) Quadrant I*(b) Quadrant II(c) Quadrant III(d) Quadrant IV(e) It lies on the x- or y-axis

Solution: Sketch a copy of ~B, labelled~B′, with its tail at the head of ~A. Then~A+ ~B has its tail at the tail of ~A and itshead at the head of ~B′.

-

6

x

y

��)~A

��~B

��

~B′

~A+ ~B


(a) Quadrant I(b) Quadrant II*(c) Quadrant III(d) Quadrant IV(e) It lies on the x- or y-axis


-

6

x

y

��)~A

��~B

��

− ~B′ ~A− ~B

Problem 769. The vectors ~A and ~B arelabelled on the figure at right. Sketch andlabel the vector ~B − ~A. What quadrantdoes ~B − ~A lie in?


Solution: Sketch a copy of − ~A, la-belled − ~A′, with its tail at the head of ~B.Then ~B − ~A has its tail at the tail of ~Band its head at the head of − ~A′.

-

6

x

y

��)~A

��~B��)

− ~A′

~B − ~A

214


(a) Quadrant I(b) Quadrant II(c) Quadrant III(d) Quadrant IV*(e) It lies on the x- or y-axis

Solution: Sketch a copy of ~B, labelled~B′, with its tail at the head of ~A. Then~A+ ~B has its tail at the tail of ~A and itshead at the head of ~B′: on the negativex-axis

-

6

x

y

��)

~A

PPPP

PPi~B

PPPP

PPi

~B′

~A+ ~B


*(a) Quadrant I(b) Quadrant II(c) Quadrant III(d) Quadrant IV(e) It lies on the x- or y-axis


-

6

x

y

��

��1 ~A��~B

��

~B′

~A+ ~B




-

6

x

y

��

��1 ~A��~B��)

− ~A′

~B − ~A

215


(a) Quadrant I(b) Quadrant II(c) Quadrant III*(d) Quadrant IV(e) It lies on the x- or y-axis


-

6

x

y

��

��1 ~A��~B

��

− ~B′

~A− ~B




-

6

x

y

��

��1 ~A

BBBBBBBN ~B

BBBBBBBM

− ~B′

~A− ~B




-

6

x

y

��

��1 ~A

BBBBBBBN ~B

BBBBBBBN

~B′

~A+ ~B

216




-

6

x

y

PPPPPPq ~A

BBBBBBBM

~B

BBBBBBBN

− ~B′~A− ~B




-

6

x

y

PPPPPPq ~A

BBBBBBBM

~B

BBBBBBBM

~B′

~A+ ~B




-

6

x

y

PPPPPPq ~A

BBBBBBBM

~BPPPP

PPi − ~A′

~B − ~A

217

6.1.5 Mixing it up

Problem 779. Five vectors are graphedon the diagram at right and described be-low. Write the letter of each vector besideits description.~E 3ı− ~B −4ı + 2~D 2ı− 3~A ı + 4~C −3ı− 3

-

6

x

y

PPPPPPq ~E

HHHH

HHH

HY

~B

JJJJJJJ ~D

��~A

��

��

��~C

Problem 780. Five vectors are graphedon the diagram at right and described be-low. Write the letter of each vector besideits description.~A 〈2, 2〉~C 〈−3, 1〉~E 〈3,−4〉~D 〈−2,−1〉~B 〈−2, 4〉

-

6

x

y

��

~A

PPPP

PPi~C

SSSSSSSSw ~E

��~D

AAAAAAAAK

~B

Problem 781. Five velocity vectors aregraphed on the diagram at right and de-scribed below. Write the letter of eachvector beside its magnitude and direction.~E 14 m/s; 0.79 rad south of east

~D 45 m/s; 1.11 rad south of east

~C 41 m/s; 1.33 rad south of west

~A 45 m/s; 0.46 rad north of west

~B 36 m/s; 0.59 rad south of west

-

6

E

N

@@R ~E

AAAAAAAAU ~D

��~C

HHHH

HHH

HY

~A

��

��

��+~B

218

Problem 782. Five velocity vectors aregraphed on the diagram at right and de-scribed below. Write the letter of eachvector beside its magnitude and direction.~E 22 mph; 64◦ north of east~C 50 mph; 37◦ south of east~A 36 mph; 34◦ south of west~D 32 mph; 18◦ north of east~B 28 mph; 45◦ south of east

-

6

E

N

��

~E

ZZZZZZZZ~~C

��

��

��+~A

��

��1 ~D

@@@@R~B

Problem 783. The vector ~A is labelledon the figure at right. Sketch and labelthe vector − ~A.

Solution: − ~A should be the samelength as ~A, in exactly the opposite di-rection. -

6

x

y

�

~A

�− ~A

Problem 784. The vector ~A is labelledon the figure at right. Sketch and labelthe vector − ~A.

Solution: − ~A should be the samelength as ~A, in exactly the opposite di-rection. -

6

x

y

PPPP

Pi~A

PPPPPq

− ~A

219

Problem 785. Which of the vectors inthe graph corresponds to ı− 3?

(a) ~A (b) ~B


Solution: Since the tail is at the ori-gin, the tip must be at the point (1,−3)(the 3rd quadrant). The magnitude ofthe y-component is greater than the x-component, so the answer must be C.

-

6

x

y

BBBBBBBN ~C

PPPPPPq

~APP

PPPPi

~D

�� ~B

Problem 786. Five vectors are graphedon the diagram at right and described be-low. Write the letter of each vector besideits description.~A = 〈2, 2〉~C = 〈−3, 1〉~E = 〈3,−4〉~D = 〈−2,−1〉~B = 〈−2, 4〉

-

6

x

y

��

~A

PPPP

PPi~C

SSSSSSSSw ~E

��~D

AAAAAAAAK

~B

Problem 787. Which of the vectors inthe graph corresponds to −6ı− 3?

(a) ~A (b) ~B


Solution: Only ~D has negative x- andy-components.

-

6

x

y

��

~A

PPPP

PPi~C

��~D

AAAAAAAAK

~B

220

6.2 Geometric Vector Addition and Subtraction

Problem 788. The vectors ~A and ~B arelabelled on the figure at right. Sketch andlabel the vector ~A+ ~B.


-

6

x

y

-

~A

6~B

~B′

�

~A+ ~B

Problem 789. The vectors ~A and ~B arelabelled on the figure at right. Sketch andlabel the vector ~A− ~B.


-

6

x

y

-~A

6~B

− ~B′JJJJJJ~A− ~B


-

6

x

y

�~A

6

~B

− ~B′�

~A− ~B

221

Problem 791. The vectors ~X, ~Y , and~Z are labelled on the figure at right.Which of the following equations is true?

(a) ~Z = ~X − ~Y *(b) ~Z = ~Y − ~X

(c) ~Z = ~X + ~Y (d) ~Z = − ~X − ~Y(e) None of these

Solution: Notice that ~Y +(− ~X) = ~Z.

-

6

x

y

�~Y

?~X

JJJJJJ]

~Z

Problem 792. The vectors ~X, ~Y , and ~Zare labelled on the figure at right. Con-sider the vector −~Z. From the figure it isclear that

−~Z = ± ~X ± ~Y .

Which of the following equations is true?

(a) −~Z = − ~X + ~Y (b) −~Z = ~X + ~Y

(c) −~Z = − ~X − ~Y (d) −~Z = ~X − ~Y(e) None of these

Solution: ~X + (−~Y ) = −~Z.

-

6

x

y

�~Y

?~X

JJJJJJ]

~Z



-

6

x

y

��*

~A

HHHH

HY~B

~B′6

~A+ ~B

222



-

6

x

y

��

~A

AAAAAK

~B

− ~B′

-

~A− ~B



-

6

x

y

��~A

��)

~B

~B′

@@

@@I~A+ ~B



-

6

x

y

PPPPPPq ~A

SSSSSSSSw ~B

− ~B′

6~A− ~B

223



-

6

x

y

BBBBBBBBN~A

��

~B

~B′

@@@@@R~A+ ~B



-

6

x

y

��9~A

@@

@@@I

~B

− ~B′

��

~A− ~B

Problem 799. The vectors ~A, ~B, and ~Care labelled on the figure at right. Sketchand label the vector ~A+ ~B + ~C.

Solution: Sketch a copy of ~B, labelled~B′, with its tail at the head of ~A. Sketcha copy of ~C, labelled ~C ′, with its tail atthe head of ~B′. Then ~A+ ~B+ ~C will haveits tail at the tail of ~A and its head at thehead of ~C ′.

-

6

x

y

SSSSSSSw

~A

6

~B

BBBBBBBBBM

~C

~B′6

~C ′

��

~A+ ~B + ~C

224

Problem 800. The vectors ~A, ~B, and ~Care labelled on the figure at right. Sketchand label the vector ~A+ ~B − ~C.

Solution: Sketch a copy of ~B, labelled~B′, with its tail at the head of ~A. Sketcha copy of −~C, labelled −~C ′, with its tailat the head of ~B′. Then ~A + ~B − ~C willhave its tail at the tail of ~A and its headat the head of −~C ′.

-

6

x

y

�� ~A

@@

@@I

~B

HHH

HHH

HHH

HHY~C

@I

~B′−~C ′

��

��1

~A+ ~B − ~C

225

6.3 Position vs. Displacement Vector Problems

Problem 801. On the graph below, draw and label the position vectors ~r1 and ~r2 corre-sponding to the points P1 and P2 on the graph, and the displacement vector ∆~r = ~r2−~r1 .

-

6

x

y

rP1

rP2

��

~r1

BBBBBBM

~r2

XXXXXX

Xy ∆~r

Problem 802. At exactly 12:00, a spi-der climbs onto the tip of a clock’s minutehand, where it remains for the next hour.On the axes at right, sketch and label thespider’s displacement vectors:~A at 12:15~B at 12:30~C at 12:45~D at 1:00

Solution: Don’t forget that you’remeasuring displacement from the 12:00position, not from the center of the clock.

-

6

x

y

@@@@R ~A

?~B

��

��~C

~D = ~0r

226

Problem 803. At exactly 12:00, aspider climbs onto the tip of a clock’sminute hand, where it remains for thenext hour. Which of the vectors on thefigure at right is in the direction of thespider’s displacement vector at 12:15?

(a) ~C *(b) ~D

(c) ~G (d) ~H(e) None of these

-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F

Problem 804. At exactly 12:00, aspider climbs onto the tip of a clock’sminute hand, where it remains for thenext hour. Which of the vectors onthe figure at right is in the direction ofthe spider’s displacement vector at 12:30?

(a) The zero vector ~0 (b) ~A

(c) ~B *(d) ~E(e) None of these

-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F


(a) ~B *(b) ~F


-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F

227


*(a) The zero vector ~0 (b) ~A

(c) ~B (d) ~E(e) None of these

-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F

Problem 807. At exactly 12:00, aspider climbs onto the tip of a clock’sminute hand, where it remains for thenext hour. If the origin is at the centerof the clock, which of the vectors on thefigure at right is the spider’s positionvector at 12:15?

(a) ~B *(b) ~C


-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F


(a) The zero vector ~0 (b) ~A

*(c) ~E (d) ~F(e) None of these

-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F

228


(a) ~B (b) ~F

*(c) ~G (d) ~H(e) None of these

-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F


(a) The zero vector ~0 *(b) ~A

(c) ~B (d) ~E(e) None of these

-

6

x

y

~0

6

~A

-~C

?~E

�~G �

��

~B

@@

@@@I

~H

��

��

~D

@@@@@R

~F

229

6.4 Finding Components of Vectors

Problem 811. The vector ~A has magnitude 5.9 and direction θ = 0.34 rad. Find the x-and y-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = (5.9)(cos 0.34 rad) = 5.6

Ay = ‖ ~A‖ sin θ = (5.9)(sin 0.34 rad) = 2.0

Problem 812. The vector ~A has magnitude 83 and direction θ = 1.98 rad. Find the x-and y-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = (83)(cos 1.98 rad) = −33

Ay = ‖ ~A‖ sin θ = (83)(sin 1.98 rad) = 76

Problem 813. The vector ~A has magnitude 41 and direction θ = 3.51 rad. Find the x-and y-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = (41)(cos 3.51 rad) = −38

Ay = ‖ ~A‖ sin θ = (41)(sin 3.51 rad) = −15

Problem 814. The vector ~A has magnitude 7.7 and direction θ = 5.66 rad. Find the x-and y-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = (7.7)(cos 5.66 rad) = 6.3

Ay = ‖ ~A‖ sin θ = (7.7)(sin 5.66 rad) = −4.5

Problem 815. The vector ~A has magnitude 0.88 and direction 37◦. Find the x- andy-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = 0.88 cos 37◦ = 0.70

Ay = ‖ ~A‖ sin θ = 0.88 sin 37◦ = 0.53

Problem 816. The vector ~A has magnitude 1.28 and direction 98◦. Find the x- andy-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = 1.28 cos 98◦ = −0.18

Ay = ‖ ~A‖ sin θ = 1.28 sin 98◦ = 1.27

Problem 817. The vector ~A has magnitude 110 and direction 219◦. Find the x- andy-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = 110 cos 219◦ = −85

Ay = ‖ ~A‖ sin θ = 110 sin 219◦ = −69

Problem 818. The vector ~A has magnitude 37 and direction 304◦. Find the x- andy-components of ~A.

Solution: Ax = ‖ ~A‖ cos θ = 37 cos 304◦ = 21

Ay = ‖ ~A‖ sin θ = 37 sin 304◦ = −31

230

Problem 819. ~A = 〈5.1, 8.8〉. Find the magnitude and direction of ~A. Give the directionin radians.

Solution: ‖ ~A‖ =√Ax

2 + Ay2 =√

5.12 + 8.82 = 10.2

Since Ax > 0 and Ay > 0, θ is in the first quadrant. Hence:

θ = tan−1

∣∣∣∣AyAx∣∣∣∣ = tan−1 8.8

5.1= 1.05 rad

Problem 820. ~A = 〈16,−41〉. Find the magnitude and direction of ~A. Give thedirection in radians.


2 + Ay2 =

√162 + (−41)2 = 44

Since Ax > 0 and Ay < 0, θ is in the fourth quadrant. Hence:

θ = 2π − tan−1

∣∣∣∣AyAx∣∣∣∣ = 2π − tan−1 41

16= 5.08 rad

Problem 821. ~A = 〈−0.81,−1.13〉. Find the magnitude and direction of ~A. Give thedirection in degrees.


2 + Ay2 =

√(−0.81)2 + (−1.13)2 = 1.39

Since Ax < 0 and Ay < 0, θ is in the third quadrant. Hence:

θ = 180◦ + tan−1

∣∣∣∣AyAx∣∣∣∣ = 180◦ + tan−1 1.13

0.81= 234◦

Problem 822. ~A = 〈−216, 82〉. Find the magnitude and direction of ~A. Give thedirection in degrees.


2 + Ay2 =

√(−216)2 + 822 = 231

Since Ax < 0 and Ay > 0, θ is in the second quadrant. Hence:

θ = 180◦ − tan−1

∣∣∣∣AyAx∣∣∣∣ = 180◦ − tan−1 82

216= 159◦

Problem 823. ~A = 4.8ı + 7.0. Find the magnitude and direction of ~A. Give thedirection in radians.


2 + Ay2 =√

4.82 + 7.02 = 8.5

Since Ax > 0 and Ay > 0, θ is in the first quadrant. Hence:

θ = tan−1

∣∣∣∣AyAx∣∣∣∣ = tan−1 7.0

4.8= 0.97 rad

Problem 824. ~A = −120ı − 220. Find the magnitude and direction of ~A. Give thedirection in radians.


2 + Ay2 =

√(−120)2 + (−220)2 = 251

Since Ax < 0 and Ay < 0, θ is in the third quadrant. Hence:

θ = π + tan−1

∣∣∣∣AyAx∣∣∣∣ = π + tan−1 220

120= 4.21 rad

231

Problem 825. ~A = −29ı + 12. Find the magnitude and direction of ~A. Give thedirection in degrees.


2 + Ay2 =

√(−29)2 + 122 = 31

Since Ax < 0 and Ay > 0, θ is in the second quadrant. Hence:

θ = 180◦ − tan−1

∣∣∣∣AyAx∣∣∣∣ = 180◦ − tan−1 12

29= 158◦

Problem 826. ~A = 0.16ı − 0.23. Find the magnitude and direction of ~A. Give thedirection in degrees.


2 + Ay2 =

√0.162 + (−0.23)2 = 0.28

Since Ax > 0 and Ay < 0, θ is in the fourth quadrant. Hence:

θ = 360◦ − tan−1

∣∣∣∣AyAx∣∣∣∣ = 360◦ − tan−1 0.23

0.16= 305◦

6.5 Algebraic Vector Addition and Subtraction

Problem 827. ~A = 3ı− 2 and ~B = 4ı + 5. Find ~A+ ~B.

Solution: ~A+ ~B = (3 + 4)ı + (−2 + 5) = 7ı + 3

Problem 828. ~A = −ı + 7 and ~B = 2ı + 3. Find ~A− ~B.

Solution: ~A− ~B = (−1− 2)ı + (7− 3) = −3ı + 4

Problem 829. ~A = 〈8, 3〉 and ~B = 〈1, 2〉. Find ~A+ ~B.

Solution: ~A+ ~B = 〈8 + 1, 3 + 2〉 = 〈9, 5〉

Problem 830. ~A = 〈2, 6〉 and ~B = 〈−4, 5〉. Find ~A− ~B.

Solution: ~A− ~B = 〈2− (−4), 6− 5〉 = 〈6, 1〉

Problem 831. ~A = 〈4,−1〉 and ~B = 〈−5,−3〉. Find ~A+ ~B.

(a) 〈−9,−4〉 (b) 〈1, 2〉(c) 〈3,−8〉 *(d) 〈−1,−4〉(e) None of these

Solution: 〈4,−1〉+ 〈−5,−3〉 = 〈4 + (−5),−1 + (−3)〉 = 〈−1,−4〉

Problem 832. ~A = 〈−2, 5〉 and ~B = 〈3,−3〉. Find ~A+ ~B.

(a) 〈5, 8〉 *(b) 〈1, 2〉(c) 〈−5, 8〉 (d) 〈−6,−15〉(e) None of these

Solution: Add like components: ~A+ ~B = 〈−2, 5〉+〈3,−3〉 = 〈−2 + 3, 5− 3〉 = 〈1, 2〉.

232

Problem 833. ~A = 〈−2, 5〉 and ~B = 〈7,−3〉. Find ~A− ~B.

(a) 〈5, 2〉 (b) 〈−9, 2〉*(c) 〈−9, 8〉 (d) 〈5,−2〉(e) None of these

Solution: Add like components: ~A − ~B = 〈−2, 5〉 − 〈7,−3〉 = 〈−2− 7, 5 + 3〉 =〈−9, 8〉.

Problem 834. ~A has magnitude 7.2 and direction 1.1 rad. ~B has magnitude 5.5 anddirection 2.9 rad. Find ~A+ ~B. Give your answer in component form: 〈Cx, Cy〉.Solution: We begin by putting ~A and ~B in component form.

~A =⟨‖ ~A‖ cos θA, ‖ ~A‖ sin θA,

⟩= 〈(7.2)(cos 1.1 rad), (7.2)(sin 1.1 rad)〉

~B =⟨‖ ~B‖ cos θB, ‖ ~B‖ sin θB,

⟩= 〈(5.5)(cos 2.9 rad), (5.5)(sin 2.9 rad)〉

~A+ ~B = 〈(7.2)(cos 1.1 rad) + (5.5)(cos 2.9 rad), (7.2)(sin 1.1 rad) + (5.5)(sin 2.9 rad)〉= 〈−2.1, 7.7〉

Problem 835. ~A has magnitude 0.83 and direction 22◦. ~B has magnitude 1.42 anddirection 157◦. Find ~A− ~B. Give your answer in terms of unit vectors: Cxı + Cy .

Solution: We begin by putting ~A and ~B in unit-vector form:

~A = ‖ ~A‖ cos θA ı + ‖ ~A‖ sin θA = 0.83 cos 22◦ ı + 0.83 sin 22◦

~B = ‖ ~B‖ cos θB ı + ‖ ~B‖ sin θB = 1.42 cos 157◦ ı + 1.42 sin 157◦

~A− ~B = (0.83 cos 22◦ − 1.42 cos 157◦) ı + (0.83 sin 22◦ − 1.42 sin 157◦)

= 2.08ı− 0.24

Problem 836. ~A has magnitude 115 and direction 3.8 rad. ~B has magnitude 303 anddirection 5.1 rad. Find the magnitude and direction of ~A+ ~B.

Solution: This will require three steps: putting ~A and ~B into component form; addingthe two vectors; and then finding the magnitude and direction of the resultant.

~A =⟨‖ ~A‖ cos θA, ‖ ~A‖ sin θA,

⟩= 〈(115)(cos 3.8 rad), (115)(sin 3.8 rad)〉

~B =⟨‖ ~B‖ cos θB, ‖ ~B‖ sin θB,

⟩= 〈(303)(cos 5.1 rad), (303)(sin 5.1 rad)〉

~A+ ~B = 〈(115)(cos 3.8 rad) + (303)(cos 5.1 rad), (115)(sin 3.8 rad) + (303)(sin 5.1 rad)〉= 〈Cx, Cy〉= 〈24,−351〉

233

You should not use these rounded component values in calculating magnitude and direc-tion. However, you’ll need to know their signs in order to make sure your angle is in theright quadrant. Since the x-component is positive and the y-component is negative, θ isin the fourth quadrant.

‖ ~A+ ~B‖ =√Cx

2 + Cy2 = 352

θ = 2π − tan−1

∣∣∣∣CyCx∣∣∣∣ = 4.78 rad

Problem 837. ~A has magnitude 57 and direction 43◦. ~B has magnitude 22 and direction117◦. Find the magnitude and direction of ~A− ~B.

This will require three steps: putting ~A and ~B in component form; finding ~C = ~A− ~B;and calculating the magnitude and direction of ~C.

~A = ‖ ~A‖ cos θA ı + ‖ ~A‖ sin θA = 57 cos 43◦ ı + 57 sin 43◦

~B = ‖ ~B‖ cos θB ı + ‖ ~B‖ sin θB = 22 cos 117◦ ı + 22 sin 117◦

~A− ~B = (57 cos 43◦ − 22 cos 117◦) ı + (57 sin 43◦ − 22 sin 117◦)

= Cxı + Cy

= 52ı + 19

You should not use these rounded component values in calculating magnitude and direc-tion of ~C. However, you’ll need their signs to make sure you have θ in the right quadrant.Since both components are positive, θ is in the first quadrant.

‖ ~A− ~B‖ =√Cx

2 + Cy2 = 55

θ = tan−1 CyCx

= 20◦

Problem 838. If ~A = 〈3.3, 5.7〉, what is the magnitude of ~A? Round your answer tothe nearest 0.1.


Solution: ‖A‖ =√A2x + A2

y =√

(3.3)2 + (5.7)2 = 6.6

Problem 839. If ~A = 〈3, 4〉, what is the magnitude of ~A?


Solution: ‖A‖ =√A2x + A2

y =√

(3)2 + (4)2 = 6.6

234

Problem 840. If ~A = 3ı− 4, what is the magnitude of ~A?

(a) -1 (b) 2(c) 4 *(d) 5(e) None of these

Solution: ‖ ~A‖ =√

(3)2 + (−4)2 = 5

235

6.6 Concept vector and scalar questions

The first three problems deal with trigonometric functions.

Problem 841. cos(5 m/s) = ?

Solution: We can only take the cosine of a pure number, so this question doesn’tmake sense.

Problem 842. Consider the expression: sin(ω∗ t∗) = ?, where t∗ has units of time. Whatmust the units of ω∗ be in order for the argument of the sine function to make sense?

Solution: [ω∗] = 1T

.

Problem 843. Why is an angle a dimensionless number when measured in radians?

Solution: The angle in radians is the ratio of the arc intercepted by the angle to theradius of the circle:

angle in radians =arc length

radius

The arc length and the radius are both measured in units of length; so length in thenumerator and denominator cancel, leaving no units.

Problem 844. True or false: If we add the scalars 1 m + 1 m, we get 2 m.

Solution: True. We add scalars as we do regular numbers: 1 + 1 = 2.

Problem 845. True or false: If we add two vectors, each with length 1 m, we get avector with length 2 m.

Solution: False. Suppose the first vector is 1 m long in the positive x-direction, andthe second is 1 m long in the negative x-direction. Then the sum of the two vectors iszero. Unless the two vectors are pointing in exactly the same direction, we can’t find themagnitude of their sum by adding the magnitudes of the two vectors.

Problem 846. Can you find a vector ~v = 〈v1, v2〉 such that ~v 6= 0, but ‖~v‖ = 0?

Solution: No. The magnitude of ~v is

‖~v‖ =√v2

1 + v22

Since it has to be the case that v21 ≥ 0 and v2

2 ≥ 0, the only way ‖~v‖ can be zero is ifv2

1 = v22 = 0. But that means that v1 = v2 = 0, so ~v = 0.

Problem 847. Is it possible for the magnitude of a vector to be smaller than any of itscomponents?

236

Solution: No. For simplicity’s sake, let’s look at a vector with two components:~v = 〈v1, v2〉; and let’s suppose that v1 < v2. The magnitude of ~v is

‖~v‖ =√v2

1 + v22 ≥

√v2

1 + 0 = |v1| ≥ v1

Hence the magnitude of ~v is no smaller than its smallest component.Technically, we should prove this for an arbitrary vector ~v = 〈v1, . . . , vn〉, where vk is thesmallest component. But the notation would get cumbersome, and the essentials of theproof are the same.

Problem 848. True or false: Let ~v = 〈a, b〉. If ‖~v‖ > 0, then a > 0 and b > 0.

Solution: False. Suppose a = −3 and b = −4. Then

‖~v‖ =√

(−3)2 + (−4)2 =√

25 = 5

In this example, ‖~v‖ > 0, and a < 0 and b < 0.

Problem 849. If ~A = Ax ı+Ay and makes an angle θ with the positive x-axis, expresssin θ in terms of Ax and Ay.

Solution: sin θ =Ay√

A2x + A2

y

Problem 850. If ~A = Ax ı+Ay and makes an angle θ with the positive x-axis, expresscos θ in terms of Ax and Ay.

Solution: cos θ =Ax√

A2x + A2

y

Problem 851. True or false: If ~A = Ax ı + Ay , then the equation for the magnitudesatisfies A = Ax + Ay?

Solution: False. A =√Ax

2 + Ay2 6= Ax + Ay. For example, let Ax = Ay = −1.

Problem 852. If ~A = Ax ı+Ay and makes an angle θ with the positive x-axis, expressθ in terms of Ax and Ay. Assume that Ax > 0 and Ay > 0.

Solution: θ = tan−1

(AyAx

). [Since Ax > 0 and Ay > 0, θ is in the first quadrant; so

we don’t have to worry about reference angles.]

Problem 853. True or false: If ~A = Ax ı + Ay , then the equation for the magnitude

satisfies ‖ ~A‖ ≥ |Ax|. Justify your answer.

Solution: True. ‖ ~A‖ =√Ax

2 + Ay2 ≥

√Ax

2 = |Ax|

237

Problem 854. True or false: Let ~r = 〈x, y〉 = x ı + y . If ‖~r‖ > 0, then x > 0 andy > 0. Justify your answer.

Solution: False. Let ~r = 〈−1,−1〉. Then ‖~r‖ =√

2 > 0. [There are many othercounterexamples you could use.]

Problem 855. Can you find a situation where the average velocity vector is equal tothe zero vector, but the average speed is not zero? In symbols:

Can we have ~vave =∆~r

∆t= ~0 and speed =

total distance travelled

∆t6= 0 ?

Give an example, or explain why one doesn’t exist.

Solution: A car drives 100 km straight north at 100 km/h, then quickly turns aroundand drives back to its starting point at 100 km/h. The total time it takes is ∆t = 2 h.Since it has returned to its starting point, ∆~r = ~0. The total distance travelled is 200km, so the average speed is 100 km/h. [There are many other examples you could use.]

Problem 856. Consider the statements (i) and (ii). Are the statements true or false?

(i) If k is a scalar, then k + k = 2k

(ii) Let ~v = 〈a, b〉. If ‖v‖ > 0, then a > 0 and b > 0.

Choose the correct answer from below.

*(a) (i) is true; (ii) is false (b) (i) is false; (ii) is true

(c) Both statements are true (d) Both statements are false

Solution: (i) is true. (ii) is false: every nonzero vector has a positive magnitude,including those with negative components. For example, if ~v = 〈−3, 4〉, then ‖~v‖ =√

(−3)2 + 42 = 5.


(i) If ~v is a vector, then ‖2~v‖ = 2‖~v‖.

(ii) Let ~v = 〈a, b〉. If ‖v‖ = 0, then a = 0 and b = 0.


(a) (i) is true; (ii) is false (b) (i) is false; (ii) is true

*(c) Both statements are true (d) Both statements are false

238


(i) If ~v is a vector, then ‖ − ~v‖ = −‖~v‖.

(ii) Let ~v = 〈a, b〉. Then 2~v = 〈2a, 2b〉.


(a) (i) is true; (ii) is false *(b) (i) is false; (ii) is true


Solution: (ii) is true. (i) is false: the magnitude of a vector is always positive, unlessthe vector is ~0, in which case the magnitude is zero. Hence ‖ − ~v‖ = ‖~v‖.


(i) If ~v is a vector, then ‖ − ~v‖ = ‖~v‖.

(ii) Let ~v = 〈a, b〉. Then −2~v = −2a− 2b.




Solution: (i) is true. (ii) is false: if ~v = 〈a, b〉, then −2~v = 〈−2a,−2b〉.


(i) If ~u and ~v are vectors, then ‖~u+ ~v‖ = ‖~u‖+ ‖~v‖.

(ii) Let ~v = 〈a, b〉. Then ‖~v‖ =√a2 + b2.




Solution: (ii) is true. (i) is false; for example, let ~u = 〈1, 0〉 and ~v = 〈−1, 0〉. Then~u+ ~v = ~0; so ‖~u‖ = ‖~v‖ = 1, but ‖~u+ ~v‖ = 0 6= 1 + 1.


(i) If ~u is a vector and k is a scalar, then ‖k~u‖ = k‖~u‖.

(ii) Let ~v = 〈a, b〉. Then ‖~v‖ =√a+ b.



(c) Both statements are true *(d) Both statements are false

Solution: (i) is false. Let ~u = 〈1, 0〉, and let k = −2. Then ‖~u‖ = 1; k~u = 〈−2, 0〉, so‖k~u‖ = 2 6= k‖~u‖ = (−2)(1). (A true version of (i) would be: ‖k~u‖ = |k| · ‖~u‖.)(ii) is also false. ‖~v‖ =

√a2 + b2.

239


(i) If ~u is a vector, k is a scalar, and k > 0, then ‖k~u‖ = k‖~u‖.

(ii) Let ~v1 = 〈a1, b1〉 and ~v2 = 〈a2, b2〉. Then v1 + v2 = 〈a1 + a2, b1 + b2〉.



*(c) Both statements are true (d) Both statements are false

Solution: Both statements are true. Statement (i) is a specific case of the generalstatement: If ~u is a vector and k is a scalar, then ‖k~u‖ = |k| · ‖~u‖. If k > 0, then |k| = k.


(i) If ~u is a vector, then ‖ − 2~u‖ = 2‖~u‖.

(ii) Let ~v1 = 〈a1, b1〉 and ~v2 = 〈a2, b2〉. Then v1 + v2 = 〈a1 + b1, a2 + b2〉.




Statement (i) is a specific case of the general statement: If ~u is a vector and k is a scalar,then ‖k~u‖ = |k| · ‖~u‖. If k = −2, then |k| = 2.

Statement (ii) is a garbled version of the true statement: if ~v1 = 〈a1, b1〉 and ~v2 = 〈a2, b2〉,then ~v1 + ~v2 = 〈a1 + a2, b1 + b2〉.


(i) If ~u is a vector, then ~u+ (−~u) = ~0.

(ii) Let ~v1 = 〈a1, b1〉 and ~v2 = 〈a2, b2〉. Then v1 − v2 = 〈a1 − b1, a2 − b2〉.




Solution: (i) is true. (ii) is a garbled version of the true statement: if ~v1 = 〈a1, b1〉and ~v2 = 〈a2, b2〉, then v1 − v2 = 〈a1 − a2, b1 − b2〉.


(i) If ~u is a vector, then ~u− 2~u = ~u.

(ii) Let ~v1 = 〈a1, b1〉 and ~v2 = 〈a2, b2〉. Then v1 − v2 = 〈a1 − a2, b1 − b2〉.




Solution: (i) is false; a true version would be: ~u− 2~u = −~u. (ii) is true.

240

6.7 Applications of Vectors

6.7.1 Breaking vectors into components

Problem 866. An airplane is flying at a speed of V in a direction 22◦ north of east.Which expression gives the northward component of the plane’s velocity?

(a) V cos 22◦ (b) V/ cos 22◦

*(c) V sin 22◦ (d) V/ sin 22◦

(e) None of these

Problem 867. An airplane is flying at a speed of V in a direction 22◦ north of east.Which expression gives the eastward component of the plane’s velocity?

*(a) V cos 22◦ (b) V/ cos 22◦

(c) V sin 22◦ (d) V/ sin 22◦

(e) None of these

Problem 868. A car is driving at a speed of V up a hill that makes an angle of 22◦

with the horizontal. What is the vertical component of the car’s velocity?

(a) V cos 22◦ (b) V/ cos 22◦

*(c) V sin 22◦ (d) V/ sin 22◦

(e) None of these

Problem 869. An airplane is flying at a speed of 190 mph in a direction 24◦ north ofeast. What is the northward component of the plane’s velocity? Round your answer tothe nearest mph.

*(a) 77 mph (b) 101 mph(c) 161 mph (d) 175 mph(e) None of these

Solution: vy = v sin θ = (190 mph) sin 24◦ = 77 mph

Problem 870. An airplane is flying at a speed of 310 mph in a direction 77◦ north ofeast. What is the eastward component of the plane’s velocity? Round your answer tothe nearest mph.

*(a) 70 mph (b) 155 mph(c) 180 mph (d) 294 mph(e) None of these

Solution: vx = v cos θ = (310 mph) cos 77◦ = 70 mph

241

Problem 871. An airplane is flying at 214 m/s in a direction 37◦ north of east. Whatis the eastward component of its velocity? Round your answer to the nearest m/s.

(a) 129 m/s (b) 134 m/s(c) 161 m/s *(d) 171 m/s(e) None of these

Solution: vE = (214) cos 37◦ = 171

Problem 872. An airplane is flying at 220 mph in a direction 30◦ north of east. Whatis the northward component of its velocity?

Solution: You don’t need a calculator, since every good physics student knows thatsin 30◦ = 1/2. Then

vN = v sin θ = (220 mph) sin 30◦ = 110 mph

Problem 873. An airplane is flying at 220 mph in a direction 30◦ east of north. Whatis the northward component of its velocity? Round your answer to the nearest mph.Solution: vN = v cos θ = (220 mph) cos 30◦ = 191 mph

Problem 874. A hill has a slope of 0.30 radians above the horizontal. If you are drivingup that hill at 22 km/hr, how fast are you gaining elevation?

Solution: You want the vertical component of your velocity:

vy = v sin θ = (22 km/hr)(sin 0.30 rad) = 6.5 km/hr

Problem 875. A car is driving at 35 mph up a hill with a slope of 8◦ above the horizontal.What is the horizontal component of the car’s velocity? Round your answer to the nearest0.1 mph.

(a) 4.9 mph (b) 5.1 mph(c) 7.5 mph *(d) 34.7 mph(e) None of these

Solution: vx = v cos θ = (35 mph) cos 8◦ = 34.7 mph

Problem 876. A gun is fired at an elevation of 39◦ above the horizontal. The shellemerges from the muzzle at 320 m/s. What is the vertical component of the shell’svelocity? Round your answer to the nearest m/s.


Solution: vy = v sin θ = (320 m/s) sin 39◦ = 201 m/s

242

Problem 877. (Lab Problem) A spring cannon is fired at an elevation of 34◦ abovethe horizontal. The ball emerges from the cannon at 14.4 m/s. What is the horizontalcomponent of the ball’s velocity? Round your answer to the nearest 0.1 m/s.

(a) 8.1 m/s (b) 9.7 m/s*(c) 11.9 m/s (d) 12.2 m/s(e) None of these

Solution: vx = v cos θ = (14.4 m/s) cos 34◦ = 11.9 m/s

Problem 878. You are on an airplane flying upward at an angle of θ above the horizontal,at 110 m/s. You have a cold, and if you gain altitude faster than 25 m/s, your eardrumswill explode. What is the maximum safe value for θ? Round your answer to the nearestdegree.

(a) 12◦ *(b) 13◦

(c) 14◦ (d) 16◦

(e) None of these

Solution: You know v and vy; you want to know θ.

vy = v sin θ ⇒ sin θ =vyv

⇒ θ = sin−1[vyv

]= sin−1

[25 m/s

110 m/s

]= 13◦

Problem 879. (Lab Problem) A spring cannon is fired at an elevation of 34◦ abovethe horizontal. The ball emerges from the cannon at 14.4 m/s. What is the horizontalcomponent of the ball’s velocity? Round your answer to the nearest 0.1 m/s.


Solution: vx = v cos θ = (14.4 m/s) cos 34◦ = 11.9 m/s

Problem 880. An bird is flying at 20 m/s in a direction 30◦ south of west. If we take thex-axis pointing east and the y-axis pointing north, then what is the southward componentof its velocity ~v = 〈vx, vy〉? Round your answer to the nearest m/s.

(a) 10√

3 m/s (b) 10 m/s

*(c) −10√

3 m/s (d) −10 m/s(e) None of these

Solution: The tip of the vector lies in the 3rd quadrant, so the y-component (thesouthward component) is negative. vy = vS = v sin θ = −20 sin(30◦) m/s = −10 m/s

Problem 881. Your Civil War cannon has a muzzle velocity of v0 = 1220 ft/s. If it isfired at an elevation of 6◦ above the horizontal, what is the vertical component v0y of thevelocity? Round your answer to the nearest ft/s.

Solution: v0y = v0 sin θ = (1220 ft/s) sin 6◦ = 128 ft/s

243

Problem 882. An airplane is flying at 330 km/hr, in a direction 0.19 rad above thehorizontal. At what speed is the airplane gaining altitude?

Solution: We want the vertical component of the airplane’s velocity vector:

vy = v sin θ = (330 km/hr)(sin 0.19 rad) = 62 km/hr

Problem 883. A train is moving at 68 mph in a direction 22◦ north of west. What isthe train’s speed westward?

Solution: We want the westward component of the train’s velocity vector:

vW = v cos θ = (68 mph)(cos 22◦) = 63 mph

Problem 884. (Position) Pirates have buried their treasure on campus, and you haveacquired a copy of their map. (Like everyone else, pirates need to be careful about that“Reply All” button.) It tells you to start at the main entrance to the physics buildingand take 120 steps in a direction 0.38 rad north of east; from there, you should go another190 steps in a direction 0.33 rad south of east. How far from the entrance is the treasure,and in what direction?

Solution: We will call the first displacement vector ~A and the second ~B. We needto find the magnitude and direction of ~C = ~A + ~B. We begin by writing the vectors incomponent form:

~A =⟨‖ ~A‖ cos θA, ‖ ~A‖ sin θA

⟩= 〈120 · cos 0.38 rad, 120 · sin 0.38 rad〉

~B =⟨‖ ~B‖ cos θB, ‖ ~B‖ sin θB

⟩= 〈190 · cos(−0.33 rad), 190 · sin(−0.33 rad)〉

Remember that cos(−θ) = cos θ and that sin(−θ) = − sin θ. Then the components of~C = ~A+ ~B are:

CE = 120 · cos 0.38 rad + 190 · cos 0.33 rad = 291 steps

CN = 120 · sin 0.38 rad− 190 · sin 0.33 rad = −17 steps

Since CE is positive and CN is negative, the direction is in the fourth quadrant: that is,south of east. We calculate

‖~C‖ =√CE

2 + CN2 = 292 steps

θ = tan−1

∣∣∣∣CNCE∣∣∣∣ south of east = 0.06 rad south of east

Problem 885. (Position) You have discovered the Lost Belgian Mine, which runsdirectly northward into a mountainside. From the opening, you go 47 m in a direction0.28 rad below the horizontal. From there, you go another 66 m in a direction 0.19 radbelow the horizontal. At this point, what is your horizontal distance from the opening,

244

and how far are you above or below it?

Solution: We begin by finding the components of the two vectors:


⟩= 〈47 m · cos(−0.28 rad), 47 m · sin(−0.28 rad)〉


⟩= 〈66 m · cos(−0.19 rad), 66 m · sin(−0.19 rad)〉

Remember that cos(−θ) = cos θ and that sin(−θ) = − sin θ. Then the components of~C = ~A+ ~B are:

Cx = 47 m · cos 0.28 rad + 66 m · cos 0.19 rad = 110 m

Cy = −47 m · sin 0.28 rad− 66 m · sin 0.19 rad = −25 m

Your horizontal distance from the entrance is 110 m, and you are 25 m below it.

Problem 886. A goose is directly east of an airplane. The airplane is flying at a speedof 160 mph, in a direction 15◦ north of east. The goose is flying at a speed of 45 mph,in a direction θ north of west. What must θ be in order for the goose to collide with theairplane? Give your answer in degrees.

Solution: Since the goose is directly east of the airplane, its northbound componentof velocity must match the airplane’s if they are to collide. That means

(45 mph)(sin θ) = (160 mph)(sin 15◦)

Solving for θ gives us

θ = sin−1

((160 mph)(sin 15◦)

45 mph

)= 67◦

Problem 887. You are tracking a submarine by determining its position when it sendsradio messages. When you first detect the sub, it is 57 km west and 24 km north of you.When it sends its next message 37 min later, it is 51 km west and 13 km north of you.What are the magnitude and direction of the sub’s average velocity during that time?(Give the magnitude in km/hr, and the direction in radians, relative to east or west:e.g. θ rad north of east, with 0 < θ < π/2.)

Solution: Don’t forget to convert 37 min to hours. A look at the numbers showsthat the submarine is moving to the east and south. The eastbound component of thevelocity is:

vE =57 km− 51 km

37 min· 60 min

1 hr≈ 9.7 km/hr

The southbound component of the velocity is:

vS =24 km− 13 km

37 min· 60 min

1 hr≈ 17.8 km/hr

245

(We present these intermediate results because it’s a good idea to look at them and makesure that they seem reasonable. You should not use these rounded numbers in your latercalculations.)

The magnitude of the velocity is√vE2 + vS2 = 20 km/hr

The direction is

tan−1

(vSvE

)= 1.1 rad south of east

Problem 888. A cockroach has a tiny embedded radio transponder, allowing you totrack its position and velocity. Just before your lab partner attempts to stomp on it, itis moving at 8 cm/s to the east and 5 cm/s to the south. 1.3 seconds later, it is scurryingaway at 31 cm/s to the west and 19 cm/s to the south. What are the components of theroach’s average acceleration during that time?

Solution: A look at the numbers shows that the roach is accelerating westward andsouthward. The westward component of acceleration is

aW =31 cm/s− (−8 cm/s)

1.3 s= 30 cm/s2

The southward component is

aS =19 cm/s− 5 cm/s

1.3 s= 11 cm/s2

Problem 889. A car is travelling eastward along a highway at 60 mph. It passes undera hawk that is flying northward at 42 mph. From the point of view of an observer in thecar, what are the hawk’s speed and direction? Give the direction in degrees relative tonorth, e.g. θ◦ east of north.

Solution: For an observer in the car, objects fixed to the ground are moving westwardat 60 mph. To that observer, the hawk is moving westward at vW = 60 mph andnorthward at vN = 42 mph. In that frame of reference, the hawk’s speed is

v =√vW 2 + vN 2 =

√(60 mph)2 + (42 mph)2 = 73 mph

The direction, relative to north, is

tan−1

(vWvN

)= tan−1

(60 mph

42 mph

)= 55◦ west of north

246

Problem 890. You can paddle a kayak in still water at 2.6 m/s. The Santa Cruz Riveris 55 m wide, and flows northward at 1.9 m/s. If you launch your kayak on the eastbank, and point it directly westward, what will your speed and direction be relative toan observer standing on the bank? Give your direction in radians relative to west, e.g. θrad south of west.

Solution: Relative to a fixed observer on the bank, you are moving westward atvW = 2.6 m/s and northward at vN = 1.9 m/s. In that frame of reference, your speed is

v =√vW 2 + vN 2 =

√(2.6 m/s)2 + (1.9 m/s)2 = 3.2 m/s

Your direction relative to west is

tan−1

(vNvW

)= tan−1

(1.9 m/s

2.6 m/s

)= 0.63 rad north of west

The width of the river is irrelevant to the problem.

Problem 891. An airplane can fly at 410 km/hr in still air. The pilot wishes to fly itfrom Smithpur to Jonesgorod, 530 km to the north. The wind is blowing at 55 km/hrfrom west to east. In what direction should the pilot point the plane to fly directly toJonesgorod? How long will the journey take? Give your direction in radians relative tonorth, e.g. θ rad east of north.

Solution: The pilot wants to fly directly north. He must point the airplane west ofnorth, so that his still-air velocity has a westward component that offsets the 55 km/hreastward wind. If θ is the pilot’s heading west of north, then

vW = (410 km/hr) sin θ = 55 km/hr

Solving this for θ yields

θ = sin−1

(55 km/hr

410 km/hr

)= 0.13 rad west of north

The northward component of the plane’s velocity is (410 km/hr) cos θ. To fly 530 km tothe north requires a time of

t =530 km

(410 km/hr) cos θ= 1.3 hr

Problem 892. A river has a current with a speed of 5 ft/s. You can swim at 4 ft/s instill water. At what angle to the cross-river direction do you need to swim to reach apoint directly across the river from you?

Solution: No solution. Whatever the angle is, your upstream component of velocitycan be no greater than the total magnitude of your velocity: 4 ft/s. Thus you will findyourself going downstream at 1 ft/s or more.

247

Problem 893. A train is going directly north at 70 mph. An action-movie star is atopthe train, running southward at 12 mph. Relative to someone standing on the ground,how fast is the action-movie star moving, and in what direction?

Solution: Relative to someone on the ground, the AMS is moving northward at70− 12 = 58 mph.

248

6.7.2 Introduction to force vectors

Working definition: You can think of a force as a push or a pull. Since a push orpull has both magnitude and direction, it follows that a force is a vector. 1

We’ll learn more about forces later on in the course when you learn about Newton’s Laws.In particular, Newton’s 2nd Law Fnet = ma requires computing the net force (sometimescalled the resultant force) acting on an object of mass m. Since a force is a vector,the net force is just the resultant vector. In more than one spatial dimension we findthe resultant vector by breaking the individual force vectors acting on the object intocomponent vectors and then finding the net force in each of the x and y directions.

The purposes of these exercises is to give you an early exposure to the concept of break-ing forces into their respective components and summing them (keeping proper ± signconventions) to find the net force in the the x and y directions, denoted Fnet,x and Fnet,y.

Problem 894. A rope is used to pull a blockup a ramp. The ramp makes an angle of 25◦

to the horizontal; the rope makes an angle of33◦ with the ramp. The rope is pulled with aforce ~F whose magnitude is 317 N. What isthe horizontal component of ~F? Round youranswer to the nearest newton.*(a) 168 N (b) 185 N(c) 203 N (d) 224 N(e) None of these

-x

6y

!!!!

!!!!

!!!!

!!!

25◦LLL!!

LLL��7

33◦

F = 317 N

Solution: Draw a free-body diagram andadd the angles to find the net angle θ between~F and the horizontal: θ = 25◦+33◦ = 58◦. Thehorizontal component is

Fx = 317 cos 58◦ = 168 N

The vertical component is

Fy = 317 sin 58◦ = 269 N -x

6y

!!!!

!

25◦��7

~F

33◦

θ = 58◦

1Actually, this is not obvious and must be proven mathematically.

249


to the horizontal; the rope makes an angle of33◦ with the ramp. The rope is pulled witha force ~F whose magnitude is 317 N. Whatis the vertical component of ~F? Round youranswer to the nearest newton.(a) 218 N (b) 242 N*(c) 269 N (d) 296 N(e) None of these -

x

6y

!!!!

!!!!

!!!!

!!!

25◦LLL!!

LLL��7

33◦

F = 317 N


Fx = 317 cos 58◦ = 168 N


Fy = 317 sin 58◦ = 269 N -x

6y

!!!!

!

25◦��7

~F

33◦

θ = 58◦


to the horizontal; the rope makes an angle of31◦ with the ramp. The rope is pulled with aforce ~F whose magnitude is 428 N. What isthe horizontal component of ~F? Round youranswer to the nearest newton.(a) 209 N (b) 232 N*(c) 258 N (d) 283 N(e) None of these

-x

6y

!!!!

!!!!

!!!!

!!!

22◦LLL!!

LLL��7

31◦

F = 428 N


Fx = 428 cos 53◦ = 258 N


Fy = 428 sin 53◦ = 342 N -x

6y

!!!!

!

22◦��7

~F

31◦

θ = 53◦

250

Problem 897. Two forces ~F1 and ~F2 are acting on aparticle, as shown at right. ~F1 makes an angle of 35◦

with the x-axis and has magnitude 87 N. ~F2 makesan angle of 18◦ with the y-axis and has magnitude181 N. Use the component method to find the netforce in the x-direction. Round your answer to thenearest newton. (The picture is not drawn to scale.)

(a) 12 N (b) 13 N*(c) 15 N (d) 16 N

y

xu��

�3F1 = 87 N

35◦

F1x

F1y

AAAAAAAAK

F2 = 181 N

18◦F2y

F2x

Solution: Let ~F = ~F1 + ~F2. Then

Fx = F1x + F2x = (87 N) cos 35◦ − (181 N) sin 18◦ = 15 N


with the x-axis and has magnitude 87 N. ~F2 makesan angle of 18◦ with the y-axis and has magnitude181 N. Use the component method to find the netforce in the y-direction. Round your answer to thenearest newton. (The picture is not drawn to scale.)

(a) 200 N *(b) 222 N(c) 244 N (d) 269 N

y

xu��

�3F1 = 87 N

35◦

F1x

F1y

AAAAAAAAK

F2 = 181 N

18◦F2y

F2x


Fy = F1y + F2y = (87 N) sin 35◦ + (181 N) cos 18◦ = 222 N

251


with the x-axis and has magnitude 87 N. ~F2 makesan angle of 18◦ with the y-axis and has magnitude181 N. Find the magnitude of the net force on theparticle. Round your answer to the nearest newton.(The picture is not drawn to scale.)

(a) 200 N *(b) 223 N(c) 245 N (d) 269 N

y

xu��

�3F1 = 87 N

35◦

F1x

F1y

AAAAAAAAK

F2 = 181 N

18◦F2y

F2x


Fx = F1x + F2x = (87 N) cos 35◦ − (181 N) sin 18◦

Fy = F1y + F2y = (87 N) sin 35◦ + (181 N) cos 18◦

F =√F 2x + F 2

y

=[((87 N) cos 35◦ − (181 N) sin 18◦)2 + ((87 N) sin 35◦ + (181 N) cos 18◦)2

]1/2= 223 N


with the x-axis and has magnitude 87 N. ~F2 makesan angle of 18◦ with the y-axis and has magnitude181 N. If ~Fnet is the net force on the particle, findthe angle that ~Fnet makes with the x-axis. Roundyour answer to the nearest degree. (The picture isnot drawn to scale.)

(a) 83◦ *(b) 86◦

(c) 89◦ (d) 92◦

y

xu��

�3F1 = 87 N

35◦

F1x

F1y

AAAAAAAAK

F2 = 181 N

18◦F2y

F2x


Fx = F1x + F2x = (87 N) cos 35◦ − (181 N) sin 18◦ ≈ 15 N

Fy = F1y + F2y = (87 N) sin 35◦ + (181 N) cos 18◦ ≈ 222 N

Since Fx and Fy are both positive, ~F is in the first quadrant. Hence

θ = tan−1

(FyFx

)= tan−1

((87 N) sin 35◦ + (181 N) cos 18◦

(87 N) cos 35◦ − (181 N) sin 18◦

)= 86◦

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Problem 901. You are pulling a crate of physics books down the hallway, using a ropethat makes an angle of 34◦ to the horizontal. If you are pulling on the rope with a force of440 N, what is the horizontal component of the force? Round your answer to the nearestnewton.

(a) 373 N (b) 233 N*(c) 365 N (d) 246 N(e) None of these

Solution: The horizontal component of the force is

Fh = F cos θ = (440 N) cos 34◦ = 365 N

Problem 902. You are pulling a crate of physics books down the hallway, using a ropethat makes an angle of 26◦ to the horizontal. If you are pulling on the rope with a force of290 N, what is the horizontal component of the force? Round your answer to the nearestnewton.

(a) 127 N (b) 188 N(c) 221 N *(d) 261 N(e) None of these

Solution: The horizontal component of the force is

Fh = F cos θ = (290 N) cos 26◦ = 261 N

Problem 903. Your car is stuck in a muddy road, and you and a friend are trying topull it out with ropes. You are pulling with a force of 820 N in a direction 0.45 rad eastof north; your friend is pulling with a force of 1040 N in a direction 0.32 rad west ofnorth. What is the northward force on the car?

Solution: We will use a coordinate system in which north is the positive x-directionand west is the positive y-direction. In that system, the angle at which you are pullingis −0.45 rad; the angle at which your friend is pulling is 0.32 rad.

We will need to write the two force vectors in component form, then find their sum, thenfind the magnitude and direction of that sum. The two force vectors are:


⟩= 〈820 N · cos(−0.45) rad, 820 N · sin(−0.45) rad〉


⟩= 〈1040 N · cos 0.32 rad, 1040 N · sin 0.32 rad〉

Remember that cos(−θ) = cos θ. Then the northerly component of ~C = ~A+ ~B is:

CN = 820 N · cos 0.45 rad + 1040 N · cos 0.32 rad = 1730 N

We have rounded this number to 10 N, since that appears to be the accuracy of theoriginal data.

253

Problem 904. A boat is held in the middle of a river by two ropes, one going to eitherbank. Each rope makes an angle of 79◦ to the upstream direction of the river. Thecurrent exerts a downstream force of 550 lbs on the boat. How much force does eachrope have to exert to hold the boat in place? (It is the same for both ropes.)

Solution: There are two ropes, each exerting a force with a magnitude of F . Theupstream component of each force vector is F cos 79◦. The two ropes working togethermust offset the downstream force of the current:

2F cos 79◦ = 550 lbs

Solving this for F give us

F =550 lbs

2 cos 79◦= 1440 lbs

Problem 905. You are riding your bicycle straight north. The wind is exerting a forceof 48 lbs on you, in a direction 39◦ east of south. What is the southward component ofthe wind’s force?

Solution: The southward component is

FS = F cos θ = (48 lbs)(cos 39◦) = 37 lbs

Problem 906. A dietitian is pulling a crate of romaine lettuce, using a rope that makesan angle of 0.41 rad to the horizontal. To move the crate requires a horizontal force of240 N. What is the force with which the dietitian needs to pull on the rope?

Solution: The number we want is F , the magnitude of the force vector. The horizontalforce is

Fx = F cos θ = F (cos 0.41 rad) = 240 N

Solving this for F give us

F =Fx

cos θ=

240 N

cos 0.41 rad= 262 N

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Part III

Kinematics

255

7 One-dimensional linear kinematics

7.1 Concept questions: kinematics

Problem 907. Is it possible for an object to have nonzero velocity and zero speed? Givean example, or explain why it’s not possible.

Solution: No. Speed is the magnitude of the velocity vector; and only the zero vectorhas a magnitude of zero.

Problem 908. Is it possible for an object to have nonzero speed and zero velocity? Givean example, or explain why it’s not possible.

Solution: No. Speed is the magnitude of the velocity vector; and the zero vector hasa magnitude of zero.

Problem 909. Is it possible for an object to have zero average velocity and nonzeroaverage speed? Give an example, or explain why it’s not possible.

Solution: Yes. Suppose you walk 100 ft north at 4 mph, then turn around and walk100 ft south at 4 mph, returning to your starting point. Your average speed is 4 mph,since that’s the speed at which you were walking the whole time. (We’ll neglect the veryshort time it took you to turn around.) However, your average velocity is zero, becauseyou returned exactly to your starting point.

Problem 910. Is it possible for an object to have zero average speed and nonzero averagevelocity? Give an example, or explain why it’s not possible.

Solution: No. Speed is always greater than or equal to zero, so the only way to havean average speed of zero is for the speed to be zero the whole time. That means that thevelocity must also be zero the whole time; so the average velocity must be zero.

Problem 911. You are given a ticket for driving at 37 mph in a school zone with aspeed limit of 15 mph. At your hearing, you inform the judge that since you were goingin reverse, your speed was −37 mph; and since −37 < 15, you were going below thespeed limit and should not be penalized. Unfortunately for you, the policeman has takenphysics. What does he tell the judge?

Solution: The policeman informs the judge that your velocity vector may have beennegative; but that speed, the magnitude of the velocity vector, must always be greaterthan or equal to zero. Thus although your velocity was negative, your speed was +37mph, and greater than the legal limit.

256

Problem 912. You drive in a straight line from your home to your physics classroom.Your driveway is 30 m long; you drive down it at 5 m/s. You then drive 300 m downSmith Street at 15 m/s. After that, you drive 4.8 km down Jones Road to your classroomat 30 m/s. Once you get there, you realize that it’s Saturday, and that you don’t haveto come to class. You turn around and return home along the same route: at 40 m/salong Jones Road; at 20 m/s along Smith Street; and at 10 m/s up your driveway, backto where you started. What is your average velocity for the round trip?

Solution: Remember the distinction between speed and velocity. You returned toyour starting point; so your net displacement for the whole trip is zero; so your averagevelocity is zero.

If we’d asked for your average speed, this problem would have been a lot more work.

Problem 913. Assume that the direction in which a car faces is positive. Give anexample of a situation in which the car has:

(a) positive velocity and positive acceleration(b) positive velocity and negative acceleration(c) negative velocity and positive acceleration(d) negative velocity and negative acceleration(e) positive velocity and zero acceleration(f) zero velocity and negative acceleration

Solution:(a) The car is moving forward and speeding up.(b) The car is moving forward, but the driver is braking to slow down.(c) The car is moving backward and slowing down. Remember that “positive accelera-tion” means accelerating in the positive direction; it doesn’t mean “speeding up, regard-less of the direction you’re going”.(d) The car is moving backward and speeding up. Remember that “negative accelera-tion” means accelerating in the negative direction, not “slowing down, regardless of thedirection you’re going”.(e) The car is moving forward at a constant speed.(f) This is a little more complicated. The car was driving up a hill when the engine died;the car coasted forward until it stopped for an instant, then started rolling backward. Atthe instant that it stopped, the velocity was zero; during the whole time after the enginedied, the acceleration was negative.

We could not have come up with a situation in which the velocity was zero and theacceleration was nonzero for any positive amount of time. Such a situation can only lastfor an instant. The reason for this is that acceleration is the change in velocity, and ifvelocity is not changing (because it’s zero!), then the acceleration must also be zero.

Here’s perhaps a better example not involving a car. Throw a ball straight up into the air.The acceleration that the ball experiences is due to the gravitational force between theball and the earth. If we take the positive y-direction as up, then the acceleration alwayspoints downward towards the earth in the negative y- direction. When the ball reaches

257

its maximum height it must stop, hence it has zero velocity, before it turns around andfalls back to earth. At no point along the balls trajectory did the earth quit pulling onthe ball, and so the ball’s acceleration never vanished. In fact, it remained negative andconstant.

Problem 914. You and your lab partner are designing a fountain for the front of thephysics building. You have a pressurized tank with a nozzle that squirts water out at 5.2m/s. You claim that if that nozzle is pointed upward, it will produce a beautiful sprayof water. Your lab partner disagrees: he says that since the acceleration due to gravityis 9.8, and 5.2 < 9.8, the water will barely dribble out of the nozzle and run back downthe sides of the pipe. Is he right? Why or why not?

Solution: Your lab partner is wrong. He is trying to compare two quantities thathave different dimensions: v0 = 5.2 m/s, and g = 9.8 m/s2. You can only write equationsand inequalities when all terms have the same dimensions.

Problem 915. A car is being driven erratically along a straight stretch of highway. Thegraph below shows its position x as a function of time t. For each interval betweenconsecutive points (A-B, B-C, etc.), describe whether the velocity of the car is positive,negative, or zero; and whether the acceleration is positive, negative, or zero.

-

6

t

x

sA��

��sB

sC��sD

sE

sF

sGAAAAAAsH

Solution: We will do these out of order, starting with the easiest one:

E-F: The position of the car does not change during this time. Therefore the car has avelocity of zero. Since the velocity does not change during the time period, the acceler-ation is also zero.

A-B: The position as a function of time is a straight line. That means that the car ismoving at a constant velocity. Since the value of x increases with time, the velocity ispositive. Since the velocity is constant, the acceleration is zero.

258

C-D: Again, the position as a function of time is a straight line, so the velocity is con-stant; the value of x increases with time, so the velocity is positive; and since the velocityis constant, the acceleration is zero. Notice that the slope of the line from C to D isgreater than the slope from A to B. That means that the velocity between C and D isgreater than the velocity between A and B.

G-H: The position as a function of time is a straight line, so the velocity is constant. Thevalue of x decreases with time, so the velocity is negative. Since the velocity is constant,the acceleration is zero.

B-C: The position as a function of time is not a straight line, so the velocity is notconstant. That means some kind of acceleration is occurring. The value of x increaseswith time over the interval, so the velocity is positive. The velocity increases: since theslope from C to D is greater than the slope from A to B, the CD velocity is greater thanthe AB velocity. Hence the acceleration is positive.

D-E: The position as a function of time is not a straight line, so the velocity is notconstant. That means that there is a nonzero acceleration. The value of x increaseswith time, so the velocity is positive. The velocity decreases from the positive value oninterval CD to the zero velocity on interval EF; so the acceleration is negative.

F-G: The position as a function of time is not a straight line, so the velocity is notconstant. That means that there is a nonzero acceleration. The value of x decreaseswith time, so the velocity is negative. The velocity decreases from zero on interval EF tosome negative value on interval GH, so there is negative acceleration. (Again, rememberthat “negative acceleration” doesn’t mean “slowing down”; it means accelerating in thenegative direction, which can mean slowing down while moving forward or speeding upwhile moving backward.)

7.2 Qualitative kinematics: descriptions

Problem 916. A car is facing in the positive direction. It is moving forward at 65 mph;the driver has just seen a deer on the road and applied the brakes. Which of the followingdescribes the car’s situation?

(a) positive velocity; positive acceleration(b) positive velocity; zero acceleration*(c) positive velocity; negative acceleration(d) negative velocity; positive acceleration(e) none of these

Solution: The car is moving forward, so the velocity is positive; it is slowing down(that is, accelerating in the backward direction), so the acceleration is negative.

259

Problem 917. A car is facing in the positive direction. It is rolling backward down ahill, and its speed is increasing. Which of the following describes the car’s situation?

(a) positive velocity; positive acceleration(b) positive velocity; negative acceleration(c) negative velocity; positive acceleration*(d) negative velocity; negative acceleration(e) none of these

Solution: The car is moving backward, so the velocity is negative. It is speeding up inthe backward direction, so the acceleration is negative. Careful! “Negative acceleration”and “slowing down” are not the same thing. In this case, negative acceleration meansspeeding up in the negative direction.

Problem 918. A car is facing in the positive direction. It is rolling backward downa hill; the drive has just applied the brakes to slow it down. Which of the followingdescribes the car’s situation?

(a) positive velocity; zero acceleration(b) positive velocity; negative acceleration*(c) negative velocity; positive acceleration(d) negative velocity; negative acceleration(e) none of these

Solution: The car is moving backward, so the velocity is negative. It is slowing down(that is, its velocity is becoming more positive), so the acceleration is positive. “Positiveacceleration” is not the same thing as “speeding up”.

Problem 919. A car is facing in the positive direction. It is moving forward, and thedriver is speeding up to merge onto the freeway. Which of the following describes thecar’s situation?

*(a) positive velocity; positive acceleration(b) positive velocity; zero acceleration(c) zero velocity; positive acceleration(d) negative velocity; zero acceleration(e) none of these

Solution: The car is moving forward, so the velocity is positive. It is speeding up inthe forward direction, so the acceleration is positive.

260

Problem 920. A car is facing in the positive direction. The driver is going at a steadyspeed of 55 mph on a highway. Which of the following describes the car’s situation?

*(a) positive velocity; zero acceleration(b) positive velocity; negative acceleration(c) positive velocity; positive acceleration(d) zero velocity; positive acceleration(e) none of these

Solution: The car is moving forward, so the velocity is positive. It is moving at aconstant speed, so the acceleration is zero.

Problem 921. A car is facing in the positive direction. It was going up a hill when theengine failed; the car slowed down as it rolled up the hill, then started rolling backward.Which of the following describes the car’s situation at the point where it started rollingbackward?

(a) positive velocity; positive acceleration(b) negative velocity; negative acceleration(c) zero velocity; positive acceleration*(d) zero velocity; negative acceleration(e) none of these

Solution: At the moment that the car stopped rolling forward and started rollingbackward, its velocity was zero. Since its speed was changing from forward to backward,the acceleration was negative.

Problem 922. In this problem, use upward as the positive direction. A rocket is flyingstraight upward; its speed increases as it rises. Which of the following describes therocket’s situation?

*(a) positive velocity; positive acceleration(b) positive velocity; negative acceleration(c) zero velocity; positive acceleration(d) zero velocity; negative acceleration(e) none of these

Solution: The rocket is moving upward, so its velocity is positive. It is speeding upin the upward direction, so its acceleration is positive.

261

Problem 923. In this problem, use downward as the positive direction. A ball has beendropped from the top of a building; it has not yet reached the ground. Which of thefollowing describes the ball’s situation?

*(a) positive velocity; positive acceleration(b) positive velocity; negative acceleration(c) zero velocity; negative acceleration(d) negative velocity; positive acceleration(e) none of these

Solution: The ball is moving downward, so the velocity is positive. It is acceleratingdownward under the influence of gravity, so the acceleration is negative.

Problem 924. In this problem, use downward as the positive direction. A ball has beendropped from the top of a building, has hit the ground and bounced, and is now movingupward. Which of the following describes the ball’s situation?

(a) positive velocity; positive acceleration(b) positive velocity; negative acceleration(c) positive velocity; zero acceleration*(d) negative velocity; positive acceleration(e) none of these

Solution: The ball is moving upward (the negative direction), so the velocity isnegative. Gravity is slowing it down as it moves upward, so the acceleration is positive.

Problem 925. In this problem, use downward as the positive direction. An elevatorhas just left the 39th floor, heading downward and speeding up as it goes. Which of thefollowing describes the elevator’s situation?

*(a) positive velocity; positive acceleration(b) positive velocity; negative acceleration(c) negative velocity; negative acceleration(d) negative velocity; positive acceleration(e) none of these

Solution: The elevator is moving downward (the positive direction), so its velocity ispositive. It is speeding up in the downward direction, so its acceleration is positive.

262

Problem 926. In this problem, use downward as the positive direction. An elevator isgoing down from the 39th floor to the 3rd. It is now at the level of the fourth floor, and isslowing down as it approaches the third. Which of the following describes the elevator’ssituation?

(a) positive velocity; positive acceleration*(b) positive velocity; negative acceleration(c) negative velocity; negative acceleration(d) negative velocity; positive acceleration(e) none of these

Solution: The elevator is moving downward (the positive direction), so the velocityis positive. It is slowing down as it moves downward, so the acceleration is negative.

Problem 927. In this problem, use downward as the positive direction. An elevator ison its way from the 89th floor to the 4th. It has reached its maximum downward speed,and has not yet started slowing down. Which of the following describes the elevator’ssituation?

(a) positive velocity; negative acceleration*(b) positive velocity; zero acceleration(c) negative velocity; negative acceleration(d) negative velocity; positive acceleration(e) none of these

Solution: The elevator is moving downward (the positive direction), so the velocityis positive. It is moving at a constant speed, so the acceleration is zero.

Problem 928. In this problem, use downward as the positive direction. An elevator hasjust left the 3rd floor, heading upward and speeding up as it goes. Which of the followingdescribes the elevator’s situation?

(a) positive velocity; positive acceleration(b) positive velocity; negative acceleration*(c) negative velocity; negative acceleration(d) negative velocity; zero acceleration(e) none of these

Solution: The elevator is moving upward (the negative direction), so its velocity isnegative. It is speeding up in the negative direction, so its acceleration is negative.

263

Problem 929. If the upward direction is positive, which situation involves positivevelocity and positive acceleration?

(a) An elevator moving upward at a constant speed(b) An elevator speeding up as it moves downward*(c) An elevator speeding up as it moves upward(d) An elevator slowing down as it moves downward(e) none of these

Solution: For velocity to be positive, the elevator must be moving upward. Foracceleration to be positive, it must be speeding up as it goes upward.

Problem 930. If the upward direction is positive, which situation involves negativevelocity and negative acceleration?

(a) An elevator moving upward at a constant speed*(b) An elevator speeding up as it moves downward(c) An elevator speeding up as it moves upward(d) An elevator slowing down as it moves downward(e) none of these

Solution: In order for velocity to be negative, the elevator must be moving downward.In order for acceleration to be negative, it must be speeding up in the downward direction.

Problem 931. If the upward direction is positive, which situation involves positivevelocity and zero acceleration?

*(a) An elevator moving upward at a constant speed(b) An elevator speeding up as it moves downward(c) An elevator speeding up as it moves upward(d) An elevator slowing down as it moves downward(e) none of these

Solution: In order for velocity to be positive, the elevator must be moving upward.In order for acceleration to be zero, it must be moving at a constant speed.

Problem 932. If the upward direction is positive, which situation involves negativevelocity and positive acceleration?

(a) An elevator moving upward at a constant speed(b) An elevator speeding up as it moves downward(c) An elevator speeding up as it moves upward*(d) An elevator slowing down as it moves downward(e) none of these

Solution: In order for velocity to be negative, the elevator must be moving downward.In order for acceleration to be positive, it must be slowing down as it moves downward.

264

Problem 933. If north is the positive direction, which situation involves positive velocityand negative acceleration?

(a) A car speeding up as it moves southward(b) A car moving southward at a constant speed*(c) A car slowing down as it moves northward(d) A car slowing down as it moves southward(e) none of these

Solution: In order for the velocity to be positive, the car must be moving northward.In order for the acceleration to be negative, it must be slowing down as it moves in thepositive direction.

Problem 934. If north is the positive direction, which situation involves negative ve-locity and negative acceleration?

*(a) A car speeding up as it moves southward(b) A car moving southward at a constant speed(c) A car slowing down as it moves northward(d) A car slowing down as it moves southward(e) none of these

Solution: In order for velocity to be negative, the car must be moving southward.In order for acceleration to be negative, the car must be speeding up as it moves in thenegative direction.

Problem 935. If north is the positive direction, which situation involves negative ve-locity and positive acceleration?

(a) A car speeding up as it moves southward(b) A car moving southward at a constant speed(c) A car slowing down as it moves northward*(d) A car slowing down as it moves southward(e) none of these

Solution: In order for the velocity to be negative, the car must be moving southward.In order for the acceleration to be positive, it must be slowing down as it moves in thenegative direction.

Problem 936. If north is the positive direction, which situation involves negative ve-locity and zero acceleration?

(a) A car speeding up as it moves southward*(b) A car moving southward at a constant speed(c) A car slowing down as it moves northward(d) A car slowing down as it moves southward(e) none of these

Solution: In order for the velocity to be negative, the car must be moving southward.In order for acceleration to be zero, it must be moving at a constant speed.

265

Problem 937. If the positive direction is downward, which situation involves negativevelocity and negative acceleration?

(a) A rocket slowing down as it moves upward(b) A rocket moving downward at a constant speed(c) A rocket moving upward at a constant speed*(d) A rocket speeding up as it moves upward(e) none of these

Solution: In order for the velocity to be negative, the rocket must be moving upward.In order for the acceleration to be negative, it must be speeding up as it moves in thenegative direction.

Problem 938. If the positive direction is downward, which situation involves positivevelocity and zero acceleration?

(a) A rocket slowing down as it moves upward*(b) A rocket moving downward at a constant speed(c) A rocket moving upward at a constant speed(d) A rocket speeding up as it moves upward(e) none of these

Solution: In order for the velocity to be positive, the rocket must be moving down-ward. In order for the acceleration to be zero, it must be moving at a constant speed.

Problem 939. If the positive direction is downward, which situation involves negativevelocity and zero acceleration?

(a) A rocket slowing down as it moves upward(b) A rocket moving downward at a constant speed*(c) A rocket moving upward at a constant speed(d) A rocket speeding up as it moves upward(e) none of these

Solution: In order for the velocity to be negative, the rocket must be moving upward.In order for the acceleration to be zero, it must be moving at a constant speed.

Problem 940. If the positive direction is downward, which situation involves negativevelocity and positive acceleration?

*(a) A rocket slowing down as it moves upward(b) A rocket moving downward at a constant speed(c) A rocket moving upward at a constant speed(d) A rocket speeding up as it moves upward(e) none of these

Solution: In order for the velocity to be negative, the rocket must be moving upward.In order for the acceleration to be positive, it must be slowing down as it moves in thenegative direction.

266

7.3 Qualitative kinematics: from graph

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Figure 1: A car is being driven erratically along a straight stretch of highway. The graphshows its position x as a function of the time t.

Problem 941. For the interval between points A and B in figure 1, the car’s velocity is:

*(a) positive (b) negative (c) zero

Solution: Since x increases between A and B, the velocity is positive.

Problem 942. For the interval between points B and C in figure 1, the car’s velocity is:


Solution: Since x increases between B and C, the velocity is positive.

Problem 943. For the interval between points C and D in figure 1, the car’s velocity is:


Solution: Since x increases between C and D, the velocity is positive.

Problem 944. For the interval between points D and E in figure 1, the car’s velocity is:


Solution: Since x increases between D and E, the velocity is positive.

Problem 945. For the interval between points E and F in figure 1, the car’s velocity is:

(a) positive (b) negative *(c) zero

Solution: Since x does not change between E and F, the velocity is zero.

Problem 946. For the interval between points F and G in figure 1, the car’s velocity is:

(a) positive *(b) negative (c) zero

Solution: Since x decreases between F and G, the velocity is negative.

267

Problem 947. For the interval between points G and H in figure 1, the car’s velocity is:


Solution: Since x decreases between G and H, the velocity is negative.

Problem 948. For the interval between points A and B in figure 1, the car’s accelerationis:


Solution: The graph of x versus t is a straight line between A and B; so the velocityis constant. That means that acceleration is zero.

Problem 949. For the interval between points B and C in figure 1, the car’s accelerationis:


Solution: The slope of the graph is greater between C and D than it is between A andB. This means that the velocity between C and D is greater than the velocity betweenA and B; so positive acceleration occurs between B and C.

Problem 950. For the interval between points C and D in figure 1, the car’s accelerationis:


Solution: The graph of x versus t is a straight line between C and D; so the velocityis constant. Thus the acceleration is zero.

Problem 951. For the interval between points D and E in figure 1, the car’s accelerationis:


Solution: The velocity between C and D is positive; the velocity between E andF is zero. Thus during the interval between D and E, the velocity decreases; so theacceleration is negative.

Problem 952. For the interval between points E and F in figure 1, the car’s accelerationis:


Solution: The velocity is constant between E and F, so the acceleration is zero.

Problem 953. For the interval between points F and G in figure 1, the car’s accelerationis:


Solution: The velocity between E and F is zero. The velocity between G and H isnegative. Hence negative acceleration occurs between F and G.

268

Problem 954. For the interval between points G and H in figure 1, the car’s accelerationis:


Solution: The graph of x versus t is a straight line between G and H; so the velocityis constant. Hence the acceleration is zero.

-

6

t

x

sA

sB s

C@@@sD s

E

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sH

Figure 2: A car is being driven erratically along a straight stretch of highway. The graphshows its position x as a function of the time t.

Problem 955. For the interval between points A and B in figure 2, the car’s velocity is:


Solution: The value of x does not change between A and B; so the velocity is zero.

Problem 956. For the interval between points B and C in figure 2, the car’s velocity is:


Solution: Since x decreases between B and C, the velocity is negative.

Problem 957. For the interval between points C and D in figure 2, the car’s velocity is:


Solution: Since x decreases between C and D, the velocity is negative.

Problem 958. For the interval between points D and E in figure 2, the car’s velocity is:


Solution: Since x decreases between D and E, the velocity is negative.

Problem 959. For the interval between points E and F in figure 2, the car’s velocity is:


Solution: Since x increases between E and F, the velocity is positive.

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Problem 960. For the interval between points F and G in figure 2, the car’s velocity is:


Solution: Since x increases between F and G, the velocity is positive.

Problem 961. For the interval between points G and H in figure 2, the car’s velocity is:


Solution: Since x increases between G and H, the velocity is positive.

Problem 962. For the interval between points A and B in figure 2, the car’s accelerationis:


Solution: The velocity is constant between A and B; so the acceleration is zero.

Problem 963. For the interval between points B and C in figure 2, the car’s accelerationis:


Solution: The velocity is zero between A and B, and negative between C and D.Hence negative acceleration occurs between B and C.

Problem 964. For the interval between points C and D in figure 2, the car’s accelerationis:


Solution: The graph of x versus t is a straight line between C and D, so the velocityis constant. Thus the acceleration is zero.

Problem 965. For the interval between points D and E in figure 2, the car’s accelerationis:


Solution: The velocity is negative between C and D, and positive between E and F.Thus positive acceleration occurs between D and E.

Problem 966. For the interval between points E and F in figure 2, the car’s accelerationis:


Solution: The velocity is negative between D and E, and positive between F and G.Thus positive acceleration occurs between E and F.

Problem 967. For the interval between points F and G in figure 2, the car’s accelerationis:


Solution: The graph of x versus t is a straight line between F and G, so the velocityis constant. Hence the acceleration is zero.

270

Problem 968. For the interval between points G and H in figure 2, the car’s accelerationis:


Solution: The slope of the line of x versus t decreases between G and H; so thevelocity decreases; so negative acceleration occurs.

271

7.4 Introducing the fundamental one-dimensional kinematic equa-tions

Assumptions: Unless instructed otherwise, assume that air resistance does not occur;and that projectiles (cannonballs, golf balls, arrows, etc.) are launched from ground level.

All of the following problems in this problem set can be solved using one of the followingthree fundamental equations:

In practice there are three different sets of subscripts used on physical variables:

1. (subscripts: i for initial values, f for final values) Example: initial position xi; finalposition xf

2. (subscripts: zero (pronounced nought) for initial values, no subscript for final val-ues) Example: initial position x0; final position x

3. (subscripts: 1 for initial values, 2 for final values) Example: initial position x1; finalposition x2

The general fundamental kinematic equations in the first type of subscript notation are:

Fundamental Equation 1: vf = vi + a(tf − ti)

Fundamental Equation 2: v2f = v2

i + 2a(xf − xi)

Fundamental Equation 3: xf = xi + vi(tf − ti) +1

2a(tf − ti)2

where a is acceleration, v is velocity, x is position, t is time, and the subscript i and fdenotes initial and final values, respectively. Typically, we take the initial time ti = 0since, in practice, when you time an experiment you always start with the stopwatch setto zero.

Warning: These equations are only valid if the acceleration is constant. Assume accel-eration is constant unless stated otherwise.

The general fundamental kinematic equations in the second type of subscript notationare:

272

Fundamental formulas for motion in 1-D

Equation 1: v = v0 + at (use when: position not in problem statement)

Equation 2: v2 = v20 + 2a(x− x0) (use when: time not in problem statement)

Equation 3: x = x0 + v0t+1

2at2 (use when: want position as a fcn of time)

The general fundamental kinematic equations in the third type of subscript notation are:

Fundamental formulas for motion in 1-D

Equation 1: v2 = v1 + at2

Equation 2: v22 = v2

1 + 2a(x2 − x1)

Equation 3: x2 = x1 + v1t2 +1

2at22

This final form has the potential of leading to much confusion. For this reason we willnot use this form. But you should be aware that it can be found in textbooks everywhere.

7.4.1 Dimensional consistency of the fundamental equations

Problem 969. Verify that the 1st fundamental equation is dimensionally consistent.

Solution: Start by rewriting the equation as ∆v = a∆t. Taking the units of theright-hand side: [a∆t] = (L/T2)T]= L/T, the units for velocity.

Problem 970. Verify that the 2nd fundamental equation is dimensionally consistent.

Solution: Start by rewriting the equation as v2f − v2

i = 2a∆x. Taking the units of theright and left-hand side yields [v2

f − v2i ] = [2a∆x]], which leads to (L/T)2 = (L/T2)L.

Problem 971. Verify that the 3rd fundamental equation is dimensionally consistent.

Solution: Start by rewriting the equation as

∆x = vi∆t+1

2a(∆t)2 ÷∆t−−−−−−→ ∆x

∆t= vi +

1

2a∆t

As we’ve already seen, a∆t has units of velocity. Thus the equation is dimensionally selfconsistent.

273

7.5 Quantitative kinematics: horizontal

Problem 972. Beginning at rest, a car accelerates down a straight road at 1.03 m/s2

for 6.7 s. What is the car’s final speed? Round your answer to the nearest 0.1 m/s.

*(a) 6.9 m/s (b) 7.6 m/s(c) 8.4 m/s (d) 9.2 m/s(e) None of these

Solution: We know acceleration and time (and initial velocity, which is zero); wewant the final velocity.

v = v0 + at = 0 m/s + (1.03 m/s2)(6.7 s) = 6.9 m/s

Problem 973. Beginning at rest, a car accelerates down a straight road at 2.18 m/s2

for 14.8 s. What is the car’s final speed? Round your answer to the nearest 0.1 m/s.

(a) 6.8 m/s *(b) 32.3 m/s(c) 35.2 m/s (d) 70.3 m/s(e) None of these

Solution: We know acceleration and time (and initial velocity, which is zero); wewant the final velocity.

v = v0 + at = 0 m/s + (2.18 m/s2)(14.8 s) = 32.3 m/s

Problem 974. A boat can accelerate at 2.9 m/s2. If it begins at rest, how long must itaccelerate before it reaches a speed of 14.4 m/s? Round your answer to the nearest 0.1s.

*(a) 5.0 s (b) 5.5 s(c) 6.0 s (d) 6.6 s(e) None of these

Solution: We know the initial velocity (which is zero), the acceleration, and the finalvelocity. We want to know the time.

v = v0 + at = at ⇒ t =v

a=

14.4 m/s

2.9 m/s2 = 5.0 s

274

Problem 975. A boat is initially going northward at 7.4 m/s. It accelerates at 0.36m/s2 northward for 22 s. At the end of this time, how far north has it travelled? Roundyour answer to the nearest 10 m.


Solution: We know the initial position (zero), the initial velocity, the acceleration,and the time. We want to know the final position.

x = x0 + v0t+1

2at2 = (7.4 m/s)(22 s) +

1

2(0.36 m/s2)(22 s)2 = 250 m

Problem 976. A car can accelerate at 1.31 m/s2. If it begins at rest, how long must itaccelerate before it reaches a speed of 20.4 m/s? Round your answer to the nearest 0.1s.

*(a) 15.6 s *(b) 17.1 s(c) 18.8 s (d) 20.7 s(e) None of these

Solution: We know the initial velocity (which is zero), the acceleration, and the finalvelocity. We want to know the time.

v = v0 + at = at ⇒ t =v

a=

20.4 m/s

1.31 m/s2 = 15.6 s

Problem 977. Beginning at rest, a car accelerates forward for 5.2 s, reaching a speed of15.0 m/s. What is the car’s acceleration? Round your answer to the nearest 0.1 m/s2.

(a) 2.6 m/s2 *(b) 2.9 m/s2

(c) 3.2 m/s2 (d) 3.5 m/s2

(e) None of these

Solution: We know the initial speed (which is zero), the time, and the final speed.We want to know the acceleration.

v = v0 + at = at ⇒ a =v

t=

15.0 m/s

5.2 s= 2.9 m/s2

275

Problem 978. A ship is initially moving northward at 12.1 m/s. It accelerates northwardat 1.91 m/s2 for 4.9 s. What is its final speed? Round your answer to the nearest 0.1m/s.


Solution: We know the initial velocity, the acceleration, and the time. We want toknow the final velocity.

v = v0 + at = 12.1 m/s + (1.91 m/s2)(4.9 s) = 21.5 m/s

Problem 979. A car is initially moving northward at 5.9 m/s. It accelerates northwardat 0.88 m/s2 for 12.1 s. What is its final speed? Round your answer to the nearest 0.1m/s.


Solution: We know the initial velocity, the acceleration, and the time. We want toknow the final velocity.

v = v0 + at = 5.9 m/s + (0.88 m/s2)(12.1 s) = 16.5 m/s

Problem 980. A ship is initially moving eastward at 4.2 m/s. It accelerates over thecourse of 27 s, at the end of which it is moving eastward at 10.3 m/s. What is the ship’sacceleration? Round your answer to the nearest 0.01 m/s2.

(a) 0.20 m/s2 *(b) 0.23 m/s2

(c) 0.25 m/s2 (d) 0.27 m/s2

(e) None of these

Solution: We know the initial velocity, the final velocity, and the time. We want toknow the acceleration.

v = v0 + at ⇒ a =v − v0

t=

10.3 m/s− 4.2 m/s

27 s= 0.23 m/s2

276

Problem 981. A train can accelerate at 1.04 m/s2. It is initially travelling eastward at10.0 m/s. How long must it accelerate before it reaches an eastward speed of 30.8 m/s?Round your answer to the nearest 0.1 s.

(a) 18.0 s (b) 19.1 s*(c) 20.0 s (d) 24.2 s(e) None of these

Solution: We know the initial velocity, the final velocity, and the acceleration. Wewant to know the time.

v = v0 + at ⇒ t =v − v0

a=

30.8 m/s− 10.0 m/s

1.04 m/s2 = 20.0 s

Problem 982. A train can accelerate from a complete stop to a speed of 31 m/s in 145s. What is the train’s acceleration? Assume constant acceleration. Round your answerto two significant figures.

(a) 0.15 m/s2 *(b) 0.21 m/s2

(c) 0.48 m/s2 (d) 0.68 m/s2

(e) None of these

Solution: Write down what information you are given, what you want, and whichfundamental equation you need to use.

Given:

vi = 0 m/s

vf = 31 m/s

∆t = 145 s

Want: a Which equation: Fundamental equation (1)

Solving for a in equation 1: a =vf − vi

∆t=

31

145= .21. Here we’ve used the fact that a = a.

Problem 983. A ship is initially moving northward at 16.7 m/s. It accelerates southwardat 2.24 m/s2 for 3.8 s. What is its final speed? Round your answer to the nearest 0.1m/s.


Solution: Careful! Notice that the initial velocity is northward, but that the ac-celeration is southward. We will take the northward direction to be positive, so theacceleration is negative. We know initial velocity, acceleration, and time; we want toknow final velocity.

v = v0 + at = 16.7 m/s + (−2.24 m/s2)(3.8 s) = 8.2 m/s

The final velocity is positive, which means that the ship is moving northward. However,the problem only asks for speed, which doesn’t include the direction.

277

Problem 984. A car is initially moving northward at 9.5 m/s. It accelerates southwardat 3.34 m/s2 for 5.9 s. What is its final speed? Round your answer to the nearest 0.1m/s.


Solution: Careful! Notice that the initial velocity is northward, but that the ac-celeration is southward. We will take the northward direction to be positive, so theacceleration is negative. We know initial velocity, acceleration, and time; we want toknow final velocity.

v = v0 + at = 9.5 m/s + (−3.34 m/s2)(5.9 s) = −10.2 m/s

The final velocity is negative, which means that the car is moving southward. Theproblem only asks for speed, which doesn’t include the direction; so 10.2 m/s is thecorrect answer.

Problem 985. A train is initially rolling backward at 6.9 m/s. It accelerates forward at0.58 m/s2. How long must it accelerate before it is moving forward at 9.2 m/s? Roundyour answer to the nearest second.

(a) 22 s (b) 25 s*(c) 28 s (d) 31 s(e) None of these

Solution: We will use forward as the positive direction, so the train’s initial velocitywill be negative. We know initial velocity, final velocity, and acceleration; we want toknow time.

v = v0 + at ⇒ t =v − v0

a=

9.2 m/s− (−6.9 m/s)

0.58 m/s2 = 28 s

Problem 986. A ship is initially moving backward at 5.6 m/s. It accelerates over thecourse of 63 s, at the end of which it is moving forward at 11.0 m/s. What is the ship’sacceleration? Round your answer to the nearest 0.01 m/s2.

(a) 0.09 m/s2 (b) 0.11 m/s2

(c) 0.24 m/s2 *(d) 0.26 m/s2

(e) None of these

Solution: We will use forward as the positive direction, so the ship’s initial velocitywill be negative. We know initial velocity, final velocity, and time; we want to knowacceleration.

v = v0 + at ⇒ a =v − v0

t=

11.0 m/s− (−5.6 m/s)

63 s= 0.26 m/s2

278

Problem 987. Beginning at rest, a car accelerates down a straight road at 3.3 m/s2 for9.8 s. At the end of this time, how far is the car from its starting point? Round youranswer to the nearest 10 m.


Solution: We know acceleration, time, and initial velocity and position (both ofwhich are zero). We want to know final position.

x = x0 + v0t+1

2at2 =

1

2at2 =

1

2(3.3 m/s2)(9.8 s)2 = 160 m

Problem 988. Beginning at rest, a car accelerates down a straight road at 5.6 m/s2 for2.9 s. At the end of this time, how far is the car from its starting point? Round youranswer to the nearest meter.


Solution: We know acceleration, time, and initial velocity and position (both ofwhich are zero). We want to know final position.

x = x0 + v0t+1

2at2 =

1

2at2 =

1

2(5.6 m/s2)(2.9 s)2 = 24 m

Problem 989. A boat is initially going northward at 15.3 m/s. It accelerates at 0.93m/s2 northward for 8.3 s. At the end of this time, how far north has it travelled? Roundyour answer to the nearest 10 m.



x = x0 + v0t+1

2at2 = (15.3 m/s)(8.3 s) +

1

2(0.93 m/s2)(8.3 s)2 = 160 m

279

Problem 990. A boat is initially going northward at 8.8 m/s. It accelerates at 3.1 m/s2

northward for 9.9 s. At the end of this time, how far north has it travelled? Round youranswer to the nearest 10 m.



x = x0 + v0t+1

2at2 = (8.8 m/s)(9.9 s) +

1

2(3.1 m/s2)(9.9 s)2 = 240 m

Problem 991. A train is initially rolling southward at 7.9 m/s. It accelerates northwardat 0.41 m/s2 for 15 s. At the end of this time, where is it relative to its starting point?Round your answer to the nearest meter.

(a) 65 m south *(b) 72 m south(c) 65 m north (d) 72 m north(e) None of these

Solution: We will take northward to be the positive direction, so the initial velocitywill be negative and the acceleration will be positive. We know the initial position (zero),the initial velocity, the acceleration, and the time; we want to know the final position.

x = x0 + v0t+1

2at2 = (−7.9 m/s)(15 s) +

1

2(0.41 m/s2)(15 s)2 = −72 m

Since the value of x is negative, it is south of the starting point.

Problem 992. A boat is initially drifting southward at 4.2 m/s. It accelerates northwardat 0.84 m/s2 for 12 s. At the end of this time, where is it relative to its starting point?Round your answer to the nearest meter.

(a) 10 m south (b) 12 m south*(c) 10 m north (d) 12 m north(e) None of these

Solution: We will take northward to be the positive direction, so the initial velocitywill be negative and the acceleration will be positive. We know the initial position (zero),the initial velocity, the acceleration, and the time; we want to know the final position.

x = x0 + v0t+1

2at2 = (−4.2 m/s)(12 s) +

1

2(0.84 m/s2)(12 s)2 = 10 m

Since the value of x is positive, it is north of the starting point.

280

Problem 993. A car is initially 30 m south of an intersection and moving northward at6.0 m/s. It accelerates northward at 2.1 m/s2 for 8.3 s. At the end of this time, where isit with respect to the intersection? Round your answer to the nearest meter.

(a) 92 m south (b) 104 m south*(c) 92 m north (d) 104 m north(e) None of these

Solution: We will take northward to be the positive direction, so the initial positionis negative and the initial velocity and acceleration are positive. We know the initialposition, initial velocity, acceleration, and time; we want to know the final position.

x = x0 + v0t+1

2at2 = −30 m + (6.0 m/s)(8.3 s) +

1

2(2.1 m/s2)(8.3 s)2 = 92 m

Since x is positive, it is north of the intersection.

Problem 994. Beginning at rest, a ship accelerates at 0.45 m/s2 until it has gone 150m. At the end of this time, how fast is the ship moving? Round your answer to thenearest m/s.

(a) 9 m/s (b) 10 m/s*(c) 12 m/s (d) 13 m/s(e) None of these

Solution: We know the initial position and initial velocity (both zero), the accelera-tion, and the final position. We want to know the final velocity. We have a formula thatwe use when time isn’t in the problem statement.

v2 = v20 + 2a(x− x0) = 2ax

⇒ v =√

2ax =[2(0.45 m/s2)(150 m)

]1/2= 12 m/s

Problem 995. Beginning at rest, a train accelerates at 0.13 m/s2 until it has gone 220m. At the end of this time, how fast is the train moving? Round your answer to thenearest 0.1 m/s.


Solution: We know the initial position and initial velocity (both zero), the accelera-tion, and the final position. We want to know the final velocity. We have a formula thatwe use when time isn’t in the problem statement.

v2 = v20 + 2a(x− x0) = 2ax

⇒ v =√

2ax =[2(0.13 m/s2)(220 m)

]1/2= 7.6 m/s

281

Problem 996. Beginning at rest, a boat accelerates at 0.13 m/s2 until it has reached aspeed of 5.6 m/s. How far has the boat travelled in this time? Round your answer tothe nearest meter.

(a) 88 m (b) 98 m(c) 109 m *(d) 121 m(e) None of these

Solution: We know the initial position and initial velocity (both zero), the accelera-tion, and the final velocity. We want to know the final position. We have a formula thatwe use when time isn’t in the problem statement.

v2 = v20 + 2a(x− x0) = 2ax ⇒ x =

v2

2a=

(5.6 m/s)2

2(0.13 m/s2)= 121 m

Problem 997. A train is moving at 31 m/s when the engineer is informed that there isa puppy on the track in front of him. If the train is capable of decelerating at 0.80 m/s2,how far does it go before it comes to a stop? Round your answer to the nearest 10 m.


Solution: We will take the forward direction to be positive, so the acceleration willbe negative; and we will take the train’s initial position as x0 = 0. We know the initialposition, the initial velocity, the final velocity (zero), and the acceleration; we want toknow the final position. We have a formula to use when time is not in the statement ofthe problem.

v2 = v20 + 2a(x− x0) ⇒ v2

0 = −2ax

⇒ x =v2

0

−2a=

(31 m/s)2

−2(−0.80 m/s2)= 600 m

Problem 998. A train can accelerate from a complete stop to a speed of 31 m/s in 145s. What is the train’s acceleration? Assume constant acceleration. Round your answerto two significant figures.

(a) 0.15 m/s2 *(b) 0.21 m/s2

(c) 0.48 m/s2 (d) 0.68 m/s2

(e) None of these

Solution: Write down what information you are given, what you want, and whichfundamental equation you need to use.

Given:

vi = 0 m/s

vf = 31 m/s

∆t = 145 s

Want: a Which equation: Fundamental equation (1)

282

Solving for a in equation 1: a =vf − vi

∆t=

31

145= .21. Here we’ve used the fact that a = a.

Problem 999. A car is initially moving at 18.9 m/s. It accelerates forward at 2.09 m/s2

until it has reached a speed of 29.3 m/s. How far has it gone in this time? Round youranswer to the nearest meter.


Solution: We know the initial position (zero), the initial velocity, the final velocity,and the acceleration. We want to know the final position. We have a formula that wecan use when time isn’t in the problem statement.

v2 = v20 + 2a(x− x0) = v2

0 + 2ax

⇒ x =v2 − v2

0

2a=

(29.3 m/s)2 − (18.9 m/s)2

2(2.09 m/s2)= 120 m

Problem 1000. During World War I, the Germans used the Paris Gun to shell thatcity from 75 miles away. The barrel of the gun was 28 m long, and the shells had amuzzle velocity of 1600 m/s. Assuming that the acceleration of the shell was constantover the whole length of the barrel, what was that acceleration? Round your answer tothe nearest 1000 m/s2.

(a) 46,000 m/s2 (b) 50,000 m/s2

(c) 55,000 m/s2 (d) 61,000 m/s2

(e) None of these

Solution: We know the initial position and velocity (both zero), and the final positionand velocity. We want to know the acceleration. We have a formula to use when time isnot included in the problem statement.

v2 = v20 + 2a(x− x0) = 2ax

⇒ a =v2

2x=

(1600 m/s)2

2(28 m)= 46, 000 m/s2

283

Problem 1001. A ship has a maximum speed in still water of 13.0 m/s. To be operatedlegally on the Rhine River, it must be able to come to a complete stop in 350 m. Whatacceleration would the ship need to do this? Round your answer to the nearest 0.01 m/s2.

(a) 0.20 m/s2 (b) 0.22 m/s2

*(c) 0.24 m/s2 (d) 0.27 m/s2

(e) None of these

Solution: We know the initial velocity, the final velocity (zero), the initial position(zero), and the final position. We want to know the acceleration. We have a formulathat we can use when time isn’t in the problem statement.

v2 = v20 + 2a(x− x0)

⇒ v20 = −2ax

⇒ a = − v20

2x= −(13.0 m/s)2

2(350 m)= −0.24 m/s2

The negative sign is because we’ve taken the direction of the ship’s movement to bepositive; to stop it, the acceleration must be in the negative direction. Since the problemdoesn’t ask for the direction of the acceleration, it’s enough to give the magnitude: 0.24m/s2.

Problem 1002. A radio-controlled airplane needs to reach a speed of 20 m/s to take off.Assuming a constant acceleration and a runway length of 20 m, what is the minimumacceleration that the plane needs if it starts from rest?

(a) 1 m/s2 (b) 5 m/s2

(c) 10 m/s2 (d) 20 m/s2

(e) None of these

Solution: Given: vi = 0 (the plane starts from rest), vf = 20 m/s, and ∆x = 20 m.Want: amin. We don’t know time, nor do we want it, so we’ll use fundamental kinematicequation 2.

v2f = v2

i + 2a∆x = 0 + 2a∆x ⇒ amin =v2f

2∆x=

202

2(20)m/s2 = 10 m/s2 ≈ g .

Thus, the plane must be able to pull one g in order to take off.

Problem 1003. A certain bi-plane needs to reach a speed of 50 m/s to take off.Assuming the plane starts from rest with a constant acceleration of 2.5 m/s2, what isthe minimum runway length the jet needs to takeoff?


Solution: Given: vi = 0 (the plane starts from rest), vf = 50 m/s, and a = 5/2 m/s2.

284

Take the initial position to be xi = 0. Want: x. We don’t know time, nor do we want it,so we’ll use fundamental kinematic equation 2.

v2f = v2

i + 2ax = 0 + 2ax ⇒ x =v2f

2a=

502

5m = 500m .

Problem 1004. A young physics student buys a pre-made remote-controlled air plane.The instructions say that the plane needs a minimum speed of 10 m/s to takeoff. Theinstructions also give the thrust-to-weight ratio of the plane when it’s at full throttle.From this information he computes the the maximum acceleration to be 3 m/s2.Allowing for friction between the wheels and the runway the student estimates that themaximum acceleration on the ground is 2.5 m/s2. Assuming the plane starts from restand has a constant acceleration 2.5 m/s2 (he keeps the throttle wide open at takeoff),what is the minimum runway length the jet needs to takeoff?


Solution: Given: vi = 0 (the plane starts from rest), vf = 10 m/s, and a = 5/2 m/s2.Take the initial position to be xi = 0. Want: x. We don’t know time, nor do we want it,so we’ll use fundamental kinematic equation 2.

v2f = v2

i + 2ax = 0 + 2ax ⇒ x =v2f

2a=

100

5m = 20m .

Problem 1005. A typical airline jet needs to reach a speed of 360 km/h to take off.Assuming a constant acceleration and a short runway length of 1.0 km, what is theminimum acceleration that the jet needs if it starts from rest? Write your answer interms of g, where g ≈ 10 m/s2.

Solution: The jet begins at x0 = 0, with initial speed v0 = 0. Hence the relationshipbetween the speed and the distance travelled is: v2 = 2ax. Solving for a, we get:

a =v2

2x=

[(360 km/h)(1000 m/km)

3600 s/h

]2

· 1

2(1 km)(1000 m/km)= 5.0 m/s2 ≈ g

2.

285

Problem 1006. An overloaded transport jet needs to reach a speed of 500 km/h totake off. Assuming a constant acceleration and a runway length of 2.0 km, what is theminimum acceleration that the jet needs if it starts from rest? Write your answer interms of g, where 1g = 9.8 m/s2. Round your answer to the nearest 0.01g.

Solution: The jet begins at x0 = 0, with initial speed v0 = 0. Hence the relationshipbetween the speed and the distance travelled is: v2 = 2ax. Solving for a, we get:

a =v2

2x=

[(500 km/h)(1000 m/km)

3600 s/h

]2

· 1

2(2 km)(1000 m/km)= 4.8 m/s2

[Don’t use this rounded value of a in further calculations; we only present it becauseit’s a good idea to check intermediate results to make sure they seem reasonable.] Theanswer we want is

a = ag

9.8 m/s2 = 0.49g

286

Problem 1007. (This problem requires solving a system of equations) A police-man is lurking behind a billboard beside the highway when someone speeds by him at38 m/s. The policeman immediately accelerates at a constant rate of 6.7 m/s2 until hehas caught up with the speeder. How long does it take for the policeman to catch up?Round your answer to the nearest second.

*(a) 11 s (b) 12 s(c) 13 s (d) 15 s(e) None of these

Solution: We will use the formula for position as a function of time: x = x0+v0t+12at2.

We will use the position of the billboard as x0 = 0. We need to use the formula twice:once for the policeman, once for the speeder. For the speeder, x0 = 0, v0,s = 38 m/s, andas = 0. For the policeman, x0 = 0, v0,p = 0, and ap = 6.7 m/s2. Hence their positionsare:

xs = x0 + v0,st+1

2ast

2 = v0,st

xp = x0 + v0,pt+1

2apt

2 =1

2apt

2

When the policeman catches up with the speeder, their positions will be the same: soxs = xp. Equating the expressions:

v0,st =1

2apt

2 ⇒ apt2 − v0,st = 0 ⇒ t (apt− 2v0,s) = 0

There are two solutions:

t = 0 and t =2v0,s

ap=

2(38 m/s)

6.7 m/s2 = 11 s

The first solution refers to the fact that both were at the billboard at t = 0; the secondsolution is the time at which the policeman catches the speeder.

Note: In general, the velocity of the police officer and the speeder at the positionxfinal = xs = xp will not be the same vs,f 6= vp,f .

287

7.6 Quantitative kinematics: vertical

Note: By convention, we will use y in place of x to denote position.

Problem 1008. An astronaut drops a watermelon from the top of a high cliff on theMoon, where the surface gravity is 1.63 m/s2. After 3.3 s, how fast is the watermelonmoving? Round your answer to the nearest 0.1 m/s.

(a) 3.9 m/s (b) 4.4 m/s(c) 4.8 m/s *(d) 5.4 m/s(e) None of these

Solution: We will take the positive direction to be downward, which will save us hav-ing to worry about negative signs. We know the initial velocity (zero), the acceleration,and the time; we want to know the final velocity.

v = v0 + at = at = (1.63 m/s2)(3.3 s) = 5.4 m/s

Problem 1009. Using a stopwatch, you determine that a television dropped from arooftop takes 2.7 s to reach the ground. How fast is it going when it hits the ground?Round your answer to the nearest m/s.


Solution: We will take the positive direction and the acceleration to be downward.We know the initial velocity (zero), the acceleration, and the time. We want to know thefinal velocity.

v = v0 + at = at = (9.8 m/s2)(2.7 s) = 26 m/s

Problem 1010. An astronaut hurls a watermelon straight downward from the top ofa high cliff on the Moon, where the surface gravity is 1.63 m/s2. The initial speed ofthe watermelon is 11.8 m/s downward. After 2.4 s, how fast is the watermelon moving?Round your answer to the nearest 0.1 m/s.


Solution: We will take the positive direction to be downward, so the acceleration andthe initial velocity will be positive. We know the initial velocity, the acceleration, andthe time; we want to know the final velocity.

v = v0 + at = 11.8 m/s + (1.63 m/s2)(2.4 s) = 15.7 m/s

288

Problem 1011. While walking in the desert, you discover an old mine. To find outhow deep it is, you drop a rock down the shaft. Using the stopwatch on your cell phone,you time how long it takes before you hear the rock hit the bottom. If it takes 3.2 s forthe rock to reach the bottom, how deep is the shaft? Round your answer to the nearestmeter.


Solution: We will take the positive direction to be downward, so that the accelerationand the final position will be positive. We will take the initial position to be zero. Thenwe know the initial position, the initial velocity (zero), the acceleration, and the time.We want to know the final position.

y = y0 + v0t+1

2at2 =

1

2at2 =

1

2(9.8 m/s2)(3.2 s)2 = 50 m

Problem 1012. A banker drops a sack of pennies from the top of a tall building. After1.5 s, how far has the sack of pennies fallen? Round your answer to the nearest 0.1 m.

(a) 9.9 m *(b) 11.0 m(c) 12.1 m (d) 13.3 m(e) None of these


y = y0 + v0t+1

2at2 =

1

2at2 =

1

2(9.8 m/s2)(1.5 s)2 = 11.0 m

Problem 1013. An astronaut drops a watermelon from the top of a tall cliff on Mercury,where the surface gravity is 3.7 m/s2. After 2.4 s, how far has the watermelon fallen?Round your answer to the nearest 0.1 m.

*(a) 10.7 m (b) 11.7 m(c) 12.9 m (d) 14.2 m(e) None of these


y = y0 + v0t+1

2at2 =

1

2at2 =

1

2(3.7 m/s2)(2.4 s)2 = 10.7 m

289

Problem 1014. A dietitian drops a head of iceberg lettuce from a bridge. How longdoes it take for the lettuce to reach the river 31.2 m below? Round your answer to thenearest 0.1 s.


Solution: We will take the positive direction to be downward, which will make theacceleration and the final position positive. We will take the initial position to be zero.Then we know the initial position and velocity (both zero), the acceleration, and thefinal position; we want to know the time.

y = y0 + v0t+1

2at2 =

1

2at2

⇒ t =

√2y

a=

[2(31.2 m)

9.8 m/s2

]1/2

= 2.5 s

Problem 1015. You are on the surface of Ganymede, where the acceleration due togravity is 1.35 m/s2. You throw an egg straight upward at 25 m/s. How long does ittake for the egg to reach the ground? Round your answer to the nearest second.


Solution: We will use upward as the positive direction and ground level as y0 = 0. Weknow the initial and final position (both zero), the initial velocity, and the acceleration.We want to know the time.

y = y0 + v0t+1

2at2 ⇒ 2v0t+ at2 = 0 ⇒ t(2v0 + at) = 0

t = 0 or t =−2v0

a=−2(25 m/s)

−1.35 m/s2 = 37 s

The first solution refers to the fact that at t = 0, you threw the egg upward from groundlevel; the second, t = 37 s, is the time at which the egg strikes the ground on its wayback down.

290

Problem 1016. An astronaut on Planet X drops a hammer from a cliff 6.8 m high. Thehammer hits the ground after 1.3 s. What is the surface gravity of Planet X? Roundyour answer to the nearest 0.1 m/s2.

*(a) 8.0 m/s2 (b) 8.9 m/s2

(c) 9.7 m/s2 (d) 10.7 m/s2

(e) None of these

Solution: We will take the positive direction to be downward. We will take the topof the cliff to be y0 = 0. We know the initial position and velocity (both zero), the finalposition, and the time. We want to know the acceleration.

y = y0 + v0t+1

2at2 =

1

2at2

⇒ a =2y

t2=

2(6.8 m)

(1.3 s)2= 8.0 m/s2

Problem 1017. A physics student angrily hurls his cell phone downward from a bridgeat 19.9 m/s. After 2.2 s, how far below the bridge will the phone be? Round your answerto the nearest meter.


Solution: We will take down to be the positive direction, and the bridge to be y0 = 0.Then we know the initial position, the initial velocity, and the time; we want to knowthe final position.

y = y0 + v0t+1

2at2 = (19.9 m/s)(2.2 s) +

1

2(9.8 m/s2)(2.2 s)2 = 67 m

Problem 1018. A banker drops a sack of pennies from the top of a building 18.8 mtall. How fast is the sack moving when it strikes the ground? Round your answer to thenearest 0.1 m/s.


Solution: We will take the top of the building to be y0 = 0, and downward as thepositive direction. We know the initial and final position, the initial velocity (zero), andthe acceleration. We want the final velocity. We have a formula to use when time is notpart of the problem statement.

v2 = v20 + 2a(y − y0) = 2ax

⇒ v =√

2ay =

√2(9.8 m/s2)(18.8 m) = 19.2 m/s

291

Problem 1019. If we ignore air drag, what is the speed of a hail stonethat falls from rest at the top of a cumulus cloud that is 10,000 m high?(a) 126 m/s (b) 13.2 m/s(c) 67.6 m/s (d) 443 m/s(e) None of these

Solution: We will take the positive direction to be downward in the direction ofmotion and take the height of 10 km to be the initial position y0. This means that thevelocity is positive, and the acceleration is positive ay = g. Let’s write down what we’regiven, what we want, and which equation we need to use.

Given:

vf = 0 (The hail stone is initially at rest at the maximum height)

a = g = 9.8 m/s2

y0 = 0

yf = 104 m

Notice that we must convert the height to meters.

Want: vf = ? (final velocity)

Which equation: equation (2)

Notice that we don’t know time; nor do we want time, so we can’t use fundamentalequations (1), or (3).

Substituting v0 = 0, y0 = 0, and a = g into equation (2) and solving for v0 we arrive at

v2f = 0 + 2gyf

√−−−−−−→ vf =

√2gyf

evaluate−−−−→ v0 =√

2(9.8m/s2)(104m)

≈ 443 m/s (that’s almost 1000 mph!)

Problem 1020. A ball is thrown straight upward from ground level, y = 0. At whatspeed must the ball be thrown if it is to reach a height of ymax = 10 meters above theground? Round your answer to the nearest m/s.

Solution: v =√

2gymax =√

2(9.8 m/s)(10 m) = 14 m/s

Problem 1021. You are standing on a bridge over a canyon. To find out how deep thecanyon is, you drop a rock off the bridge and time its fall. The rock hits the river at thecanyon bottom after 7.2 s. How deep is the canyon? Round your answer to the nearestmeter.


Solution: Given: yi = 0 (take y-axis pointing downward in the direction of motion),vi = 0, ay = g, and tf = 7.2 s. Want: yf . Use fundamental equation 3: y = 1

2gt2fall.

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7.6.1 The difference between average speed and average velocity in one-dimension

Problem 1022. (Speed vs. Velocity) An object is launched straight up into the airfrom the ground with an initial vertical velocity of 30 m/s. The object rises to a heightof approximately 45 m above the ground in 3 seconds; it then falls back to the ground in3 more seconds, impacting with a speed of 30 m/s. Approximately what is the averagespeed of the object during its time in the air?


Solution: The definition of average speed is

average speed =total distance travelled

total time

=distance up + distance down

time to go up + time to go down

=2distance up

2time up(since yup = ydown = 45 m; tup = tdown = 3 s)

=45 m

3 s= 15 m/s

Problem 1023. (Speed vs. Velocity) An object is launched straight upward fromground level with an initial vertical velocity of about 30 m/s. In roughly 3 seconds, theobject reaches a maximum height of approximately 45 m above the ground; it takesanother 3 seconds to fall back down, striking the ground at 30 m/s. If vave is the averagevelocity of the object during its 6-second flight, then the magnitude of the averagevelocity vave is:

*(a) 0 m/s (b) 5 m/s(c) 15 m/s (d) 30 m/s(e) None of these

Solution: Recall the definition of average velocity the change in location (the displace-ment) over the change in time. It is a vector quantity, which in one spatial dimension is aquantity that can be positive or negative denoting which direction the object is travelingwith respect to the x-axis. Let (ti, yi) be the initial time and position, respectively, and

293

(tf , yf ) be the final time and position.

average velocity =displacement

change in time

=yf − yitf − ti

=yi − yitf − ti

(Since the ball starts and stops at the same point yi = yf )

= 0 (since ∆y = yf − yi = 0 and ∆t = tf − ti > 0)

294

7.7 Similarity problems

The Idea: With all similarity problems there is always two similar problems (hence thename!) that are modeled by the same equation/equations. The two similar situations aresometimes called experiments, and the equation/equations that describes the situation(the model that governs the behavior of the particle) is/are referred to as the governingequation/equations. Some of the variables/parameters have the same value in both exper-iments. These parameters are known as common parameters. The variables/parametersthat are situation dependent (i.e., different between the two experiments) are labelledby subscripts: a subscript 1 is given to the parameters from the first experiment anda subscript 2 is given to the parameters from the second experiment. The parametersthat are in common to both experiments are the link between the two experiments. Wetypically don’t give these common parameters subscripts.

Problem-Solving Algorithm: Write down what your given, what you want, and de-termine the governing equation that describes both situations. Identify all of the vari-ables/parameters that are in common to both situations and put them on the right-handside (RHS) of the equation, and put all of the variables/parameters that are situation-dependent on the LHS of the equation. Then equate the two situations through thecommon RHS of the equation of the re-written governing equation. This is your commonlink between the two situations!

Problem 1024. (Similarity Problem) A train has a constant forward acceleration ofa. Starting at rest, it reaches a speed of v1 at time t1. At time t2 = 2t1, how fast is thetrain moving?

(a)√

2 v1 *(b) 2v1

(c) 2√

2 v1 (d) 4v1

(e) None of these

Solution: We have two similar situations that can both be modeled by one governingequation. In both situations we have an object (the train) starting from rest vi = 0under constant acceleration a = const. In both situations we are given time and in thesecond situation we want the final velocity. The equation describing the motion (i.e.,the governing equation), is fundamental kinematic equation 1 vf = vi + atf (we’re notgiven distance, nor do we want it). Taking ti = 0, vi = 0, a = const, and dropping the fsubscripts yields

v = at (The governing equation) ,

where we recognize vi = 0 and a as common parameters (i.e., parameters that are incommon to both situations).We’ll now use a to relate the velocity and time from each of the similar situations. Re-writing the governing equation, by putting the common parameter a on the right andthe situation-dependent parameters on the left yields:

v1

t1= a =

v2

t2

· t2−−−−−→ v2 = v1

(t2t1

)=

(2t1t1

)v1 = 2v1 .

295

Problem 1025. (Similarity Problem) A physics student drops a cell phone off a veryhigh bridge. At time t1 after being released, the phone has fallen a distance y1. How farhas the phone fallen at time t2 = 3t1?

(a)√

3 y1 (b) 3y1

(c) 3√

3 y1 *(d) 9y1

(e) None of these

Solution: We will take the positive direction as downward, and the bridge level asy0 = 0. We know that the initial velocity is v0 = 0. Hence y = y0 + v0t + 1

2at2 = 1

2gt2

is our governing equation for both cases. We are given t1, y1, and t2. At time t = t1,y1 = 1

2gt21, at time t = t2 = 3t1, y2 = 1

2gt22, and g/2 is a constant that is in common to

both situations. Re-writing the governing equationy

t2=g

2(notice that the right-hand

side of the equation is constant). We want y2, thus equating the two equations gives

y1

t21=g

2=y2

t22

· t22−−−−−→ y2 = y1

(t2t1

)2

=

(3t1t1

)2

y1 = 9y1 .

Problem 1026. (Similarity Problem) An object dropped from a building with heighty1 takes time t1 to reach the ground. How long would it take for an object dropped froma building with height y2 = 2y1?

*(a)√

2 t1 (b) 2t1(c) 2

√2 t1 (d) 4t1

(e) None of these

Solution: We will take downward as the positive direction, and the top of the buildingas y0 = 0. The initial velocity is zero v0 = 0. Then fundamental equation (3) reduces toy = 1

2gt2. Solving this equation for time gives the governing equation for the fall time as

a function of height.

y =1

2gt2

· 2g−−−−−→ t2 =

2

gy

√−−−−−−→ tfall =

√2y

g.

We’re given: y1, t1, and y2. We want: t2. Re-writing the governing equation with commonparameters on the RHS of the equation and the situation-dependent parameters on theLHS of the equation yields

÷√y−−−−−−→ tfall

y=

√2

g.

Equating the two situations via the common parameters yields

tfall,1√y1

=

√g

2=tfall,2√y2

·√y2−−−−−−→ tfall,2 = tfall,1

√y2

y1

√y2=2y1−−−−−−−−→ tfall,2 =

√2tfall,1 .

296

Problem 1027. (Similarity Problem) An object dropped from a building with heighty1 strikes the ground with speed v1. If an object is dropped from a building with heighty2 = 4y1, how fast is it moving when it hits the ground?

*(a) 2v1 (b) 4v1

(c) 8v1 (d) 16v1

(e) None of these

Solution: We will take the positive direction as downward, and the top of the buildingas y0 = 0. The initial velocity is zero v0 = 0. Since time isn’t in the statement of theproblem, we use the fundamental formula (2): v2 = v2

0 +2a(y−y0); in this case, v2 = 2gy.Taking the square root of this equation gives v as a function of y. The governing equationis v =

√2gy. Note that 2g is the same for both situations. This will be are common

parameters. Putting all of the parameters that are situation dependent on the left-handside of the equation, and the common parameters on the right-hand side of the governingequation yields:

v =√

2gy

÷√y−−−−−−→ v

√y

=√

2g (similarity form of governing equation)

Substituting the variables into this equation and using the common link (the RHS of thegoverning equation) to equate the two situations yields

v1√y1

=√

2g =v2√y2

·√y2−−−−−−→ v2 = v1

√y2

y1

= v1

√4y1

y1

= 2v1

Problem 1028. (Similarity Problem) An object dropped from a certain height onPlanet X takes time tX to reach the ground. An object dropped from the same heighton Planet Y takes time tY to fall. Which equation describes the relation between thesurface gravities gX and gY of the two planets?

(a)gYgX

=tYtX

(b)gYgX

=t2Yt2X

(c)gYgX

=tXtY

*(d)gYgX

=t2Xt2Y

(e) None of these

Solution: An object falling in a gravity of g for a time of t falls for a distance ofx = 1

2gt2. The distance fallen, x, is the same on both planets. Hence

gXt2X = 2x = gY t

2Y = ⇒ gY

gX=t2Xt2Y

297

Problem 1029. (Similarity Problem) Your friend’s hot rod can go from 0 to 60 mphin 6 seconds. Your father’s minivan can only do 0 to 60 mph in 12 seconds. Assumingconstant acceleration for each vehicle, what is the ratio of the hot rod’s acceleration ahrto the minivan’s acceleration amv?

(a)ahramv

=1

2(b)

ahramv

=1√2

(c)ahramv

=√

2 *(d)ahramv

= 2

(e) None of these

Solution: Since each vehicle starts from rest v0 = 0 and accelerates until it reaches thefinal velocity v = 60 mph, we have ∆v = v− v0 = v = 60 m/hr. Because of the constantacceleration assumption, we can use the equation v = at and the common parameter vto find a relationship between the ratio of accelerations and the times:

ahr thr = v = amv tmv

÷amv thr−−−−−−−−→ ahramv

=tmvthr

=2 thrthr

= 2

Notice that by taking a ratio, we avoided having to convert the acceleration a =v

t=

60 mph

6sinto ft/s2 or some other “standard” measure of acceleration.

Problem 1030. (Similarity Problem) A spherical balloon has a radius of r whenfilled with 0.50 gallons of water. If you fill the balloon with 1.0 gallons of water, whatwill its radius be? Your answer should be exact.

Solution: The volume of a sphere is proportional to the cube of the radius. If theradius of the one-gallon balloon is R, then:

0.50 gal

r3=

1.0 gal

R3⇒ R3

r3=

1.0

0.50= 2.0

Hence R3 = 2.0 r3; so R = 3√

2 r.

Problem 1031. (Similarity Problem) An unloaded train has an acceleration of au.Starting from a standstill, it can accelerate to a speed of v in time tu. A fully loaded trainhas an acceleration of al = 0.4au. How long does it take the loaded train to acceleratefrom zero to v?

Solution: Since the initial velocity is zero, we can write: v = autu for the unloadedtrain. For the loaded train, the acceleration is al = 0.40au. Let tl be the time that ittakes for the loaded train to accelerate to v. Then

altl = v = autu÷al−−−−−→ tl =

autual

=tuau

0.40au= 2.5 tu

298

Problem 1032. (Similarity Problem) The Smith Building is twice as tall as the JonesBuilding. A sack of pennies dropped from the top of the Jones Building takes time tJ toreach the ground. How long does it take for a similar sack dropped from the top of theSmith Building? Your answer should be exact. (Hint: Use the top of the building as theorigin, and let downward be the positive y-direction.)

Solution: Since y0 = 0 and v0 = 0, distance fallen as a function of time is: y = 12gt2.

Since the Smith Building is twice as tall as the Jones Building, yS = 2yJ . Hence:

yJ =1

2gtJ

2 ⇒ yS =1

2gtS

2 = 2yJ = gtJ2 ⇒ tS

2 = 2tJ2 ⇒ tS =

√2 tJ

Problem 1033. (Similarity Problem) The Smith Building is three times as high asthe Jones Building. A watermelon dropped from the top of the Smith Building takes tsseconds to hit the ground. How long does it take for a watermelon dropped from the topof the Jones Building? Your answer should be exact.

Solution: We will use the downward direction as positive, and take the top of eachbuilding to be x = 0. Then x0 = 0 and v0 = 0, so we can write:

x =1

2gt2 ⇒ t =

√2x

g

Thus the falling time is proportional to the square root of the height of the building. Weuse the subscripts s and j to designate Smith and Jones:

ts√xs

=tj√xs/3

=

√3tj√xs

⇒ tj =

√3

3ts

Problem 1034. (Similarity Problem) Two identical balls are launched vertically intothe air with the same initial velocity v0 . One of them is launched from the earth; theother from the surface of Planet X, which has 1/10 of the earth’s gravitational attraction(gearth = 10gX). The ball thrown on the earth reaches a maximum height of HE; theball launched on Planet X reaches a maximum height of HX . What is the relationshipbetween HE and HX? (Assume that air drag has no effect.)

Solution: We can use the formula: v2 = v20 + 2a(x−x0). The balls are launched from

ground level, so we can write x0 = 0; the maximum height x = H is reached when v = 0.Hence the formula becomes: 0 = v0

2 + 2aH. Since v0 is the same on both planets, wecan write: 2gearthHE = 2gXHX. Solving for HE yields

HE =gXHX

gearth

=gXHX

10gX

=HX

10

Problem 1035. (Similarity Problem) Two identical balls are launched vertically intothe air with the same initial velocity v0 . One of them is launched from the earth; theother from the surface of Planet X, which has 1/10 of the earth’s gravitational attraction(gE = 10gX). The ball thrown on the earth reaches its maximum height after time tE;the ball launched on Planet X reaches its maximum height after time tX . What is therelationship between tE and tX? (Assume that air drag has no effect.)

299

Solution: For the two similar experiments we are given: vE,i = vX,i = v0 (initialvelocity), vE,f = vX,f = 0 (final velocity), and gE = 10gX), where we have used thefact when the ball reaches its maximum height, then its velocity is zero. The commonparameters are: the initial and final velocities. Using the fundamental kinematic equation1 (vf = vi − gt) with the y axis pointing upward yields:

gEtE = vE,i = vX,i = gXtX ⇒ tEtX

=gXgE

.

300

7.8 Lab-application problems: One-Dimensional Kinematics

This is a mixture of problem types based on actual equipment found in instructionalphysic labs everywhere.

7.8.1 Ball-drop apparatus

Problem 1036. (Lab Problem) In a ball-drop apparatus, a ball is held in place abovea drop pad. When a knob is turned, the ball is released to fall onto the pad; an electricalcircuit through the drop apparatus and the pad is used to measure the time it takes forthe ball to fall. If the ball is initially 0.85 m above the pad, how long should it take forthe ball to fall to the pad? Round your answer to the nearest 0.01 s.

(a) 0.37 s m *(b) 0.42 s(c) 0.46 s (d) 0.50 s(e) None of these

Solution: We will take the positive direction to be downward, which will make theacceleration and the final position positive. We will take the initial position to be zero.Then we know the initial position and velocity (both zero), the acceleration, and thefinal position; we want to know the time.

y = y0 + v0t+1

2at2 =

1

2at2

⇒ t =

√2y

a=

[2(0.85 m)

9.8 m/s2

]1/2

= 0.42 s

7.8.2 Spring-cannon apparatus

Problem 1037. (Lab Problem) In this problem, use upward as the positive direction.A spring cannon has been pointed straight up, and a ball has been fired from it. Theball has not yet reached the highest point in its flight. Which of the following describesthe ball’s situation?

(a) positive velocity; positive acceleration*(b) positive velocity; negative acceleration(c) positive velocity; zero acceleration(d) zero velocity; negative acceleration(e) none of these

Solution: The ball hasn’t reached its highest point, so it is still moving upward.Hence the velocity is positive. It is slowing down as it moves upward, so its accelerationis negative.

301

Problem 1038. (Lab Problem) In this problem, use upward as the positive direction.A spring cannon has been pointed straight up, and a ball has been fired from it. Theball has reached the highest point in its flight, and its speed is increasing as it falls backtoward the ground. Which of the following describes the ball’s situation?

(a) negative velocity; positive acceleration*(b) negative velocity; negative acceleration(c) negative velocity; zero acceleration(d) zero velocity; positive acceleration(e) none of these

Solution: The ball is moving downward, so its velocity is negative. It is speeding upin the downward direction, so its acceleration is negative.

Problem 1039. (Lab Problem) A physics student shoots a ball straight upward froma spring cannon, with an initial speed of 15.1 m/s. How long does it take for the ball toreach the highest point in its flight? Round your answer to the nearest 0.1 s.


Solution: We will take the positive direction to be upward. This means that theinitial velocity is positive, and the acceleration is negative. We know the initial velocity,the final velocity (zero at the highest point in the flight), and the acceleration; we wantto know the time.

v = v0 + at ⇒ t =v − v0

a=−15.1 m/s

−9.8 m/s2 = 1.5 s

Problem 1040. (Lab Problem) A physics student on Pluto shoots a ball straightupward from a spring cannon, with an initial speed of 9.1 m/s. The ball reaches thehighest point in its flight after 11.2 s. What is Pluto’s surface gravity? Round youranswer to the nearest 0.01 m/s2.

(a) 0.59 m/s2 (b) 0.66 m/s2

(c) 0.73 m/s2 *(d) 0.81 m/s2

(e) None of these

Solution: We will take the positive direction to be downward, so the initial velocitywill be negative and the acceleration will be positive. We know the initial velocity, thefinal velocity (zero), and the time; we want to know the acceleration.

v = v0 + at ⇒ a =v − v0

t=−v0

t=

9.1 m/s

11.2 s= 0.81 m/s2

302

Problem 1041. (Lab Problem) A student shoots a ball straight upward from a springcannon. The ball reaches the highest point in its flight after 1.2 s. What was the ball’sinitial speed? Round your answer to the nearest 0.1 m/s.


Solution: We will take the positive direction to be upward, so the accelerationwill be negative and the initial velocity positive. We know the final velocity (zero), theacceleration, and the time; we want to know the initial velocity.

v = v0 + at ⇒ v0 = v − at = −at = (9.8 m/s2)(1.2 s) = 11.8 m/s

Problem 1042. (Lab Problem) A physics student shoots a ball from a spring cannonstraight upward at 15.0 m/s. After 1.1 s, how high will the ball be? Round your answerto the nearest 0.1 m.


Solution: We will take upward to be the positive direction; so the acceleration willbe negative. We will take ground level to be y0 = 0. We know initial position, initialvelocity, acceleration, and time; we want to know final position.

y = y0 + v0t+1

2at2 = (15.0 m/s)(1.1 s) +

1

2(−9.8 m/s2)(1.1 s)2 = 10.6 m

Problem 1043. (Lab Problem) A physics student shoots a ball from a spring cannonstraight upward at 15.0 m/s. After 2.3 s, how high will the ball be? Round your answerto the nearest 0.1 m.


Solution: We will take upward to be the positive direction; so the acceleration willbe negative. We will take ground level to be y0 = 0. We know initial position, initialvelocity, acceleration, and time; we want to know final position.

y = y0 + v0t+1

2at2 = (15.0 m/s)(2.3 s) +

1

2(−9.8 m/s2)(2.3 s)2 = 8.6 m

303

Problem 1044. (Lab Problem) A physics student shoots a ball from a spring cannonstraight upward. After 1.2 s, the ball is 15.0 m high. What was the initial velocity of theball?


Solution: We will take upward to be the positive direction; so the acceleration will benegative. We will take ground level to be y0 = 0. We know initial position, final position,acceleration, and time; we want to know initial velocity.

y = y0 + v0t+1

2at2 = v0t+

1

2at2

⇒ 2y = 2v0t+ at2

⇒ v0 =2y − at2

2t=

2(15.0 m)− (−9.8 m/s2)(1.2 s)2

2(1.2 s)= 18.4 m/s

304

Problem 1045. (Lab Problema) A physics student shoots a ball straight upward froma spring cannon. Using a meter stick, the student measures the ball’s maximum height.If the height was found to be 53 cm, what was the ball’s initial velocity as it left themuzzle of the cannon? Round your answer to the nearest 0.1 m/s.


Solution: We will take the positive direction to be upward in the direction of motion.This means that the initial velocity is positive, and the acceleration is negative a = −g.Let’s write down what we’re given, what we want, and which equation we need to use.

Given:

vf = 0 (The ball is instantaneous at rest at the maximum height)

a = −g = −9.8 m/s2

ymax = 53 cm = .53 m

y0 = 0 (We’ll that the mouth of the cannon as our origin)

Notice that we must convert the height to meters (or acceleration to cm/s2).

Want: v0 = ? (initial velocity)

Which equation: equation (2)

Notice that we don’t know time; nor do we want time, so we can’t use fundamentalequations (1), or (3).

Substituting v = 0, a = −g into equation (2) and solving for v0 we arrive at

02 = v20 − 2gymax

+ 2gymax−−−−−→ v20 = 2gymax

√−−−−−−→ v0 =

√2gymax

evaluate−−−−→ v0 =√

2(9.8m/s2)(.53m)

= 3.2 m/s

aThis is a very accurate method for determining the initial velocity of the ball, especially when youshoot the spring cannon at a high angle of inclination (θ0 ≥ 60◦).

305

Problem 1046. (Lab Problem) A physics student is designing a spring cannon. Hewants to fire a ball straight upward from ground level; and he wants the ball to have anupward speed of 10.2 m/s when it reaches a height of 6.8 m above the ground. Whatmust the ball’s initial speed be? Round your answer to the nearest 0.1 m/s.


Solution: We will take upward as the positive direction and ground level as y0 = 0.We know the initial and final position, the acceleration (negative), and the final velocity.We want the initial velocity. We have a formula we can use when time isn’t in thestatement of the problem.

v2 = v20 + 2a(y − y0) = v2

0 + 2ay

⇒ v0 =√v2 − 2ay =

√(10.2 m/s)2 − 2(−9.8 m/s2)(6.8 m) = 15.4 m/s

Problem 1047. (Lab Problem) A physics student shoots a ball straight upward froma spring cannon, with an initial velocity of 18.2 m/s. At what time will the ball reach aheight of 10.4 m on its way up? Round your answer to the nearest 0.01 s.

(a) 0.51 s (b) 0.57 s(c) 0.63 s *(d) 0.71 s(e) None of these

Solution: We will take upward as the positive direction and ground level as y0 = 0.Then acceleration is negative. We know initial position and velocity, final position, andacceleration; we want to know time.

y = y0 + v0t+1

2at2 = v0t+

1

2at2

⇒ at2 + 2v0t− 2y = 0

We will use the quadratic formula to solve this for t. As often happens with quadraticequations, we will get two solutions. This is because the ball will pass through mostpoints on its trajectory twice: once on the way up, and again on the way back down.The problem asks when the ball reaches y on the way up; so we want the smaller of thetwo times.

t =−2v0 ±

√4v2

0 + 8ay

2a=−v0 ±

√v2

0 + 2ay

a

=−18.2 m/s±

√(18.2 m/s)2 + 2(−9.8 m/s2)(10.4 m)

−9.8 m/s2

= 0.71 s or 3.01 s

306

Problem 1048. (Lab Problem) A physics student shoots a ball straight upward froma spring cannon, with an initial velocity of 14.3 m/s. At what time will the ball reach aheight of 8.8 m on its way back down, after it has reached its maximum height? Roundyour answer to the nearest 0.01 s.

(a) 1.83 s *(b) 2.04 s(c) 2.24 s (d) 2.46 s(e) None of these

Solution: We will take upward as the positive direction and ground level as y0 = 0.Then acceleration is negative. We know initial position and velocity, final position, andacceleration; we want to know time.

y = y0 + v0t+1

2at2 = v0t+

1

2at2

⇒ at2 + 2v0t− 2y = 0

We will use the quadratic formula to solve this. As often happens with quadratic equa-tions, we will get two solutions. This is because the ball will pass through most points onits trajectory twice: once on the way up, and again on the way back down. The problemasks when the ball reaches y on the way down; so we want the larger of the two times.

t =−2v0 ±

√4v2

0 + 8ay

2a=−v0 ±

√v2

0 + 2ay

a

=−14.3 m/s±

√(14.3 m/s)2 + 2(−9.8 m/s2)(8.8 m)

−9.8 m/s2

= 0.88 s or 2.04 s

Problem 1049. A student shoots a ball straight upward from a spring cannon. Theball reaches the highest point in its flight after 1.5 s. What was the ball’s initial speed?Round your answer to the nearest meter per second.


Solution: We will take the positive direction to be upward, so the accelerationwill be negative and the initial velocity positive. We know the final velocity (zero), theacceleration, and the time; we want to know the initial velocity.

v = v0 + at ⇒ v0 = v − at = −at = (9.8 m/s2)(1.5 s) = 15 m/s

307

7.9 Mixing It Up

Problem 1050. Harry Potter’s magic flying broomstick is described as having an accel-eration of 150 miles per hour in 10 seconds. What is this acceleration in ft/s2? What isit in terms of g?

Solution: We’ll begin by converting miles per hour to ft/s:

150 mi

hr· 5280 ft

1 mi· 1 hr

3600 s= 220 ft/s

The broomstick attains this speed in 10 s; so we divide by 10 s to get acceleration:

a =220 ft/s

10 s= 22 ft/s2

Since g = 32 ft/s2,

a =22 ft/s2

32 ft/s2 = 0.69 g

Problem 1051. Using a stopwatch, you determine that a television dropped from arooftop takes 2.7 s to reach the ground. How fast is it going when it hits the ground?Give your answer in m/s.

Solution: For constant acceleration,

v = v0 + at = 0 + (9.8 m/s2)(2.7 s) = 26 m/s

Problem 1052. An experimental aircraft is capable of accelerating horizontally at 4.2g. How long does it take for the aircraft to reach a horizontal speed of 5600 ft/s?

Solution: For constant acceleration, v = at; so t = v/a.We need to convert the acceleration to units of ft/s2:

a = 4.2 g = (4.2)(32 ft/s2) = 134.4 ft/s2

Now that we have consistent units, we can calculate the time:

t =v

a=

5600 ft/s

134.4 ft/s2 = 42 s

Notice that we’ve rounded this answer to two significant digits, since that’s all we hadin the statement of the problem.

Problem 1053. You are on the surface of the Moon, where the acceleration due togravity is 1.62 m/s2. You throw an egg straight upward at 25 m/s. How long does ittake for the egg to reach the ground?

308

Solution: We can use the formula: x = x0 + v0t + 12at2. We know x0 = 0, since

we’re throwing the egg from ground level; v0 = 25 m/s; a = −1.62 m/s2 (taking theupward direction as positive); and x = 0, since we want to know the time at which theegg reaches ground level. Plugging these numbers into the formula gives us

0 = −0.81t2 + 25t = −t(0.81t− 25)

Solving this yields

t = 0 s and t =25

0.81= 31 s

The first solution refers to the fact that at t = 0, you were throwing the egg up from thesurface. The second solution is the time at which the egg reaches x = 0 on its way backdown. Hence the answer is: 31 s.

Problem 1054. A train is moving at 32 m/s when the engineer is informed that a schoolbus full of orphans and puppies is on the tracks ahead of him. If the train is capable ofdecelerating at 0.80 m/s2, how long does it take for the train to come to a halt? How fardoes it go after the brakes are applied?

Solution: We’ll use: v = v0 + at, with v = 0 m/s, v0 = 32 m/s and a = −0.80 m/s2.(We’ve taken the forward direction of the train as positive.) Then

0 = 32− 0.80t so t =32 m/s

0.8 m/s2 = 40 s

Now that we know the time, we can use the formula: x = x0 + v0t + 12at2. We’ll use

x0 = 0, v0 = 32 m/s, a = −0.80 m/s2, and t = 40 s. Then

x = 0 + (32 m/s)(40 s)− 1

2(0.8 m/s2)(40 s)2 = 640 m

Problem 1055. Your lab partner accidentally rides his canoe over Horseshoe Falls, whichis 53 m high. How long does it take for him to reach the bottom? What is his downwardspeed when he gets there?

Solution: We’ll take the edge of the falls as x = 0, and the downward directionas positive. Then we can use the formula: x = x0 + v0t + 1

2at2, with x0 = 0, v0 = 0,

a = g = 9.8 m/s2, and x = 53 m. We get

53 = 0 + 0t+1

2(9.8 m/s2)t2 = 4.9t2

Hence t2 = 53/4.9; so t =√

53/4.9 = 3.3 s.

Since v0 = 0, we can write v = at = 9.8√

53/4.9 = 32 m/s.

309

Problem 1056. A ship has a maximum speed in still water of 8.8 m/s. To be operatedlegally on the Rhine River, it must be able to come to a complete stop in 350 m. Whatacceleration would the ship need to do this?

Solution: Again, we’ll need two formulas to solve this:

x = x0 + v0t+1

2at2 v = v0 + at

We know that x = 350 m, x0 = 0, v = 0, and v0 = 8.8 m/s. We want to calculate a.We’ll begin by solving the second equation for t:

0 = v0 + at ⇒ t = −v0

a

We substitute this into the first equation:

x = 0− v02

a+v0

2

2a= −v0

2

2a

We solve this for a:

a = −v02

2x= − (8.8 m/s)2

(2)(350 m)= −0.11 m/s2

Problem 1057. A ball is thrown straight up from ground level, which we take to bey = 0.(a) At what speed must it be thrown to reach a height ymax meters above the ground?(b) How long does it take the ball to reach the maximum height tup?

Solution: (a) Use fundamental equation 2 with vf = 0, since when the ball reachesits highest point the its velocity is zero.

0 = v2f = v2

i − 2gymax ⇒ vi =√

2gymax

(b) Use fundamental equation one vf = vi + gt with vf = 0. Then tup = −vi

g=√

2ymax

g

310

8 One-dimensional rotational kinematics

8.1 Angular speed, period, and frequency

Problem 1058. A Ferris wheel has a radius of 12 m. If a point on the outer rim ismoving at 4 m/s, what is its angular speed?

*(a) 1/3 rad/s (b) 3 rad/s(c) 2π/3 rad/s (d) 6π rad/s(e) None of these

Solution: v = rω ⇒ ω =v

r=

4 m/s

12 m=

1

3rad/s.

Problem 1059. A reflector is attached to the spoke of a bicycle wheel, 20 cm from thecenter. If the wheel is turning at 18 rad/s, how fast is the reflector moving? Round youranswer to two significant figures.

(a) 110 cm/s *(b) 360 cm/s(c) 640 cm/s (d) 1100 cm/s(e) None of these

Solution: v = rω = (20 cm)(18 rad/s) = 360 cm/s

Problem 1060. A bug is clinging to the tip of a wind turbine blade, 45 meters from thecenter. The blade turns through a complete circle every 7.5 s. What is the bug’s angularspeed? Round your answer to two significant figures.

*(a) 0.84 rad/s (b) 6.0 rad/s(c) 19 rad/s (d) 38 rad/s(e) None of these

Solution: If T is the period (the time that it takes to make a complete circle), then

ω =2π rad

T=

2π rad

7.5 s= 0.84 rad/s

The 45-m radius of the circle is irrelevant to this problem.

Problem 1061. The orbit of Jupiter’s moon Callisto has a radius of 1.9× 109 m. Themoon makes a complete circle around Jupiter in 1.4 × 106 s. What is Callisto’s orbitalspeed? Round your answer to two significant figures.


Solution: v =2πr

T=

2π · 1.9× 109 m

1.4× 106 s= 8500 m/s

311

Problem 1062. Planet X orbits its star at a distance of 1.5× 1011 m. Its orbital speedis 3.0× 104 m/s. How long does it take for the planet to make a complete circle aroundthe star? Round your answer to two significant figures.

(a) 2.3× 107 s (b) 2.5× 107 s(c) 2.8× 107 s *(d) 3.1× 107 s(e) None of these

Solution: In making a complete circle, the planet goes a distance of 2πr; if it does soat speed v, then the time that it takes is

T =2πr

v=

2π · 1.5× 1011

3.0× 104 m= 3.1× 107 s

Problem 1063. An airplane propeller is turning at a rate of 40 revolutions per second.What is the propeller’s angular speed? Round your answer to two significant figures.

(a) 6.4 rad/s (b) 13 rad/s(c) 130 rad/s *(d) 250 rad/s(e) None of these

Solution: 1 revolution = 2π rad; so

ω =(40 rev)(2π rad/rev)

1 s= 250 rad/s

Problem 1064. A Ferris wheel takes 40 s to make a complete circle. What is its angularspeed? Round your answer to two significant figures.

(a) 0.025 rad/s *(b) 0.16 rad/s(c) 6.4 rad/s (d) 13 rad/s(e) None of these

If T is the time that it takes to make a complete circle, then

ω =2π rad

T=

2π rad

40 s= 0.16 rad/s

Problem 1065. An electric fan turns at a rate of 24 revolutions per second. How longdoes it take for the fan to turn through 60 radians? Round your answer to two significantfigures.


Solution: Careful! We need to convert revolutions to radians. Since there are 2πradians in a revolution, if f is the angular speed in revolutions/s, then 2πf is the angularspeed in rad/s.

θ = ωt ⇒ t =θ

ω=

θ

2πf=

60 rad

(2π rad/rev)(24 rev/s)= 0.40 s

312

Problem 1066. A wind turbine turns with an angular speed of 1.1 rad/s. How longdoes it take for the turbine to make 10 complete revolutions? Round your answer to twosignificant figures.

(a) 9.1 s (b) 17 s(c) 31 s *(d) 57 s(e) None of these

Solution: We need to convert revolutions to radians, at 2π rad/rev. Then

θ = ωt ⇒ t =θ

ω=

(10 rev)(2π rad/rev)

1.1 rad/s= 57 s

Problem 1067. (Similarity Problem) Two bugs are clinging to a spoke of a turningbicycle wheel. The first bug is 10 cm from the center of the wheel; the second bug is 20cm from the center. If the first bug is experiencing an angular speed of ω1, what is thesecond bug’s angular speed?

(a) ω1/2 *(b) ω1

(c)√

2ω1 (d) 2ω1

(e) None of these

Solution: Both bugs experience the same angular speed, which is that of the wheelitself.

Problem 1068. (Similarity Problem) Two bugs are clinging to a spoke of a turningbicycle wheel. The first bug is 10 cm from the center of the wheel; the second bug is 20cm from the center. If the first bug is moving at a speed of v1, what is the speed of thesecond bug?

(a) v1/2 (b) v1

(c)√

2 v1 *(d) 2v1

(e) None of these

Solution: Both bugs experience the same angular velocity ω, which is that of thewheel. Then v = ωr ⇒ v1 = (10 cm)ω and v2 = (20 cm)ω = 2v1

Problem 1069. (Similarity problem) A physics professor is riding on a moving merry-go-round. At a distance of r1 > 0 from the center, his angular speed is ω1. At whatdistance from the center will his angular speed be 2ω1?

(a)√

2 r1 (b) 2r1

(c) 4r1 *(d) No solution(e) None of these

Solution: No matter where the professor goes on the merry-go-round, his angularvelocity will be the same: that of the merry-go-round itself. Since the merry-go-round ismoving and r1 > 0, ω1 6= 0. Hence there is no distance at which his angular velocity willbe 2ω1 6= ω1.

313

Problem 1070. (Similarity problem) A physics professor is riding on a moving merry-go-round. At a distance of r1 > 0 from the center, his speed is v1. At what distance fromthe center will his speed be 2v1?

(a)√

2 r1 *(b) 2r1

(c) 4r1 (d) No solution(e) None of these

Solution: The professor’s angular velocity is always ω, the angular velocity of themerry-go-round. In the equation v = ωr, v and r change between the two situations; ωdoesn’t. Separating the changing from the unchanging parameters, we get

ω =v

r=v1

r1

=2v1

r2

⇒ r2 =2v1r1

v1

= 2r1

8.2 Fundamental equation 1

Problem 1071. A ceiling fan is initially at rest. When it is switched on, it experiencesan angular acceleration of 2.2 rad/s2. How long does it take before the fan is turning at7.5 rad/s? Round your answer to two significant figures.

(a) 1.6 s *(b) 3.4 s(c) 7.5 s (d) 17 s(e) None of these

Solution: We know ω0 = 0, ω, and α. We want to know t. Use the first fundamentalkinematic equation:

ω = ω0 + αt = αt ⇒ t =ω

α=

7.5 rad/s

2.2 rad/s2 = 3.4 s

Problem 1072. An airplane propeller is initially turning at 220 rad/s. When the engineis switched off, it takes the propeller 18 seconds to come to a complete stop. What is thepropeller’s angular acceleration? Round your answer to two significant figures.

(a) −8.9 rad/s2 (b) −9.9 rad/s2

(c) −11 rad/s2 *(d) −12 rad/s2

(e) None of these

Solution: We know ω0, ω = 0 rad/s, and t. We want to know α. Use the firstfundamental kinematic equation:

ω = 0 = ω0 + αt ⇒ α = −ω0

t= −220 rad/s

18 s= −12 rad/s2

314

Problem 1073. A wheel is initially at rest. It experiences an angular acceleration of13 rad/s2 for 22 s. At the end of this time, what is its angular speed? Round your answerto two significant figures.

*(a) 290 rad/s (b) 950 rad/s(c) 3100 rad/s (d) 6300 rad/s(e) None of these

Solution: We know ω0 = 0, α, and t. We want to know ω. Use the first fundamentalkinematic equation.

ω = ω0 + αt = αt = (13 rad/s2)(22 s) = 290 rad/s

Problem 1074. A wind turbine is turning clockwise at 0.85 rad/s. The wind strength-ens, causing the turbine to experience a clockwise angular acceleration of 0.044 rad/s2.After 15 seconds, how fast is the turbine turning? Round your answer to two significantfigures.

(a) 0.56 rad/s *(b) 1.5 rad/s(c) 4.1 rad/s (d) 6.1 rad/s(e) None of these

Solution: We know ω0, α, and t. We want to know ω. Use the first fundamentalkinematic equation. Careful! Notice that the problem specifies “clockwise”. When yousee a direction specified, you should check the problem with extra care to make sure thatall of your signs are right. There are no negative quantities in this particular problem,but there might be on the exam...

ω = ω0 + αt = 0.85 rad/s + (0.044 rad/s2)(15 s) = 1.5 rad/s

Problem 1075. A disc is spinning at an initial angular speed of 10 rad/s. It is subjectedto an angular acceleration of 2 rad/s2 for 5 seconds. At the end of this time, how fast isthe disc spinning?

(a) 10 rad/s *(b) 20 rad/s(c) 35 rad/s (d) 60 rad/s(e) None of these

Solution: We know ω0, α, and t; we want to know ω. Use the first fundamentalkinematic equation.

ω = ω0 + αt = 10 rad/s + (2 rad/s2)(5 s) = 20 rad/s

315

Problem 1076. A propeller is initially turning at 20 rad/s. When the power to it isincreased, it speeds up, reaching a speed of 48 rad/s after 4 seconds. What was itsangular acceleration?

*(a) 7 rad/s2 (b) 12 rad/s2

(c) 80 rad/s2 (d) 192 rad/s2

(e) None of these

Solution: We know ω0, ω, and t. We want to know α. Use the first fundamentalkinematic equation.

ω = ω0 + αt ⇒ α =ω − ω0

t=

48 rad/s− 20 rad/s

4 s= 7 rad/s2

Problem 1077. A fan is initially turning at 120 rad/s. When it is switched off, itexperiences an angular acceleration of −5 rad/s2. After 10 seconds, how fast is it turning?

(a) 24 rad/s (b) 50 rad/s*(c) 70 rad/s (d) 96 rad/s(e) None of these

Solution: We know ω0, α, and t. We want to know ω. Use the first fundamentalkinematic equation.

ω = ω0 + αt = 120 rad/s + (−5 rad/s2)(10 s) = 70 rad/s

Problem 1078. A circular saw is initially turning at 450 rad/s. It is subjected to anangular acceleration of −30 rad/s2 until it is turning at 300 rad/s. How long does it takebefore it reaches this speed?

(a) 5/3 s *(b) 5 s(c) 10/3 s (d) 10 s(e) None of these

Solution: We know ω0, ω, and α. We want to know t. Use the first fundamentalkinematic equation. Watch your signs!

ω = ω0 + αt ⇒ t =ω − ω0

α=

300 rad/s− 450 rad/s

−30 rad/s2 = 5 s

316

Problem 1079. A windmill is turning at 10 rad/s in a clockwise direction. The windsuddenly shifts, producing an angular acceleration of 0.5 rad/s2 in a counterclockwisedirection. After 12 sec, what is the windmill’s angular velocity?

(a) 6 rad/s counterclockwise (b) 16 rad/s counterclockwise(c) 6 rad/s clockwise *(d) 4 rad/s clockwise(e) None of these

Solution: For no particularly good reason, we’ll take clockwise as the positive direc-tion. We know ω0, α, and t. We want to know ω. Use the first fundamental kinematicequation.

ω = ω0 + αt = 10 rad/s + (−0.5 rad/s2)(12 s) = 4 rad/s

Since ω is positive, it’s clockwise.

Problem 1080. A flywheel is turning at 60 rad/s in a counterclockwise direction. It isthen subjected to an angular acceleration of 5 rad/s2 in a clockwise direction. After 20seconds, what is the flywheel’s angular velocity?

(a) 30 rad/s counterclockwise *(b) 40 rad/s clockwise(c) 100 rad/s clockwise (d) 40 rad/s counterclockwise(e) None of these

Solution: We’ll arbitrarily choose clockwise as the positive direction. We know ω0,α, and t. We want to know ω. Use the first fundamental kinematic equation.

ω = ω0 + αt = −60 rad/s + (5 rad/s2)(20 s) = 40 rad/s

Since ω is positive, it’s turning clockwise.

Problem 1081. Starting from rest, a wheel is subjected to an angular acceleration of3.5 rad/s2 for 5 seconds; after that, it turns at a constant angular speed. At that constantspeed, how long does it take for the wheel to make one revolution? Round your answerto two significant figures.


Solution: We’ll need to do this in two parts: first, calculate the final angular velocity;then calculate the period (the time that it takes for the wheel to make one revolution).We know ω0 = 0, α and t. We can get ω from the first fundamental kinematic equation,then calculate the period using 2π = ωT .

ω = ω0 + αt = αt

2π = ωT ⇒ T =2π

ω=

2π

αt=

2π

(3.5 rad/s2)(5 s)= 0.36 s

317

Problem 1082. A shaft is initially turning at 50 revolutions per second. When a brakeis applied to it, it slows down and comes to a halt after 5.0 seconds. What angularacceleration does the shaft experience while it is slowing down? Round your answer totwo significant figures.

(a) −3.2 rad/s2 (b) −6.4 rad/s2

(c) −31 rad/s2 *(d) −63 rad/s2

(e) None of these

Solution: We need to convert revolutions per second to radians per second, using 2πrad/revolution. Then we’ll know ω0, ω = 0, and t; we want to know α. Use the firstfundamental kinematic equation. We’ll use f (for “frequency”) to signify the revolutionsper second.

ω = 0 = ω0 + αt ⇒ α = −ω0

t= −2πf0

t= −(2π rad/rev)(50 rev/s)

5.0 s= −63 rad/s2

Problem 1083. A merry-go-round is turning at a constant rate, with a period (thetime that it takes for it to make one complete revolution) of 8.0 seconds. The operatorincreases the power to it, applying an angular acceleration of 0.25 rad/s for 5.0 seconds.At the end of this time, what is the merry-go-round’s period? Round your answer to twosignificant figures.


Solution: We will need to convert back and forth between period and angular speed.Since a complete circle is 2π radians, the conversion is ω = 2π/T , where T is the period.When we’ve done that with the initial period T0, we’ll know ω0, α, and t; we’ll use thoseto calculate the final angular speed ωf , which we’ll then convert to the final period Tf .From the first fundamental kinematic equation,

ωf = ω0 + αt =2π

T0

+ αt

We convert ωf to Tf :

Tf =2π

ωf=

2π2π

T0

+ αt=

2πT0

2π + αtT0

=2π(8 s)

2π + (0.25 rad/s2)(5.0 s)(8.0 s)= 3.1 s

318

Problem 1084. A Ferris wheel is turning at a rate of one revolution every 10 seconds.The operator applies a brake to it for 8 seconds, after which it is turning at a rate of onerevolution every 15 seconds. What angular acceleration does the Ferris wheel experiencewhile it is slowing down? Round your answer to two significant figures.

(a) −0.011 rad/s2 *(b) −0.026 rad/s2

(c) −0.064 rad/s2 (d) −0.16 rad/s2

(e) None of these

Solution: We will need to convert back and forth between period and angular speed.Since a complete circle is 2π radians, the conversion is ω = 2π/T , where T is the period.When we’ve done that with the initial period T0 and the final period Tf , we’ll know ω0,ωf , and t; we’ll use these to calculate α. From the first fundamental kinematic equation:

ωf = ω0 + αt

⇒ α =ωf − ω0

t=

2π

Tf− 2π

T0

t=

2π(T0 − Tf )T0Tf t

=2π(10 s− 15 s)

(10 s)(15 s)(8 s)= −0.026 rad/s2


Problem 1085. Starting from rest, a flywheel is subjected to an angular acceleration of2.5 rad/s2. After it has turned through 11 radians, how fast is it turning? Round youranswer to two significant figures.

*(a) 7.4 rad/s (b) 8.2 rad/s(c) 9.0 rad/s (d) 9.9 rad/s(e) None of these

Solution: We know ω0 = 0, θ0 = 0, θ, and α. We want to know ω. We neither knownor care about the time. Use the second fundamental kinematic equation.

ω2 − ω20 = 2α(θ − θ0)

⇒ ω2 = 2αθ ⇒ ω =√

2αθ =

√2(2.5 rad/s2)(11 rad) = 7.4 rad/s2

319

Problem 1086. A windmill is initially held at rest by a brake. When the brake isreleased, it is subjected to an angular acceleration of 3 rad/s2. What angle will it turnthrough before it reaches an angular speed of 12 rad/s?

(a) 4 rad (b) 12 rad(c) 16 rad *(d) 24 rad(e) None of these

Solution: We know ω0 = 0, α, θ0 = 0, and ω. We want to know θ. We neither knownor care about time. Use the second fundamental kinematic equation.

ω2 − ω20 = 2α(θ − θ0) ⇒ ω2 = 2αθ

⇒ θ =ω2

2α=

(12 rad/s)2

2(3 rad/s2)= 24 rad

Problem 1087. A physics instructor is riding on a merry-go-round, which is initiallyturning at 3 rad/s. He ignites a rocket that he has attached to the merry-go-round,producing an angular acceleration of 8 rad/s2. How fast will the merry-go-round beturning after it has turned through 15 rad? Round your answer to the nearest rad/s.

(a) 12 rad/s (b) 13 rad/s(c) 15 rad/s *(d) 16 rad/s(e) None of these

Solution: We know ω0, θ0 = 0, α, and θ. We want to know ω. We neither know norcare about the time. Use the second fundamental kinematic equation.

ω2 − ω20 = 2α(θ − θ0) = 2αθ

⇒ ω =√ω2

0 + 2αθ =

√(3 rad/s)2 + 2(8 rad/s2)(15 rad) = 16 rad/s

Problem 1088. A wheel is turning at a rate of 12 rad/s. When a brake is applied toit, it experiences an angular acceleration of α. If the wheel turns through 9 rad beforecoming to a halt, what is α?

(a) −1/3 rad/s2 (b) −4/3 rad/s2

(c) −2 rad/s2 *(d) −8 rad/s2

(e) None of these

Solution: We know ω0, ω = 0, θ0 = 0, and θ. We want to know α. We neither knownor care about t. Use the second fundamental kinematic equation.

ω2 − ω20 = 2α(θ − θ0) ⇒ −ω2

0 = 2αθ

⇒ α = −ω20

2θ= −(12 rad/s)2

2(9 rad)= −8 rad/s2

320

Problem 1089. A propeller is turning at 30 rad/s. The power to it is increased, sothat after it has turned through 20 radians, it is turning at 45 rad/s. What angularacceleration did the propeller experience? Round your answer to 2 significant figures.

(a) 21 rad/s2 (b) 23 rad/s2

(c) 25 rad/s2 *(d) 28 rad/s2

(e) None of these

Solution: We know ω0, ω, θ0 = 0, and θ. We want to know α. We neither know norcare about t. Use the second fundamental kinematic equation.

ω2 − ω20 = 2α(θ − θ0) = 2αθ

⇒ α =ω2 − ω2

0

2θ=

(45 rad/s)2 − (30 rad/s)2

2(20 rad)= 28 rad/s2

Problem 1090. A shaft is initially turning at 10 rad/s. A brake is applied, producingan angular acceleration of −0.5 rad/s2. What angle does the shaft turn through beforeit has slowed down to 6 rad/s?

(a) 36 rad (b) 45 rad(c) 56 rad *(d) 64 rad(e) None of these

Solution: We know ω0, ω, α, and θ0 = 0. We want to know θ. We neither know norcare about t. Use the second fundamental kinematic equation.

ω2 − ω20 = 2α(θ − θ0) = 2αθ

⇒ θ =ω2 − ω2

0

2α=

(6 rad/s)2 − (10 rad/s)2

2(−0.5 rad/s2)= 64 rad

Problem 1091. A windmill is initially spinning at a constant rate. The wind abruptlydies, slowing the windmill down at a rate of 0.25 rad/s2. After the wind dies, the windmillturns through 30 rad before stopping. What was the windmill’s initial angular speed?Round your answer to two significant figures.


Solution: Watch your signs! We want ω0 to be positive, so α will be negative. Weknow α, θ0 = 0, θ, and ω = 0. We want to know ω0. We neither know nor care about t.Use the second fundamental kinematic equation.

ω2 − ω20 = 2α(θ − θ0) ⇒ ω2

0 = −2αθ

ω =√−2αθ =

√−2(−0.25 rad/s2)(30 rad) = 3.9 rad/s

321

Problem 1092. A steam turbine is initially turning at a constant rate. The steampressure to it is increased, producing a positive angular acceleration of 20 rad/s2. Afterit has turned through 30 rad, it has an angular speed of 50 rad/s. What was its initialangular speed? Round your answer to two significant figures.


Solution: We know θ0 = 0, θ, α, and ω. We want to know ω0. We neither know norcare about t. Use the second fundamental kinematic equation.

ω2 − ω20 = 2α(θ − θ0) = 2αθ ⇒ ω2 − 2αθ = ω2

0

⇒ ω0 =√ω2 − 2αθ =

√(50 rad/s)2 − 2(20 rad/s2)(30 rad) = 36 rad/s

Problem 1093. A wind turbine is initially spinning at a constant rate, making onerevolution every 4.5 seconds. The wind increases, subjecting the turbine to an angularacceleration of 0.2 rad/s. After the turbine has completed 3 revolutions, it is once againturning at a constant rate. What is its new period of revolution? Round your answer tothe nearest 0.1 s.


Solution: We need to convert between the period T (the time that it takes for theturbine to make one complete circle) and the angular speed ω. Since one complete circleis 2π rad, ωT = 2π. We know the initial period T0 (from which we can get ω0), theangular acceleration, θ0 = 0, and θ = (3 rev)(2π rad/rev) = 6π rad. We want the finalperiod Tf , which we can get from the final angular speed ωf . We neither know nor careabout t. Use the second fundamental kinematic equation.

ω2f − ω2

0 = 2α(θ − θ0) = 2αθ

⇒ ω2f = ω2

0 + 2αθ

⇒ 4π2

T 2f

=4π2

T 20

+ 2αθ =4π2 + 2αθT 2

0

T 20

⇒ T 2f =

4π2T 20

4π2 + 2αθT 20

⇒ Tf =2πT0√

4π2 + 2αθT 20

=2π(4.5 s)√

4π2 + 2(0.2 rad/s2)(6π rad)(4.5 s)2

= 2.0 s

322

Problem 1094. A Ferris wheel is turning at a rate of one revolution every 8.0 seconds.The operator applies a brake while the wheel makes two revolutions; at the end of thistime, it is turning at a rate of one revolution every 15.0 sec. What is the magnitude ofthe angular acceleration produced by the brake? Round your answer to two significantfigures. α?

*(a) 0.018 rad/s2 (b) 0.11 rad/s2

(c) 6.4 rad/s2 (d) 40 rad/s2

(e) None of these

Solution: We need to convert between the period T (the time that it takes to makeone complete revolution) and the angular speed ω. Since one revolution is 2π radians,ωT = 2π. We know the inital period T0, from which we can get ω0; the final period, Tf ,from which we can get ωf ; and the number of revolutions turned, from which we can getthe angle ∆θ = (2 revolutions)(2π rad/rev) = 4π rad. We want to know α. We neitherknow nor care about t. Use the second fundamental kinematic equation.

2α∆θ = ω2f − ω2

0 =4π2

T 2f

− 4π2

T 20

=4π2(T 2

0 − T 2f )

T 2f T

20

⇒ α =2π2(T 2

0 − T 2f )

T 2f T

20 ∆θ

=2π2[(8.0 s)2 − (15.0 s)2]

(15.0 s)2(8.0 s)2(4π rad)= −0.018 rad/s2

Since the question asks for the magnitude of α, we use the absolute value: 0.018 rad/s2.


Problem 1095. A fan is initially turning at 12 rad/s. The power to the fan is increased,applying a rotational acceleration of 3 rad/s2. What angle does the fan turn throughover the next 2 sec?

*(a) 30 rad (b) 36 rad(c) 42 rad (d) 45 rad(e) None of these

Solution: We know θ0 = 0, ω0, α, and t. We want to know θ. Use the thirdfundamental kinematic equation.

θ = θ0 + ω0t+1

2αt2 = (12 rad/s)(2 s) +

1

2(3 rad/s2)(2 s)2 = 30 rad

323

Problem 1096. Starting at rest, a wheel experiences an angular acceleration of 2 rad/s2.How much time elapses before the wheel has turned through 16 rad?

(a) 2 s (b)√

8 s*(c) 4 s (d) 8 s(e) None of these

Solution: We know θ0 = 0, ω0 = 0, α, and θ. We want to know t. Use the thirdfundamental kinematic equation.

θ = θ0 + ω0t+1

2αt2 =

1

2αt2 ⇒ t =

√2θ

α=

[2(16 rad)

2 rad/s2

]1/2

= 4 s

Problem 1097. A shaft is turning at 15 rad/s. A brake is applied, producing an angularacceleration of −0.25 rad/s2. What angle does the shaft turn through over the next 4seconds?

(a) 49 rad (b) 52 rad*(c) 58 rad (d) 62 rad(e) None of these

Solution: We know θ0 = 0, ω0, α, and t. We want to know θ. Use the thirdfundamental kinematic equation.

θ = θ0 + ω0t+1

2αt2 = (15 rad/s)(4 s) +

1

2(−0.25 rad/s2)(4 s)2 = 58 rad

Problem 1098. A propeller is spinning at 40 rad/s. When the power is switched off, itexperiences a constant angular acceleration α. In the next 3 seconds, the propeller turnsthrough 60 rad. What is α? Round your answer to the nearest rad/s2.

(a) −10 rad/s2 (b) −11 rad/s2

(c) −12 rad/s2 *(d) −13 rad/s2

(e) None of these

Solution: We know θ0 = 0, θ, ω0, and t. We want to know α. Use the thirdfundamental kinematic equation.

θ = θ0 + ω0t+1

2αt2 = ω0t+

1

2αt2

⇒ α =2(θ − ω0t)

t2=

2[60 rad− (40 rad/s)(3 s)]

(3 s)2= −13 rad/s2

324

Problem 1099. A windmill is turning at 12 rad/s. When the wind abruptly dies, itslows down at a rate of 3 rad/s2. How long does it take for the windmill to turn through16 rad? Round your answer to two significant figures.


Solution: We know θ0 = 0, ω0, α, and θ. We want to know t. Use the thirdfundamental kinematic equation. Watch your signs; since the windmill is slowing down,we’ll take α as negative.

θ = θ0 + ω0t+1

2αt2 = ω0t+

1

2αt2 ⇒ α

2t2 + ω0t− θ = 0

Since the solution choices don’t include any integers, it’s likely that we won’t be able tofactor this. We’ll use the quadratic formula.

t =−ω0 ±

√ω2

0 − 4(α/2)(−θ)2(α/2)

=−ω0 ±

√ω2

0 + 2αθ

α

=−12 rad/s±

√(12 rad/s)2 + 2(−3 rad/s2)(16 rad)

−3 rad/s2

= 1.7 s or 6.3 s

There are two positive solutions; we use the smaller one. Why? The equation doesn’t“know” that the windmill will slow down and remain stopped; as far as it’s concerned,the angular acceleration of −3 m/s2 will continue forever. If that were the case, thenthe windmill would turn through 16 rad as it slowed down; would come to a halt andthen reverse its direction; and would then speed up in that backward direction, passingthrough the 16-rad mark a second time as it turned backward.

Problem 1100. A turbine is turning at some constant angular velocity. The power to itis increased, producing an angular acceleration of 2.0 rad/s2. Over the next 10 seconds,the turbine turns through 300 radians. What was its original angular velocity? Roundyour answer to two significant figures.


Solution: We know α, θ0 = 0, θ, and t. We want to know ω0. Use the thirdfundamental kinematic equation.

θ = θ0 + ω0t+1

2αt2 = ω0t+

1

2αt2

⇒ ω0 =2θ − αt2

2t=

2(300 rad)− (2 rad/s2)(10 s)2

2(10 s)= 20 rad/s

325

Problem 1101. A winch is turning clockwise at 3 rad/s. The operator reverses the powerto it, producing a counterclockwise angular acceleration of 0.50 rad/s2 for 10 seconds.At the end of this time, what is the net angle through which the winch has turned?

*(a) 5 rad clockwise (b) 5 rad counterclockwise(c) 55 rad clockwise (d) 55 rad counterclockwise(e) None of these

Solution: We know θ0 = 0, ω0, α, and t. We want to know θ. Use the thirdfundamental kinematic equation. We’ll use clockwise as the positive direction, so α willbe negative.

θ = θ0 + ω0t+1

2αt2 = (3 rad/s)(10 s) +

1

2(−0.50 rad/s2)(10 s)2 = 5 rad

Since this is positive, it’s 5 rad clockwise.

Problem 1102. A fan is turning at some constant angular velocity. It is subjected to aclockwise angular acceleration of 0.4 rad/s2 for 6 seconds, during which it turns 8 radiansclockwise. What was its initial angular velocity? Round your answer to two significantfigures.

*(a) 0.13 rad/s clockwise (b) 0.50 rad/s clockwise(c) 0.13 rad/s counterclockwise (d) 0.50 rad/s counterclockwise(e) None of these

Solution: We know θ0 = 0, θ, α, and t. We want to know ω0. Use the fourthfundamental kinematic equation. We’ll use clockwise as the positive direction.

θ = θ0 + ω0t+1

2αt2 = ω0t+

1

2αt2

⇒ ω0 =2θ − αt2

2t=

2(8 rad)− (0.4 rad/s2)(6 s)2

2(6 s)= 0.13 rad/s

Since this number is positive, it’s 0.13 rad/s clockwise.

326

Problem 1103. A windmill is initially turning at a rate of one revolution per 4.8 seconds.The wind strengthens, applying an angular acceleration of 0.2 rad/s2. Over the next8 seconds, what angle does the windmill turn through? Round your answer to twosignificant figures.

(a) 6.4 rad (b) 10 rad*(c) 17 rad (d) 27 rad(e) None of these

Solution: We need to convert from the period T (the time that it takes for thewindmill to make one revolution) to the angular speed. Since one revolution equals 2πrad, ωT = 2π. Then we know T0, from which we can get ω0; θ0 = 0; α; and t. We wantto know θ. Use the third fundamental kinematic equation.

θ = ω0t+1

2αt2 =

2πt

T0

+1

2αt2 =

2π(8 s)

4.8 s+

(0.2 rad/s2)(8 s)2

2= 17 rad

Problem 1104. A merry-go-round is turning at a rate of one revolution per 5.5 sec. Theoperator applies a brake, producing an angular acceleration α. In the next 4.0 seconds,the merry-go-round turns half a revolution. What is the magnitude of α? Round youranswer to two significant figures.

(a) 0.098 rad/s2 *(b) 0.18 rad/s2

(c) 0.59 rad/s2 (d) 0.93 rad/s2

(e) None of these

Solution: We need to convert from revolutions to radians, and from the period T (thetime that it takes to make one revolution) to the angular speed. Since one revolution is2π rad, ωT = 2π. We know the initial period T0, from which we can calculate ω0; θ0 = 0;θf = 1/2 revolution = π rad; and t. We want to know α. Use the third fundamentalkinematic equation.

θf = ω0t+1

2αt2

⇒ α =2(θf − ω0t)

t2=

2(θf − 2πt/T0)

t2=

2(θfT0 − 2πt)

T0t2

=2[(π rad)(5.5 s)− 2π(4.0 s)]

(5.5 s)(4.0 s)2= −0.18 rad/s2

Since we’re asked for the magnitude, we use the absolute value: 0.18 rad/s2.

327

Problem 1105. A water wheel is turning at a rate of one revolution per 7.5 seconds.The current to the wheel increases, applying an angular acceleration of 0.5 rad/s. Howlong does it take for the wheel to turn through two revolutions? Round your answer totwo significant figures.


Solution: We need to convert the period T (the time that it takes to make onerevolution) to the angular speed ω. Since one revolution is 2π rad, ωT = 2π. We knowthe initial period T0, from which we can calculate ω0; α; and θ = 4π rad. We want toknow t. Use the third fundamental kinematic equation.

θ = ω0t+1

2αt2 ⇒ 1

2αt2 + ω0t− θ = 0

⇒ t =−ω0 ±

√ω2

0 − 4(α/2)(−θ)2(α/2)

=−ω0 ±

√ω2

0 + 2αθ

α=−2π ±

√4π2 + 2αθT 2

0

αT0

=−2π ±

√4π2 + 2(0.5 rad/s2)(4π)(7.5 s)2

(0.5 rad/s2)(7.5 s)= 5.6 s

There’s also a negative root, but we ignore it.

Problem 1106. An evil physics instructor tries to fail as many students as possibleby speeding up the clock during an exam. He secretly installs a device that applies aconstant angular acceleration α to the minute hand. At the beginning of the test, thehand is moving at the correct speed; because of the angular acceleration, it makes acomplete circle in only 40 minutes. What is the value of α? Round your answer to twosignificant figures.

(a) 8.7× 10−8 rad/s2 *(b) 7.3× 10−7 rad/s2

(c) 6.6× 10−5 rad/s2 (d) 4.2× 10−4 rad/s2

(e) None of these

Solution: We need to convert revolutions and minutes and hours to radians andseconds. We can easily determine ω0, θ, and t. We want α. Use the first fundamentalkinematic equation.

θ = ω0t+1

2αt2 ⇒ α =

2(θ − ω0t)

t2

We can calculate ω0 from the fact that at the correct speed, a minute hand circlesthe clock in T0 = 1 hr = 3600 s; so ω0 = 2π/T0. We need to convert t to seconds:t = 40 min = 2400 s. The minute hand makes a complete circle, so θ = 2π rad. Hence

α =2(θ − 2πt/T0)

t2=

2(θT0 − 2πt)

t2T0

=2[2π(3600 s)− 2π(2400 s)]

(2400 s)2(3600 s)= 7.3× 10−7 rad/s2

328

9 Two-dimensional kinematics

9.1 The basics of velocity and acceleration in two dimensions

9.1.1 The difference between average speed and average velocity in a plane

Problem 1107. (Speed vs. Velocity) A race car goes around a track at a constantspeed of 200 mph (It has cruise control). What is the car’s average speed and velocityover one lap on the race track?

Solution: The average speed is 200 mph, but the average velocity is zero! The initialand final positions are the same, so

∆x = xf − xi = xi − xi = 0 ⇒ vave =∆x

∆t.

The point of this exercise is to emphasize that one must be careful when using long-timeaverages with vector quantities. The average velocity may give you a result that youwere not intending to get.

Problem 1108. (Speed vs. Velocity) If we throw a ball straight up in the air andit comes back down to the same spot, what is the average velocity of the ball over onecomplete trip up and back down?

Solution: The average velocity is zero. The initial and final positions are the same,so

∆y = yf − yi = yi − yi = 0 ⇒ vave =∆y

∆t.

The point of this exercise is to emphasize that one must be careful when using averagevelocity, because you may get a result that you were not intending to get.

329

9.1.2 Average acceleration in a plane

Problem 1109. A circular room has a radius of 4 m and a small window facing north.A fly moving at 6 m/s flies directly south through the window and into the room. Overthe next 20 s, he flies three times counterclockwise around the perimeter of the room,then flies directly north out the window at 6 m/s. What are the magnitude and directionof the fly’s average acceleration between the time he enters the room and the time heleaves?

Solution: Since we asked for the magnitude and direction of the average acceleration,we must’ve meant the acceleration vector. (Unlike “velocity” and “speed”, we don’t havea convenient way to distinguish between the acceleration vector and its magnitude.) Theonly data that matter are the fly’s initial and final velocity, and the time over which itchanged: from 6 m/s southward to 6 m/s northward, over 20 s. The difference betweenthe final and initial velocity is 12 m/s northward; the change occurred over 20 s, so theaverage acceleration is

12 m/s

20 s= 0.6 m/s2 northward

9.1.3 Relative velocity in a plane

Problem 1110. A river has a current with a speed of 5 ft/s. You can swim at 4 ft/sin still water. At what angle to the cross-river direction do you need to swim to reach apoint directly across the river from you?

Solution: No solution. Whatever the angle is, your upstream component of velocitycan be no greater than the total magnitude of your velocity: 4 ft/s. Thus you will findyourself going downstream at 1 ft/s or more.

Problem 1111. A train is going directly north at 70 mph. An action-movie star is atopthe train, running southward at 12 mph. Relative to someone standing on the ground,how fast is the action-movie star moving, and in what direction?

Solution: Relative to someone on the ground, the AMS is moving northward at70− 12 = 58 mph.

330

9.1.4 Graphical interpretation of velocity and acceleration

Problem 1112. On the graph below, draw the velocity vectors at the points P1 and P2

on the curve. Assume the particle is moving with uniform speed from left to right.

-

6

x

y

rP1

rP2

��

��

Solution: Since the particle is moving with uniform speed, your vectors shouldhave the same length. Since it’s moving from left to right, they should have positivex-components—that is, they should point up and to the left, not down and to the right.

Problem 1113. The graph atright shows the top view of thepath of a particle moving along ahorizontal surface. The particle ismoving from the lower left to theupper right. Which of the labelledvectors is the velocity vector atpoint P?

(a) ~A *(b) ~B

(c) ~C (d) ~D(e) None of these -

6

x

y

sP��~B

��

~D

@@@I

~A

@@@R~C

Solution: The velocity vector is tangent to the curve in the direction of motion.

331

Problem 1114. The graph atright shows the path of a particlemoving from the lower left to theupper right. Which of the labelledvectors is the acceleration vectorat point P?

(a) ~A (b) ~B


Solution: The accelerationvector always points into the turn(on the concave side of the curve).

-

6

x

y

sP��~B

��

~D

@@@I

~A

@@@R~C

332

9.2 Uniform circular motion

Acceleration in uniform circular motion:

arad =v2

r,

where v is the tangential speed of the object and r is the radius.

Speed in uniform circular motion:

v = v =circumference of circle

total time=

2πr

Tperiod

= 2π r f ,

where the frequency f is defined as f ≡ 1

Tperiod

.

Problem 1115. A freeway on-ramp has a curve with a radius of 59 m. You attemptto drive around the curve at 16 m/s. What is the centripetal acceleration on your car?Round your answer to the nearest 0.1 m/s2.

*(a) 4.3 m/s2 (b) 4.8 m/s2

(c) 5.3 m/s2 (d) 5.8 m/s2

(e) None of these

Solution: a =v2

r=

(16 m/s)2

59 m= 4.3 m/s2

Problem 1116. A ceiling fan has blades 81 cm long. A spider is clinging to the tip ofone of the blades, which are turning so that the spider moves at 10.2 m/s. What is thespider’s radial acceleration? Round your answer to the nearest m/s2.

(a) 105 m/s2 (b) 117 m/s2

*(c) 128 m/s2 (d) 143 m/s2

(e) None of these

Solution: Don’t forget to convert 81 cm to 0.81 m.

a =v2

r=

(10.2 m/s)2

0.81 m= 128 m/s2

333

Problem 1117. A toddler’s parents have placed him on a merry-go-round, 2.7 m fromthe center. If the toddler experiences a radial acceleration greater than 1.2 m/s2, it willfall off, prompting the angry parents to sue the merry-go-round operator. How fast canthe operator allow the toddler to go without fear of legal action? Round your answer tothe nearest 0.1 m/s.


Solution: We know r and a; we want to know v.

a =v2

r⇒ v =

√ar =

√(1.2 m/s2)(2.7 m) = 1.8 m/s

Problem 1118. A curve in a highway has a radius of 93 m. A truck will roll over ifit experiences a centripetal acceleration greater than 1.9 m/s2. How fast can the truckdrive around the curve? Round your answer to the nearest m/s.


Solution: We know a and r; we want to know v.

a =v2

r⇒ v =

√ar =

√(1.9 m/s2)(93 m) = 13 m/s

Problem 1119. You are designing a highway that will include a curve. Cars will skidoff the curve if they experience a centripetal acceleration greater than 2.2 m/s2. Youwant cars to be able to drive the curve safely at 20 m/s. What must the radius of thecurve be? Round your answer to the nearest 10 m.


Solution: We know a and v; we want to know r.

a =v2

r⇒ r =

v2

a=

(20 m/s)2

2.2 m/s2 = 180 m

334

Problem 1120. An airplane is moving at 330 m/s when the pilot realizes that he isgoing in the wrong direction and needs to make a U-turn. If the pilot experiences aradial acceleration greater than 49 m/s2, he will lose consciousness and crash the plane.What is the minimum safe radius of the turn? Round your answer to the nearest 100 m.


Solution: We know a and v; we want to know r.

a =v2

r⇒ r =

v2

a=

(330 m/s)2

49 m/s2 = 2200 m

Problem 1121. A bug is standing atop the blade of a ceiling fan, which is turning at1.7 revolutions per second. The bug will lose its grip on the fan and go sailing off intothe room if it experiences a radial acceleration greater than 91 m/s2. How far from thecenter of the fan can the bug go without losing its hold? Round your answer to thenearest centimeter.


Solution: We can’t use our formula directly, since we know a but we don’t know ror v. However, there is a relationship between r and v, and we can substitute that intoour formula. If the bug is standing at a distance r from the center of the fan, then everytime the fan makes a complete revolution, the bug travels a distance of 2πr. We knowthat the fan is turning at a frequency of f = 1.7 revolutions/second. Hence the bug’sspeed is v = 2πrf . Now we can use the formula:

a =v2

r=

4π2r2f 2

r= 4π2rf 2 ⇒ r =

a

4π2f 2=

91 m/s2

4π2(1.7 s−1)2= 0.80 m = 80 cm

Problem 1122. A windmill has three blades, each 37 m long; the blades make a completecircle every 5.2 s. A bug is clinging to the tip of one of the blades. How much radialacceleration does the bug experience? Ignore the effect of gravity.

Solution: The radius of the circle traced by the bug is r = 37 m. The bug makesa complete circle in t = 5.2 s. Hence the bug’s speed is v = 2πr/t. Its centripetalacceleration is

ac =v2

r=

4π2r2

rt2=

4π2r

t2=

4π2(37 m)

(5.2 s)2= 54 m/s2

335

Problem 1123. A windmill has three blades, each 8.3 m long; the blades make a com-plete circle every 4.1 s. A bug is clinging to the tip of one of the blades. How much radialacceleration does the bug experience? Ignore the effect of gravity. Round your answerto the nearest m/s2.

(a) 10 m/s2 (b) 0.1 m/s2

(c) 6 m/s2 *(d) 19 m/s2

(e) None of these

Solution: The bug is moving in a circle with radius r = 8.3 m, going around it onceevery T = 4.1 s. The circumference of the circle is 2πr; so the bug’s speed is v = 2πr/T .The centripetal acceleration is

a =v2

r=

4π2r2

T 2r=

4π2r

T 2=

4π2(8.3 m)

(4.1 s)2= 19 m/s2

Problem 1124. A ceiling fan is turning at a rate of 1.8 revolutions per second. A spideris clinging to the blade of the fan. If he experiences centripetal acceleration greater than2.7 g, he will lose his grip on the blade and go sailing off. How far from the center of thefan can the spider safely go? Give your answer in centimeters.

Solution: Since we want the answer in centimeters, we will convert: g = 980 cm/s2.

If the spider is at a distance r from the center of the fan, which is turning at f revolutionsper second, his speed is

v = 2πrf

The centripetal acceleration is

ac =v2

r=

4π2r2f 2

r= 4π2rf 2

We know ac and f , and we want to find r:

r =ac

4π2f 2=

(2.7)(980cm/s2)

4π2(1.8/s)2= 21 cm

Problem 1125. A highway has a curve with radius of curvature 72 m. A truck will skidoff the highway if it experiences a radial acceleration of more than 2.0 m/s2. How fastcan the truck safely go on the curve?

Solution: The centripetal acceleration is

ac =v2

r

Solving this for v yields

v =√acr =

√(2.0 m/s2)(72 m) = 12 m/s

336

9.2.1 Similarity problems

For all vehicles-going-around-circular-curve problems: Although it is not obviousat this point, the vehicle’s ability to stay on the road depends on the centripetal forceprovided by the friction and the tires. It turns out that there is a maximum centripetalforce that the tires can supply to the road to keep the car from slidding. There is adirect correlation between force and acceleration, so it follows that there is a maximumcentripetal acceleration for which the car can remain on the road. If this centripetalacceleration is exceeded, then the car will begin to slid.

Problem 1126. (Similarity Problem) Two bugs are clinging to the second hand ofa very large clock: one 3 ft from the center of the clock, the other 6 ft from the center.If the first bug experiences a radial acceleration of a, what radial acceleration does thesecond bug experience?

Solution: The centripetal acceleration of a body in circular motion is

a =v2

r

Be careful! Don’t forget that the two bugs are moving at different speeds. Since thesecond bug is twice as far from the center of the clock, he is moving twice as fast as thefirst bug. The second bug’s centripetal acceleration is

(2v)2

2r=

2v2

r= 2a

Problem 1127. (Similarity Problem) A highway has a curve with radius r1. Largetrucks must drive the curve at a speed of v1 or less to keep from skidding off the road.What would the radius of the curve have to be in order for trucks to drive it safely atαv1, where α > 0 is a real number?

(a)√α r1 *(b) α2r1

(c) α√α r1 (d) αr1

(e) None of these

Solution: In the first case we are given v1 is the maximum speed of the truck ona curve with radius r1. In the second case let v2 be the maximum speed of the truckon a curve with radius r2. We want to solve for r2 in the case v2 = αv1. The equationgoverning uniform circular motion is ac = v2/r. Since the centripetal acceleration mustbe the same for both cases, we can relate the two cases to one another via the centripetalacceleration ac as follows:

v21

r1

= ac =v2

2

r2

⇒ r2 = r1

(v2

v1

)2

= r1

(αv1

v1

)2

= α2r1

337

Problem 1128. (Similarity Problem) A highway has a curve with a radius of r1.Large trucks must drive the curve at a speed of v1 or less to keep from skidding off theroad. If the highway were rebuilt so that the curve had twice radius, how fast couldtrucks safely drive around the curve?

*(a)√

2 v1 (b) 2v1

(c) 2√

2 v1 (d) 4v1

(e) None of these

Solution: We are given: r1, v1, and r2 = 2r1, and we want v2. Since the centripetalacceleration ac must be the same for both situations the governing equation is alreadyin the standard similarity form: v2/r = ac. Equating the two situations through thisequation, and using the centripetal acceleration as the common link gives:

v21

r1

= ac =v2

2

r2

·r2−→ v22 = v2

1

r2

r1

= 2v21

√−−−−−−→ v2 =

√2v1

Problem 1129. (Similarity Problem) A highway has a curve with radius r1. Largetrucks must drive the curve at a speed of 20 mph or less to keep from skidding off theroad. What would the radius of the curve have to be in order for trucks to drive it safelyat 60 mph?

(a)√

3 r1 (b) 3r1

(c) 3√

3 r1 *(d) 9r1

(e) None of these

Solution: Let v1 = 20 mph be the maximum speed of the truck on a curve with radiusr1. The governing equation is in standard similarity form v2/r = ac. The maximumcentripetal acceleration that a truck can experience without sliding off the road is ac =v2

1/r1. Now we want to know how large r2 must be for a truck going at a speed ofv2 = 60 mph = 3v1 to experience an acceleration of ac. Equating the two situationsthrough the common parameter in the governing equation yields

v21

r1

= ac =v2

2

r2

flip eqn−−−−−−−→ r1

v21

=r2

v22

·v22−−−−−→ r2 = r1

(v2

v1

)2

= 9r1

Problem 1130. (Similarity Problem) A highway has a quarter-circle curve with aradius of r. Large trucks must take the curve at a speed of 25 mph or less to avoidskidding off the road. What would the radius of the curve have to be in order for trucksto drive around it safely at 50 mph?

Solution: The centripetal acceleration of a body in circular motion is

a =v2

r

We want to double the value of v and change the value of the radius so that a remainsthe same. Thus we need to increase the radius to 4r.

STOP

338

9.3 Projectile motion

If ~A = Axı+Ay , then from the definition of the trig functions about the right triangle:

A =

∥∥∥ ~A∥∥∥ =√A2x + A2

y

θ0 = tan−1

(AyAx

)

9.3.1 General equations for projectile motion

Let the initial time be t0 = 0, the initial position be (x0, y0), and the initial velocity be~v0 = v0xı+ v0y . Typically we take the initial position to be (x0, y0) = (0, h), where h isthe height of the mechanism doing the launching.

The general equations governing motion in 2-D are the set of fundamental equations in1-D repeated in each of the x and y directions. Substituting (x0, y0) = (0, h) into thefundamental equations (1) and (3) in each coordinate direction, and taking ax = 0 (sincethere is no force acting in the horizontal direction) yields

velocity x-component: vx = v0x (9.1a)

position x-component: x = v0xt (9.1b)

If we take our y-axis pointing upward, then ay = −g and we have

velocity y-component: vy = v0y − gt (9.2a)

position y-component: y = h+ v0yt−1

2gt2 (9.2b)

If the initial velocity is given in terms of magnitude ‖~v0‖ = v0, and direction θ0, asmeasured in the counterclockwise direction from the x-axis, then we must convert thisinformation into information about the components of the vector in order to use theabove formulas.{

v0x = v0 cos θ0 (x-component of initial velocity)

v0y = v0 sin θ0 (y-component of initial velocity)

339

9.3.2 Velocity and acceleration of a projectile

Problem 1131. The following four diagrams show the trajectory of a cannonball. Whichdiagram correctly shows the acceleration vectors at three points along the trajectory?

-

6

x

y(a)

rr r

6 ~0?

-

6

x

y(b)

rr r

��7

-

QQs

-

6

x

y(c)

rr r

ZZ~

? �

-

6

x

y*(d)

rr r

?

??

(e) None of theseSolution: At every point, the acceleration is 9.8 m/s2, directed straight down. Yourvectors should all be of the same length. The answer is (d).

Problem 1132. The following four diagrams show the trajectory of a cannonball. Whichdiagram correctly shows the velocity vectors at three points along the trajectory?

-

6

x

y(a)

rr r

6 ~0?

-

6

x

y*(b)

rr r

��7

-

QQs

-

6

x

y(c)

rr r

ZZ~

? �

-

6

x

y(d)

rr r

?

??

(e) None of these

Solution: Only figure (b) has vectors that are tangent to the curve.

340

Problem 1133. Draw the acceleration vector due to gravity at the points A, B, and Calong the curve of the trajectory of the cannonball.

-

6

x

y rArB

rC?

?

?

Solution: At every point, the acceleration is 9.8 m/s2, directed straight down. Yourvectors should all be of the same length.

341

9.3.3 Special Case I: The Half-Parabola

Note: For the case of the half-parabola the initial height is the maximum height andthe ball is fired horizontally. Setting h = ymax > 0 and θ0 = 0, which leads to v0y = 0and v0x = v0, in the general equations reduces them to

velocity x-component: vx = v0 (9.3a)

position x-component: x = v0t (9.3b)


velocity y-component: vy = −gt (9.4a)

position y-component: y = ymax −1

2gt2 (9.4b)

Note: For this case, it would also make sense to take the y-axis pointing downward.Then the time to fall is just a one-dimensional problem. The time to fall, denoted bytfall, is found from the fundamental equation 3, with the y-axis pointing downward sothat ay = g. Once again v0y = 0, so the equation becomes

ymax =1

2gt2fall ⇒ tfall =

√2ymax

g.

Problem 1134. A skateboarder inadvertently rides off the edge of a flat roof at 8.2 m/s.The roof is 23 m high. How long does it take for the skateboarder to reach the ground?Round your answer to the nearest 0.1 s.


Solution: Since the skateboarder is riding off a flat roof, we assume that his angleof launch is θ0 = 0. Hence there will be no vertical component of his initial velocity.That means that the time it takes for him to reach the ground is the same as if he weredropped from roof level:

t =

√2y

g=

[2(23 m)

9.8 m/s2

]1/2

= 2.2 s

The skateboarder’s initial velocity has nothing to do with this problem.

342

Problem 1135. A bicyclist rides off a flat roof at 13.4 m/s. The roof is 3.6 m abovethe ground. How far from the edge of the building does the bicyclist land? Round youranswer to the nearest meter.


Solution: Since the bicyclist is riding off a flat roof, we assume that the elevation isθ0 = 0. Then the vertical component of the initial velocity is v0y = 0, and the horizontalcomponent is v0x = v = 13.4 m/s. The horizontal velocity does not change, since there’sno horizontal acceleration. Hence the horizontal distance that the bicycle travels is theinitial horizontal velocity times the time it takes for the bicycle to fall to the ground:

x = v0xt = v0x

√2y

g= (13.4 m/s)

[2(3.6 m)

9.8 m/s2

]1/2

= 11 m

Problem 1136. You fire your potato gun horizontally from shoulder height, which is140 cm above the ground. The potato strikes the ground 15 m away from you. Whatwas the initial speed of the potato? Round your answer to the nearest m/s.


Solution: Since you fired the potato gun horizontally, the angle of elevation is θ0 = 0.This means that there is no vertical component to the initial velocity. The potato’s initialvelocity was all horizontal; and since there’s no horizontal acceleration, the horizontalvelocity is constant. The time it takes for the potato to reach the ground is t =

√2y/g,

where y = 1.4 m is the initial height. (Don’t forget to convert 140 cm to 1.4 m.) Duringthis time, the potato travels a horizontal distance of x = 15 m. Hence the initial velocitywas

v0 = v0x =x

t=

x√2y/g

=

√g

2yx =

√9.8 m/s2

2(1.4 m)(15 m) = 28 m/s

343

Problem 1137. You fire your potato gun horizontally from shoulder height, which is150 cm above the ground. The potato strikes the ground 11 m away from you. Whatwas the initial speed of the potato? Round your answer to the nearest m/s.


Solution: Since you fired the potato gun horizontally, the angle of elevation is θ0 = 0.This means that there is no vertical component to the initial velocity. The potato’s initialvelocity was all horizontal; and since there’s no horizontal acceleration, the horizontalvelocity is constant. The time it takes for the potato to reach the ground is t =

√2y/g,

where y = 1.5 m is the initial height. (Don’t forget to convert 150 cm to 1.5 m.) Duringthis time, the potato travels a horizontal distance of x = 11 m. Hence the initial velocitywas

v0 = v0x =xmax

tfall

=xmax√2y/g

=

√g

2yxmax =

√9.8 m/s2

2(1.5 m)(11 m) = 20 m/s

Problem 1138. Your potato gun has a muzzle velocity of 27 m/s. You shoot it horizon-tally from a window 12 m above the ground. How far from the building does the potatostrike the ground? Round your answer to the nearest meter.


Solution: Since you’re shooting the gun horizontally, the vertical component of theinitial velocity is v0y = 0, and the horizontal component is v0x = v = 27 m/s. The rangexmax is the product of the horizontal velocity v0x and the time tfall that it takes for thepotato to reach ground level (i.e., xmax = v0x tfall, where tfall is the time it takes the objectto fall from its initial position at y = y0 to its final position on the ground at y = 0). Tofind tfall, we use the formula y = y0 + v0yt− 1

2gt2. We use y0 = 12 m, y = 0, and t = tfall.

The formula simplifies to: 0 = y0 − 12gt2fall. Solving for tfall yields

tfall =

√2y0

g

Then the range is

x = v0xtfall = v0x

√2y0

g=

(27 m/s)√

2 · 12 m√9.8 m/s2

= 42 m

344

Problem 1139. Your potato gun has a muzzle velocity of 10 m/s. You shoot it horizon-tally from a window 5 m above the ground. How far from the building does the potatostrike the ground? Approximate gravity as g = 10 m/s2. Round your answer to thenearest meter.


Solution: Since you’re shooting the gun horizontally, the vertical component of theinitial velocity is v0y = 0, and the horizontal component is v0x = v = 27 m/s. The rangexmax is the product of the horizontal velocity v0x and the time tfall that it takes for thepotato to reach ground level (i.e., xmax = v0x tfall, where tfall is the time it takes the objectto fall from its initial position at y = y0 to its final position on the ground at y = 0). Tofind tfall, we use the formula y = y0 + v0yt− 1

2gt2. We use y0 = 12 m, y = 0, and t = tfall.

The formula simplifies to: 0 = y0 − 12gt2fall. Solving for tfall yields

tfall =

√2y0

g

Then the range is

x = v0xtfall = v0x

√2y0

g= (10 m/s)

√2 · 5 m

10 m/s2 = 10 m

Problem 1140. (Similarity Problem) You shoot your potato gun horizontally fromshoulder height, which is 150 cm. The potato lands 25 m away from you. You wouldlike the potato to go 75 m, so you climb up a ladder and fire the gun horizontally fromthere. How high off the ground must the gun be?

Solution: This is a similarity problem. Since the gun is fired horizontally, this is ahalf-parabola case. Step 1 is always write down what you’re given, what you want, andany assumptions.

Step 1:

Given:

h1 = 150 cm = 1.5 m (convert to meters right away)

xmax,1 = 25 m

xmax,2 = 75 m = 3xmax,1

Notice that we must convert the height to meters (or acceleration to cm/s2).

Want: h2 = ? (initial velocity)

345

Assumptions: The potato gun always produces the same initial speed for every potato.This is a big assumption that is probably not true.

Step 2: Determine the equation that governs the motion of the particle in both exper-iments. Then determine which of the variables in the governing equation are held fixedfor both experiments (we call these the common parameters) and which variables arevaried. Put all of the equations that vary on the left-hand side of the governing equation(LHS) and all of the parameters that are held fixed over the two experiments on the RHSof the governing equation. Lastly, solve for the desired variable.

Governing equation: We need an equation that gives xmax in terms of height. We’vealready derived such an equation from our fundamental equations:

xmax = v0

√2h

g

Common parameters: v0, g, 2

Next, we rearrange the governing equation and put all of the parameters that are variedover the experiments on the LHS of the governing equation and all of the parametersthat are held fixed over the two experiments on the RHS of the governing equation.

xmax = v0

√2h

g

·√h−−−−−→ xmax√

h= v0

√2

g

Now we equate the results from the two experiments through the common parameters:

xmax,1√h1

= v0

√2

g=xmax,2√h2

flip equation−−−−−−−−−−→√h1

xmax,1

=

√h2

xmax,2

( )2−−−−−−−→ h1

x2max,1

=h2

x2max,2

·x2max,2−−−−−−−→ h2 =

(xmax,2

xmax,1

)2

h1 =

(3xmax,1

xmax,1

)2

h1 = 32 h1

Discussion: The range of a projectile fired horizontally is proportional to the time thatit takes it to fall from its initial height to the ground. This time is proportional to thesquare root of the initial height. Since you want your potato’s range to be three times

346

greater, you must increase the fall time by a factor of 3; so you must increase the initialheight by a factor of 32 = 9. Therefore you must fire the gun from a height of

9 · 3/2m = 27/2 m = 13.5m ≈ 41 ft.

You are going to need a very tall ladder. Try the local fire department...

Problem 1141. A movie stuntman wants to ride a motorcycle off a flat roof, sail acrossa gap 24 m wide, and land on another roof 3.6 m below the first one. How fast does heneed to be moving when he leaves the first roof?


Solution: The stuntman is going off a flat roof, with no mention of a ramp; so weassume that the elevation is θ0 = 0. There is no vertical component of initial velocity,and v0 = v0x.The time it takes for the motorcycle to fall y = 3.6 m is t =

√2y/g. During this time,

the motorcycle needs to cover a horizontal distance of x = 24 m. Hence the initial(horizontal) velocity must be

v =x

t=

x√2y/g

=

√g

2yx =

√9.8 m/s2

2(3.6 m)(24 m) = 28 m/s

Problem 1142. While an unmanned rover is exploring the surface of Planet X, it goesoff the edge of a cliff at 11.2 m/s. The cliff is 3.1 m high; the rover strikes the ground18.8 m from the base of the cliff. What is the surface gravity of Planet X? Round youranswer to the nearest 0.1 m/s2.

*(a) 2.2 m/s2 (b) 2.4 m/s2

(c) 2.6 m/s2 (d) 2.9 m/s2

(e) None of these

Solution: Since no elevation is mentioned, we assume that θ0 = 0 and thus thatv0y = 0 and v0x = v0. We can calculate the time of fall from the horizontal velocity andthe horizontal distance that the rover travels as it falls: t = x/v0. During this time t, therover falls a distance of y.

y =1

2gXt

2 ⇒ gX =2y

t2=

2y

(x/v0)2=

2yv20

x2=

2(3.1 m)(11.2 m/s2)

(18.8 m)2= 2.2 m/s2

347

Problem 1143. A ball rolls off the edge of a horizontal table 1.32 m high. It strikesthe floor at a point 4.45 m horizontally away from the edge of the table. What was theball’s speed at the moment that it left the tabletop? Round your answer to the nearest0.1 m/s.

Solution: The ball’s initial velocity is horizontal; so if the table’s height is H, thenthe time it takes for the ball to fall to the ground is

t =

√2H

g

During that time, it traverses a horizontal distance of X; so its horizontal speed is

v0 =X

t=X√g

√2H

=(4.45 m)

√9.8 m/s2

√2 · 1.32 m

= 8.6 m/s

348

9.3.4 Special Case II: The Full Parabola

Note: For the case of the full-parabola the initial height is zero and the ball is fired atan angle of incline 0 < θ0 ≤ π

2. Setting h = 0 and θ0 6= 0, which leads to v0x 6= 0 (except

if θ0 = π2) and v0y 6= 0, in the general equations reduces them to

velocity x-component: vx = v0x (9.5a)

position x-component: x = v0xt (9.5b)


velocity y-component: vy = v0y − gt (9.6a)

position y-component: y = v0yt−1

2gt2 (9.6b)

Problem 1144. Your Civil War musket has a muzzle velocity of 290 m/s. If you fire itat an elevation of 6.2◦, what is the maximum height reached by the bullet? Round youranswer to the nearest meter.


Solution: We have the initial velocity, elevation, and gravitational acceleration; wewant to know the maximum height. We have a formula for this:

ymax =1

2

v20

gsin2 θ0 =

(290 m/s)2(sin 6.2◦)2

2(9.8 m/s2)= 50 m

Problem 1145. A golfer hits a ball from ground level with an initial speed of 42 m/sat an elevation of 19◦. How far from the initial location does the ball hit the ground?Round your answer to the nearest 10 m.


Solution: We have the initial velocity, the elevation, and the gravitational accelera-tion. We want to know the range. We have a formula for this:

xmax =v2

0

gsin(2θ0) =

(42 m/s)2 sin(2 · 19◦)

9.8 m/s2 = 110 m

349

Problem 1146. A golfer hits a ball from ground level with an initial speed of 57 m/sat an elevation of 34◦. How far from the initial location does the ball hit the ground?Round your answer to the nearest 10 m.


Solution: We have the initial velocity, the elevation, and the gravitational accelera-tion. We want to know the range. We have a formula for this:

xmax =v2

0

gsin(2θ0) =

(57 m/s)2 sin(2 · 34◦)

9.8 m/s2 = 310 m

Problem 1147. You fire your potato gun from ground level at an angle of 36◦ abovethe horizontal. The potato strikes the ground 47 m away from you. What was the initialspeed of the potato? Round your answer to the nearest m/s.


Solution: We know the elevation, the gravitational acceleration, and the range; wewant to know the initial velocity. Use the formula for range:

xmax =v2

0

gsin(2θ0) ⇒ v0 =

√gxmax

sin(2θ0)=

[(9.8 m/s2)(47 m)

sin(2 · 36◦)

]1/2

= 22 m/s

Problem 1148. You are holding the neck of a champagne bottle at an elevation of 41◦

when you pop the cork. The highest point in the cork’s flight is 2.6 m above its initialheight. What was the cork’s initial speed? Round your answer to the nearest m/s.


Solution: We know the elevation, the gravitational acceleration, and the maximumheight; we want to know the initial velocity. Use the formula for maximum height:

ymax =1

2

v20

gsin2 θ0 ⇒ v0 =

√2gymax

sin θ0

=

√2(9.8 m/s2)(2.6 m)

sin 41◦= 11 m/s

350

Problem 1149. You are holding the neck of a champagne bottle at an elevation of 34◦

when you pop the cork. The highest point in the cork’s flight is 1.9 m above its initialheight. What was the cork’s initial speed? Round your answer to the nearest m/s.


Solution: We know the elevation, the gravitational acceleration, and the maximumheight; we want to know the initial velocity. Use the formula for maximum height:

ymax =1

2

v20

gsin2 θ0 ⇒ v0 =

√2gymax

sin θ0

=

√2(9.8 m/s2)(1.9 m)

sin 34◦= 11 m/s

Problem 1150. A motorcyclist would like to ride up a short ramp on a rooftop, flyacross a gap between two buildings, and land on a second rooftop at the same heightas the first. The ramp is inclined at 32◦; the buildings are 18 m apart. How fast doesthe motorcyclist have to be going when he makes the jump? Round your answer to thenearest m/s.


Solution: We know the gravitational acceleration, the elevation, and the range; wewant to know the initial velocity. Use the formula for range:

xmax =v2

0

gsin(2θ0) ⇒ v0 =

√gxmax

sin(2θ0)=

[(9.8 m/s2)(18 m)

sin(2 · 32◦)

]1/2

= 14 m/s

Problem 1151. A crossbow fires its arrow at 105 m/s. If the arrow is aimed at anelevation of 12◦ above the horizontal, what is the horizontal distance that it will travelbefore striking the ground? Round your answer to the nearest 10 m.


Solution: We know the initial velocity, the gravitational acceleration, and the eleva-tion; we want to know the range. We have a formula for this:

xmax =v2

0

gsin(2θ0) =

(105 m/s)2 sin(2 · 12◦)

9.8 m/s2 = 460 m

351

Problem 1152. A Napoleon cannon has a muzzle velocity of 450 m/s. If the cannon isfired at an elevation of 5.1◦, how much time elapses before the ball strikes the ground?Round your answer to the nearest 0.1 s.


Solution: We know the elevation, initial velocity, and gravitational acceleration. Wewant to know the flight time. We have a formula for this:

txmax =2v0 sin θ0

g=

2(450 m/s) sin 5.1◦

9.8 m/s2 = 8.2 s

Problem 1153. Space pirates have kidnapped you from your physics lab and are holdingyou for ransom on another planet. Luckily, they have overlooked your cell phone andyour spring cannon. The spring cannon has a muzzle velocity of 8.6 m/s; when you fireit at an elevation of 45◦, the ball strikes the ground 41 m away. (The pirates are keepingyou in a rather large room.) What is the surface gravity of the planet on which you’rebeing held? Round your answer to the nearest 0.1 m/s2.

(a) 1.6 m/s2 *(b) 1.8 m/s2

(c) 2.0 m/s2 (d) 2.2 m/s2

(e) None of these

Solution: We know the initial velocity, elevation, and range; we want to know thegravitational acceleration, which we’ll call gX . Use the formula for range:

xmax =v2

0

gXsin(2θ0) ⇒ gX =

v20

xmax

sin(2θ0) =(8.6 m/s)2 sin(2 · 45◦)

41 m= 1.8 m/s2

Problem 1154. Your potato gun has a muzzle velocity of 23 m/s. You want to shoota potato and have it strike the ground 3.5 seconds later. At what elevation do you haveto fire the gun? Round your answer to the nearest degree.

(a) 43◦ *(b) 48◦

(c) 53◦ (d) 58◦

(e) None of these

Solution: We know the initial velocity, the gravitational acceleration, and the timeof flight. We want to know the angle of elevation. Use the formula for flight time:

txmax =2v0 sin θ0

g⇒ sin θ0 =

gtxmax

2v0

⇒ θ0 = sin−1

[gtxmax

2v0

]= sin−1

[(9.8 m/s2)(3.5 s)

2(23 m/s)

]= 48◦

352

Problem 1155. You have designed a tennis-ball cannon with a muzzle velocity of 62m/s. You want to land a tennis ball in a swimming pool 210 m away. At what angle tothe horizontal do you need to aim the gun? Round your answer to the nearest degree.

(a) 11◦ (b) 13◦

(c) 15◦ *(d) 16◦

(e) None of these

Solution: We know the initial velocity, the gravitational acceleration, and the range.We want the angle of elevation. Use the formula for range:

xmax =v2

0

gsin(2θ0) ⇒ sin(2θ0) =

gxmax

v20

⇒ 2θ0 = sin−1

[gxmax

v20

]⇒ θ0 =

1

2sin−1

[gxmax

v20

]=

1

2sin−1

[(9.8 m/s2)(210 m)

(62 m/s)2

]= 16◦

Problem 1156. You have designed a tennis-ball cannon with a muzzle velocity of 62m/s. You want to land a tennis ball in a swimming pool 210 m away. At what angle tothe horizontal do you need to aim the gun? Your answer should be less than π/4 radians.(a) 1.08 rad (b) 0.13 rad(c) 0.97 rad (d) 0.28 rad(e) None of these

Solution: The horizontal range of a projectile is the product of the flight time and thehorizontal component of velocity. The horizontal component of velocity is vx = v0 cos θ.For a projectile fired from ground level, the flight time is t = 2v0 sin θ/g. Then thehorizontal range is

xmax =2v0

2 cos θ sin θ

g

We know x = 210 m, v0 = 62 m/s, and g = 9.8 m/s2. We want to find θ. The presenceof both sin θ and cos θ in the formula creates some difficulty. Happily, we remember thetrigonometric identity:

sin(2θ) = 2 sin θ cos θ

Substituting this produces

xmax =v0

2 sin(2θ)

g

Now we can solve for θ:

θ =1

2sin−1

(gxmax

v02

)=

1

2sin−1

((9.8 m/s2)(210 m)

(62 m/s)2

)= 0.28 rad

353

Problem 1157. A dietitian is standing at the edge of a cliff. He throws a piece of cheeseupward and outward, so that it reaches a maximum height of 12 m above him, then fallsdown into the canyon 65 m below him and 40 m from the edge of the cliff. Betweenthe time he releases the cheese and the time that it strikes the bottom, where is theacceleration at a minimum?

Solution: There is no maximum or minimum acceleration: during the entire timethat the cheese is in the air, it is experiencing acceleration of 9.8 m/s2 downward. Thevarious numbers describing the trajectory are irrelevant to the problem.

Problem 1158. The surface gravity on Mars is 0.4 times the surface gravity on theEarth. A golfer can hit a ball so that it strikes the ground 200 m from him on Earth. Ifhe hits the ball with the same speed and at the same angle on Mars, how far will it go?

Solution: The range of a projectile is the product of the horizontal velocity andthe time that it takes for the projectile to reach the ground. The horizontal velocity isvx = v0 cos θ. If the projectile is launched from ground level, the time is t = 2v0 sin θ/a,where a is the acceleration due to gravity. Hence the range is

x =2v0

2

acos θ sin θ

On Earth, a = g; on Mars, a = 0.4g. The values of v0 and θ are the same on bothplanets. Hence

g · xE = 0.4g · xM so xM =xE0.4

=200 m

0.4= 500 m

354

Problem 1159. Your physics teacher is trying to get a planeload of “prescription” drugsacross the border. As an extra-credit project, he has assigned you and your lab partnerto shoot down the radar blimp. The blimp flies at a height of 10,000 ft. You fire acannon at an angle of θ to the horizontal, and the maximum height attained by the shellis 5000 ft. Your lab partner says that if you increase the angle to 2θ, the shell will gohigh enough to hit the blimp. Is he correct? Why or why not?

Solution: Your lab partner is wrong. The obvious counterexample is: let θ = 90◦.Then 2θ = 180◦; but a shell fired at 180◦ will never rise above ground level, so will reacha maximum height of zero.More generally, let’s look at ymax as a function of θ. The time from the initial release ofthe projectile to its maximum height is the time that it takes for its vertical velocity toreach zero:

t =v0 sin θ

g

The maximum height is the distance that a body travels in that time if its initial velocityis v0 sin θ and it experiences an acceleration of −g:

ymax = v0t sin θ − gt2

2= v0 sin θ

(v0 sin θ

g

)− gv0

2 sin2 θ

2g2=v0

2 sin2 θ

2g

Clearly, a simple relation like “2θ ⇒ 2ymax” isn’t going to hold.

Problem 1160. You are trying to put out a fire 150 m away with a water cannon. Themanual informs you that if you aim the water cannon at an elevation of 23◦ above thehorizontal, it will have a range of 150 m. Unfortunately, there is a building between youand the fire, and at an elevation of 23◦ the water will hit it. Is there another angle ofelevation θ that you could use to get a range of 150 m? Give θ, or explain why it doesn’texist.

Solution: A projectile launched from ground level at an elevation of θ will have thesame range as a similar projectile launched at an elevation of 90◦ − θ. In this case, sinceyou can’t use an elevation of 23◦, an elevation of 90◦− 23◦ = 67◦ should work (provided,of course, that the obstructive building isn’t too high).

355

Problem 1161. If a champagne bottle is held with its neck elevated 78◦ above thehorizontal, the popped cork will reach a maximum altitude of 4.8 m. Is there anotherangle θ between 0◦ and 90◦ for which the cork will reach the same maximum altitude?Give θ, or explain why it doesn’t exist.

Solution: There is no such angle θ.

Let’s calculate the maximum altitude ymax as a function of θ. The time from the poppingof the cork to its reaching its maximum height is the time that it takes for its verticalvelocity to reach zero:

t =v0 sin θ

g

The maximum height is the distance that a body travels in that time if its initial velocityis v0 sin θ and it experiences an acceleration of −g:


2= v0 sin θ

(v0 sin θ

g

)− gv0

2 sin2 θ

2g2=v0

2 sin2 θ

2g

Between 0 and 90◦, sin θ is a nonnegative and strictly increasing function of θ. Thatmeans that within that range, there are no two values of θ that will produce the samevalue of ymax.

In conclusion, for projectile motion there are two θ values that will give the same distance,but only one that will give the same height.

Problem 1162. The Paris Gun had a muzzle velocity of 1600 m/s. If the gun was firedat an elevation of 45◦ above the horizontal, what was the maximum altitude reached bythe shell? Give your answer in kilometers.

Solution: The vertical component of the shell’s velocity as a function of time isvy = v0 sin θ − gt. The shell reaches its maximum altitude when the vertical componentof velocity is zero; so at time t = v0 sin θ/g. The maximum altitude is the vertical distancetravelled in that time t by a projectile with initial vertical speed v0 sin θ, subjected to anacceleration of −g:


2= v0 sin θ

(v0 sin θ

g

)− gv0

2 sin2 θ

2g2=v0

2 sin2 θ

2g

Substituting our values for v0, θ, and g, we get:

ymax =(1600 m/s)2 sin2 45◦

2(9.8 m/s2)= 65, 000 m = 65 km

356

Problem 1163. A crossbow fires its arrow at 345 ft/s. If the arrow is aimed at anelevation of 12◦ above the horizontal, what is the horizontal distance that it will travelbefore striking the ground?

Solution: The horizontal range of a projectile is the product of the horizontalcomponent of velocity and the time that the projectile is in the air. If the projectileis fired from ground level with a velocity of v0 and an elevation of θ, the time in the air is2v0 sin θ/g. The horizontal component of velocity is vx = v0 cos θ. Hence the horizontalrange is

x =2v0

2

gsin θ cos θ =

2(345 ft/s)2

32 ft/s2 sin 12◦ cos 12◦ = 1500 ft

Problem 1164. Barbora Spotakova won a gold medal in the 2008 Olympics by throwinga javelin 71.42 m. Assuming that Spotakova threw the javelin at an angle of π/4 radiansto the horizontal, what was its speed when initially thrown?

Solution: The horizontal range of a projectile is the product of the horizontalcomponent of velocity and the time that the projectile is in the air. If the projectileis fired from ground level with a velocity of v0 and an elevation of θ, the time in the air is2v0 sin θ/g. The horizontal component of velocity is vx = v0 cos θ. Hence the horizontalrange is

x =2v0

2

gsin θ cos θ

We know x, θ, and g. Solving for v0, we get

v0 =( gx

2 sin θ cos θ

)1/2

=

((9.8 m/s2)(71.42 m)

2 sin π4

cos π4

)1/2

= 26 m/s

Problem 1165. The Napoleon cannon of the Civil War had a muzzle velocity of 1485ft/s. If the cannon was fired at an elevation of 5.0◦ above the horizontal, how much timewould elapse between the firing and the ball’s striking the ground?

Solution: If a projectile is fired from ground level with initial velocity v0 and elevationθ, its flight time is

2v0 sin θ

g=

2(1485 ft/s) sin 5◦

32 ft/s2 = 8.1 s

357

Problem 1166. Using surgical tubing and ingenuity, you have constructed a water-balloon launcher in a window 22 m above the ground. Your device launches a waterballoon at a speed of 19 m/s, and at an elevation of 0.25 radians above the horizontal.At what horizontal distance from the building will the water balloon strike the ground?

Solution: The horizontal range is the product of the flight time and the horizontalvelocity. The horizontal velocity is vx = v0 cos θ. The flight time is the time that it takesfor the projectile to reach the ground. The projectile’s height above ground level is:

y = y0 + v0t sin θ − gt2

2

We need to find t for which y = 0. Using the quadratic formula yields

t =−v0 sin θ ±

√(v0 sin θ)2 + 4(g/2)y0

−2(g/2)=−v0 sin θ ±

√v0

2 sin2 θ + 2gy0

−g

=(−19 m/s)(sin 0.25 rad)±

√(19 m/s)2(sin 0.25 rad)2 + 2(9.8 m/s2)(22 m)

−9.8 m/s2

We want to use the positive value, which is about 2.65 s. (You should not use thisrounded value in subsequent calculations; we present it only because it’s a good idea tocheck intermediate results and make sure they seem reasonable.) The range is

x = vxt = v0t cos θ = (19 m/s)(cos 0.25 rad)t = 49 m

358

Problem 1167. A bicyclist unfamiliar with the terrain is riding along a level sidewalkwhen he goes over the edge of a staircase. He hits the staircase 3.7 m below the edge,and 3.7 m horizontally from the edge. How fast was he going when he went over theedge?

Solution: We will use the bicyclist’s vertical travel to calculate his flight time; andhis flight time and horizontal travel to calculate his initial horizontal speed.

We take the height of the sidewalk as y0 = 0. We want to know how long it takes for thebicyclist to reach a height of y = −3.7 m. Since his initial velocity was horizontal, hisheight as a function of time is

y = −gt2

2

Solving for t:

t =

√−2y

g=

√2(3.7 m)

9.8 m/s2 = 0.87 s

During this time, the bicyclist travels a horizontal distance of x = 3.7 m. Thus hishorizontal velocity is

vx =x

t=

x√g

√−2y

=(3.7 m)

√9.8 m/s2

√2 · 3.7 m

= 4.3 m/s

Problem 1168. You fire a cannon at an angle of θ to the horizontal, with a muzzlevelocity of v0. The ball lands at a horizontal distance of x from the cannon. For yoursecond shot, you increase the powder charge in the cannon so that the muzzle velocitywill be 2v0. How far from the cannon will the ball land?

Solution: The range of a projectile is the product of the horizontal velocity andthe time that it takes for the projectile to reach the ground. The horizontal velocityis vx = v0 cos θ. If the projectile is fired from ground level, the time is t = 2v0 sin θ/g.Hence the range is

x =2v0

2

gcos θ sin θ

If we increase the muzzle velocity v0 by a factor of 2, we increase the range by a factorof 22 = 4; so the new range is 4x.

359

9.4 Lab application problems: Two-Dimensional Kinematics

This is a mixture of problem types based on actual equipment found in instructionalphysic labs everywhere.

9.4.1 Spring-cannon apparatus

Problem 1169. (Lab Problema) You fire a spring cannon horizontally from a fixedheight hcannon = ymax. It strikes the ground at a distance xmax from the base of themouth of the muzzle. Find a formula for the ball’s initial velocity as a function of theinitial height ymax and range xmax. The initial velocity is taken to be the velocity of theball as it leaves the mouth of the cannon.

(a) xmax

√2gymax (b) xmax

√2ymax

g

(c) xmax

√2g

ymax

*(d) xmax

√g

2ymax

(e) None of these

aThis is a common technique for determining initial velocity, based on measured lengths rather thanon measured time. It is good for estimating the initial velocity when the spring cannon is shot at smallto moderate angles of incidence (θ0 ≤ 45◦).

Solution: Since the ball is fired horizontally, v0x = v0. Using fundamental kinematic

equation (1) in the x-direction we have xmax = v0 tfall, where tfall =

√2ymax

gis the time

for the ball to fall a distance ymax. Thus,

xmax = v0 tfall÷tfall−−−−−−→ v0 =

xmax

tfall

= xmax

√g

2ymax

360

Problem 1170. (Lab Similarity Problem) You fire a spring cannon horizontally froma fixed height. The initial velocity of the ball is v0,1; it strikes the ground at a distancexmax,1 from the base of the mouth of the muzzle. If you want the ball to travel twice thehorizontal distance, what must the initial velocity be?

(a)√

2 v0,1 *(b) 2v0,1

(c) 2√

2 v0,1 (d) 4v0,1

(e) None of these

Solution: If a projectile is fired horizontally, the horizontal range is the initial velocitytimes the time that it takes for the projectile to fall to the ground: xmax = v0tfall. Sincethe cannon is fired horizontally, the fall time is the same as if we dropped the ball straightdown from rest from the height of the cannon hcannon. The fall time is then

tfall =

√2hcannon

g

·v0−−−−−→ xmax = v0

√2hcannon

g.

This is our governing equation. Since g and hcannon are the same in both situations (i.e.,these are our common parameters) we re-arrange the governing equation as

tfall =

√2hcannon

g

÷v0−−−−−→ xmax

v0

=

√2hcannon

g.

Using the standard form of the governing equation to equate the two situations throughthe common parameters gives

xmax,1

v0,1

=

√2hcannon

g=xmax,2

v0,2

⇒ v0,2 =xmax,2

xmax,1

v0,1 = 2v0,1 .

361

Problem 1171. (Lab Similarity Problem) If you fire a spring cannon horizontallyfrom a height of hcannon,1, the ball travels a horizontal distance of xmax,1 before hitting thefloor. How high must the spring cannon be to get the ball to travel twice the horizontaldistance before hitting the ground?

(a)√

2hcannon,1 (b) 2hcannon,1

(c) 2√

2hcannon,1 *(d) 4hcannon,1

(e) None of these

Solution: If a projectile is launched horizontally, the horizontal distance that ittravels is the initial velocity times the time that it takes for it to fall to the ground:xmax = v0tfall = v0

√2hcannon/g. This is our governing equation. Re-arranging the terms

to put the common factors on the right and the parameters that are situation dependenton the left gives:

xmax = v0

√2hcannon

g

÷√hcannon−−−−−−→ xmax√

hcannon

= v0

√2

g

( )2−−−→ x2max

hcannon

=2v2

0

g.

The last step is to use the governing equation in the similarity standard form to determinethe relationship between the heights of the two similar problems:

x2max,1

hcannon,1

=2v2

0

g=

x2max,2

hcannon,2

flip equation−−−−−−−−−−→ hcannon,1

x2max,1

=hcannon,2

x2max,2

·x2max,2−−−−−−−→ hcannon,2 =

(xmax,2

xmax,1

)2

hcannon,1 = 4hcannon,1

Problem 1172. (Lab Problem) You fire a spring cannon from ground level at anelevation of 23◦. The ball emerges from the cannon at a speed of 14.2 m/s. How longdoes it take before the ball strikes the ground? Round your answer to the nearest 0.1 s.


Solution: We know the initial velocity, the elevation, and the surface gravity; wewant to know the flight time. We have a formula for this:

txmax =2v0 sin θ0

g=

2(14.2 m/s) sin 23◦

9.8 m/s2 = 1.1 s

362

Problem 1173. (Lab Problem) A high-powered spring cannon discharges its ball at16.6 m/s. If the cannon is fired at an elevation of 77◦, what is the maximum heightreached by the ball? Round your answer to the nearest meter.


Solution: We have the initial velocity, elevation, and gravitational acceleration; wewant to know the maximum height. We have a formula for this:

ymax =1

2

v20

gsin2 θ0 =

(16.6 m/s)2(sin 77◦)2

2(9.8 m/s2)= 13 m

Problem 1174. (Lab Problem) In your physics lab, you are using a new high-poweredspring cannon with a muzzle velocity of 19 m/s. The ceiling of the lab is 2.8 m above thelevel of the cannon. What is the maximum elevation at which you can fire the cannonwithout hitting the ceiling? Round your answer to the nearest degree.

(a) 16◦ (b) 18◦

(c) 21◦ *(d) 23◦

(e) None of these

Solution: We know the initial velocity, the maximum height, and the gravitationalacceleration. We want to know the angle of elevation. Use the formula for maximumheight:

ymax =1

2

v20

gsin2 θ0 ⇒ sin2 θ0 =

2gymax

v20

⇒ sin θ0 =

√2gymax

v0

⇒ θ0 = sin−1

[√2gymax

v0

]= sin−1

√

2(9.8 m/a2)(2.8 m)

19 m/s

= 23◦

363

Problem 1175. (Lab Similarity Problem) You fire a spring cannon at a fixed eleva-tion θ0 6= 0. The spring cannon uses a compressed spring to launch the ball. The cannonhas three settings with the first being the lowest and the third being the strongest. Usingthe first setting (one click) you find that the ball emerges from the muzzle with a speedof v0,1; it strikes the ground at a distance xmax,1 from the cannon. You then increase theenergy stored in the spring by using the second setting (two clicks). The muzzle velocityof the ball is now measured to be v0,2 = α v0,1, where α is a positive constant. What isthe new range of the ball?

(a)√αxmax,1 (b) αxmax,1

(c) α√αxmax,1 *(d) α2 xmax,1

(e) None of these

Solution: Use the formula (the governing equation) for range:

xmax =v2

0

gsin(2θ0)

÷v20−−−−−→ xmax

v20

=sin(2θ0)

g(the common factors) ,

where we’ve re-arranged the governing equation to be in the standard form for a similarityequation with the common factors on the right-hand side and the situation-dependentvariables on the left-hand side. If we increase the initial velocity from v0,1 to v0,2 = α v0,,1,we get

xmax,2

v20,2

=sin(2θ0)

g=xmax,1

v20,1

·v20,2−−−−−−→ xmax,2 =

(v0,2

v0,1

)2

xmax,1 = α2xmax,1

An alternative approach is to make a direct substitution for v0,2 in terms of v0,1 in theoriginal equation:

xmax,2 =(αV )2

gsin(2θ0) = α2

(V 2

gsin(2θ0)

)= α2xmax,1

Thus, if we double the initial speed we quadruple the distance traveled by the ball.Actually, in the real world the faster the ball travels, the larger the air drag.

364

Problem 1176. (Lab Similarity Problem) You fire a spring cannon at a fixed eleva-tion θ0 6= 0. The ball emerges from the muzzle with a speed of v0,1; it reaches a maximumheight of ymax. You would like to increase the muzzle velocity so that the ball can doubleits height. What must the new muzzle velocity be?

*(a)√

2 v0,1 (b) 2 v0,1

(c) 2√

2 v0,1 (d) 4 v0,1

(e) None of these

Solution: Use the formula for maximum height as the governing equation for the twosituations.

ymax =1

2

v20

gsin2 θ0

÷v20−−−−−→ ymax

v20

=sin2 θ0

2g(common factors)

We want to find v0,2 such that the maximum height is ymax,2 = 2ymax,1.

ymax,2

v20,2

=sin2 θ0

2g=ymax,1

v20,1

flip equation−−−−−−−−−−→v2

0,2

ymax,2

=v2

0,1

ymax,1

·ymax,2−−−−−−−→ v20,2 =

(ymax,2

ymax,1

)v2

0,1

√−−−−−−→ v0,2 = v0,1

√ymax,2

ymax,1

=√

2 v0,1

Problem 1177. (Derive Problem) A ball rolls off the edge of a horizontal table ofheight H. It strikes the floor at a horizontal distance X away from the edge of the table.What was the ball’s speed at the moment that it left the tabletop? Derive a formulathat involves the given data: H and X.

Solution: The ball’s initial velocity is horizontal; so if the table’s height is H, thenthe time it takes for the ball to fall to the ground is

tfall =

√2H

g

During that time, it traverses a horizontal distance of X; so its horizontal speed is

v0 =X

tfall

= X

√g

2H

Problem 1178. (Derive Problem) A child whirls a stone in a horizontal circle at aheight of H above the ground, on the end of a string with length L. The string breaksand the stone flies off horizontally, striking the ground at a horizontal distance of X fromwhere it was when the string broke. What was the centripetal acceleration of the stonebefore the string broke? Your answer should be a formula of the form ac = ac(H,L,X, g),

365

where g is the acceleration due to gravity.

Solution: To find the centripetal acceleration, we need to know the speed v of thestone while it was circling. If we knew the time t that it took for the stone to cover thehorizontal distance X, we could calculate: v = X/t. We are not given t, but we knowthat it’s how long it takes for the stone to fall to the ground from an initial height of H.Thus we’re going to proceed in three steps: use H and g to determine t; use X and t todetermine v; and use v and L to determine the centripetal acceleration ac.

The time t is:

t =

√2H

g

The speed v is:

v =X

t=X√g

√2H

The centripetal acceleration ac is:

ac =v2

L=X2g

2HL

Problem 1179. (Lab Derive Problem) A ball is fired horizontally out of a springcannon with an initial velocity of v0; the cannon is a height h above the ground. The ballstrikes the ground at a horizontal distance xmax from the muzzle of the cannon. Find v0

as a function of hcannon, xmax, and g.Note: The purpose of this problem is to find the initial velocity of the ball as it leavesthe muzzle of the spring cannon. Using this initial velocity one can then point the cannonin any direction and predict the maximum height and the range of the projectile.

Solution: Since the cannon is fired horizontally, the initial vertical velocity of theball is zero; and we’re assuming that the horizontal component of the velocity remainsv0 for the entire flight. We will use hcannon and g to determine the flight time (the timeit takes the object to fall a distance hcannon) tfall; then use tfall and xmax to determine v0.

The time it takes for the ball to fall from a height of hcannon to the ground is

tfall =

√2hcannon

g

The horizontal component of the ball’s velocity is

v0 =xmax

tfall

= xmax

√g

2hcannon

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9.5 Advanced-Level Problems (Kinematics)

General algorithm for physics derivation problems

Step 1: (Setup) Write down what you are given, what you want, and list any assumptionsthat you need to make to solve the problem: for example, “the pulley and string aremassless”, “the surface is frictionless”, etc. Draw any relevant pictures and labelthem appropriately.

Step 2: Starting from the general equations for projectile motion written on the formulasheet, substitute the values for the initial position, velocity, etc. That is, find theset of equations that are particular to this problem.

Step 3: Solve the resulting equations.

Derive problems will be graded out of 10 points. The setup step is worth between 3 and4 points. The point breakdown for the setup is given below.

Given: (1 point) Write down what you are given.

Want: (1 point) Write down what you want (what you’re solving for).

Assumptions: (1 point) Write down any assumptions and key physical laws/equations that youare going to use. Notice that I am including writing down the relevant equationsand physical laws in this step.

Here are some examples:(1.) Assume the surface is made of ice that can be treated as a frictionless surface.You should include this even if it is in the problem statement because it is a goodhabit to get into. Such an assumption of frictionless in a real problem could leadto dubious results.(2.) Use fundamental equation 2 and Newton’s second law.

Picture: (1 point) Draw a picture, if one is not drawn for you. If you are resolving forces ona system, then draw a free-body diagram for each object in the system that you’reanalyzing, based on the picture. This should include a coordinate system for eachobject.. (If a picture has been provided for you, then you will not receive credit forthis step).

How the rest of the problem is graded is too problem specific to provided an explicitalgorithm. Here is a general guide line:

• Use the free-body diagram/diagrams to determine the net force acting on eachobject in the system, in each component direction.

• Apply the correct physical law/laws and solve the problem. This is the hard part!

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Problem 1180. (Lab Derive Problem) Consider the situation shown in the figure 3below. Take the mouth of the cannon to be at the point (x0, y0) = (0, h). Assume thatyou know the speed v0 and direction θ of the initial velocity ~v0 = v0 〈cos θ, sin θ〉. Derivea formula for the vertex of the trajectory (xymax, ymax) and the range point (xmax, 0) interms of the given information.Note: The purpose of this problem is to find the vertex and the range of a projectile.

Figure 3: Spring cannon

Solution: See the file chapter4-projectile-motion-general-case under the physics lec-ture notes webpage.

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Problem 1181. In one of your physics labs you are given the challenge of shooting a ballfrom a spring cannon into a bucket. Your job is to figure out where to place the centerof the bucket in order to launch the ball into the bucket! Also, you are only allowed oneattempt to put the ball in the bucket.

You are given the following information (see figure 5):

• The ball is released from the mouth of the cannon and is located at the initialposition (x0, y0) = (0, Hcannon).

• The bucket standsHbucket units above the table with the center of the bucket locatedxmax units in the horizontal direction from the mouth of the cannon. Also, as thepicture suggests, Hbucket < Hcannon.

• The mouth of the bucket is just large enough to allow the ball to be captured inthe bucket provided that the ball lands close enough to the point (xmax, Hbucket).

• The initial speed v0 = ‖~v0‖ and direction θ0 that the ball leaves the mouth of thespring cannon.

Figure 4: Spring cannon and bucket

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Problem 1182. (Lab Derive Problem [Put the ball in the bucket])In one of your physics labs you are given the challenge of shooting a ball from a springcannon into a bucket. Specifically, you need to work out the angle the cannon should beaimed in order to launch the ball into the bucket! You are only allowed one attempt toput the ball in the bucket and you cannot move the bucket. You are also told that thecannon has plenty enough power to launch the ball into the bucket.

You are given the following information (see figure 5):

• The bucket standsHbucket units above the table with the center of the bucket locatedxmax units in the horizontal direction from the mouth of the cannon.

• The mouth of the bucket is just large enough to allow the ball to be captured inthe bucket provided that the ball lands close enough to the point (xmax, Hbucket).

• You are instructed to take the mouth of the cannon to be at the point (x0, y0) =(0, Hcannon). By using a swivel mount at the mouth of the cannon, the device isdesigned in such a way as to keep the height of the ball the same regardless of thedirection θ that the cannon is aimed. Thus Hcannon is independent of θ.

• The initial speed v0 = ‖~v0‖ that the ball leaves the cannon.

Derive a formula for θ in terms of the given information.

Figure 5: Spring cannon and bucket

Solution: This is identical to the problem solved in class except that we must reducethe height of the cannon by the height of the bucket. Let h = Hcannon −Hbucket > 0.

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Problem 1183. (Lab Derive Problem [Put the ball in the moving bucket])The Idea: In one of your physics labs you are given the challenge of shooting a ball froma spring cannon into a bucket mounted on top of a glider moving at a constant velocityalong an air track that is aligned in the direction of the cannon.

The Apparatus: A horizontal air track is sitting atop a table. The track passes undera spring cannon that is attached to a mount that holds the cannon directly above the airtrack. From a top view the cannon is aimed along the track in the direction of motionof the glider. From a side view, the cannon makes an angle 0 ≤ θ0 ≤ π/2 with respectto the horizontal air track (see figure 6 below).

A bucket sits atop a glider moving down the air track at constant speed Vglider. Whenthe bucket passes directly below the the cannon, the cannon is fired and the ball emergesat the muzzle of the cannon with a velocity of V0. The apparatus is designed so that thecenter of mass of the ball, denoted by x, is directly over the center of the bucket xcenter

as the ball leaves the muzzle. Take this time to be the initial time t0 = 0 and take thelocation along the air track to be x0 = 0.

Figure 6: Side view of the spring cannon firing ball into a bucket mounted on top of aglider

The Setup: You are given the following information

• You are instructed to take the mouth of the cannon to be at the point (x0, y0) =(0, Hcannon). By using a swivel mount at the mouth of the cannon, the device isdesigned in such a way as to keep the height of the ball the same regardless of thedirection θ0 that the cannon is aimed. Thus Hcannon is independent of θ0.

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• The bucket stands Hbucket units above the table with the center of the bucketlocated a distance xcenter = Vglider t from the mouth of the cannon in the horizontaldirection along the track. Thus the ordinates of the center of the bucket, the target,are (xcenter, Hbucket).

• The mouth of the bucket is large enough to offset the small effects of air drag so thatthe ball can be captured in the bucket provided that the ball lands close enough tothe point (xcenter, Hbucket).

• Assume Hbucket < Hcannon.

• The initial speed of the ball as it leaves the mouth of the cannon is ‖~v0‖ = V0 > 0.

• Assume that you know the speed of the glider Vglider and the initial speed of theball as it leaves the mouth of the cannon V0.

The Questions:

(a) Starting from the general equations for projectile motion with x0 = 0:{x(t) = v0xt (the ball’s x-component:)

y(t) = h+ v0yt− 12gt2 (the ball’s y-component:)

substitute the values for the initial position (0, h), velocity ~v0 = V0〈cos(θ0), sin(θ0)〉, etc.That is, find the set of equations that are particular to this problem.

(b) Taking the initial speed of the ball as leaves the mouth of the cannon V0 to be knownfrom a previous experiment, what is the range of speeds for the glider Vglider > 0 forwhich there is a solution. That is, for what range of speeds will the ball land in thebucket, provided it is aimed properly.

(c) In order for the ball to land in the bucket, what must the cannon’s angle of elevationθ0 be?

(d) At what horizontal distance xmax from the mouth of the cannon will the ball land inthe bucket?

Potential future problem: What would happen if the air track were not level?

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Part IV

Newton’s Laws

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10 Introduction to Newton’s Laws

10.1 Concept questions involving Newton’s three laws

Problem 1184. (Derived quantities) Determine the expression for the unit of force(the Newton N) in terms of the fundamental dimensions mass, length, and time from theequation expressing Newton’s 2nd Law F = ma.

Solution: Using our take-the-dimensions-of-the-quantity operator [ ] on the equationgives

[F ] = [ma] = ML

T 2= kg

m

s2⇒ 1 Newton = kg

m

s2

Problem 1185. An object is subjected to a nonzero constant net force ~F . Which ofthe following properties of the object will be constant?

(a) Position (b) Speed(c) Velocity *(d) Acceleration(e) None of these

Problem 1186. (Similarity Problem) Consider the following two experiments. Inthe first experiment you push a block along a level surface at a constant velocity V .To overcome the friction between the block and the surface, you must apply a constanthorizontal force Fpush,1. In the second experiment you push the block at a constant speed2V across the same surface. Assuming that the friction is the same as it was the firstexperiment, which of the following choices best describes the new push force Fpush,2 interms of the first push force?

*(a)A constant force Fpush,1

(b) A constant force 2Fpush,1

(c) A force that increases from Fpush,1 to 2Fpush,1

(d) A force of zero, since the block does not accelerate.

Solution: For the block to move at constant speed, by Newton’s 1st law the netforce on it must be zero. Doing a force balance in the horizontal direction we find:0 = Fnet = Fpush,1 − f ⇒ Fpush,1 = f , where f is the frictional force. Thus whenthe block is moving at speed V , the pushing force Fpush,1 is equal to the frictional force.When the table is moving at 2V , the frictional force is the same; so the pushing forcemust again be Fpush,1. This is a subtle problem because students generally choose (b).The key to understanding this problem is to realize that that the force that it takes tomove the block at a constant speed is different that the force it takes to accelerate itfrom rest to a certain speed. Also the time that it would take. These other issues haveto do with the concepts of work and power, concepts that we have not yet discussed.

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Problem 1187. You are standing on a spring scale in an elevator that is moving at aconstant speed V . Before the elevator started moving, the scale registered your weightas W . It now registers a weight of Wapp. What is the relationship between Wapp and W?

(a) Wapp < W*(b) Wapp = W(c) Wapp > W

(d) Not enough information; it depends on the direction of ~V .

Solution: The elevator is moving at constant speed, so it’s not accelerating upwardor downward. Your apparent weight is the same as your true weight.

Problem 1188. You come into your physics lab and findthree books stacked on a table, as shown at right. Whatis the net force on the middle book?(a) 10 N downward (b) 20 N upward(c) 15 N downward *(d) 0 N

Solution: The book is motionless, so it’s not accel-erating in any direction. Hence the net force on it iszero.

20 N

15 N

10 N

Problem 1189. You are piloting your spaceship through interstellar space where thereis no gravity and no air, when your timing belt breaks and your engine abruptly quits.This occurs at time T , when you are travelling at speed V . Which of the graphs belowdescribes your motion starting at time T?

6v

V

-tT

(a)6

v

V

-tT

*(b)6

a

-tT

(c)6

a

-tT

(d)

Solution: There is no gravity, no friction, and no air resistance in this situation; sowe assume that once the engine has stopped, there are no forces acting on the spaceship.Thus it keeps going at the same velocity forever. Graph (b) shows that. Graph (a) showsa changing speed; and graphs (c) and (d) show nonzero acceleration.

Problem 1190. A piano with a weight of 4500 N is resting on a frictionless horizontalice-covered parking lot. You push on the piano with a constant horizontal force of F .Which of the following statements is true?

*(a) The piano will accelerate, no matter how small F is.(b) The piano will not accelerate unless F > 4500 N.(c) The piano will move with constant velocity, because F is constant.(d) The piano will move as long as you apply F ; once the force is removed, it will stop.

Solution: Since the lot is level and there is no friction, the only horizontal force onthe piano is your pushing force F . Hence as long as F > 0, there is a net force on thepiano, so it will accelerate. Answer (d) is wrong because the piano, once moving, willcontinue to move at a constant velocity once there is no horizontal force on it.

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Problem 1191. You come into your physics lab and findthree weights suspended by thin strings from the ceiling,as shown at right. Which of the following statements istrue about the tensions T1, T2, and T3 in the strings?

(a) T1 = T2 = T3

*(b) T1 > T2 > T3

(c) T3 > T2 > T1

(d) T3 > T1 > T2 6 kg

T3

1 kg

T2

4 kg

T1

Solution: The tension in each string is equal to the weight of the masses below it.Hence T1 = 11 kg · g; T2 = 7 kg · g; and T3 = 6 kg · g.

Problem 1192. Your physics instructor asks you to help him push an atomic bomb intothe classroom. The bomb weighs 8000 N. You push on it with a horizontal force of 400N. Which of the following statements is true?

(a) If the atomic bomb moves across the floor, then you feel it pushing back onyou with a force of less than 400 N.

*(b) You feel the atomic bomb pushing back on you with a force of 400 N, whetherit moves or not.

(c) If the atomic bomb does not move, you feel it pushing back on you with a forceof 8000 N.

(d) You feel the atomic bomb pushing back on you with a force of 8000 N, whetherit moves or not.

Solution: This is Newton’s third law. If you’re pushing on the bomb with a force of400 N, then the bomb is pushing on you with an equal force.

Problem 1193. You are pushing two crates across alevel frictionless floor, as shown at right. You apply ahorizontal force of 300 N to crate A, which is heavierthan crate B. Which of the following statements is true?

-300 N

A B

(a) Crate A pushes on crate B with a force of 300 N; crate B pushes on crate A witha force of 300 N.

(b) Crate A pushes on crate B with more force than crate B pushes on crate A.*(c) Crate A and crate B push on one another with forces that are equal to one

another and less than 300 N.(d) The two crates will not move if their weight is greater than 300 N.

Solution: We can rule out answer (b) by Newton’s third law; and we can ruleout (d), since the floor is level and frictionless: any horizontal force will move thecrates. The two crates together experience a force of 300 N; so they accelerate ata = (300 N)/(MA + MB). The force on crate B is the mass of crate B times thisacceleration:

FB = MB · a =MB

MA +MB

(300 N) < 300N

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Problem 1194. A spaceship is very far away from stars, planets, or other sources ofgravitational force. It is moving at high speed through the intergalactic void when itsrockets stop firing. Which of the following describes the spaceship’s future course?

(a) It will immediately stop, throwing the cargo and crew violently forward.

(b) It will begin slowing down, eventually coming to a complete stop in the vacuumof space.

(c) It will move at constant speed for a while, but will then start slowing down.

*(d) It will keep on going forever at constant speed.

(e) None of these

Problem 1195. You are idly playing with a ball on a level floor. You give the ball alight push, and it rolls about halfway across the floor before slowing down and comingto a stop. Why does the ball slow and stop?

(a) Because you weren’t pushing it.

(b) Because speed is proportional to force.

*(c) Because there must have been some force on it opposing the direction of itsmotion.

(d) Because the net force on the ball was zero.

(e) None of these

Problem 1196. Two objects with masses M and m, where M > m, are on a levelfrictionless surface. If a force F is applied to the smaller object, it will experience anacceleration of a. If that same force F is applied to the larger object, what will happento that object?

(a) It will experience an acceleration greater than a.(b) It will experience an acceleration of a.*(c) It will experience an acceleration less than a.(d) It will move only if the force F is greater than some minimum value.(e) None of these

Problem 1197. An object is moving northward with increasing speed. What do weknow about the forces on the object?

(a) There is a single force on the object, directed northward.*(b) There is a net force on the object, directed northward.(c) There may be several forces on the object, but the largest is directed northward.(d) We know nothing about the forces on the object.(e) None of these

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Problem 1198. An object is moving northward. What do we know about the forceson the object?

(a) There is a single force on the object, directed northward.(b) There is a net force on the object, directed northward.(c) There may be several forces on the object, but the largest is directed northward.*(d) We know nothing about the forces on the object.(e) None of these

Problem 1199. Which of the following objects is not experiencing a net force directednorthward?

(a) An object moving southward with decreasing speed.(b) An object moving northward with increasing speed.(c) An object that is instantaneously at rest, and that then begins moving northward.*(d) An object moving northward with constant speed.(e) None of these

Problem 1200. A rock is at rest on a level surface. The force that the surface exerts onthe rock has magnitude FSR; the force that the rock exerts on the surface has magnitudeFRS. What do we know about the two forces?

(a) FSR < FRS *(b) FSR = FRS (c) FSR > FRS

(d) We don’t know anything about the relative sizes of the forces.(e) None of these

Problem 1201. An object with a mass of 20kg is suspended from two ropes, as shown atright. The magnitude of the force exerted byeach rope on the object is 139 N; the magnitudeof the force of gravity on the object is 196 N.What is the magnitude of the net force on theobject?

(a) 47.4 N (b) 33.5 N(c) 13.9 N *(d) 0 N(e) None of these

\\\\\\\\��

20 kg

Solution: Since the system is static, the sum of the forces in each component directionis zero. Thus the net force is zero.

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Problem 1202. An astronaut with a mass of 100 kg is trying to move a very smallasteroid with a mass of 400 kg. The astronaut pushes on the asteroid with a force of120 N. What force does the asteroid exert on the astronaut? Round your answer to thenearest newton.


Solution: This doesn’t require any computation: only Newton’s third law. Theasteroid must exert the same magnitude of force, but in the opposite direction, as theastronaut by Newton’s 3rd Law.

Since the astronaut exerts a force of 120 N on the asteroid, the asteroid exerts a force of120 N on the astronaut. The masses of the astronaut and the asteroid have nothing todo with the problem.

Problem 1203. You are pushing a crate with mass m across a floor. You are pushingforward with a force whose magnitude is Fpush; the force of friction is directed backward,with magnitude f . If the crate is moving at a constant velocity, which of the followingis true?

*(a) Fpush = f (b) Fpush > f(c) Fpush = mg + f (d) Fpush > mg + f(e) None of these

Solution: Since the crate is moving at a constant velocity, the net horizontal forceon it must be zero. Hence Fpush = f .

Problem 1204. An object with a mass of 12kg is suspended from the ceiling by two ropesof equal length, each making an angle of 74◦ tothe horizontal. Each rope has a tension of 61N. What is the magnitude of the net force onthe object? Round your answer to the nearestnewton.

*(a) 0 N (b) 59 N(c) 118 N (d) 122 N(e) None of these

12 kgBBBBBBB

��

74◦ 74◦

Solution: You don’t have to do any calculations on this. Since the object isn’tmoving, it has a constant velocity of zero; so the net force on it is zero.

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10.2 Free-body diagrams

Problem 1205. A book is given a push along a frictionless horizontal tabletop. Afterit has been pushed, it slides along the tabletop and off the edge.

(a) What are the forces acting on the book while it is being pushed along the table?What are the reaction forces to those forces?

Solution: Sketch a free-body diagram showing action-reaction pairs. While the bookis being pushed, it experiences a horizontal force from the hand doing the pushing; itpushes back on the hand with equal force. It experiences two vertical forces adding up tozero: the downward force of gravity, and the normal force of the table pushing upward.The reaction forces are the gravitational force pulling the Earth upward toward the book,and the force of the book pushing downward on the table.

(b) What are the forces acting on the book while it is sliding along the table after beingpushed? What are the reaction forces to those forces?

Solution: As in part (a), only with no horizontal forces.

(c) What are the forces acting on the book while it is falling from the table to the floor?What are the reaction forces to those forces?

Solution: The only force acting on the book is the Earth’s gravity pulling it downward.The reaction force is the gravitational force pulling the Earth upward toward the book.

Problem 1206. Two crates are connected by athin horizontal rope and sitting on a horizontalfloor. You are pulling them by a second horizon-tal rope attached to one of the crates. There isfriction between the crates and the floor. Draw afree-body diagram for each of the crates.

Solution: We isolate each crate as a separatesystem and draw a free-body diagram for thatsystem. In this course, we will usually assumethat the rope is massless and doesn’t stretch; soT1 = T2.

� m1m2

u6N1

?m1g

--T1 + f1�

Pu6

N2

?m2g

-f2�

T2

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10.3 Pushing and pulling objects on horizontal frictionless sur-faces

Problem 1207. Two blocks are connected by athin rope and sitting on a horizontal frictionlesssurface. You are pulling them by a second ropeattached to the smaller block. You pull on yourrope with a horizontal force of P ; the tension inthe rope connecting the blocks is T . Which of thefollowing is true of P and T?

� P 37 kg T 89 kg

(a) P < T (b) P = T*(c) P > T (d) Not enough information; it could be any of these

Solution: P is giving a certain acceleration to the combined mass of both crates. Tis giving the same acceleration to the 89-kg crate alone. Thus T is smaller than P ; infact,

T =89 kg

37 kg + 89 kg· P

Problem 1208. You are pushing two blocks across ahorizontal frictionless surface, as shown at right. Youapply a horizontal force of 240 N to the smaller block.How much force does the smaller block exert on the largerone?

-240 N

20 kg 60 kg

(a) 60 N (b) 120 N*(c) 180 N (d) 240 N

Solution: You are applying a net horizontal force of 240 N to a total mass of20 + 60 = 80 kg. By Newton’s second law, the two crates together will accelerate at(240 N)/(80 kg) = 3 m/s2. Again by Newton’s second law, the net horizontal force onthe larger crate must be F = ma = (60 kg)(3 m/s2) = 180 N.

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11 Applications of Newton’s Laws to Linear Motion

11.1 Newton’s laws in one-dimension

11.1.1 One-dimensional kinematics and Newton’s second law

The idea behind these problems is add one more equation to our three fundamentalkinematic equations:

{Fnet = ma}+

Fund. Eqn 1: v = v0 + at

Fund. Eqn 2: v2 = v20 + 2a(x− x0)

Fund. Eqn 3: x = x0 + v0t+ 12at2

This builds our system of possible fundamental equations to 4 equations:Fund. Eqn 1: v = v0 + at

Fund. Eqn 2: v2 = v20 + 2a(x− x0)

Fund. Eqn 3: x = x0 + v0t+ 12at2

Fund. Eqn 4: Fnet = ma

with the catch that when mass or force is involved in a problem then fundamental equa-tion 4 is always used together with one of the other three equations. That is, we arealways solving a system of equations: F = ma plus one of the fundamental kinematicequations.

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Problem 1209. A block with a mass of 132 kg is pulled with a horizontal force of ~Facross a rough floor. The coefficient of friction between the floor and the block is 0.51.If the block is moving at a constant velocity, what is the magnitude of ~F? Round youranswer to the nearest newton.


Solution: Since the block’s velocity is constant, the net force on it is zero. The twoforces acting in the horizontal direction are the pull, directed forward with magnitude F ;and the frictional force fk, directed backward with magnitude µmg. Hence:

F = fk = µmg = (0.51)(132 kg)(9.8 m/s2) = 660 N

Problem 1210. A 5.5-kg block is initially at rest on a frictionless horizontal surface.It is pulled with with a constant horizontal force of 3.8 N. How long must it be pulledbefore its speed is 5.2 m/s?(a) 2.5 s (b) 5.0 s*(c) 7.5 s (d) 10.0 s(e) None of these

Solution: Start by using Newton’s 2nd law for 1-d motion to determine a. Once a isknown, this becomes a kinematics problem and we can apply one of our 3 fundamentalequations.

F = ma÷m−−→ a =

F

m=

3.8 kg m/s2

5.5 kg= 0.69 m/s2.

We now know the acceleration a; so write down what information you are given, whatyou want, and which fundamental equation you need to use.

Given:

vi = 0 m/s

vf = 5.2 m/s

a = .69 m/s2

ti = 0

Want: t Which Equation: Fundamental equation (1)

Solving for t in equation 1: t =vf − via

=5.2 m/s

.69 m/s2 = 7.5 s.

Problem 1211. A 523-kg experimental rocket sled can be accelerated from rest to 1620km/h in 1.82 s. What is the net force required to produce this motion? Assume theapplied force is constant.*(a) 1.29× 105 N (b) 6.34× 103 N(c) 2.74× 106 N (d) 1.29× 104 N(e) None of these

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Solution: Start by down what we’re given and what we want:

Given:

m = 5 kg

vi = 0 m/s

vf = 1620 kmh

(1 h

3600 s

) (1000 m1 km

)= 450m/s

ti = 0

tf = 1.82 s

Want: F = ?

We can find the net force using Newton’s 2nd law for 1-d motion F = ma, but we’ll needto use 1-d kinematics to determine a. To find a we can apply one of our 3 fundamentalequations. Since we know both velocities and time, but there is no x in the problem weshould apply equation (1). Put another way, since the acceleration is constant,

a = aave

=∆v

∆t

=vf − vitf − ti

=450 m/s

1.82 s= 247 m/s2

×m−−−−−→ F = ma = (523 kg)(247 m/s2) = 1.29× 105 N .

Problem 1212. A box has a mass of 160 kg. It is sitting on an icy sidewalk, whichconstitutes a horizontal frictionless surface. A horizontal force of 32 N is applied to thebox. What is the box’s acceleration?

(a) 5 m/s2 *(b) 0.2 m/s2

(c) 6.4 m/s2 (d) 0.16 m/s2

(e) None of these

Solution: Let +x be the direction of the force and acceleration. The net force on thebox is 32 N. Since F = ma, we can solve for acceleration:

a =Fnet

m=

32 N

160 kg= 0.2 m/s2

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Problem 1213. A box has a mass of 60 kg. It is sitting on an icy sidewalk, whichconstitutes a horizontal frictionless surface. A horizontal force of 30 N is applied to thebox. What is the box’s acceleration?

(a) 15 m/s2 (b) 5 m/s2

(c) 2 m/s2 *(d) 0.5 m/s2

(e) None of these

Solution: The only horizontal force on the box is the push of 30 N. Since F = ma,

a =F

m=

30 N

60 kg= 0.5 m/s2

Problem 1214. You are trying to move a 150-lb crate across a floor by pulling on a ropeinclined at 17◦ to the horizontal. The coefficient of static friction is 0.52; the coefficientof kinetic friction is 0.35. You pull on the rope until the crate starts moving. What isthe crate’s initial acceleration?

Solution: See the solution for Problem 25 in Chapter 5 Exercises on my website.

Problem 1215. A 10-kg block is initially at rest on a frictionless horizontal surface. Itis pulled with with a constant horizontal force of 10 N until it reaches a final velocity of10 m/s. How far does it move in this time?(a) 5 m (b) 10 m(c) 100 m *(d) 50 m(e) None of these


F = ma÷m−−→ a =

F

m=

10 kg m/s2

10 kg= 1.0 m/s2.

We now know the acceleration a; so write down what information you are given, whatyou want, and which fundamental equation you need to use. Notice that time does notappear explicitly in the problem, and we are not asked to find the time.

Given:

vi = 0 m/s

vf = 10 m/s

a = 1.0 m/s2

ti = 0

Want: ∆x Which Equation: Fundamental equation (2)

Solving for ∆x in equation 2: ∆x =v2f − v2

i

2a=

102 (m/s)2

2(1.0) m/s2 = 50 m.

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Problem 1216. A 1.0-kg block is initially at rest on a frictionless horizontal surface.It is pulled with with a constant horizontal force of 1.0 N. How long must it be pulledbefore its speed is 1.0 m/s?(a) 0.25 s (b) 0.5 s(c) .75 s *(d) 1.0 s(e) None of these


F = ma÷m−−−−−→ a =

F

m=

1.0 kg m/s2

1.0 kg= 1.0 m/s2.

We now know the acceleration a; so write down what information you are given, whatyou want, and which fundamental equation you need to use.

Given:

vi = 0 m/s

vf = 1.0 m/s

a = 1.0 m/s2

ti = 0

Want: t Which Equation: Fundamental equation (1)

Solving for t in equation 1: t =vf − via

=1.0 m/s

1.0 m/s2 = 1.0 s.

386

Problem 1217. A crate of auto parts is sitting on a horizontal frictionless frozen lake.A worker applies a constant horizontal force of 750 N to the crate. Starting from rest,the crate moves 20 m in the first 4 s. What is the mass of the crate?(a) 250 kg *(b) 300 kg(c) 350 kg (d) 400 kg(e) None of these

Solution: If you sketch a free-body diagram, you’ll see that the only horizontal forceis the constant 750 N. That is therefore the net horizontal force. There’s no net verticalforce.

Given: Fnet = 750 N; vi = 0 m/s; vf = ? ; ∆x = 20 m; and ∆t = 4 s.

Want: m = ?

If we try to solve for m directly using Newton’s second law, we get stuck:

Fnet = ma÷a−−−−−→ m =

Fnet

a=

750 N

a(11.1)

We need to use one of our three fundamental kinetic equations to find a. Since we know∆x, vi = 0, and t, we can use

∆x = vit+1

2at2 =

at2

2

· 2/t2−−−−→ a =2∆x

t2(11.2)

Substituting equation (11.5) into equation (11.4) gives us

m =Fnett

2

2∆x=

(750 N)(4 s)2

2(20 m)= 300 kg

Problem 1218. A ball with a mass of 0.4 kg is at rest on top of a tee when it is hit witha bat moving horizontally. The bat is in contact with the ball for 20 ms; when the ballleaves the bat, it is moving horizontally at 50 m/s. If the bat applies a constant forceFnet to the ball during the whole time that they are in contact, what is the magnitudeof Fnet?

(a) 400 N *(b) 1000 N(c) 2500 N (d) 4000 N(e) None of these

Solution: We know the ball’s mass m, its initial speed v0 and final speed vf , andthe time over which it accelerates ∆t. We want to know the force Fnet. We can useNewton’s second law: Fnet = ma. We don’t know a, but we can find it from the equationvf = vi + a∆t, in which we know everything but a:

vf = vi + a∆t ⇒ a =vf − vi

∆t

Fnet = ma =m(vf − vi)

∆t=

(0.4 kg)(50 m/s− 0 m/s)

0.02 s= 1000 N

387

Problem 1219. A 10.0-kg block is initially at rest on a frictionless horizontal surface.It is pulled with with a constant horizontal force of 10.0 N.(a) How long must it be pulled before its speed is 100 m/s? Round your answer to thenearest second(b) How far does it move in this time? Round your answer to the nearest meter.

Solution: Since F = ma, the acceleration of the block is

a =F

m=

10.0 N

10.0 kg= 1 m/s2

(a) Since the block starts at v0 = 0, speed as a function of time is v = at; so

t =v

a=

100 m/s

1 m/s2 = 100 s

(b) Since v0 = 0 and x0 = 0, position as a function of time is

x =1

2at2 =

(1 m/s2)(100 s)2

2= 5000 m

Problem 1220. A 1-kg ketchup bottle is slid across a smooth counter at a local diner.You notice that the bottle slid 1.5 meters and took 2 seconds to come to rest. Assuminga constant net force owing to the friction on the table, what was the initial velocity ofthe bottle? Round your answer to the nearest 0.1 m/s.

(a) 5 m/s (b) 0.5 m/s*(c) 1.5 m/s (d) 0.15 m/s(e) None of these

Solution: If the force on the bottle was constant, then so was the acceleration a. If welet x0 = 0, then position as a function of time is: x = v0t+

12at2. Since the acceleration is

constant and slows the bottle from v0 to v = 0 in time t = 2 s, we can write: a = −v0/t.Substituting that expression for a into the equation of position, we get

x = v0t−1

2

v0

tt2 =

v0t

2

Solving for v0 yields

v0 =2x

t=

2(1.5 m)

2 s= 1.5 m/s

[Notice that we didn’t need the mass of the bottle.]

388

Problem 1221. A block of mass M is pulled along a horizontal frictionless surface by abar of mass m that is attached to the block and horizontal to the surface . A horizontalforce ~P (P for pull) is applied to the free end of the bar (see figure below).(a) Find the acceleration of the bar-block system (i.e., the bar and the block takentogether).(b) What is the force that the bar exerts on the block?

Solution: (a) The bar-block system has mass M +m. The acceleration of the systemis the force applied to it divided by its mass:

a =P

M +m

(b) The force on the block is its mass times its acceleration. Its mass is M and itsacceleration is that of the whole bar-block system, so

F = Ma =MP

M +m

11.1.2 Weight

Remember weight is defined as w = mg, where m is mass and g is the magnitude of theacceleration due to gravity. g is always positive.

Problem 1222. The acceleration due to gravity at the surface of Neptune’s moon Tritonis 2.41 m/s2. If a rock weighs 106 N on Triton, what is its mass? Round your answer tothe nearest kilogram.

(a) 26 kg *(b) 44 kg(c) 255 kg (d) 431 kg(e) None of these

Solution: w = mgT

÷ gT−−−→ m =

w

gT

=106 N

2.41 m/s2 = 44 kg

Problem 1223. Triton’s gravity is 2.41 m/s2. If a rock weighs 482 N on Triton, whatis its mass? Round your answer to the nearest kilogram.


Solution: w = mgT

÷gT−−−−−−→ m =

w

gT

=482 N

2.41 m/s2 = 200 kg.

389

Problem 1224. The acceleration due to gravity at the surface of Neptune is 11.15 m/s2.If a rock weighs 266 N on Neptune, what is its mass? Round your answer to the nearest0.1 kg.

(a) 27.1 kg (b) 2965.9 kg(c) 0.4 kg *(d) 23.9 kg(e) None of these

Solution: On Neptune,

WN = mgN ⇒ m =WN

gN=

266 N

11.15 m/s2 = 23.9 kg

Problem 1225. The acceleration due to gravity at the surface of Planet X is 12.2 m/s2.If a rock weighs 307 N on Planet X, what is its mass? Round your answer to the nearestkilogram.

*(a) 25 kg (b) 247 kg(c) 382 kg (d) 3745 kg(e) None of these

Solution: On Planet X,

WX = mgX ⇒ m =WX

gX=

307 N

12.2 m/s2 = 25 kg

Problem 1226. An astronaut weighs 790 N on Earth. On Planet X, he weighs 640 N.What is astronaut’s mass on Planet X? Round your answer to the nearest kilogram.


Solution: Remember that the mass of an object is the same wherever it may be. Theonly way to change mass is by adding material to or taking it away from the object. Weknow that on Earth, wE = mgE; and we know both wE and gE. Hence

wE = mgE÷gE−−→ m =

wEgE

=790 N

9.8 m/s2 = 81 kg

390

Problem 1227. An astronaut weighs 790 N on Earth. On Planet X, he weighs 640 N.What is the astronaut’s mass on Earth? Round your answer to the nearest kilogram.


Solution: Remember that the mass of an object is the same wherever it may be. Theonly way to change mass is by adding material to or taking it away from the object. Weknow that on Earth, wE = mgE; and we know both wE and gE. Hence

wE = mgE÷gE−−→ m =

wEgE

=790 N

9.8 m/s2 = 81 kg

Problem 1228. (Similarity Problem) An astronaut weighs 790 N on Earth. OnPlanet X, he weighs 640 N. What is the gravitational acceleration on Planet X? Roundyour answer to the nearest 0.1 m/s2.

(a) 5.8 m/s2 (b) 6.4 m/s2

(c) 7.1 m/s2 *(d) 7.9 m/s2

(e) None of these

Solution:Step 1: Write down what you’re given, what you want, and any assumptions:Given: wE = mgE = 790 N; wX = mgX = 640 N

In addition we know gE = 9.8 m/s2, but don’t know mWant: gX = ?

Step 2: Solve. We can eliminate the common parameter m by solving both equations form and setting the expressions equal (m is our common link):

wEgE

= m =wXgX

× gXgEwE−−−−→ gX =

wXgEwE

=(640 N)(9.8 m/s2)

790 N= 7.9 m/s2

391

Problem 1229. A uniform chain is 2.0 m long and hasa mass of 20 kg. It hangs from the ceiling and supportsa lamp with a mass of 60 kg. What is the tension in thetop link of the chain? For ease of calculation, assume thatg = 10 m/s2.


wL

u12wcu

12wcu

u6

Ttop

?wL

?wc

Solution: Sketch a picture of the entire system; then a free-body diagram of thesystem consisting of the top link. The top link supports the entire weight of the chainwc, plus the weight of the lamp wL. Since the system is in static equilibrium, there is noacceleration, so no net force; so

Ttop = wc + wL = (mc +mL)g = (20 kg + 60 kg)(10 m/s2) = 800 N

Problem 1230. A uniform chain is 2.0 m long and hasa mass of 20 kg. It hangs from the ceiling and supportsa lamp with a mass of 60 kg. What is the tension in themiddle link of the chain? For ease of calculation, assumethat g = 10 m/s2.


wL

u12wcu

12wcu

u6

Tmiddle

?wL

?12wc

Solution: Sketch a picture of the entire system; then a free-body diagram of thesystem consisting of the middle link. The middle link supports half of the weight of thechain, 1

2wc, plus the weight of the lamp wL. Since the system is in static equilibrium,

there is no acceleration, so no net force; so

Ttop =1

2wc + wL = (

1

2mc +mL)g = (10 kg + 60 kg)(10 m/s2) = 700 N

392

Problem 1231. A uniform chain is 2.0 m long and hasa mass of 20 kg. It hangs from the ceiling and supportsa lamp with a mass of 60 kg. What is the tension in thebottom link of the chain? For ease of calculation, assumethat g = 10 m/s2.


wL

u12wcu

12wcu

u6

Tbottom

?wL

Solution: Sketch a picture of the entire system; then a free-body diagram of thesystem consisting of the bottom link. The bottom link supports the weight of the lampwL, but none of the weight of the chain. Since the system is in static equilibrium, thereis no acceleration, so no net force; so

Tbottom = wL = mLg = (60 kg)(10 m/s2) = 600 N

Problem 1232. (Force-to-weight ratio) A softball is at rest on top of a tee whenit is hit with a bat moving horizontally. The bat is in contact with the ball for 20 ms;when the ball leaves the bat, it is moving horizontally at 50 m/s. If the bat appliesa constant force Fnet to the ball during the whole time that they are in contact, whatis the force-to-weight ratio Fnet/wball for the ball? For ease of calculation, assume thatg = 10 m/s2.


Solution: We know the ball’s initial speed v0 and final speed vf , and the timeover which it accelerates ∆t. We want to know the ratio of net force to weight Fnet/w.Since we don’t know the ball’s mass, we can’t know the net force or the weight; but incalculating the ratio, the mass cancels out:

Fnet

wball

=ma

mg=a

g

We know enough to calculate the acceleration:

vf = vi + at ⇒ a =vf − vi

tFnet

wball

=a

g=vf − vigt

=50 m/s− 0 m/s

(10 m/s2)(0.02 s)= 250

393

Problem 1233. (Determining the weight of an object without a scale) You havejust landed on Planet X and realized that you’ve left your scale behind. However, youhave a 5.0 kg ball, and a spring cannon that can launch the ball straight upward at aspeed of 15 m/s. You launch the ball upward from ground level; it strikes the groundagain 3.0 s after launching. What is the magnitude of gravity on Planet X? Use this todetermine the weight of the object. Round your answer to the nearest newton.


Solution: We know m = 5.0 kg; vi = 15 m/s; and ∆t = 3.0 s. We want wX = mgX ,where gX is the gravitational acceleration on Planet X. We are assuming that there isno air resistance, so the ball returns to the ground at the same speed at which it waslaunched, and it takes as much time going up as it does falling back down: vf = −vi =−15 m/s, and ∆t = 2tymax, where tymax is the time that the ball takes to reach itsmaximum height after being launched.The velocity as a function of time is v(t) = vi− gXt. At the top of the trajectory (i.e. att = tymax), the velocity is zero. Hence

v(tymax) = 0 = vi − gXtymax ⇒ gX =vi

tymax

∆t=2tymax−−−−−−→ gX =2vi∆t

⇒ wX = mgX =2mvi∆t

=2(5.0 kg)(15 m/s)

3.0 s= 50 N

394

Problem 1234. You are standing on a 100 m tall cliff on Planet X. You drop a 5 kgstone off the cliff, and notice that it takes 10 seconds for it to hit the ground below.What is the magnitude of gravity g

Xon planet X? You may assume that you can ignore

air drag.(a) 1.0 m/s2 *(b) 2.0 m/s2

(c) 5.0 m/s2 (d) 10.0 m/s2

(e) None of these

Solution: This is just a kinematics problem! The weight of the stone is irrelevant.By Newton’s 2nd law for 1-d motion with weight as our force:

mgX

= w = F = ma÷m−−−−−→ g

X= a .

We can now apply one of our 3 fundamental equations. Start by writing down whatinformation you are given, what you want, and which fundamental equation you needto use. We’ll take our initial position to be the origin yi = 0, and the final position atthe bottom of the cliff as our final position yf = 100 m. Since this is ”drop problem”the natural direction for the y-axis is in the direction of motion, so we’ll take the y-axispointing downward.

Given:

yi = 0 m

yf = 100 m

vi = 0 m/s

ti = 0 s

tf = 10 s

Want: a = gX

Which Equation: Equation (3)

Solving for a in equation 3 with v0 = 0 and y0 = 0 gives:

yf =1

2g

Xt2fall ⇒ g

X=

2yft2fall

=2 (100) m

102 s2= 2.0 m/s2 .

395

11.1.3 Apparent weight problems (The Elevator Equation)

The apparent weight of a mass that is accelerating (upward or downward) with the y-axistaken pointing upward is

wapp = w

(1 +

ayg

)

Problem 1235. A block is hanging from a spring scale that isattached to the ceiling of an elevator. Under which of the followingcircumstances will the reading on the scale be less than the weightof the block?

(a) The elevator is moving downward and slowing down.*(b) The elevator is moving upward and slowing down.(c) The elevator is moving downward at constant speed.(d) The elevator is moving upward at constant speed.

u

?wb

6Tu

Solution: Sketch a free-body diagram for the block, as at right. There are twoforces acting on the block: the weight wb pulling down, and the tension in the scale Tupulling upward. If Tu < wb, then the net force is downward; so the block is experiencingdownward acceleration. This can happen if the elevator is moving upward and slowingdown, or if it is moving downward and speeding up. Only the former option is in theanswers.

Problem 1236. A block is hanging from a spring scale that is attached to the ceiling ofan elevator. What will the reading on the scale be if the elevator cable breaks?

Solution: The vertical forces on the block are the weight pulling downward, and thetension in the scale pulling up. If the cable breaks and the elevator is in free fall, thenthe net force on the block is equal to its weight; so the tension in the scale is zero. Hencethe scale will register a reading of zero.

Problem 1237. An object is hung from a spring scale attached to the ceiling of anelevator. The scale reads 100 N when the elevator is sitting still. What is the readingon the scale if the elevator is moving downward with a constant speed of 10 m/s?Approximate g as 10 m/s2.(a) 0 N (b) 50 N*(c) 100 N (d) 200 N(e) None of these

Solution: Since the elevator is moving at constant speed, the acceleration a is 0 m/s2.From Newton’s 2nd law it follows that the net force is just the force due to gravity, whichwas present when the elevator was sitting still. Thus the reading on the scale is 100 N.

396

Problem 1238. What is the apparent weight of a 150 N block if it is acceleratingupward at 1 m/s2? For ease of calculation, assume that g = 10 m/s2.


Solution: Use the apparent weight formula.

Problem 1239. What is the apparent weight of a 150 N block if it is acceleratingdownward at 1 m/s2? For ease of calculation, assume that g = 10 m/s2.

(a) 20 N (b) 135 N(c) 150 N (d) 165 N(e) None of these

Solution: Use the apparent weight formula.

Problem 1240. What is the apparent weight of a 150 N block if it is moving upwardat a constant velocity of 1 m/s? For ease of calculation, assume that g = 10 m/s2.


Solution: Constant velocity ⇒ No acceleration ⇒ wapp = w.

Problem 1241. What is the weight of a 150 N block if it is accelerating downward at1 m/s2? For ease of calculation, assume that g = 10 m/s2.


Solution: The weight of an object is the force exerted on it by gravity, and that’s notaffected by its upward or downward acceleration.

397

Problem 1242. An elevator weighing 6200 lb is pulled upward by a cable, with anupward acceleration of 3.2 ft/s2. What is the tension in the cable? Round your answerto the nearest 100 lb.

(a) 218,200 lb (b) 19,800 lb*(c) 6800 lb (d) 600 lb(e) None of these

Solution: A free-body diagram shows the force of weight pointing straight downwardwith magnitude W = mg, and the tension T pointing straight upward. The sum of thetwo is the net force, which produces an upward acceleration of a. Hence Newton’s secondlaw becomes: T −W = T −mg = ma. We solve for tension: T = m(g + a). We knowthat W = mg, so m = W/g. Hence

T = m(g + a) =W (g + a)

g= W

(1 +

a

g

)= (6200 lb)

(1 +

3.2 ft/s2

32 ft/s2

)= 6800 lb

Problem 1243. A bucket weighing 500 N is lifted withan upward acceleration of 2 m/s2 by a uniform chainweighing 200 N. What is the tension in the top link of thechain? For ease of calculation, assume that g = 10 m/s2.

(a) 360 N (b) 720 N*(c) 840 N (d) 1080 N(e) None of these wb

u12wcu

12wcu

u6

Ttop

?wb

?wc

Solution: Sketch a picture of the entire system; then a free-body diagram of thesystem consisting of the top link. The top link supports the weight of the bucket wb plusthe entire weight of the chain wc. Since the system is accelerating upward, it experiencesa net upward force. We will use Newton’s second law:

Ttop − wb − wc = Fnet = msystema = (mb +mc)a

⇒ Ttop = wb + wc + (mb +mc)a

w=mg−−−→ = wb + wc +(wb + wc)a

g

= (wb + wc)g + a

g= (500 N + 200 N)

10 m/s2 + 2 m/s2

10 m/s2 = 840 N

398

Problem 1244. A bucket weighing 500 N is lifted withan upward acceleration of 2 m/s2 by a uniform chainweighing 200 N. What is the tension in the middlelink of the chain? For ease of calculation, assume thatg = 10 m/s2.


wb

u12wcu

12wcu

u6

Tmid

?wb

?12wc

Solution: Sketch a picture of the entire system; then a free-body diagram of thesystem consisting of the middle link. The middle link supports the weight of the bucketwb plus half the weight of the chain, 1

2wc. Since the system is accelerating upward, it

experiences a net upward force. We will use Newton’s second law:

Tmid − wb −wc2

= Fnet = msystema =(mb +

mc

2

)a

⇒ Tmid = wb +wc2

+(mb +

mc

2

)a

w=mg−−−→ = wb +wc2

+(wb + wc/2)a

g

=(wb +

wc2

) g + a

g=

(500 N +

200 N

2

)10 m/s2 + 2 m/s2

10 m/s2 = 720 N

Problem 1245. A bucket weighing 500 N is lifted withan upward acceleration of 2 m/s2 by a uniform chainweighing 200 N. What is the tension in the bottomlink of the chain? For ease of calculation, assume thatg = 10 m/s2.


wb

u12wcu

12wcu

u6

Tbottom

?wb

Solution: Sketch a picture of the entire system; then a free-body diagram of thesystem consisting of the bottom link. The bottom link supports the weight of the bucketwb, but none of the weight of the chain wc. Since the system is accelerating upward, itexperiences a net upward force. We will use Newton’s second law:

Tbottom − wb = Fnet = msystema = mba

⇒ Tbottom = wb +mbaw=mg−−−→ = wb +

wba

g

= wbg + a

g= (500 N)

10 m/s2 + 2 m/s2

10 m/s2 = 600 N

399

Problem 1246. You are standing on a spring scale on an elevator.When the elevator is not moving, the scale registers a weight of 600N. When the elevator starts moving, it has an upward accelerationof 2 m/s2. What weight does the scale register while the elevator isaccelerating? For ease of calculation, assume that g = 10 m/s2.

(a) 360 N (b) 500 N*(c) 720 N (d) 1200 Nv(e) None of these

u6

Ns

?w

Solution: Sketch a free-body diagram, as at right. The forces working on the system(you) are the normal force of the scale Ns, which is the weight registered on the scale;and your weight. Since you are accelerating, there is a nonzero net force. Use Newton’ssecond law:

Fnet = Ns − w = ma

⇒ Ns = w +ma

w=mg−−−→ Ns = w +wa

g=

(1 +

a

g

)w =

(1 +

2 m/s2

10 m/s2

)(600 N) = 720 N

Problem 1247. You are standing on a spring scale on an elevator. When the elevatoris not moving, the scale registers a weight of 730 N. When the elevator starts moving, ithas an upward acceleration of 1.6 m/s2. What weight does the scale register while theelevator is accelerating? Round your answer to the nearest 10 N.


Solution: There are two vertical forces acting on you: your weight w = mg = 730 N,pulling down; and the normal force of the scale, Nscale = wapp, your apparent weight,pushing upward. The combination of these two is the net force, which gives you anupward acceleration of a = 1.6 m/s2. Taking the upward direction as positive, this givesus

Nscale − w = Fnet = ma ⇒ Nscale = w +ma

We solve w = mg to get: m = w/g. Substituting this gives

Nscale = w +ma = w +wa

g= w

(1 +

a

g

)= (730 N)

(1 +

1.6 m/s2

9.8 m/s2

)= 850 N

400

Problem 1248. Someone has left a spring scale on an elevator. While the elevatoris not moving, you get on the scale and see that your weight is 810 N. When theelevator starts moving, you see that the scale registers 890 N. What is the accelerationof the elevator? Round your answer to the nearest 0.1 m/s2; use a positive sign if theacceleration is upward, a negative sign if the acceleration is downward.

*(a) 1.0 m/s2 (b) 10.8 m/s2

(c) −1.0 m/s2 (d) −10.8 m/s2

(e) None of these

Solution: If your mass is m, then gravity is pulling you downward with a force of yourweight: −w = −mg = −810 N. When the elevator is accelerating, the scale reading isthe normal force of the floor pushing up on you. This upward force is Fscale = 890 N. Thenet vertical force on you is Fnet = Fscale−mg = 80 N. Since Fnet = ma, your accelerationis a = Fnet/m. We can get m from your weight: m = w/g. Hence

a =Fnet

m=Fnetg

w=

(80 N)(9.8 m/s2)

810 N= 1.0 m/s2

Problem 1249. Someone has left a spring scale in an express elevator. While theelevator is standing still, you get on the scale and see that you weigh 142 lb. You ridethe elevator nonstop to the top floor of the building. In the middle of the ride, you aregoing upward at at a constant speed of 44 ft/s. What weight does the scale register atthis point? Round your answer to the nearest pound.

(a) 103 lb *(b) 142 lb(c) 154 lb (d) 195 lb(e) None of these

Solution: Since the elevator is moving at a constant speed, it is experiencing noacceleration. The weight registered by the scale is the same as it would be if the elevatorwere standing still: 142 lb.

Problem 1250. A lamp with a mass of 7.0 kg is hanging from athin cord attached to the ceiling of an elevator. When the elevator isaccelerating upward at 3 m/s2, what is the tension in the cord? Forease of calculation, assume that g = 10 m/s2.


u6

T

?w = mg

Solution: Sketch a free-body diagram for the lamp, as at right. The forces workingon the system are the tension of the cord T , and the lamp’s weight w = mg. Since thesystem is accelerating, there is a nonzero net force. Use Newton’s second law:

Fnet = T −mg = ma

⇒ T = mg +ma = m(g + a) = (7.0 kg)(10 m/s2 + 3 m/s2) = 91 N

401

Problem 1251. A block with weight w is hanging from a springscale that is attached to the ceiling of an elevator. If the elevatorhas an upward acceleration of a > 0 and the reading on the scale isT , what is w as a function of a and T?

(a) w =Tg

g − a(b) w =

T (g − a)

g

*(c) w =Tg

g + a(d) w =

T (g + a)

g

(e) None of these

u?w

6T

Solution: Sketch a free-body diagram of the block, as at right. There are two verticalforces acting on the block: the weight w pulling down, and the tension in the scale Tpulling up. The block is experiencing an upward acceleration of a. Using Newton’ssecond law and the fact that w = mg:

T − w = ma =wa

g

⇒ T = w

(1 +

a

g

)= w

(g + a

g

)⇒ w =

Tg

g + a

Problem 1252. A block with weight w is hanging from a springscale that is attached to the ceiling of an elevator. If the elevatorhas an upward acceleration of a > 0, what is the reading on thescale T?

(a) T = wa

g(b) T = w

g

a

(c) T = w(

1 +g

a

)(d) T = w

(1 +

a

g

)(e) None of these

u?w

6T

Solution: Sketch a free-body diagram of the block, as at right. There are two verticalforces acting on the block: the weight w pulling down, and the tension in the scale Tpulling up. The block is experiencing an upward acceleration of a. Using Newton’ssecond law and the fact that w = mg:

T − w = ma =wa

g

⇒ T = w

(1 +

a

g

)Quick way: Notice the the tension in the string is the force that the scale reads. Butthat is just the apparent weight. Now apply the apparent weight formula.

402

11.1.4 Spring Problems

Problem 1253. (Derived quantities) Determine the units of the spring constant kfrom the equation for the ideal spring force: Fspring = −kx, where F is force, k is thespring constant, and x is the displacement from the equilibrium position. Write theanswer in terms of Newtons.

Solution: Using our take-the-dimensions-of-the-quantity operator [ ] on the mathe-matical expression for the spring force gives

[F ] = [−kx] = [k][x]÷[x]−−−−−−→ [k] =

[F ]

[x]=

N

m

There is no special name for a Newton divided by a meter.

Problem 1254. A rubber band has an unstretched length of 8 cm and a force constantof 4 N/cm. What is the force required to stretch it to a length of 12 cm?


Solution: The rubber band is being stretched by ∆x = 12 cm − 8 cm = 4 cm. Theforce required to stretch it is F = k∆x = (4 N/cm)(4 cm) = 16 N.

Problem 1255. A spring is hanging from the ceiling. Its unstretched length is 12 cm.When a block with a mass of 6 kg is hanging from the end, it stretches to a length of15 cm. What is the spring constant of the spring? For ease of calculation, assume thatg = 10 m/s2.

(a) 4 N/m (b) 20 N/m(c) 400 N/m *(d) 2000 N/m(e) None of these

Solution: Don’t forget to convert centimeters to meters. When a mass of m is hangingfrom the spring, the force on the spring is w = mg. By Hooke’s law, F = k∆y; so

k =F

∆y=

mg

yf − y0

=(6 kg)(10 m/s2)

0.15 m− 0.12 m= 2000 N/m

403

Problem 1256. A spring is hanging from the ceiling. Its unstretched length is 10 cm.When a block with a mass of 8 kg is attached to the end, it stretches to a length of 20cm. If you want to stretch the spring to a length of 30 cm, how much mass must youhang from it?

(a) 8 kg (b) 12 kg*(c) 16 kg (d) 24 kg(e) None of these

Solution: When a mass of m is hanging from the spring, the force on the spring isw = mg. By Hooke’s law, F = mg = k∆y. When we change the mass from m1 to m2,we change from ∆y1 to ∆y2; k and g don’t change. Hence

k

g=

m

∆y=

m1

∆y1

=m2

∆y2

⇒ m2 =m1∆y2

∆y1

=(8 kg)(0.30 m− 0.10 m)

0.20 m− 0.10 m= 16 kg

Problem 1257. A horizontal spring with a spring constant of 500 N/m is used to pull asled across a frictionless frozen lake. The sled has a mass of 40 kg, and is undergoing anacceleration of 0.5 m/s2. How much does the spring stretch while pulling the sled? Forease of calculation, use g = 10 m/s2.

(a) 2.5 cm *(b) 4 cm(c) 10 cm (d) 25 cm(e) None of these

Solution: Since the ice is horizontal and frictionless, we need only concern ourselveswith the horizontal forces. There is only one: the tension in the spring, pulling on thesled. Using Newton’s second law gives us the value of the force: F = ma. Using Hooke’slaw gives us the amount by which the spring stretches.

F = ma = k∆x ⇒ ∆x =ma

k=

(40 kg)(0.5 m/s2)

500 N/m= 0.04 m = 4 cm

404

Problem 1258. An unstretched spring has a length of 22.8 cm. When an object witha mass of 12.0 kg is hung from it, it stretches to a length of 35.5 cm. What is the springconstant of the spring? Round your answer to the nearest N/m.


Solution: The force applied to the spring is the weight of the object w = mg. Sincethere is no acceleration in the mass-spring system, applying Newton’s 2nd law in thevertical direction gives |Fspring| = w = mg. The spring stretches by x = 35.5 cm −22.8 cm = 0.127 m. (Since the problem asks for the spring constant in N/m, you mustconvert centimeters to meters.) Hooke’s law gives us

|Fspring| = kx = mg ⇒ k =mg

x=

(12.0 kg)(9.8 m/s2)

0.127 m= 926 N/m

Problem 1259. An unstretched spring is 62 cm long. When you hang an object with amass of 4.6 kg from it, it stretches to a length of 71 cm. What is the mass of the objectyou would have to hang from the spring to get it to stretch to a length of 90 cm? Roundyour answer to the nearest 0.1 kg.

(a) 1.8 kg (b) 5.8 kg*(c) 14.3 kg (d) 450.8 kg(e) None of these

Solution: Let m1 = 4.6 kg be the first mass, and let x1 = 71 cm− 62 cm = 9 cm bethe distance by which it stretches the spring. We can calculate the force constant of thespring:

m1g = kx1 ⇒ k =m1g

x1

We want to know what mass m2 it takes to stretch the spring by a distance of x2 =90 cm− 62 cm = 28 cm.

m2g = kx2 ⇒ m2 =kx2

g=m1x2

x1

=(4.6 kg)(28 cm)

9 cm= 14.3 kg

405

Problem 1260. A bungee cord obeys Hooke’s law, with a spring constant of 850 N/m.It is attached to a crate with a mass of 170 kg and pulled at an angle of 19◦ to thehorizontal, in order to move the crate across a horizontal frictionless floor. If the crateaccelerates at 0.81 m/s2, how much does the bungee cord stretch? Round your answerto the nearest centimeter.

(a) 16 cm *(b) 17 cm(c) 159 cm (d) 168 cm(e) None of these

Solution: Let T be the tension in the bungee cord. Since it makes an angle of 19◦ tothe horizontal, the horizontal component of the force is Fh = T cos 19◦. This is the onlyhorizontal force acting on the crate, and it is accelerating the crate at a = 0.81 m/s2. ByNewton’s second law, T cos 19◦ = ma. By Hooke’s law, T = kx, where x is the amountby which the bungee cord stretches. Hence

kx cos 19◦ = ma ⇒ x =ma

k cos 19◦=

(170 kg)(0.81 m/s2)

(850 N/m) cos 19◦= 0.17 m = 17 cm

Problem 1261. A bungee cord obeys Hooke’s law, with a spring constant of 620 N/m.It is attached to a crate with a mass of 220 kg and pulled at an angle of 24◦ to thehorizontal, in order to move the crate across a horizontal frictionless floor. If the crateaccelerates at 0.72 m/s2, how much does the bungee cord stretch? Round your answerto the nearest centimeter.


Solution: Let T be the tension in the bungee cord. Since it makes an angle of 19◦ tothe horizontal, the horizontal component of the force is Fh = T cos 24◦. This is the onlyhorizontal force acting on the crate, and it is accelerating the crate at a = 0.72 m/s2. ByNewton’s second law, T cos 24◦ = ma. By Hooke’s law, T = kx, where x is the amountby which the bungee cord stretches. Hence

kx cos 24◦ = ma ⇒ x =ma

k cos 24◦=

(220 kg)(0.72 m/s2)

(620 N/m) cos 24◦= 0.28 m = 28 cm

406

Problem 1262. (Lab Problem) A spring is hanging from the ceiling in the physicslab. When there is no weight hanging from it, the spring is 42 cm long; it has a springconstant of 8300 N/m. You hang a lead block with a mass of 87 kg from the spring,stretching it. What is the length of the spring with the weight hanging from it? Roundyour answer to the nearest centimeter.


Solution: We know the unstretched length of the spring Lu = 42 cm = 0.42 m; thespring constant k = 8300 N/m; and the mass m = 87 kg.

If y is the change in length of the spring, we want the stretched length: Ls = Lu + y.

Use the equation |Fspring| = ky to find y. We assume that the system is static, so thereis no acceleration; so the two forces on the weight match: mg = |Fspring| = ky. Divideby k:

y =mg

k⇒ Ls = Lu +

mg

k= 0.42 m +

(87 kg)(9.8 m/s2)

8300 N/m= 0.52 m = 52 cm

Problem 1263. (Lab Problem) An unstretched spring is 62 cm long. When youhang an object with a mass of 4.6 kg from it, it stretches to a length of 71 cm. What isthe mass of the object you would have to hang from the spring to get it to stretch to alength of 90 cm? Round your answer to the nearest 0.1 kg.

(a) 1.8 kg (b) 5.8 kg*(c) 14.3 kg (d) 450.8 kg(e) None of these

Solution: Let m1 = 4.6 kg be the first mass, and let x1 = 71 cm− 62 cm = 9 cm bethe distance by which it stretches the spring. We can calculate the force constant of thespring:

m1g = kx1 ⇒ k =m1g

x1

We want to know what mass m2 it takes to stretch the spring by a distance of x2 =90 cm− 62 cm = 28 cm.

m2g = kx2 ⇒ m2 =kx2

g=m1x2

x1

=(4.6 kg)(28 cm)

9 cm= 14.3 kg

Problem 1264. (Derive Problem) Two air-track gliders of mass m1 and m2 withm1 < m2 are connected by a spring whose mass is negligible compared to the mass ofthe gliders. The gliders sit on a horizontal frictionless air-track and are constrained toone-dimensional motion. Treat the problem as a one-dimensional problem by assumingthat the force of gravity is exactly cancelled by the upward blowing air of the air track.

The system: Take the pair of gliders together with the spring to be our system underinvestigation. There are no external forces acting on the system in the direction of thetrack.

407

The initial condition: The system is initially at rest with the spring compressed andheld in place by a thin massless string connecting the two gliders. At time t = 0, thestring is cut setting the mass-spring system in motion.

The final condition:(a) If at time tf > 0 the acceleration of mass m2 is a2, then what is the force on m2 bythe spring, denoted Fm2s?

(b) What is the acceleration of mass m1 at time tf?

Comment: After the string is cut, the energy stored in the spring sets the system inmotion. The motion comes from internal forces, there are no external forces acting onthe mass-spring system.

Solution: See solutions to chapter 3 homework on my website (Exercises #11 ).

408

11.2 Newton’s laws in two-dimensions

11.2.1 Static equilibrium problems

Problem 1265. Two nails are driven into a wall atdifferent heights, and a weight is hung from two thinstrings fastened to the nails, as shown at right. Bothstrings make the same angle θ to the horizontal, butthe first string is longer than the second. If T1 is thetension in the longer string and T2 is the tension inthe shorter, what is the relationship between T1 andT2?

M

��

rθ

T1ZZ

ZZrθ

T2

(a) T1 < T2 *(b) T1 = T2

(c) T1 > T2 (d) Not enough information: could be any of these

Solution: Since the weight isn’t moving, the net force on it must be zero. Inparticular, the net horizontal force must be zero. There are two horizontal forces actingon it: the horizontal component of T2 pulling to the left, and the horizontal componentof T1 pulling to the right. These horizontal forces must be equal to one another:

T1x = T2x ⇒ T1 cos θ = T2 cos θ ⇒ T1 = T2

The difference in the lengths of the strings doesn’t matter; what’s important is that theirangles are the same.

409

Problem 1266. A rope is stretched betweentwo poles. The rope is initially horizontal. Af-ter an 82 N block is hung from the middle ofthe rope each side of the rope makes an angleof 9.5◦ with the horizontal. What is the tensionin the rope? Round your answer to the nearest10 N.


Solution: Begin by sketching a free-bodydiagram for the block, as at right. There aretwo horizontal forces acting on the block: thehorizontal components of the tension in the tworopes, T cos θ. These forces are equal and op-posite.

hhhhhhh

hh((((

(((((

82 N

�

9.5◦

@R

9.5◦

T T

u?w

��

��1T

θ -

6

T cos θ

T sin θ

PPPP

PPPPi

T

θ�

6

T cos θ

T sin θ

There are three vertical forces acting on the block: the force due to gravity (i.e., theweight w) pulling downward, and the vertical components of the tension in the two ropespulling upward, each with magnitude Ty = T sin θ. Since the crate is in static equilibrium,the net force on it is zero; so

2T sin θ = w ⇒ T =w

2 sin θ=

82 N

2 sin 9.5◦= 250 N

410

Problem 1267. In an attempt to keep bearsfrom eating his food, a hiker stretches a ropebetween two trees and hangs the food from themiddle of the rope. When the food is hungfrom the rope, each side of the rope makes anangle of 11.4◦ with the horizontal. The rope willbreak if its tension exceeds 220 N. What is themaximum weight of food wmax that the hikercan hang without breaking the rope? Roundyour answer to the nearest newton.


Solution: Begin by sketching a free-body di-agram for the food, as at right. There are twohorizontal forces acting on the food: the hor-izontal components of the tension in the tworopes, T cos θ. These forces are equal and op-posite.

hhhhhhh

hh((((

(((((

wmax

�

11.4◦

@R

11.4◦

T T

u?wmax

��

��1T

θ -

6

T cos θ

T sin θ

PPPP

PPPPi

T

θ�

6

T cos θ

T sin θ

There are three vertical forces acting on the food: the weight wmax pulling downward,and the vertical components of the tension in the two ropes pulling upward, each withmagnitude Ty = T sin θ. Since the food is in static equilibrium, the net force on it is zero;so

wmax = 2T sin θ = 2(220 N) sin 11.4◦ = 87 N

Problem 1268. In an attempt to keep bearsfrom eating his food, a hiker stretches a ropebetween two trees and hangs the food from therope. The food weighs 82 N; each side of therope makes an angle of 9.5◦ with the horizontal.What is the tension in the rope? Round youranswer to the nearest 10 N.


hhhhhhh

hh(((

((((((

82 N

�

θ = 9.5◦

@R

θ = 9.5◦

T T

Solution: The food is not accelerating in any direction, so the net force on it is zero.In particular, the net vertical force on it is zero. The vertical forces on it are the weight,w = 82 N, pulling downward; and the vertical component of the tension of the rope oneither side, pulling upward. If the tension is T , its vertical component on either side ofthe food is Ty = T sin θ. If we take upward as the positive direction, the net vertical force

411

is:

2T sin θ − w = 0+w−−−−−→ 2T sin θ = w

÷2 sin θ−−−−−−−→ T =w

2 sin θ(11.3)

evaluate−−−−−−−−→ T =82 N

2 sin 9.5◦= 250 N

Notice that from equation (11.3), that as θ → 0, sin θ → 0, and so T →∞.

Problem 1269. Two ropes are attached to theceiling, and their free ends are tied together.A block with weight w is hung from the pointwhere the ropes are joined. Rope A makes anangle of 51◦ to the horizontal; rope B makes anangle of 68◦ to the horizontal. The tension inrope B is TB = 138 N. What is the weight w ofthe block? Round your answer to the nearestnewton.

\\\\\\\\

A

51◦

��

B

68◦

w

Solution: Since the weight isn’t accelerating to the left or to the right, we knowthat the net horizontal force on it is zero. The only horizontal forces acting on it are thehorizontal components of the tensions in the ropes. Thus these two must be equal:

TA cos 51◦ = TB sin 68◦ ⇒ TA =TB cos 68◦

cos 51◦

Since the weight isn’t accelerating up or down, the net vertical force on it is zero. Thevertical forces are the weight w pulling down, and the vertical components of TA and TBpulling up. Hence:

w = TA sin 51◦ + TB sin 68◦ =TB cos 68◦ sin 51◦

cos 51◦+ TB sin 68◦

Since TB = 138 N,

w = (138 N) [cos 68◦ tan 51◦ + sin 68◦] = 192 N

412

Problem 1270. A hanging mass with weightw is suspended by two thin ropes, labelled Aand B. Rope A makes an angle of θA withthe horizontal; rope B makes an angle of θBwith the horizontal. The tension in rope A isTA; the tension in rope B is TB. Which of thefollowing equations represents the condition forstatic equilibrium of the hanging-mass systemin the horizontal direction?

wZ

ZZ

ZZZ

ZZZZ θA

TA

��θB

TB

*(a) TA cos θA = TB cos θB (b) TA sin θA = TB sin θB

(c) TA cos θB = TB cos θA (d) TA sin θB = TB sin θA

(e) None of these

Solution: The hanging mass is not moving, so the net force on it must be zero. Thenet horizontal force is the sum of the horizontal components of the tensions in the tworopes: TA cos θA, pulling to the left; and TB cos θB, pulling to the right. Since the nethorizontal force must be zero, TA cos θA = TB cos θB.

Problem 1271. A hanging mass with weightw is suspended by two thin ropes, labelled Aand B. Rope A makes an angle of θA withthe horizontal; rope B makes an angle of θBwith the horizontal. The tension in rope A isTA; the tension in rope B is TB. Which of thefollowing equations represents the condition forstatic equilibrium of the hanging-mass systemin the vertical direction?

wZ

ZZ

ZZZ

ZZZZ θA

TA

��θB

TB

(a) w = TA cos θA + TB cos θB (b) w =TA

cos θA+

TBcos θB

*(c) w = TA sin θA + TB sin θB (d) w =TA

sin θA+

TBsin θB

(e) None of these

Solution: The hanging mass is not moving, so the net force on it must be zero.The vertical forces acting on it are the weight w, pulling downward; and the verticalcomponents of the tensions in the two ropes, TA sin θA and TB sin θB, both pulling upward.Since the net vertical force must be zero, w = TA sin θA + TB sin θB.

413

Problem 1272. A block with weight w = 130N is suspended by two thin ropes, labelled Aand B. Rope A makes an angle of θA = 37◦

with the horizontal; rope B makes an angle ofθB = 54◦ with the horizontal. The tension inrope A is TA; the tension in rope B is TB. FindTA; round your answer to the nearest newton. w

ZZZ

ZZZ

ZZZZ θA

TA

��θB

TB


Solution: The block is not moving, so the net force on it must be zero. The net forcein the horizontal direction is the sum of the horizontal components of the two tensions:TA cos θA pulling leftward; and TB cos θB pulling rightward. The net force in the verticaldirection is the sum of the weight w, pulling downward; and the vertical components ofthe two tensions, TA sin θA and TB sin θB, pulling upward. Thus we get two equationsthat we can solve for the two variables TA and TB:

TA cos θA = TB cos θB

TA sin θA + TB sin θB = w

We want to solve for TA, so we will begin by solving the first equation for TB, thensubstitute into the second equation.

TA cos θA = TB cos θB÷ cos θB−−−−→ TB =

TA cos θAcos θB

substitute−−−−−→ TA sin θA +TA cos θA sin θB

cos θB= w

× cos θB−−−−→ TA sin θA cos θB + TA cos θA sin θB = w cos θBfactor−−−→ TA(sin θA cos θB + cos θA sin θB) = w cos θB

divide−−−→ TA =w cos θB

sin θA cos θB + cos θA sin θBtrig identity−−−−−−→ TA =

w cos θBsin(θA + θB)

=(130 N) cos 54◦

sin 37◦ cos 54◦ + cos 37◦ sin 54◦

= 76 N

414

Problem 1273. A block with weight w = 130N is suspended by two thin ropes, labelled Aand B. Rope A makes an angle of θA = 37◦

with the horizontal; rope B makes an angle ofθB = 54◦ with the horizontal. The tension inrope A is TA; the tension in rope B is TB. FindTB; round your answer to the nearest newton. w

ZZZ

ZZZ

ZZZZ θA

TA

��θB

TB


Solution: The block is not moving, so the net force on it must be zero. The net forcein the horizontal direction is the sum of the horizontal components of the two tensions:TA cos θA pulling leftward; and TB cos θB pulling rightward. The net force in the verticaldirection is the sum of the weight w, pulling downward; and the vertical components ofthe two tensions, TA sin θA and TB sin θB, pulling upward. Thus we get two equationsthat we can solve for the two variables TA and TB:

TA cos θA = TB cos θB

TA sin θA + TB sin θB = w

We want to solve for TB, so we will begin by solving the first equation for TA, thensubstitute into the second equation.

TA cos θA = TB cos θB÷ cos θA−−−−→ TA =

TB cos θBcos θA

substitute−−−−−→ TB cos θB sin θAcos θA

+ TB sin θB = w

× cos θA−−−−→ TB cos θB sin θA + TB sin θB cos θA = w cos θAfactor−−−→ TB(cos θB sin θA + sin θB cos θA) = w cos θA

divide−−−→ TB =w cos θA

cos θB sin θA + sin θB cos θAtrig identity−−−−−−→ TA =

w cos θAsin(θA + θB)

=(130 N) cos 37◦

cos 54◦ sin 37◦ + sin 54◦ cos 37◦

= 104 N

415

Problem 1274. (Lab Derive Problem) Youwalk into the lab and find a block hanging fromthe ceiling suspended by two threads as shownin the figure to the right. Using a scale, youfind the weight of the block wblock. Using a pro-tractor, you find the angle between the ceilingand the left-hand string to be θL. Similarly, θRis the angle between the ceiling and the right-hand thread. From this information determinethe tensions in the threads: TL and TR.

wblock

ZZZ

ZZZ

ZZZZ θL

TL

��θR

TR

Solution: The block is not moving, so the net force on it must be zero. The net forcein the horizontal direction is the sum of the horizontal components of the two tensions:TL cos θL pulling leftward; and TR cos θR pulling rightward. The net force in the verticaldirection is the sum of the weight of the block wblock, pulling downward; and the verticalcomponents of the two tensions, TL sin θL and TR sin θR, pulling upward. Thus we gettwo equations that we can solve for the two variables TL and TR:

TL cos θL = TR cos θR

TL sin θL + TR sin θR = wblock

The order in which we solve for the tensions is arbitrary, so we’ll solve for TR. We willbegin by solving the first equation for TL, then substitute this result into the secondequation.

TL cos θL = TR cos θR÷ cos θL−−−−−−−−→ TL =

(cos θRcos θL

)TR

substitute−−−−−→[(

cos θRcos θL

)TR

]sin θL + TR sin θR = wblock

× cos θL−−−−→ TR sin θL cos θR + TR sin θR cos θL = wblock cos θLfactor−−−→ TR(sin θL cos θR + sin θR cos θL) = wblock cos θL

trig identity−−−−−−→ TR sin(θL + θR) = wblock cos θL

divide−−−→ TR =

(cos θL

sin(θL + θR)

)wblock

We can now substitute this result back into our equation for TL and TR to find TL. Thereis a faster and more insightful way to solve for TL. By symmetry, we can reverse the rolesof the left and right angles, just call the left angle right and the right angle left (i.e., justswitch the left and right subscripts) to get

TL =

(cos θR

sin(θR + θL)

)wblock =

(cos θR

sin(θL + θR)

)wblock .

416

Problem 1275. Two blocks with weights wAand wB are held in place on a frictionless in-clined plane by two thin ropes, as shown atright. One rope connects block A to a wall atthe top of the plane; a second rope connectsblock B to block A. Both ropes are parallel tothe inclined plane, which makes an angle of αwith the horizontal. What is the tension TABin the rope that connects blocks A and B?

!!!!

!!!!

!!!!

!!!!

α

!!!

LLL!!!

LLL

wA!!

!!TAw

!!!

LLL!!!

LLL

wB!!

!!TAB

!!! x

LLL

y

(a) wB cosα *(b) wB sinα(c) (wA + wB) cosα (d) (wA + wB) sinα(e) None of these

Solution: Begin by drawing a free-bodydiagram for each block. We will take the x-axis as pointing up the slope, and the y-axisas perpendicular to the slope pointing upward:see the axes sketched on the upper figure.There are only two forces acting in the x-direction on block B: the tension TAB in thepositive direction, and the x-component of theweight wBx = wB sinα in the negative direc-tion. Since the system is in static equilib-rium, the net force on block B is zero; soTAB = wB sinα.

u��

wB

�wBx

?wBy

α

6NB

-TAB u��

wA

�wAx

?wAy

α

6NA

-TAw�TAB

417

Problem 1276. Two blocks with weights wAand wB are held in place on a frictionless in-clined plane by two thin ropes, as shown atright. One rope connects block A to a wall atthe top of the plane; a second rope connectsblock B to block A. Both ropes are parallel tothe inclined plane, which makes an angle of αwith the horizontal. What is the tension TAwallin the rope that connects block A to the wall?

!!!!

!!!!

!!!!

!!!!

α

!!!

LLL!!!

LLL

wA!!

!!TAwall

!!!

LLL!!!

LLL

wB!!

!!TAB

!!! x

LLL

y

(a) wA cosα (b) wA sinα(c) (wA + wB) cosα *(d) (wA + wB) sinα(e) None of these

Solution: Begin by drawing a free-bodydiagram for each block. We will take the x-axis as pointing up the slope, and the y-axisas perpendicular to the slope pointing upward:see the axes sketched on the upper figure.There are only two forces acting in the x-direction on block B: the tension TAB in thepositive direction, and the x-component of theweight wBx = wB sinα in the negative direc-tion. Since the system is in static equilib-rium, the net force on block B is zero; soTAB = wB sinα.

u��

wB

�wBx

?wBy

α

6NB

-TAB u��

wA

�wAx

?wAy

α

6NA

-TAwall�TAB

Now we can look at the forces acting on block A. There are three forces in the x-direction:TAB and the x-component of the weight, wAx = wA sinα, both pulling in the negativex-direction; and TAwall in the positive x-direction. (Since the ropes are thin, we assumethat TAB doesn’t change over the length of the rope.) Since the system is in staticequilibrium, the net force on block A must be zero; so

TAwall = wAx + TAB = wA sinα + wB sinα = (wA + wB) sinα

418

Problem 1277. Two blocks with weights wAand wB are held in place on a frictionless in-clined plane by two thin ropes, as shown atright. One rope connects block A to a wall atthe top of the plane; a second rope connectsblock B to block A. Both ropes are parallel tothe inclined plane, which makes an angle of αwith the horizontal. What is the magnitude ofthe force that the plane exerts on block B?

!!!!

!!!!

!!!!

!!!!

α

!!!

LLL!!!

LLL

wA!!

!!

!!!

LLL!!!

LLL

wB!!

!!

!!! x

LLL

y

*(a) wB cosα (b) wB sinα(c) (wA + wB) cosα (d) (wA + wB) sinα(e) None of these

Solution: Begin by drawing a free-bodydiagram for each block. We will take the x-axis as pointing up the slope, and the y-axisas perpendicular to the slope pointing upward:see the axes sketched on the upper figure.There are only two forces acting in the y-direction on block B: the normal force NB inthe positive y-direction, and the y-componentof the block’s weight, wBy = wB cosα. Sincethe block is in static equilibrium, the net forceon it is zero; so NB = wB cosα.

u��

wB

�wBx

?wBy

α

6NB

-TAB u��

wA

�wAx

?wAy

α

6NA

-TAw�TAB

419

Problem 1278. Two blocks with weights wAand wB are held in place on a frictionless in-clined plane by two thin ropes, as shown atright. One rope connects block A to a wall atthe top of the plane; a second rope connectsblock B to block A. Both ropes are parallel tothe inclined plane, which makes an angle of αwith the horizontal. What is magnitude of theforce that the plane exerts on block A?

!!!!

!!!!

!!!!

!!!!

α

!!!

LLL!!!

LLL

wA!!

!!

!!!

LLL!!!

LLL

wB!!

!!

!!! x

LLL

y

*(a) wA cosα (b) wA sinα(c) (wA + wB) cosα (d) (wA + wB) sinα(e) None of these

Solution: Begin by drawing a free-bodydiagram for each block. We will take the x-axis as pointing up the slope, and the y-axisas perpendicular to the slope pointing upward:see the axes sketched on the upper figure.There are only two forces acting in the y-direction on block A: the normal force NA inthe positive y-direction, and the y-componentof the block’s weight, wAy = wA cosα. Sincethe block is in static equilibrium, the net forceon it is zero; so NA = wA cosα.

u��

wB

�wBx

?wBy

α

6NB

-TAB u��

wA

�wAx

?wAy

α

6NA

-TAw�TAB

Problem 1279. Two forces with magnitudes F1 and F2 are acting on an object. Whichof the following inequalities always hold for the magnitude of the net force Fnet on theobject?

(a) F1 ≤ Fnet ≤ F2 (b) (F1 − F2)/2 ≤ Fnet ≤ (F1 + F2)/2

*(c) |F1 − F2| ≤ Fnet ≤ F1 + F2 (d) F 21 − F 2

2 ≤ F 2net ≤ F 2

1 + F 22

(e) None of these

Problem 1280. An object with a mass of 20kg is suspended from two ropes, as shown atright. The magnitude of the force exerted byeach rope on the object is 139 N; the magnitudeof the force of gravity on the object is 196 N.What is the magnitude of the net force on theobject?

(a) 47.4 N (b) 33.5 N(c) 13.9 N *(d) 0 N(e) None of these

\\\\\\\\��

20 kg

Solution: The system is static, so the net force is zero.

420

11.2.2 Block and pulley frictionless systems

Key Idea: Frictionless block-and-pulley systems are just one-dimensional systems indisguise! The system sits in two dimensions, but can be converted to a one-dimensionalsystem. The background space is known as the ambient space and the system can beshown to be one dimensional by mathematical arguments from a field known as Topology.We will just take these topological arguments as fact.

Problem 1281. A rope, one end of which is attached to a ceiling, passes through apulley and then goes straight upward where it is held in place by a man that is standingon the ground. A weight of 400 N is attached to the pulley. Assuming that the weightof the pulley can be neglected in comparison to the 400 N weight and that there is nofriction between the pulley and the rope, what upward force must the man exert to holdthe weight in place?


Solution: Let W = 400 N be the downward force of the weight, and let T be thetension of the rope. If you sketch a free-body diagram, it will show a downward force ofW on the pulley, an upward force of T exerted on the pulley by the rope to the ceiling,and another upward force of T exerted on the pulley by the man pulling upward on theother end of the rope. Since the pulley isn’t moving, the net force must be zero: so2T = W ; so T = W/2 = 200 N.

Problem 1282. Two crates are connected by athin rope and sitting on a frictionless horizontalfloor. You are pulling them by a second rope at-tached to the smaller crate. You pull on yourrope with a horizontal force of P ; the tension inthe rope connecting the crates is T . Which of thefollowing is true of P and T?

� P 37 kg T 89 kg

(a) P < T (b) P = T*(c) P > T (d) Not enough information; it could be any of these

Solution: P is giving a certain acceleration to the combined mass of both crates. Tis giving the same acceleration to the 89-kg crate alone. Thus T is smaller than P ; infact,

T =89 kg

37 kg + 89 kg· P

421

Problem 1283. In your physics lab, you find the ap-paratus shown at right. A block with a mass of m1 isresting on a horizontal table. A light string runs hor-izontally from the block, over a massless frictionlesspulley, and down to a hanging bucket. The coefficientof static friction between the block and the table isµs, and the coefficient of kinetic friction between theblock and the table is µk. Initially, the system doesnot move. You slowly add water to the bucket untilthe system starts moving. At this point, what is theweight w of the bucket?

w

m1

given: µsgiven: µk

(a) m1 (b) m1g(c) µsm1g (d) µkm1g

Solution: The system is initially at rest, so you must add water to the bucket until itsweight is sufficient to overcome the static friction of the block on the table. This occurswhen W = µsN = µsMg, where N is the normal force on the block from the table.

Problem 1284. (Lab Problem) In your physicslab, you find the apparatus shown at right. A blockwith a weight of 60 N is resting on a horizontal table.A light string runs horizontally from the block, overa massless frictionless pulley, and down to a hangingbucket. The coefficient of static friction between theblock and the table is µs = 0.6, and the coefficientof kinetic friction between the block and the table isµk = 0.2. Initially, the system does not move. Youslowly add water to the bucket until the system startsmoving. At this point, what is the weight w of thebucket?

w

60 N

µs = 0.6µk = 0.2

(a) 12 N *(b) 36 N(c) 48 N (d) 60 N

Solution: The system is initially at rest, so you must add water to the bucket until itsweight is sufficient to overcome the static friction of the block on the table. This occurswhen W = µs · 60 N = (0.6)(60 N) = 36 N.

422

Problem 1285. In your physics lab, you find theapparatus shown at right. A glider with a weight ofm1 is resting on a horizontal air-track with the airoff. A light string runs horizontally from the glider,over a massless frictionless pulley, and down to ahanging block with a weight of m2. The coefficientof static friction between the glider and the air-trackis µs, and the coefficient of kinetic friction betweenthe glider and the air-track is µk. The coefficients offriction are assumed known. After the glider-string-block system have been released to move freely, andbefore the hanging block has hit the floor, what isthe tension T in the string?

m2

m1

µs > µk

(a)

(m1

m1 +m2

)(1 + µk)g *(b)

(m1m2

m1 +m2

)(1 + µk)g

(c)

(m1m2

m1 +m2

)(1− µk)g (d)

(m1 +m2

m1m2

)(1 + µk)g

Solution: First, you should determine a condition between m1 and m2 that assuresus that the system moves when the glider-block system is released. The maximal staticfriction of the glider on the air-track is fk = µsN1 = µsm1g. That is, we need m2g >µsm1g ⇒ m2 > µsm1. This is our condition for motion.

Assuming the system will move, we can determine the forces on it. We will first analyzethe forces on the two separate subsystems: mass m1 and m2, and then we will connectthem by examining the combined system: glider, massless string, and hanging-mass. Wewill connect the forces on the systems by observing that because the string is taunt andassumed not to stretch, that the velocity and the acceleration of the two masses must bethe same.

423

The glider m1: If we take the x1-axis for m1 to point in thedirection of motion and the y1-axis to point upward, then applyingNewton’s second law we arrive at the system of equations:

Fnet1,x = T1 − fk = m1a1,x ,

Fnet1,y = N1 −m1g = m1a1,y = 0 ,

where we’ve used the fact that the glider does not move in thevertical direct, so ay,1 = 0 (see free-body diagram).

u�T1

?m1g

6N1

-fk

free-body diagramfor mass m1

From the second equation we have

N1 = m1g ⇒ fk = µkN1 = µkm1g .

Upon substituting this equation into the first equation we have

T1 = fk +m1a1,x = µkm1g +m1a1,x .

The hanging mass m2: Since the motion of m2 is pure vertical, we only need a y2-axis to describe the motion. We take the y2-axis pointing downward in the direction ofmotion. Applying Newton’s second law to mass m2, we arrive at the system of equations:

Fnet1,y = m2g − T2 = m2a2,y ⇒ T2 = m2g −m2a2,y .

By hypothesis, the pulley is massless and frictionless, so it only serves to redirect theforces. Thus, T1 = T2 and we have

T1 = T = µkm1g +m1a1,x ,

T2 = T = m2g −m2a2,y .

Equating these equations through T and using the fact that since the glider and hangingmass are connected by a string, they move as one and therefore have the same magnitudeof acceleration (i.e., a1,x = a2,y = a) gives the following equation for a

µkm1g +m1a = T = m2g −m2are-arraging terms−−−−−−−−−→ (m1 +m2)a = (m2 − µkm1)g

÷ (m1+m2)−−−−−−→ a =

(m2 − µkm1

m1 +m2

)g

424

We can now use a to determine T .

T = m2g −m2a

= m2g −m2

(m2 − µkm1

m1 +m2

)g

=

((m1m2 +m2

2)− (m22 − µkm1m2)

m1 +m2

)g

=

(m1m2

m1 +m2

)(1 + µk)g

As a check on our formulas that we found for acceleration and tension, let’s ask whathappens if the mass of the glider goes to zero (i.e., m1 → 0+)? Then a = g and T = 0as would be expected since the hanging mass would then be in free fall. Notice that wecannot let m2 go to zero because of the constraint m2 > µsm1 > 0.

425

Problem 1286. (Lab Problem) This problem has a fundamental mistake init, can you find it?A glider is sitting at rest on a horizontal frictionless air track. A light string is attachedto a light string that runs over a massless frictionless pulley and attaches to a hangingmass of 1 kg. The system is held in place and released from rest. Using photo gates thatare placed 1 m apart the time is measured to be 2 s. Assuming that the string remainshorizontal and taunt throughout the entire run, what is the mass of the glider? For easeof computation take g to be 10 m/s2. Round your answer to the nearest kilogram.

(a) 1 kg (b) 2 kg(c) 4 kg (d) 19 kg(e) None of these

Solution: The massless, frictionless pulley justifies the assumption that the tensionis the same throughout the string, from which it follows that there is a constanthorizontal force on the glider of mg = 10 N from the hanging weight. This force hasbeen transmitted through the string. Recall that a pulley is a machine that redirectsforce. The light string means that we can ignore the mass of the string.

The mistake is highlighted in italics: as the hanging weight falls, it acceler-ates, so it’s apparent weight increases since it’s in an accelerating frame ofreference. This leads to an increase in tension. If you sketch a free-body diagramof the glider, you’ll see that the only horizontal force is the constant 1 N. There’s no netvertical force.

Given: Fnet = 1 N; vi = 0 m/s; ∆x = 1 m; and t = 2s;

Want: m = ?

Note: We don’t know, nor do we want vf .

If we try to solve for m directly using Newton’s second law, we get stuck:

Fnet = ma÷a−−−−−→ m =

Fnet

a=

1 N

a(11.4)

We need to use one of our three fundamental kinetic equations to find a. Since we know∆x, vi = 0, and t, we can use

∆x = vit+1

2at2 =

at2

2

· 2/t2−−−−→ a =2∆x

t2(11.5)

Substituting equation (11.5) into equation (11.4) gives us

m =Fnett

2

2∆x=

(1 N)(2 s)2

2(1 m)= 2 kg

426

Problem 1287. In your physics lab, you find theapparatus shown at right. A tank of water is rest-ing on a horizontal frictionless table. A light stringruns horizontally from the tank, over a massless fric-tionless pulley, and down to a hanging block with amass of m1. You add increasing amounts of water tothe tank, each time releasing the system and observ-ing its motion or lack thereof. How does the systembehave as you increase the mass m2 of the tank? m1

m2

(a) The acceleration does not change, since there is no friction and m1 remainsconstant.

(b) The system accelerates as long as m2 < m1; when m2 = m1, the system nolonger accelerates.

(c) The system moves as long as m2 < m1; when m2 = m1, the system no longermoves.

*(d) The system always accelerates; but the acceleration gets smaller as m2 becomeslarger.

Solution: Since there is no friction, the system always accelerates, no matter howlarge m2 is. However, since the force remains the same (F = m1 · g), and since the massof the system is m1 +m2, increasing m2 increases the mass of the system and decreasesa. In fact,

a =m1 · gm1 +m2

427

Problem 1288. (Derive Lab Problem) In yourphysics lab, you find the apparatus shown at right.A block with mass m1 is resting on a horizontalair-track with the air turned off. A light stringruns horizontally from the block, over a masslessfrictionless pulley, and down to a hanging bucket.The coefficient of static friction between the blockand the track is µs, and the coefficient of kineticfriction between the block and the table is µk < µs.Initially, the system does not move. You slowly addwater to the bucket until the system just starts tomove. Call this moment in time t = t0 = 0; at thistime, call the mass of the bucket of water m2.

m2

m1

µk < µs

Note: Not drawn to scale.

Setup: For the block, take the x1-axis in the direction of motion along the table andthe y1-axis going upward. Take the y2-axis pointing downward, in the direction of mo-tion. Let the origin along the x1-axis be the initial position of the block with mass m1

(x1,i = 0). The initial velocity of mass m1 is zero (at t = 0, v1,i = 0).

(a) Find the value of m2: that is, find m2 such that the force of gravity on the hangingbucket is just enough to overcome the force of static friction on the block with mass m1.

(b) Find a formula for the initial acceleration of the system, denoted asys.

(c) Assuming that the initial position and velocity of the block with mass m1 is zero,find an equation for its position as a function of time: x1 = x1(t). Ignore the fact thatthe block will eventually crash into the pulley.

Solution: Work it out!

428

Problem 1289. (Derive Problem) In your physics lab, you find the apparatus shownbelow. A glider with mass m1 is resting on a horizontal air track. A light string runs hori-zontally from the glider, over a massless frictionless pulley, and down to a hanging bucket.

The air track is specially designed so that the coefficient of static friction between theglider and the air track is µs > 0 (a constant), and the coefficient of kinetic frictionbetween the block and the table is µk = µse

−x (a function of position) provided thatthe front of the glider is initially at rest at a location marked as the origin on the airtrack. The coefficient of static friction between the block and the table is µs (a constant),and the coefficient of kinetic friction between the block and the table is µk = µse

−x (afunction of position). Initially, the system does not move. You slowly add water to thebucket until the system just starts to move. Call this moment the initial time t0 = 0;at this time, call the mass of the bucket of water m2. Define the initial position of thecenter of mass for block m1 to be the origin and take the initial velocity to be zero, sincethe block starts from rest.

m2

m1

µk = µse−x

The Coordinate System: For the glider, take the x1-axis in the direction of motionalong the table and the y1-axis going upward. Take the y2-axis pointing downward, inthe direction of motion. Let the origin along the x1-axis be the initial position of theglider with mass m1 (x1,i = 0). Since the direction of motion is one dimensional, you cansimplify things by letting x = x1 = y2.

The Setup and Initial Conditions: Take the initial position of the glider to be at theorigin defined above (i.e., x1,i = 0) and the initial velocity of the two-mass-plus-stringsystem to be zero at time t0 = 0. Take the point of zero potential to be y2 = initialheight of the hanging mass m2. This is equivalent to taking x = 0 as the point of zeropotential.

The Problem: Find an equation for the mechanical energy of the system as a functionof time and position:

K.E.sys(x, t) + P.E.sys(x, t) = D(x, t) ,

429

where D(x, t) is a dissipation term. Ignore the fact that the glider will eventually crashinto the pulley, or the hanging mass will hit the ground. Your model is only designed todescribe the motion in the interim time.

Comments, Advice, and Hints:

Comment (1): The first step is to determine if the system will even move! To do thisyou will need to find the force exerted by the hanging weight m2 that over comes theforce of static friction. That is, find m2 such that the force of gravity on the hangingbucket plus water is just enough to overcome the force of static friction fs,max on theblock with mass m1.

Comment (2): Notice that the coefficient of kinetic friction µk is a function of position!We have never solved such a problem in class! However, we have solved a similar equationin the new lab 6.

Comment (3): Be sure to clearly label all of the intermediate steps in comments (1)-(3).If I have to go searching for them, I will take points off.

Advice: To find the kinetic and potential energy, and any possible dissipation term forthe system, as a function of position and time, you will need to first find a formula for theinitial acceleration of the system asys as a function of time t and position x and integratethe system using the initial conditions. When you get your result, ask yourself what isthe source of the kinetic energy of the system, and what is the source of the potentialenergy of the system. Also, ask yourself: is the system mechanically conservative? If not,then what are the sources of energy loss? Can you identify these terms in the equation?

Hint: To perform one integration on a differential equation of the form:

d2x

dt2+dF (x(t))

dx= constant ,

The standard trick is to multiple the equation by the velocity term v = x and integratethe equation using the chain rule together with the Fundamental Theorem of Calculus(FTC). With this hint, you can solve the resulting equation with the techniques that youlearn in Calculus 1 (a prerequisite for the course).

Solution: Work it out!

430

Problem 1290. Two identical blocks, eachwith mass m, are attached to opposite ends of athin rope that passes over a massless frictionlesspulley. The pulley is suspended from the ceilingby a thin chain. Draw a free-body diagram foreach block, and one for the block-pulley system.

Solution: The diagrams are drawn to theright of the illustration. The diagram on theleft is for an individual block; the one on theright is for the block-pulley system.

m

m

r u6Trope

?

wb = mg

u6

Tchain

?

wsys = 2mg

Problem 1291. Two identical blocks, eachwith mass m, are attached to opposite ends ofa thin rope that passes over a massless friction-less pulley. The pulley is suspended from theceiling by a thin chain. What is the tension inthe chain holding up the pulley-block system?

(a) 0 (b) mg/2(c) mg *(d) 2mg

Solution: A free-body diagram for thepulley-blocks system is sketched at right. Thereare two forces acting on the system: the weightw = 2mg pulling down; and the tension Tchain

pulling up. Since the system is in static equi-librium, Fnet = 0; so Tchain = 2mg.

m

m

r u6

Tchain

?

w = 2mg

Problem 1292. Two identical blocks, eachwith mass m, are attached to opposite ends ofa thin rope that passes over a massless friction-less pulley. The pulley is suspended from theceiling by a thin chain. What is the tension inthe rope?

(a) 0 (b) mg/2*(c) mg (d) 2mg

Solution: A free-body diagram for oneof the blocks is sketched at right. There areonly two forces acting on the block: the weightw = mg pulling downward, and the tensionTrope pulling upward. Since the system is instatic equilibrium, Fnet = 0; so Trope = mg.

m

m

r u6

Trope

?

w = mg

431

Problem 1293. In your physics lab, you find theapparatus at right. A block with mass M1 is on africtionless slope. A thin string from the block runsparallel to the slope, up to and over a massless fric-tionless pulley, then down to a second block withmass M2. Initially, the system is in static equilib-rium. If it is shaken slightly, what will the systemdo?

M2

��

��

@@M1

@@@@

@@

@@@

@@

@@@@u

(a) The first block will accelerate up the slope; the second will accelerate downward.(b) The first block will accelerate down the slope; the second will accelerate upward.*(c) The blocks will not accelerate.

Solution: Since the system is in static equilibrium and there is no friction, thedownward force of gravity on M2 equals the component of the gravitational force on M1

parallel to the slope. Vibrating the system doesn’t change this; the net force on thesystem will still be zero afterward, so it won’t accelerate in either direction.

Problem 1294. A block with mass m1 is on a fric-tionless ramp that makes an angle of θ to the hori-zontal. A thin string runs from the block, parallel tothe ramp, up and over a massless frictionless pulley,and down to a hanging block with mass m2. Thesystem is held in place, then released. If the blocksdo not move after the release, which of the followingmust be true?*(a) m1 sin θ = m2 (b) m2 sin θ = m1

(c) m1 cos θ = m2 (d) m2 cos θ = m1

(e) None of these

Solution: Sketch a free-body diagram for eachof the blocks, as at right. Since the rope is thinand the pulley is massless and frictionless, weassume that the tension in the rope is constantover its entire length. In the diagram for thesliding block, we take the x-direction as parallel tothe slope, and the y-direction as perpendicular to it.

��

��

��

��

θ

m2��

AAA��

AAA

m1

��

��

u

um2

?w2 = m2g

6T

um1

θ

?w1 = m1g

��* T

AAAAAK

N

AAAAAUw1y

��w1x

If the system does not move, then the net force on each block must be zero. For thehanging block, this means that T = w2 = m2g. For the sliding block, the net force inthe x-direction must be zero; so T = w1x = m1g sin θ. Combining these, we get

T = m1g sin θ = m2g ⇒ m1 sin θ = m2

432

Problem 1295. A block with mass m1 is on africtionless ramp that makes an angle of θ to thehorizontal. A thin string runs from the block,parallel to the ramp, up and over a masslessfrictionless pulley, and down to a hanging blockwith mass m2. The system is held in place,then released. What is the acceleration of thehanging block? Use downward as the positivedirection.

��

��

��

��

θ

m2��

AAA��

AAA

m1

��

��

u

(a)(m2 −m1 cos θ)g

m1 +m2

*(b)(m2 −m1 sin θ)g

m1 +m2

(c)(m2 −m1 cos θ)g

m2

(d)(m2 −m1 sin θ)g

m2

(e) None of these

Solution: Sketch a free-body diagram for eachof the blocks, as at right. Since the rope is thin andthe pulley is massless and frictionless, we assumethat the tension in the rope is constant over itsentire length. In the diagram for the sliding block,we take the x-direction as parallel to the slope, andthe y-direction as perpendicular to it.We know the masses m1 and m2, the angle θ, andthe gravitational acceleration g. We don’t knowthe tension in the rope T . We want to know theacceleration of the hanging block; and since the twoblocks are connected by a taut rope, the accelera-tion of the sliding block must have equal magnitude.

um2

?w2 = m2g

6T

um1

θ

?w1 = m1g

��*T

AAAAAK

N

AAAAAUw1y

��w1x

The net force acting on the hanging block in the positive (downward) direction is F2 =m2g − T . The net force acting on the sliding block in the same direction (up the slope)is F1 = T −m1g sin θ. The net force acting on the whole system is

Fnet = F1 + F2 = T −m1g sin θ +m2g − T = m2g −m1g sin θ

The mass of the system is m1 + m2; so by Newton’s second law, the acceleration of thesystem is

a =Fnet

msystem

=(m2 −m1 sin θ)g

m1 +m2

433

Problem 1296. A block with a mass of 50 kg is ona frictionless ramp that makes an angle of 30◦ to thehorizontal. A thin string runs parallel to the rampfrom the block to a massless frictionless pulley, thendown to a second block with a mass of 30 kg. Thesystem is initially held in place, then released. Whatis the acceleration of the hanging (30 kg) block?Round your answer to the nearest 0.1 m/s2. [Thepicture is not drawn to scale.]

(a) 0.6 m/s2 upward *(b) 0.6 m/s2 downward(c) 1.6 m/s2 upward (d) 1.6 m/s2 downward

30 kg

��

��

@@50 kg

@@@@

@@

@@@

@@@

@@@u

Solution: Since the pulley is massless and frictionless and since the string is massless,the tension is the same throughout the length of the string: T .

Sketch a free-body diagram of each block. The hanging block is easy: the weight w1 =m1g pulling straight down, and the tension T pulling straight up. (We’ll use the subscripts1 for the hanging block and 2 for the sliding block.)

The sliding block is more complicated. There arethree forces: the weight w2 = m2g pulling straightdown; the tension T pulling upward parallel to theslope; and the normal force N pushing upward per-pendicular to the slope. We decompose w2 into twocomponents: w2‖ = w2 sin 30◦, parallel to the slope;and w2⊥ = w2 cos 30◦, perpendicular to the slope. Afree-body sketch is shown at right.

r30◦

HHHH

HYT

?w2

HHHj w2‖�

��w2⊥

�� N

We can ignore forces on the sliding block perpendicular to the slope, since there’s nofriction. We will take forces as positive if they tend to move the system so that thehanging block moves downward and the sliding block moves upward. The net force onthe hanging block is F1 = w1 − T = m1g − T ; the net force parallel to the slope on thesliding block is F2 = T −w2‖ = T −m2g sin 30◦. Then the net force on the whole systemis

Fnet, system = F1 + F2 = m1g − T + T −m2g sin 30◦ = m1g −m2g sin 30◦

By Newton’s second law, the acceleration of the system is the net force on the systemdivided by the total mass:

a =Fnet, system

msystem

=m1g −m2g sin 30◦

m1 +m2

=g(m1 −m2 sin 30◦)

m1 +m2

=(9.8 m/s2)[30 kg− (50 kg) sin 30◦]

30 kg + 50 kg= 0.6 m/s2

Since the sign is positive, the hanging block accelerates downward.

434

11.2.3 Accelerometers

Problem 1297. (Accelerometer) A thinstring is attached to the ceiling of a car, and alead sinker with mass m is attached to the freeend of the string. The car is going at highwayspeed when the driver hits the brakes. Whilethe car is slowing down, the string makesan angle of θ to the vertical. What is themagnitude of the car’s acceleration? Take thedirection of motion to be the positive x-axisand the y-axis to point upward.

*(a) −g tan θ (b) −mg tan θ

(c) − g

tan θ(d)

mg

tan θ

(e) None of these

LLLLLLu

��θ

Solution: Sketch a free-body diagram of the object on thestring. There are two forces acting on it: gravity, pulling straightdownward with a force of w = mg, where m is the mass of theobject; and the tension T in the string, pulling upward and to theleft at an angle of θ to the vertical. The latter can be decomposedinto its x- and y-components: Tx = T sin θ and Ty = T cos θ.

Since there is no vertical acceleration, the net force in the y-direction must be zero: thus

0 = may = Fnet,y = Ty − w ⇒ T cos θ = mg

The only force in the x-direction is Tx; so it supplies all thehorizontal acceleration. Hence

max = Fnet,x = −Tx ⇒ T sin θ = −max

uAAAAAAK

T

θ

�Tx

6Ty

?w

From these, we can calculate θ:

tan θ =TxTy

=−maxmg

=−axg

⇒ ax = −g tan θ

Note: If we take the x-axis pointing in the initial direction of motion, then ax < 0.

435

Problem 1298. (Accelerometer) A thinstring is attached to the ceiling of a car, anda weight is attached to the free end of thestring. The car is going at 43 m/s when thedriver hits the brakes. The car’s accelerationhas a magnitude of 2.3 m/s2. While the caris slowing down, what angle θ does the stringmake to the vertical? Take the direction ofmotion to be the positive x-axis and the y-axisto point upward. Round your answer to thenearest degree.

(a) 11◦ (b) 12◦

*(c) 13◦ (d) 15◦

LLLLLLu

��θ

Solution: Sketch a free-body diagram of the object on thestring. There are two forces acting on it: gravity, pulling straightdownward with a force of w = mg, where m is the mass of theobject; and the tension T in the string, pulling upward and to theleft at an angle of θ to the vertical. The latter can be decomposedinto its x- and y-components: Tx = T sin θ and Ty = T cos θ.

Since there is no vertical acceleration, the net force in the y-direction must be zero: thus

0 = may = Fnet,y = Ty − w ⇒ T cos θ = mg

The only force in the x-direction is Tx; so it supplies all thehorizontal acceleration. Hence

max = Fnet,x = −Tx ⇒ T sin θ = −max

uAAAAAAK

T

θ

�Tx

6Ty

?w

From these, we can calculate θ:

tan θ =TxTy

=−maxmg

=−axg

⇒ θ = tan−1

(−axg

)= tan−1

(2.3 m/s2

9.8 m/s2

)= 13◦

Note: If we take the x-axis pointing in the initial direction of motion, then

ax < 0 ⇒ −ax > 0 ⇒ θ > 0 .

436

Problem 1299. (Accelerometer) A thinstring is attached to the ceiling of a car, and alead sinker with mass m is attached to the freeend of the string. The car is going at highwayspeed when the driver hits the brakes, produc-ing an acceleration with magnitude ax. Whilethe car is slowing down, the string makes anangle of θ to the vertical. What is θ? Take thedirection of motion to be the positive x-axis andthe y-axis to point upward.

(a) θ = tan

(g

−ax

)(b) θ = tan−1

(g

−ax

)(c) θ = tan

(−axg

)*(d) θ = tan−1

(−axg

)(e) None of these

Solution: Sketch a free-body diagram forthe sinker, as at right. There are two forcesacting on it: the tension in the string T , pullingupward and to the left; and the weight w = mg,pulling straight down. While the car is slowingdown, the sinker has a horizontal accelerationof ax and no vertical acceleration.

LLLLLLu m

��θ

uAAAAAAK

T

θ

6Ty = T cos θ

�Tx = T sin θ

?w = mg

Since there is no vertical acceleration, the two vertical forces must be equal and opposite:Ty = T cos θ = mg. The only horizontal force is Tx = T sin θ; so max = Fnet,x = −T sin θ.We can eliminate the unknowns T and m by taking the ratio of these two equations:

T cos θ = mg and T sin θ = −max

⇒ T sin θ

T cos θ=

sin θ

cos θ= tan θ =

−maxmg

=−axg

⇒ θ = tan−1

(−axg

)= − tan−1

(axg

)In the last step we used the fact that the arctan is an odd function to factor out thenegative sign.

437

11.2.4 Static and kinetic friction problems

Problem 1300. (Similarity Problem) You are pushing a crate of physics books acrossa concrete floor, with friction present. If you push with a horizontal force of F1, the crateexperiences an acceleration a1. If you increase the horizontal force to F2 = 2F1, theacceleration is a2. What is the relationship between a2 and a1?

(a) a2 = a1 (b) a1 < a2 < 2a1

(c) a2 = 2a1 *(d) a2 > 2a1

Solution: If you push the crate hard enough to move it, then the net horizontal forceon it is your pushing force minus the force of kinetic friction. In the first case, that’sF1 − fk. If you push with twice the horizontal force, it doesn’t affect the friction; so inthe second case, the net horizontal force on the crate is 2F1 − fk > 2(F1 − fk). Sincethe mass of the crate doesn’t change from the first case to the second, the accelerationis proportional to the net horizontal force; so a2 > 2a1.

Problem 1301. Two blocks, one with mass m1

and one with mass m2, are connected by a thin hor-izontal rope. A second horizontal rope is attachedto the block with mass m1 and pulled to the leftwith a force whose magnitude is P . The coefficientof kinetic friction between each block and the flooris µk. If the blocks move at a constant velocity,what is the tension T in the rope connecting thetwo blocks?

(a) m1gµk (b) (m2 +m1)gµk*(c) m2gµk (d) (m2 −m1)µk(e) None of these

Solution: Sketch a free-body diagram for eachblock, as at right. Let T be the tension in the ropeconnecting the two boxes.

� P m1T m2

u -

f2k

�

T

?w2 = m2g

6

N2

u -

f1k

-

T�

P

?w1 = m1g

6

N1

Since the floor is horizontal, N1 = m1g and N2 = m2g. Hence f1k = µkm1g andf2k = µkm2g. Since the blocks are moving at constant velocity, the net force on each iszero. For block 2, this means that

T = f2k = m2gµk

438

Problem 1302. Two blocks, one with mass m1

and one with mass m2, are connected by a thin hor-izontal rope. A second horizontal rope is attachedto the block with mass m1 and pulled to the leftwith a force whose magnitude is P . The coefficientof kinetic friction between each block and the flooris µk. If the blocks move at a constant velocity,what is P?

(a) 0 *(b) (m1 +m2)gµk(c) m1µk (d) (m2 −m1)µk(e) None of these


� P m1T m2

u -

f2k

�

T

?w2 = m2g

6

N2

u -

f1k

-

T�

P

?w1 = m1g

6

N1

Since the floor is horizontal, N1 = m1g and N2 = m2g. Hence f1k = µkm1g andf2k = µkm2g. Since the blocks are moving at constant velocity, the net force on each iszero. For block 2, this means that T = f2k = µkm2g. For block 1, it means that

P = f1k + T = µkm1g + µkm2g = (m1 +m2)gµk

Problem 1303. Two blocks, one with mass m1 =3 kg and one with mass m2 = 7 kg, are connectedby a thin horizontal rope. A second horizontal ropeis attached to the block with mass m1 and pulled tothe left with a force whose magnitude is P . The co-efficient of kinetic friction between each block andthe floor is µk = 0.4. If the blocks move at a con-stant velocity, what is P? For ease of calculation,assume that g = 10 m/s2.



� P m1T m2

u -

f2k

�

T

?w2 = m2g

6

N2

u -

f1k

-

T�

P

?w1 = m1g

6

N1

Since the floor is horizontal, N1 = m1g and N2 = m2g. Hence f1k = µkm1g andf2k = µkm2g. Since the blocks are moving at constant velocity, the net force on each iszero. For block 2, this means that T = f2k = µkm2g. For block 1, it means that

P = f1k + T = µkm1g + µkm2g = (m1 +m2)gµk = (3 kg + 7 kg)(10 m/s2)(0.4) = 40 N

439

Problem 1304. Two blocks, one with mass m1 =3 kg and one with mass m2 = 5 kg, are connectedby a thin horizontal rope. A second horizontal ropeis attached to the block with mass m1 and pulled tothe left with a force whose magnitude is P . The co-efficient of kinetic friction between each block andthe floor is µk = 0.2. If the blocks move at a con-stant velocity, what is the tension T in the ropeconnecting the two blocks? For ease of calculation,assume that g = 10 m/s2.



� P m1T m2

u -

f2k

�

T

?w2 = m2g

6

N2

u -

f1k

-

T�

P

?w1 = m1g

6

N1

Since the floor is horizontal, N1 = m1g and N2 = m2g. Hence f1k = µkm1g andf2k = µkm2g. Since the blocks are moving at constant velocity, the net force on each iszero. For block 2, this means that

T = f2k = m2gµk = (5 kg)(10 m/s2)(0.2) = 10 N

440

Problem 1305. You are towing a block on a horizontal surface, using a rope that makesan angle of θ to the horizontal. The block has a mass of m; the coefficient of kineticfriction between the block and the floor is µk. As you pull the block, you are giving theblock a forward acceleration of a. What is the tension T in the rope, expressed in termsof m, a, θ, µk, and the gravitation g?

(a) (mgµk + a) cos θ *(b)m(gµk + a)

cos θ + µk sin θ

(c)m(gµk + a)

cos θ(d) mgµk + a

(e) None of these

m ��

�*T

θ

Solution: The first step is draw a picture and a free-body diagram. The forces onthe block are the normal force, the weight, the kinetic frictional force, and the tensionthat must be broken into its x and y components, where the x-axis points to the rightand the y-axis pointing upward. See the diagram. The horizontal force from the rope isTx = T cos θ, directed forward and the vertical force is Ty = T sin θ. The frictional forceis given by fk = µkN . Notice that because of the vertical component of the tension thenormal force is not equal to the weight. Applying Newton’s second law in each of thecomponent directions: {

T cos θ − µkN = Fnet,x = max = ma

T sin θ +N −mg = Fnet,x = may = 0

This is a system of two equations and two unknowns: N and T . Solving the secondequation for N gives N = mg − T sin θ, and substituting this equation into the firstequation yields

T cos θ − µk(mg − T sin θ) = maIsolate T−−−−−→ T (cos θ + µk sin θ) = m(gµk + a)

Solve for T−−−−−−→ T =m(gµk + a)

cos θ + µk sin θ

Note 1: A common mistake is to take N = mg, giving an incorrect force of frictionfk = −µkmg, which would only be the case if there was no y component of tension.

Note 2: If the force is horizontal, then θ = 0, and the problem is straight forward.

441

Problem 1306. A large stone block has been loadedin the bed of a dump truck. The bed is slowly raiseduntil it makes an angle of θs with the horizontal; at thispoint, the block starts to slide downward. What is thecoefficient of static friction µs between the block andthe truck bed?(a) sin θs *(b) tan θs(c) cos θs (d) θs(e) None of these

Solution: Sketch a picture and then a free-bodydiagram, as at right. For the free-body diagram, wechoose the x-axis parallel to the truck bed and the y-axisperpendicular to it.

Since the block does not sink into the truck bed or riseabove it, the net force in the y-direction must be zero.Hence N = wy = w cos θs. At the point where the blockbegins to slide, the downslope force in the x-direction isequal to the force of static friction. Hence

��

��

��

��

��

θs

AAA A

AA��r?w

AAAAAK

N

��*fs

��*x

AAK

y

u��w

θs?wy = w cos θs

6N

�

wx =w sin θs

-fs = µsN

w sin θs = fs = µsN = µsw cos θs ⇒ µs =w sin θsw cos θs

=sin θscos θs

= tan θs

Problem 1307. A brick is sitting on a roof, whichmakes an angle of θk with the horizontal. If the brickis given a slight nudge, it slides down the rooftop at aconstant speed. What is the coefficient of kinetic frictionµk between the brick and the rooftop?

(a) sin θk *(b) tan θk(c) cos θk (d) θk(e) None of these

Solution: Sketch a picture and then a free-bodydiagram, as at right. For the free-body diagram, wechoose the x-axis parallel to the roof top and the y-axisperpendicular to it.

Since the brick does not sink into the roof or rise aboveit, the net force in the y-direction must be zero. HenceN = wy = w cos θk. When the brick slides down theroof without accelerating, the downslope force in the x-direction exactly matches the force of kinetic friction.Hence

��

��

��

��

��

θk

AAA A

AA��r?w

AAAAAK

N

��*

fk

��*x

AAK

y

u��w

θk?wy = w cos θk

6N

�

wx =w sin θk

-fk = µkN

w sin θk = fk = µkN = µkw cos θk ⇒ µk =w sin θkw cos θk

=sin θkcos θk

= tan θk

442

Problem 1308. A crate whose weight is w isbeing pulled up a ramp that makes an angle ofθ with the horizontal, using a thin rope thatruns parallel to the ramp. The coefficient ofkinetic friction between the crate and the rampis µk. If the crate is moving at a constant speedof v0, what is the tension in the rope? !!

!!!!

!!!!

!!!!

!!

θ

!!!!

LLLL!!

!!

LLLL

w!!

!!!!!

!!x

LLL

y

*(a) w(sin θ + µk cos θ) (b) w(sin θ − µk cos θ)(c) w(cos θ + µk sin θ) (d) w(cos θ − µk sin θ)(e) None of these

Solution: Draw a free-body diagram with the x-axisparallel to the ramp and the y-axis perpendicular to it.Since the crate does not sink into or rise above the ramp,the net force in the y-direction is zero; so N = w cos θ.Hence fk = µkN = µkw cos θ.Since the crate is moving at constant speed, the net forceon it is zero. Hence the net force in the x-direction iszero; so

T = wx + fk = w sin θ + µkw cos θ = w(sin θ + µk cos θ)

u��w

θ

?wy = w cos θ

6N

�

wx =w sin θ

�

fk =µkN

-T

Problem 1309. A crate whose mass is mc issliding down a ramp that makes an angle of θwith the horizontal. The coefficient of kineticfriction between the crate and the ramp is µk.What is the magnitude of the crate’s accelera-tion?

!!!!

!!!!

!!!!

!!!!

θ

!!!!

LLLL!!

!!

LLLL

mc!!

xLLL

y

(a) g(sin θ + µk cos θ) *(b) g(sin θ − µk cos θ)(c) g(cos θ + µk sin θ) (d) g(cos θ − µk sin θ)(e) None of these

Solution: Draw a free-body diagram with the x-axisparallel to the ramp and the y-axis perpendicular to it.Since the crate does not sink into or rise above the ramp,the net force in the y-direction is zero; so N = w cos θ.Hence fk = µkN = µkw cos θ.The net force on the crate in the downslope x-directionis Fx = wx − fk = w sin θ − µkw cos θ. We can use New-ton’s law, plus the fact that w = mcg, to calculate theacceleration.

u��w

θ

?wy = w cos θ

6N

�

wx =w sin θ

-fk = µkN

a =Fxmc

=mcg sin θ − µkmcg cos θ

mc

= g(sin θ − µk cos θ)

443

Problem 1310. Two crates, one full of avocados and the other full of beets, areconnected by a light horizontal rope. A second horizontal rope is attached to the crateof avocados and used to pull both crates across the floor at a constant speed of 1.4 m/s.The avocados have a mass of 72 kg; the beets have a mass of 93 kg. The coefficient ofkinetic friction between the crates and the floor is 0.21. What is the tension TAB on therope connecting the two crates? Round your answer to the nearest newton. (See theillustration following the answers.)


B: 93 kg A: 72 kg -TAB

Solution: In this situation, the properties of the crate of avocados have nothing todo with TAB. The exact speed of the two crates is also unimportant; all that matters isthat they’re moving at a constant velocity. The horizontal forces acting on the beets areTAB pulling forward, and the friction mBgµ pulling backward. Since the crate is movingat a constant velocity, the net horizontal force must be zero: so

TAB = mBgµ = (93 kg)(9.8 m/s2)(0.21) = 191 N

Problem 1311. A block with a mass of 9.7 kg is sliding down a plane at a constantspeed of 4.4 m/s. The plane is angled at 22.4◦ to the horizontal. What is the coefficientof kinetic friction µk of the block on the plane? Round your answer to the nearest 0.01.


Solution: Take the x-axis to be down the plane and the y-axis to be the upwardnormal to the plane. Since the speed is constant, ax = 0 ⇒ Fnet,x = 0 ⇒ the force offriction must match the component of the gravitational force parallel to the plane. Thiscomponent is: Wx = mg sin θ. The component of the gravitational force normal to theplane is: N = mg cos θ; so the frictional force is: fk = µkN = µkmg cos θ. Equating thetwo forces gives:

mg sin θ = µkmg cos θ ⇒ µk =sin θ

cos θ= tan θ = tan 22.4◦ = 0.41

The mass of the block and the value of its constant speed are irrelevant to the problem.

444



Solution: Since the block’s velocity is constant, the net force on it is zero. The twoforces acting in the horizontal direction are the pull, directed forward with magnitude F ;and the frictional force fk, directed backward with magnitude µmg. Hence:

F = fk = µmg = (0.51)(132 kg)(9.8 m/s2) = 660 N



Solution: Since the block is moving at constant velocity, the net force on it must bezero. The horizontal forces on it are the pull, directed forward with a magnitude of F ;and the frictional force, directed backward with a magnitude of µmg. Hence

F = µmg = (0.64)(84 kg)(9.8 m/s2) = 527 N

Problem 1314. You trying to move a crate across a floor by pulling horizontally on arope. The crate has a mass of 114 kg. The coefficient of static friction between the crateand the floor is 0.76; the coefficient of kinetic friction is 0.51. You gradually increase yourpull on the rope until the crate starts to move. What is the crate’s initial acceleration?Round your answer to the nearest 0.01 m/s2.

Solution: To get the crate to start moving, you must exert a horizontal force equalto the crate’s weight times the coefficient of static friction: Fpull = µsmg. Once the cratebegins to move, the force pulling backward on it is kinetic friction: Fk = µkmg. At thatpoint, the net force on it is: F = Fpull − Fk = (µs − µk)mg. The acceleration is

a =F

m= (µs − µk)g = (0.76− 0.51)9.8 m/s2 = 2.45 m/s2

445

Problem 1315. A block has mass m. It is sliding at a constant speed v0 down a planethat makes an angle of θk to the horizontal. If the coefficient of kinetic friction of theblock on the plane is µk, which equation gives the value of µk?

(a) µk = mg sin θk (b) µk = µs −1

g cos θs

2∆x

(∆t)2

(c) µk =mg

sin θk*(d) µk = tan θk

(e) None of these

Solution: The component of the gravitational force parallel to the surface of theplane is: Fp = mg sin θk. The frictional force is µkmg cos θk. Since the block is sliding atconstant velocity, the net force on it is zero; so

mg sin θk = µkmg cos θk ⇒ µk =sin θkcos θk

= tan θk

446

11.2.5 Stopping distance problems

Problem 1316. The coefficient of kinetic friction between a set oftires and the road surface is µk. If you are driving on level pavementat a speed of v0, and you lock up the brakes and skid to a halt, howfar do you travel before you stop?

*(a)v2

0

2µkg(b)

v0

µkg

(c)v0µ

2k

g2(d)

v0g2

µ2k

(e) None of these

u?

w = mg

6

N

�fk = µkN

Solution: Sketch a free-body diagram, as at right; assume that your mass is m. Sincethe road surface is level, N = w = mg, so fk = µkN = µkmg. Use Newton’s law to findyour acceleration: a = fk/m = µkg. Then we know v0, vf = 0, and a = µkg; we wantto know the distance ∆x. We have a kinetic equation that we can use in this situation.Remember that the signs of a and of ∆x are opposite; since we want ∆x to be positive,we will use a = −µkg.

v2f = v2

0 + 2a∆x ⇒ ∆x =v2f − v2

0

2a=−v2

0

−2µkg=

v20

2µkg

447

Problem 1317. The coefficient of kinetic friction between a set oftires and the road surface is µk,dry when the road is dry, and µk,wet

when the road is wet, with 2µk,wet = µk,dry. If you are drivingon level wet pavement at a speed of vwet and hit the brakes, youwill travel a distance of ∆x before you come to a stop. How fastcan you drive on dry pavement and be able to stop in the samedistance?(a) vwet *(b)

√2 vwet

(c) 2vwet (d) 4vwet

(e) None of these

u?

w = mg

6

N

�fk = µkN

Solution: Sketch a free-body diagram, as at right; assume that your mass is m. Sincethe road surface is level, N = w = mg, so fk = µkN = µkmg. Use Newton’s law to findyour acceleration: a = fk/m = µkg. Then we know ∆x, vf = 0, and a = µkg; we want toknow the initial speed v0. We have a kinetic equation that we can use in this situation.Remember that the signs of a and of ∆x are opposite.

v2f = v2

0 + 2a∆x ⇒ v20 = v2

f − 2a∆x = 2µkg∆x

In going from the wet- to the dry-pavement situation, the only things that change are µkand v0; 2g∆x remains the same. Thus we can write:

v20

µk= 2g∆x =

v2dry

µk,dry

=v2

wet

µk,wet

⇒ v2dry =

µk,dry

µk,wet

v2wet =

2µk,wet

µk,wet

v2wet = 2v2

wet

⇒ vdry =√

2 vwet

448

Problem 1318. The coefficient of kinetic friction between a setof tires and the road surface is µk,dry when the road is dry, andµk,ice = µk,dry/4 when the road is icy. If you are driving on leveldry pavement at a speed of v0 and hit the brakes, you will travela distance of ∆xdry before you come to a stop. What will yourstopping distance be if you are driving at v0 on an icy road?

(a) ∆xdry (b) 2∆xdry

*(c) 4∆xdry (d) 16∆xdry

(e) None of these

u?

w = mg

6

N

�fk = µkN

Solution: Sketch a free-body diagram, as at right; assume that your mass is m. Sincethe road surface is level, N = w = mg, so fk = µkN = µkmg. Use Newton’s law to findyour acceleration: a = fk/m = µkg. Then we know v0, vf = 0, and a = µkg; we wantto know the distance ∆x. We have a kinetic equation that we can use in this situation.Remember that the signs of a and of ∆x are opposite; since we want ∆x to be positive,we will use a = −µkg.

v2f = v2

0 + 2a∆x ⇒ ∆x =v2f − v2

0

2a=−v2

0

−2µkg=

v20

2µkg

In switching from a dry to an icy road, the only things that change are the coefficient offriction µk and the stopping distance ∆x; v0 and g do not change. Thus we can write

µk∆x =v2

0

2g= µk,ice∆xice = µk,dry∆xdry

⇒ ∆xice =µk,dry

µk,ice

∆xdry =µk,dry

µk,dry/4∆xdry = 4∆xdry

449

12 Applications of Newton’s Laws to Uniform Cir-

cular Motion

12.1 Centripetal force

12.1.1 Conceptual questions: centripetal force

Problem 1319. A satellite is in circular orbit around the Earth. What is the source ofthe centripetal force?

(a) Momentum (b) Kinetic energy*(c) Gravity (d) There is no centripetal force(e) None of these

Solution: Look for a force directed toward the center of the circle. In this case,it’s gravity. Momentum and kinetic energy aren’t forces at all; and there must be acentripetal force, since the satellite is in circular motion.

Problem 1320. A car goes around a banked curve on an icy highway. What is thesource of the centripetal force?

(a) Kinetic energy (b) The car’s engine*(c) Normal force (d) There is no centripetal force(e) None of these

Solution: The highway is described as icy, which suggests that there’s no friction.The curve is banked, so there’s a normal force directed toward the inside of the curve.Kinetic energy is not a force at all; and since the car is going around a curve, there mustbe some centripetal force.

Problem 1321. A car goes around an unbanked curve. What is the source of thecentripetal force?

(a) Momentum (b) Normal force*(c) Friction (d) There is no centripetal force(e) None of these

Solution: Momentum is not a force at all. Since the curve is unbanked, the normalforce points vertically upward, with no component pointing toward the inside of thecurve. Friction is the only force that might work; there must be a centripetal force, sincethe car is going around a curve.

450

Problem 1322. A model airplane is attached to a wire so that it flies in horizontalcircles around a central pylon. What is the source of the centripetal force?

*(a) Tension in the wire (b) Air resistance(c) Inertia (d) There is no centripetal force(e) None of these

Solution: Look for a force directed toward the center of the circle. Tension in the wireis such a force. Air resistance is directed backward against the plane’s motion. Inertia isnot a force at all. There must be a centripetal force, since the plane is moving in a circle.

Problem 1323. A bullet is fired straight up into the air; it slows down, comes to amomentary halt, then falls straight down again. What is the source of the centripetalforce?

(a) Inertia (b) Air resistance(c) Normal force *(d) There is no centripetal force(e) None of these

Solution: The bullet is moving in a straight line: first straight up, then straightdown. Since it is not moving in a curve, there is no centripetal force.

Problem 1324. You have left your coffee mug on the flat roof of your car. As you drivearound an unbanked curve, what is the source of the centripetal force on the mug?

(a) Normal force (b) Kinetic energy*(c) Static friction (d) There is no centripetal force(e) None of these

Solution: The mug is moving in a curve, so there is a centripetal force. Since theroof and the road are level, the normal force is directed straight upward. Kinetic energyis not a force at all. Static friction is the source of the centripetal force: if there were nostatic friction, the mug would slide off on the curve.

Problem 1325. Your keys are attached to the end of a thin chain. You are holdingthe end of the chain and twirling the keys in a vertical circle. What is the source of thecentripetal force on the keys at the top of the circle?

(a) Gravity (b) Tension in the chain*(c) Both gravity and tension (d) There is no centripetal force(e) None of these

Solution: The keys are moving in a circle, so there is a centripetal force. At the topof the circle, both gravity and the tension in the chain are directed straight downward,toward the center of the circle. Hence the combination of the two produces the centripetalforce.

451

Problem 1326. A car is moving down a straight and level road. The driver comes to astop sign, at which he slows down but does not stop completely, then speeds up again.What is the source of the centripetal force?

(a) Kinetic friction (b) Air resistance(c) Normal force *(d) There is no centripetal force(e) None of these

Solution: The car is moving in a straight line. Since it is not moving in a curve,there is no centripetal force.

Problem 1327. A dead cockroach is lying on its back on the flat blade of a ceiling fan,which is slowly turning in a horizontal circle. What is the source of the centripetal forceon the roach?

*(a) Static friction (b) Kinetic friction(c) Gravity (d) There is no centripetal force(e) None of these

Solution: The late roach is moving in a circle, so there must be a centripetal force.Gravity is pulling straight downward, so has no component toward the center of thecircle. There is no kinetic friction, since the roach is not moving with respect to the fanblade on which it is resting. Static friction is the force that is holding it in place as thefan turns, so is the source of the centripetal acceleration.

Problem 1328. Which of the following statements is correct?

(a) Uniform circular motion causes a constant force toward the center.

(b) Uniform circular motion is caused by a constant force toward the center.

*(c) Uniform circular motion is caused by a net force toward the center with aconstant magnitude.

(d) Uniform circular motion is caused by a net force away from the center with aconstant magnitude.

(e) None of these

452

Problem 1329. Let ~Fnet be the net force on an object moving in a circle (not necessarilywith constant speed). Which of the following statements best describes the componentof the net force in the radial direction?

*(a) The radial component of the net force is the centripetal force.

(b) The radial component of the net force is the centrifugal force.

(c) The radial component of the net force causes the object to spiral inwardstowards the center of the circular path.

(d) The radial component of the net force causes the object to spiral outwardsaway from the center of the circular path.

(e) None of these

Solution: Recall: The word Centripetal originates from the Greek word for centerseeking. By assumption, the object is moving in a circle, so (c) and (d) are wrong. Theobject is traveling in a circle because the radial component of the net force points towardsthe center of the circle. The tangential component of the net force will cause the objectto accelerate in the θ-direction (i.e., it will cause the object to speed up or slow down,but it will continue to move in a circle).

There is no such thing as centrifugal force; it is an apparent force, not a real force (i.e., itdoesn’t originate from a push or a pull). The effect of centrifugal force comes from linearinertial. For example, suppose you are a passenger in a car that is initially traveling ina straight line and then makes a hard left. Why are you thrown outward against thecar door? There is no magical force that is throwing you outward. The explanation issimple. You were initially traveling in a straight line, and by Newton’s first law (the lawof linear inertial) you want to remain moving in a straight line, but the car turned outfrom under you. You went straight and the car didn’t! How did the car mange to dothis? The friction between the tires and the road allowed the car to move in a circulararc. It should also be mentioned that the friction between the tires and the road is thesource of the very real centripetal force.

Problem 1330. Determine whether the two claims are true or false.

Claim 1: Centripetal force is a real force and its existence is brought about when objectsmove in circles.

Claim 2: Centrifugal force is a real force that points radially outwards when objectsmove in circles.

(a) Claim 1 is true; Claim 2 is true

(b) Claim 1 is true; Claim 2 is false

(c) Claim 1 is false; Claim 2 is true

*(d) Claim 1 is false; Claim 2 is false

Solution: Claim 1 is false. The centripetal force comes from a real push or pullbetween two or more objects, it does not just appear magically because you are movingin a circle. To the contrary, it is a real force that causes you to move in a circle.

Claim 2 is also false. There is no such thing as a centrifugal force.

453

12.1.2 Horizontal Motion

Problem 1331. A car is driving at aconstant speed counter-clockwise aroundthe horizontal track shown at right. Ateach of the labelled points on the track,sketch a vector representing the net forceon the car. Ignore gravity, which pointsdownward into the page. Remember tomake the lengths of your vectors roughlyproportional to the magnitude of theforce: greater forces should be indicatedby longer vectors.

Solution: Since the car never changesspeed, there is no acceleration in thedirection tangent to the track. Theacceleration vector and thus the forcemust therefore always be perpendicularto the track. At points A, B, and C,there is a centripetal force keeping thecar on the curves. The curves at A andB have about the same radius; the curveat C has a radius about twice as big.

rA

6

rB�

��

rC-

rD~Fnet = ~0

Since arad = v2/r, and since v is the same at all points, the centripetal force at C shouldhave about half the magnitude that it has at A and B. At all three points, the forcevector should point toward the inside of the curve (the concave side of the curve).Note: You don’t need to be fussy about making the vector at C exactly half as long asthe vectors at A and B; but it should clearly be shorter on your drawing.

The car is moving in a straight line at D, so it is experiencing no acceleration, hence noforce. Indicate the zero vector at that point.

454

Problem 1332. A car is travelling at a constant speed clockwise around a horizontaltrack. Which of the diagrams below best shows the force vectors on the car at two pointson the track?

(a)

-

? (b)

�

6*(c)

-

?

(d)

�

6

Solution: Since the car never changes speed, there is no acceleration in the directiontangent to the track. The acceleration vector and thus the force must always be perpen-dicular to the track. That doesn’t help us: it doesn’t allow us to rule out any of the fourdiagrams.

At each of the two points shown, the car is on a curve. There must be a centripetalforce keeping the car on the curve, and that centripetal force must be directed towardthe concave side of the curve. That lets us rule out (b) and (d), which show the forcedirected toward the convex side of the curves.

The magnitude of the centripetal force is Frad = v2/r. Since the speed of the car isconstant, Frad should be proportional to 1/r. The curve on the left of the picture is wide(large r); the curve on the top of the picture is tight (small r). Hence Frad should besmaller on the left and larger at the top. We see this in (c) but not in (a). Hence thecorrect answer is (c).

Problem 1333. A model gas airplane is flying in a horizontal circle, attached to thecentral point by a string 5 m long. The airplane has a mass of 1 kg. The string willbreak if subjected to a tension of more than 100 N. How fast can the airplane fly withoutbreaking the string? Round your answer to the nearest 0.1 m/s.

Solution: If the airplane’s speed is v, then the centripetal acceleration necessary tokeep it moving in a circle of radius r is:

arad =v2

r

We assume that the tension T in the string is supplying all of the centripetal force; soif the airplane has mass m, then T = marad. We want to find the value of v for whichT = 100 N. Thus:

T = marad =mv2

r⇒ v =

√rT

m=

[(5 m)(100 N)

1 kg

]1/2

= 22.4 m/s

455

Problem 1334. A circular curve on a highway is designed for traffic moving at 15 m/s.The radius of the unbanked curve is 100 m. What is the minimum coefficient of frictionbetween tires and the highway necessary to keep cars from sliding off the curve? Roundyour answer to the nearest 0.01.


Solution: In order for a car to remain on the curve, it must experience a centripetalacceleration of arad = v2/r. For this acceleration to come from friction, it must be thecase that

gµs ≥ arad =v2

r⇒ µs ≥

v2

rg=

(15 m/s)2

(100 m)(9.8 m/s2)= 0.23

Problem 1335. You are designing a highway interchange that includes an unbankedcircular curve. Assuming that the coefficient of static friction between tires andpavement is µs = 0.51 (pointing in the direction transverse to the motion), and thatcars will need to negotiate the curve at 20 m/s, what must the radius of the curve be?Round your answer to the nearest meter.


Solution: For a car with mass m to follow the curve, it must experience a centripetalforce of mv2/r. The maximum frictional force on such a car is mgµr. Hence

mv2

r= mgµr ⇒ r =

v2

gµr=

(20 m/s)2

(9.8 m/s2)(0.51)= 80 m

456

Problem 1336. A highway curve in north-ern Minnesota has a radius of 140 m. Thecurve is banked so that a car travelling at 25m/s will not skid sideways, even if the curve iscoated with a frictionless glaze of ice. Sketch afree-body diagram of a car under these circum-stances.Solution: See figure at right. The vector Nis the normal force, so it’s perpendicular to thehighway surface. The y-component of N mustmatch the weight w, since the car doesn’t moveup or down. The x-component of N is an un-balanced force. By Newton’s second law, thisforce must lead to an acceleration. It is thisforce that provides the centripetal accelerationarad.

��

��

��

�

θ

r

?w = mg

AAAAAAKN

θ

6Ny = mg

�Nx = marad

Problem 1337. A highway curve in northernMinnesota has a radius of 160 m. The curve isbanked so that a car travelling at 25 m/s willnot skid sideways, even if the curve is coatedwith a frictionless glaze of ice. At what angleto the horizontal is the curve banked? Roundyour answer to the nearest degree.

(a) 18◦ (b) 20◦

*(c) 22◦ (d) 24◦


��

��

��

θ

r

?w = mg

AAAAAAKN

θ

6Ny = mg

�Nx = marad

Solution: Sketch a free-body diagram. There are two forces acting on the car: theweight w = mg, pulling straight down; and the normal force of the highway N , pushingperpendicular to the road surface. If the banking angle of the highway is θ = θbank, thenN decomposes into two components: an x-component Nx = N sin θ; and a y-componentNy = N cos θ.

Since the car doesn’t move up or down, the total vertical force must be zero; so Ny = mg.Since Nx is the only horizontal force, it must account for the centripetal acceleration; soif the speed of the car is V and the radius of the curve is R, then Nx = marad = mV 2/R.Hence

tan θ =Nx

Ny

=mV 2/R

mg=V 2

Rg

⇒ θbank = tan−1

(V 2

Rg

)= tan−1

((25 m/s)2

(160 m)(9.8 m/s2)

)= 22◦

457

Problem 1338. (Derive Problem) You are a civil engineer working in the wilds ofNorth Dakota. You are designing a highway with a circular curve of radius r. Youknow that at times, the highway will be glazed with ice so that the coefficient of frictionbetween the highway and a car’s tires is essentially zero. In order for the cars to safelynegotiate the icy curve at a speed of v, the road must be banked at an angle of θbank

above the horizontal. Find a formula for θbank in terms of r, v, and g.

(a) tan−1(rgv2

)(b) tan−1

(v2rg

)*(c) tan−1

(v2

rg

)(d) tan−1

(v

r2g

)(e) None of these

Solution: The highway must exert two components of force on a car: a vertical forceequal to the car’s weight; and the centripetal force, directed horizontally. If there is nofriction, then the sum of these two forces must be normal to the surface of the highway.The vertical weight force is mg; the horizontal centripetal force is mv2/r. Hence

tan θbank =mv2/r

mg=v2

rg⇒ θbank = tan−1

(v2

rg

)

Problem 1339. A highway curve in northernMinnesota has a radius of 90 m. The curve isbanked so that a car travelling at 15 m/s willnot skid sideways, even if the curve is coatedwith a frictionless glaze of ice. At what angleto the horizontal is the curve banked? Roundyour answer to the nearest degree.

*(a) 14◦ (b) 16◦

(c) 17◦ (d) 19◦


��

��

��

�

θ

r

?w = mg

AAAAAAKN

θ

6Ny = mg

�Nx = marad

Solution: Sketch a free-body diagram. There are two forces acting on the car: theweight w = mg, pulling straight down; and the normal force of the highway N , pushingperpendicular to the road surface. If the banking angle of the highway is θ = θbank, thenN decomposes into two components: an x-component Nx = N sin θ; and a y-componentNy = N cos θ.

Since the car doesn’t move up or down, the total vertical force must be zero; so Ny = mg.Since Nx is the only horizontal force, it must account for the centripetal acceleration; soif the speed of the car is V and the radius of the curve is R, then Nx = marad = mV 2/R.Hence

tan θ =Nx

Ny

=mV 2/R

mg=V 2

Rg

⇒ θbank = tan−1

(V 2

Rg

)= tan−1

((15 m/s)2

(90 m)(9.8 m/s2)

)= 14◦

458

Problem 1340. A circular curve on a highway is designed for traffic moving at 15 m/s.The radius of the curve is 100 m. What is the correct angle of banking for the road?Round your answer to the nearest degree.

*(a) 13◦ (b) 11◦

(c) 9◦ (d) 7◦

(e) None of these

Solution: The centripetal acceleration is horizontal, and has magnitude arad = v2/r.The acceleration due to gravity is vertical and has magnitude g. The surface of thehighway at the correct angle of banking is normal to the sum of these vectors; so

tan θbank =arad

g=v2

rg⇒ θbank = tan−1

(v2

rg

)= tan−1

((15 m/s)2

(100 m)(9.8 m/s2)

)= 13◦

Problem 1341. (Similarity Problem) A car travelling at a speed of V goes aroundan unbanked curve with a radius of R. The centripetal force that keeps it on the road isthe frictional force F between the road and the tires. A few miles later, the car comesto another curve, this one with radius R/2, and tries to go around the curve at the samespeed. What must the frictional force fs be in order to keep the car on the road?

(a) F/2 (b) F/√

2

(c)√

2F *(d) 2F(e) None of these

Solution: On the first curve, F = MV 2/R, where M is the mass of the car. On thesecond curve,

fs = MV 2

R/2= M

2V 2

R= 2F .

Since this is a similarity problem, we could also solve this problem by finding the param-eters that are in common (held fixed) in both experiments. Notice that the mass of thecar M and the speed of the car V are the same in both situations (experiments).

FR = MV 2 = fs

(R

2

)· 2

R−−−−−→ fs = 2F .

Notice that the result is independent of the mass of the car.

459

Problem 1342. (Similarity Problem) A car with mass M and a pickup with mass3M both go around an unbanked curve with radius R. Each vehicle is moving at aconstant speed V . While the car is going around the curve, it experiences a centripetalforce of F . What is the centripetal force fs experienced by the pickup on the curve?

(a) F (b)√

3 F*(c) 3F (d) 9F(e) None of these

Solution: The two vehicles go around the same curve at the same speed, so theyexperience the same centripetal acceleration. The centripetal force is the mass times theacceleration. Since the pickup has three times the mass of the car, it experiences threetimes the centripetal force: fs = 3F .

Problem 1343. (Similarity Problem) A car with mass M and a pickup with mass2M both go around a curve with radius R. Each vehicle is moving at a constant speedV . While the car is going around the curve, it experiences a centripetal force of F . Whatis the centripetal force Ftruck experienced by the pickup on the curve?

(a) F/2 (b) F*(c) 2F (d) 4F(e) None of these

Solution: The two vehicles go around the same curve at the same speed, so theyexperience the same centripetal acceleration. The centripetal force is the mass times theacceleration. Since the pickup has twice the mass of the car, it experiences twice thecentripetal force: Ftruck = 2F .

Problem 1344. (Similarity Problem) An object with a mass of m is on a frictionlesshorizontal surface. The object is attached to a horizontal string with length r whoseother end is attached to a pivot fixed in the surface; the object is then set to moving ina horizontal circle at a constant speed v1. Under these circumstances, the tension in thestring is T1. If the speed of the object were doubled to a constant v2 = 2v1, what wouldbe the tension in the string T2?

(a) T1 (b)√

2T1

(c) 2T1 *(d) 4T1

(e) None of these

Solution: The tension in the string is the sole source of centripetal acceleration; sothe tension equals mac. When the speed of the object is v1, the tension is T1 = mv2

1/r.When the speed is doubled, it becomes

T2 =mv2

2

r=m(2v1)2

r=

4mv21

r= 4T1

460

Problem 1345. (Similarity Problem) An object with a mass of m is on a frictionlesshorizontal surface. The object is attached to a horizontal string with length r1 whoseother end is attached to a pivot fixed in the surface; the object is then set to moving ina horizontal circle at a constant speed v1. Under these circumstances, the tension in thestring is T1. If the speed of the object is doubled to a constant v2 = 2v1, and the lengthof the string is increased to r2 = 4r1, what will be the tension in the string T2?

*(a) T1 (b)√

2T1

(c) 2T1 (d) 4T1

(e) None of these

Solution: The tension in the string is the sole source of centripetal acceleration; so thetension equals mac. When the speed of the object is v1 and the radius is r1, the tensionis T1 = mv2

1/r1. When the speed is doubled and the radius quadrupled, the tension isunchanged:

T2 =mv2

2

r2

=m(2v1)2

4r1

=mv2

1

r1

= T1 .

461

12.1.3 Vertical Motion

Problem 1346. A block with mass M is whirled on the end of a thin rigid rod that isattached to the axel of a motor so that it moves at a constant speed in a vertical circlewith radius R. At the top of the circle, the tension in the rod is twice the weight of theblock. What is the speed of the block?

(a)√

2gR *(b)√

3gR(c) 2

√gR (d) 4

√gR

(e) None of these

Solution: At the top of the circle, two forces are pulling straight down on the block:gravity, pulling downward with the weight of the block W = Mg; and tension in therod at the top of the circle Ttop, pulling downward with twice the weight of the block.The sum of these two is the centripetal force: Frad = W + Ttop = 3W = 3Mg. SinceFrad = Marad, it follows that the centripetal acceleration is 3g. Hence

arad = 3g =v2

R⇒ v =

√3gR

Problem 1347. A brass sphere with weight W is attached to the end of a thin rigidrod; the other end of the rod is attached to the axle of a motor, which is set to turningso that the sphere moves in a vertical circle at a constant speed. When the sphere is atthe lowest point on the circle, the tension in the rod is 2W . What is the tension in therod when the sphere is at the top of the circle?

*(a) 0 (b) W/2(c) W (d) 2W(e) None of these

Solution: At the bottom of the circle, the weight is pulling downward and the tensionis pulling upward; so

Frad = Tbottom −W = 2W −W = W ⇒ Tbottom = W + Frad .

Notice that the tension on the rod when the sphere is at the bottom of the circle is thesum of the centripetal force, needed to overcome the sphere’s inertia and keep the spheremoving in a circle, and the gravitational force. This should be the maximum tension inthe rod.

At the top of the circle, the radial force is the same Frad because the object is travellingwith constant speed, but now the weight and the tension are pulling in the same direction.Thus

Frad = W = Ttop +W ⇒ Ttop = 0 .

Notice that the tension on the rod when the sphere is at the top of the circle is thedifference between the centripetal force and the gravitational force. This should be theminimum tension in the rod. Since Ttop < Tbottom, this result agrees with our intuition.

462

Problem 1348. A brass sphere with weight W is attached to the end of a thin rigidrod; the other end of the rod is attached to an axle, which is set to turning so that thesphere moves in a vertical circle at a constant speed. When the sphere is at the lowestpoint on the circle, the tension in the rod is 3W . What is the tension in the rod whenthe sphere is at the top of the circle?

(a) 0 *(b) W(c) 2W (d) 4W(e) None of these

Solution: At the bottom of the circle, the weight is pulling downward and the tensionis pulling upward; so

Frad = Tbottom −W = 3W −W = 2W+W−−−−−−→ Tbottom = W + Frad .

Notice that the tension on the rod when the sphere is at the bottom of the circle is thesum of the centripetal force, needed to overcome the sphere’s inertia and keep the spheremoving in a circle, and the gravitational force. This should be the maximum tension inthe rod.

At the top of the circle, the radial force is the same Frad because the object is travellingwith constant speed, but now the weight and the tension are pulling in the same direction.Thus

Frad = 2W = Ttop +W ⇒ Ttop = W .

Notice that the tension on the rod when the sphere is at the top of the circle is thedifference between the centripetal force and the gravitational force. This should be theminimum tension in the rod. Since Ttop < Tbottom, this result agrees with our intuition.

463

Problem 1349. A pendulum in the physics lab consists of a metal sphere with a massof 220 g attached to the end of a thin string 104 cm long. The sphere is pulled awayfrom the center with the string taut and then released to swing freely. When the spherereaches the lowest point in its arc, it is moving at 83 cm/s. At that point, what is themagnitude of the sphere’s acceleration? Round your answer to the nearest 0.01 m/s2.

*(a) 0.66 m/s2 (b) 0.73 m/s2

(c) 0.80 m/s2 (d) 0.88 m/s2

(e) None of these

Solution: At the pendulum’s lowest point, there is no acceleration due to gravity,since gravity is pulling directly against the string (see the comments below). Hence theonly acceleration to consider is the centripetal acceleration. Not forgetting to convertcentimeters to meters, we get

arad =v2

r=

(0.83 m/s)2

1.04 m= 0.66 m/s2

The mass of the sphere is irrelevant to this problem.

Comment (1): When the pendulum is at its lowest point (the bottom of the circulararc), there is no component of gravity in the tangential direction. The tension in thestring counters the gravitational force and must also bear the burden of having tochange the direction of the mass (the sphere), which possesses inertia. Remember thatby Newton’s 1st Law, the sphere wants to travel in a straight line. If the string did notpull on the mass, then it would not go in a circle!

Comment (2): If the sphere were brought to rest (no motion), then the sphere wouldhang straight down and a simple free-body diagram would show that the tension inthe string would equal the weight (no motion ⇒ no acceleration ⇒ net force is zero).The only difference between the static situation and the one involving the motion of thependulum through the lowest point is that a centripetal force is needed to change thependulum’s direction. This additional force is the centripetal force.

Problem 1350. A Ferris wheel has a radius of 79 m. At its maximum speed, it makesone revolution every 38 s. If you are riding the Ferris wheel at this speed, how fast areyou moving? Round your answer to the nearest m/s.


Solution: Since you are moving at a constant speed, your average velocity is equal toyour instantaneous velocity: v = v. In travelling a complete revolution, you go a distanceof one circumference: 2πr. If you travel this distance in a time period Tperiod, your speedis

v =2πr

Tperiod

=2π(79 m)

38 s= 13 m/s

464

Problem 1351. A Ferris wheel has a radius of 65 m. At its maximum speed, it makesone revolution every 31 s. A dog is sitting on a spring scale on one of the seats on thewheel. When the wheel is not moving, the scale registers 124 N. When the wheel isturning at its maximum speed, what weight does the scale register at the highest pointin the circle? Round your answer to the nearest newton.


Solution: At the top of the wheel, the dog is experiencing two forces: the forceof gravity pulling downward with a magnitude of w = 124 N, and the normal force ofthe scale pushing upward with a magnitude of Nscale = wapp, which will be the weightregistered by the scale. This additional force on an object, resulting from the fact thatthe object is accelerating, together with the force of gravity on the object (its weight) isknown as the object’s apparent weight, which we will denote by wapp and use from thispoint forward. The sum of these two forces will produce a net centripetal force, whichwill give rise to a downward-directed centripetal acceleration arad. Hence if the dog’smass is m, then marad = w − wapp; so wapp = w −marad.

Since we know the dog’s weight on the ground, w = 124 N, we can calculate the mass:w = mg ⇒ m = w/g.

Calculating arad requires the dog’s speed on the wheel. If the radius of the wheel is Rand the dog makes a complete circle in time T , then the speed is V = 2πR/T . Then thecentripetal acceleration is

arad =V 2

R=

4π2R

T 2

Putting all of these equations together gives us

wapp = w −marad

= w − warad

g

= w

[1− arad

g

](the elevator equation with y-axis pointing up)

= w

[1− 4π2R

gT 2

]= (124 N)

[1− 4π2(65 m)

(9.8 m/s2)(31 s)2

]= 90 N

Notice that in the elevator equation , we have ay = −arad, since ~arad always pointstowards the center of the Ferris wheel.

465

Problem 1352. A Ferris wheel has a radius of 79 m. At its maximum speed, it makesone revolution every 44 s. A dog is sitting on a spring scale on one of the seats on thewheel. When the wheel is not moving, the scale registers 227 N. When the wheel isturning at its maximum speed, what weight does the scale register at the lowest point inthe circle? Round your answer to the nearest newton.


Solution: At the bottom of the wheel, the dog is experiencing two forces: theforce of gravity pulling downward with a magnitude of w = 227 N, and the normalforce of the scale pushing upward with a magnitude of Nscale = wapp, which will be theapparent weight registered by the scale. The sum of these two forces will produce thecentripetal acceleration arad, which is directed upward. Hence if the dog’s mass is m,then marad = wapp − w; so wapp = w +marad.



arad =V 2

R=

4π2R

T 2


wapp = w +marad

= w +warad

g

= w

[1 +

arad

g


= w

[1 +

4π2R

gT 2

]= (227 N)

[1 +

4π2(79 m)

(9.8 m/s2)(44 s)2

]= 264 N

Notice that in the elevator equation we have ay = −arad since ~arad always points towardsthe center of the Ferris wheel.

466

Problem 1353. A Ferris wheel has a radius of 88 m. A dog is sitting on a spring scaleon one of the seats on the wheel. When the wheel is not moving, the scale registers 265N. When the wheel is turning at its maximum speed, the scale registers zero at the topof the circle. At this speed, how long does it take for the wheel to make one revolution?Round your answer to the nearest second.

(a) 14 s (b) 15 s(c) 17 s *(d) 19 s(e) None of these

Solution: At the top of the wheel, the dog is experiencing two forces: the forceof gravity pulling downward with a magnitude of w = mg = 265 N, and the normalforce of the scale pushing upward with a magnitude of Nscale = wapp, which will be theapparent weight registered by the scale. The sum of these two forces will produce thecentripetal acceleration arad, which is directed downward. Hence if the dog’s mass is m,then marad = w − wapp. Since the scale registers zero, wapp = 0, and marad = w = mg.We can factor out the dog’s mass (its weight is irrelevant to this problem): arad = g.

If the radius of the wheel is R and the time it takes to make one revolution is T , thenthe dog is going around at a speed of V = 2πR/T . Then

g = arad =V 2

R=

4π2R

T 2

· T2

g−−−−−−→ T 2 =4π2R

g

√−−−−−−→ T = 2π

√R

g

evaluate−−−−−−−−→ T = 2π

√88 m

9.8 m/s2 = 19 s

467

Problem 1354. A Ferris wheel has a radius of 81 m. At its maximum speed, it makesone revolution every 56 s. A dog is sitting on a spring scale on one of the seats on thewheel. When the wheel is not moving, the scale registers 119 N. When the wheel isturning at its maximum speed, what weight does the scale register at the lowest point inthe circle? Round your answer to the nearest newton.





arad =V 2

R=

4π2R

T 2


wapp = w +marad

= w +warad

g

= w

[1 +

arad

g


= w

[1 +

4π2R

gT 2

]= (119 N)

[1 +

4π2(81 m)

(9.8 m/s2)(56 s)2

]= 131 N

Notice that in the elevator equation we have ay = −arad, since ~arad always points towardsthe center of the Ferris wheel.

468

Problem 1355. A Ferris wheel has a radius of 57 m. At its maximum speed, the seatsare moving at 12.2 m/s. A dog is sitting on a spring scale on one of the seats on thewheel. When the wheel is not moving, the scale registers 303 N. When the wheel isturning at its maximum speed, what weight does the scale register at the lowest point inthe circle? Round your answer to the nearest newton.



Since we know that the dog’s true weight is w = 303 N, we can calculate the mass:w = mg ⇒ m = w/g.

We know V = 12.2 m/s and R = 57 m for the Ferris wheel, so we can calculate thecentripetal acceleration: arad = V 2/R.


wapp = w +marad

= w +warad

g

= w

[1 +

arad

g


= w

[1 +

V 2

gR

]= (303 N)

[1 +

(12.2 m/s)2

(9.8 m/s2)(57 m)

]= 384 N

Notice that in the elevator equation we have ay = −arad, since ~arad always points towardsthe center of the Ferris wheel.

469

Problem 1356. An airplane pilot is flying downward. He wants to pull out of the diveand start flying upward again by making a vertical semicircle. At the bottom of thesemicircle, his speed will be 125 m/s. If he experiences more than 8.2 g’s, he will loseconsciousness and crash the plane. What is the minimum radius of the semicircle? Roundyour answer to two significant figures.


Solution: If the pilot’s centripetal acceleration at the bottom of the circle is arad,then the apparent gravity, denoted gapp, he will experience is gapp = arad +g. We wantto find the radius of the semicircle such that arad + g = 8.2g, so arad = 7.2g. If V is hisspeed and R is the radius of the semicircle, then

arad =V 2

R= 7.2g ⇒ R =

V 2

7.2g=

(125 m/s)2

(7.2)(9.8 m/s2)= 220 m

Comment: The apparent weight and apparent gravity for a mass m are related by theequation wapp = mgapp. To derive this relationship, we start by finding the centripetalforce.

The centripetal force is Frad = N − w, where N is the normal force of the pilot’s seat.If the pilot were sitting on a scale, then the reading on the scale would be N , usuallydenoted Nscale, and Nscale = wapp. From the equation Frad = N − w we get

wapp = w + Frad = mg +marad = m(g + arad) = mgapp .

Problem 1357. A skateboard park includes a U-shaped half-pipe with a semicircularcross section whose radius is 3.4 m. A skater who starts at the top will be moving at 7.4m/s at the lowest point. If a skater weighs 160 lb, what is his apparent weight at thelowest point? Round your answer to the nearest 10 lb.


Solution: At the lowest point on the circle, the apparent weight of the skateboarderwill be wapp = w + marad, where m is the mass of the skateboarder. We don’t knowthe mass of the skater, and it would be tedious to calculate it in kg from the weight inpounds. Happily, we don’t need to do so. The true weight is w = mg; so m = w/g. Thus

wapp = mgapp = m(g + arad) =w(g + arad)

g= w

[1 +

arad

g

](The elevator equation)

Now we need to calculate arad. We know the radius of the circle R and the speed of theskater V . Hence arad = V 2/R. Then the apparent weight is

wapp = w

[1 +

V 2

Rg

]= (160 lb)

[1 +

(7.4 m/s)2

(3.4 m)(9.8 m/s2)

]= 420 lb

470

Problem 1358. You have tied a short length of rope to the handle of a bucket full ofwater, and are swinging the bucket in a vertical circle with radius 83 cm. What is theminimum speed that the bucket must be moving in order for no water to spill out at thetop of the circle? Round your answer to two significant figures.

(a) 2.6 m/s *(b) 2.9 m/s(c) 26 m/s (d) 29 m/s(e) None of these

Solution: When the bucket is moving just fast enough to keep the water in at thetop of the circle, then arad = g. To see this, notice that at the top of the circular arcFrad = Ttop + w, where Ttop is the tension in the rope at the top of the circle, and w isthe weight of the bucket of water. Now, since the speed of the bucket V is related to thecentripetal acceleration via the equation arad = V 2/R, and since we are assuming thatthe radius R is fixed, it follows that if we want to minimize V , then we must minimizearad. But to minimize arad we must minimize Frad, since the mass is assumed to be afixed constant. Finally, to minimize Frad = Ttop + w, we need to minimize the tensionin the rope, since the weight of the water is a fixed constant. Since tension is alwaysgreater than or equal to zero, it follows that to minimize V , we should take the ten-sion at the top of the rope to be zero. That is, to minimize V = Vmin, we must set Ttop = 0.

Hence if the minimum speed of the bucket is Vmin and the radius of the circle is R (andif we don’t forget to convert R from centimeters to meters), then

V 2min

R= g ⇒ Vmin =

√Rg =

√(0.83 m)(9.8 m/s2) = 2.9 m/s

Problem 1359. A block with a mass of M slides down a frictionless loop-the-loop track.The loop has a radius of R. At the top of the loop, the block is moving at a speed of V .What is the net force acting on the block at that point?

Solution: The net force is the centripetal force: the centripetal acceleration timesthe mass of the block.

Fnet = Marad =MV 2

R

471

Problem 1360. A sack of potatoes has a mass of 4.4 kg. It is hanging from a springscale that is hanging from the ceiling of a car. The car is driven around a vertical loop-the-loop track with a radius of 7.9 m. At the top of the track, the car is going at a speedof 9.6 m/s. At this point, what apparent weight does the scale show for the potatoes?Round your answer to the nearest 0.1 N.

*(a) 8.2 N (b) 9.0 N(c) 9.9 N (d) 10.9 N(e) None of these

Solution: At the top of the circle, the potatoes are experiencing two forces: gravity,pulling downward with a force of w = mg, where m = 4.4 kg is the mass of the potatoes;and the tension in the scale Tscale = wapp, the apparent weight registered by the scale.Since the car is upside down at the top of the loop, with the potatoes “hanging” upwardfrom the ceiling, Tscale is directed downward. The sum of these two forces will producethe centripetal acceleration arad, which is directed downward toward the center of theloop. Hence mg + wapp = marad; so

wapp = marad −mg = m(arad − g)

We know the car’s speed V = 9.6 m/s and the loop’s radius R = 7.9 m. Thus we cancalculate arad = V 2/R. Substituting that into the equation for wapp gives us

wapp = m(arad − g) = m

[V 2

R− g]

= (4.4 kg)

[(9.6 m/s)2

7.9 m− 9.8 m/s2

]= 8.2 N

472

Problem 1361. The college’s Wellness Committee has decreed that overweight facultymembers will not receive pay raises. To get his raise, a certain physics instructor mustget his weight from 200 lbs down to 180 lbs. Rather than giving up pizza and beer, hedecides to employ trickery. He stands on a spring scale in the back of a moving van andhas it driven around a vertical loop-the-loop track with a radius of 8.7 m. At the top ofthe loop, he has a friend take a picture with the scale showing an apparent weight of 180lbs. At what speed does the van need to be going at the top of the loop? Round youranswer to the nearest 0.1 m/s.


Solution: At the top of the loop, the pudgy instructor will experience two forces:gravity, pulling downward with a force ofW = mg = 200 lbs, wherem is his mass; and thenormal force of the scale, pushing downward with a magnitude Nscale = Wapp = 180 lbs,the apparent weight. (Since the van is upside down at the top of the loop, the scale ispushing the instructor downward.) The sum of these two forces produces the centripetalacceleration arad, which is directed downward toward the center of the loop. HenceW +Wapp = marad.

It would be nice if we could avoid having to calculate m. Happily, we can. Notice thatm = W/g. Substituting this expression into the equation W +Wapp = marad gives

W

garad = W +Wapp

· gW−−−−−→ arad =

(W +Wapp)g

W=

(1 +

Wapp

W

)g

Using our given information, we can now compute arad.

Wapp

W=

180 lbs

200 lbs= 0.9 ⇒ arad = 1.9g .

Now that we know arad we can use the formula for centripetal acceleration. We know theradius R = 8.7 m of the loop; we want to know the speed v. Setting the two expressionsfor arad equal to one another will give us v

v2

R= arad =

(1 +

Wapp

W

)g

·R−−−−−→ v2 =

(1 +

Wapp

W

)gR

√−−−−−→ v =

√(1 +

Wapp

W

)gR

evaluate−−−−−−−−→v =[(1.9)(9.8 m/s2)(8.7 m)

]1/2= 12.7 m/s

473

12.2 Gravitation and circular orbits

12.2.1 Gravation

Problem 1362. A wrench has a mass of 1.43 kg. It has been left outside of a spaceshipat a distance of 1.61 × 107 m from the center of a planet whose mass is 5.98 × 1024 kg.What is the gravitational force on the wrench? Round your answer to the nearest 0.01N.(a) 1.78 N (b) 1.98 N*(c) 2.20 N (d) 2.42 N

Solution: Use the formula for gravitational force:

Fg =Gm1m2

r2=

(6.674× 10−11 N m2/kg2)(1.43 kg)(5.98× 1024 kg)

(1.61× 107 m)2= 2.20 N

Problem 1363. A very small asteroid with a mass of 32.5 kg is at a distance of 7.71×106

m from the center of a planet whose mass is 2.12 × 1021 kg. What is the gravitationalforce on the asteroid? Round your answer to three significant figures.

(a) 6.96× 10−2 N *(b) 7.74× 10−2 N(c) 8.51× 10−2 N (d) 9.36× 10−2 N

Solution: Use the formula for gravitational force:

Fg =Gm1m2

r2=

(6.674× 10−11 N m2/kg2)(32.5 kg)(2.12× 1021 kg)

(7.71× 106 m)2= 7.74× 10−2 N

Problem 1364. An astronaut with a mass of 73.3 kg is standing on the surface of aplanet whose radius is 1.15 × 106 m and whose mass is 1.31 × 1022 kg. What is theastronaut’s weight on the planet? Round your answer to the nearest 0.1 N.

(a) 39.3 N (b) 43.6 N*(c) 48.5 N (d) 53.3 N

Solution: The weight of an object is the gravitational force acting on it. We cantherefore use the formula for gravitational force:

Fg =Gm1m2

r2=

(6.674× 10−11 N m2/kg2)(73.3 kg)(1.31× 1022 kg)

(1.15× 106 m)2= 48.5 N

474

Problem 1365. An unmanned space probe has a mass of 129 kg. It is sent to the surfaceof a planet with a radius of 2.48 × 107 m and a mass of 1.02 × 1026 kg. What is theprobe’s weight on the planet? Round your answer to the nearest 10 N.

(a) 1040 N (b) 1160 N(c) 1290 N *(d) 1430 N

Solution: The weight of an object is the gravitational force acting on it. We cantherefore use the formula for gravitational force:

Fg =Gm1m2

r2=

(6.674× 10−11 N m2/kg2)(129 kg)(1.02× 1026 kg)

(2.48× 107 m)2= 1430 N

Problem 1366. A space probe has a mass of 31.2 kg. It is sent to the surface of PlanetX, whose radius is 2.11 × 106 m. On the planet’s surface, the probe’s weight is 109 N.What is the mass of Planet X? Round your answer to three significant figures.

(a) 1.70× 1023 kg (b) 1.89× 1023 kg(c) 2.10× 1023 kg *(d) 2.33× 1023 kg

Solution: The weight of an object is the force of gravity on it. We know thegravitational force, the mass of the probe mp, and the radius of the planet; we want toknow the mass of the planet mX . We can use the formula for gravitational force andsolve for mX .

Fg = w =GmpmX

r2

·(r2/Gmp)−−−−−−→ mX =wr2

Gmp

mX =(109 N)(2.11× 106 m)2

(6.674× 10−11 N m2/kg2)(31.2 kg)= 2.33× 1023 kg

Problem 1367. An astronaut has a mass of 67.3 kg. On the surface of Planet X, hisweight is 418 N. The radius of Planet X is 4.87× 106 m. What is the mass of Planet X?Round your answer to three significant figures.

(a) 1.99× 1024 kg *(b) 2.21× 1024 kg(c) 2.43× 1024 kg (d) 2.67× 1024 kg

Solution: The weight of an object is the force of gravity on it. We know thegravitational force, the mass of the astronaut ma, and the radius of the planet; we wantto know the mass of the planet mX . We can use the formula for gravitational force andsolve for mX .

Fg = w =GmamX

r2

·(r2/Gma)−−−−−−→ mX =wr2

Gma

mX =(418 N)(4.87× 106 m)2

(6.674× 10−11 N m2/kg2)(67.3 kg)= 2.21× 1024 kg

475

12.2.2 Orbits

Problem 1368. You want to calculate the period of a satellite in a circular orbit arounda planet. Which of the following do you not need to know for your calculation?

(a) The mass of the planet *(b) The radius of the planet(c) The radius of the orbit (d) The gravitational constant G(e) None of these

Solution: The formula for orbital period is

T =2πr3/2

√GM

where T is the period, r is the radius of the orbit, G is the gravitational constant, andM is the mass of the primary (in this case, the planet). The radius of the planet doesnot matter.

Problem 1369. Astronomers discover a small planet in a circular orbit around a distantstar. They are able to determine the distance of the planet from the star, and the periodof the planet’s orbit. Which of the following can they calculate based on these?

(a) The mass of the planet (b) The radius of the planet*(c) The mass of the star (d) The radius of the star(e) None of these

Solution: The formula for the orbital period is

T =2πr3/2

√GM

where T is the period, r is the radius of the orbit, G is the gravitational constant, and Mis the mass of the primary (in this case, the star). We know T , r, and G; the only othervariable in the formula is M . We can solve the formula for M and calculate it. The massof the planet and the radius of the planet and the star do not come into the formula atall.

476

Problem 1370. A satellite in circular orbit around the Earth has a period of 90 minutes.A second satellite in circular orbit around Planet X has a period of 180 minutes. If themass of the Earth is ME and the mass of Planet X is MX , which planet’s mass is greater?

(a) ME < MX *(b) Not enough information: need radii of orbits(c) ME > MX (d) Not enough information: need masses of satellites(e) None of these


T =2πr3/2

√GM

where T is the period, r is the radius of the orbit, G is the gravitational constant, andM is the mass of the primary. We can solve this for M :

T =2πr3/2

√GM

( )2−−→ T 2 =4π2r3

GM

·M/T 2

−−−−→ M =4π2r3

GT 2

We see that M depends on r as well as on T . Since we don’t know anything about r, wecan’t conclude anything about M .

Problem 1371. A satellite with a mass of 24 kg in circular orbit at a distance of 10,000km from the center of Planet X has a period of 2 hours. Which of the following satellitescircling Planet X would also have a period of 2 hours?

(a) A satellite with a mass of 24 kg at a distance of 20,000 km*(b) A satellite with a mass of 96 kg at a distance of 10,000 km(c) A satellite with a mass of 3 kg at a distance of 20,000 km(d) A satellite with a mass of 3 kg at a distance of 5,000 km(e) None of these


T =2πr3/2

√GM

where T is the period, r is the radius of the orbit, G is the gravitational constant, andM is the mass of the primary (in this case, of Planet X). Hence T depends on r and Malone. Since both satellites are circling the same planet, M is the same; so if r is thesame, then T is the same as well. The mass of the satellite doesn’t enter into the formulaat all.

477

Problem 1372. Russell’s celestial teapot was observed to be in a circular orbit aroundthe Earth at a distance of 1.48 × 107 m from the center of the planet. The mass of theEarth is 5.97× 1024 kg. What is the teapot’s speed? Round your answer to the nearest10 m/s.

(a) 3780 m/s (b) 4200 m/s(c) 4670 m/s *(d) 5190 m/s

Solution: For an object in circular orbit, gravitation provides the centripetal force.Thus we can use the formula for gravitational force and the one for centripetal force:

Fg = Frad =GmtmE

r2=mtv

2

r

·(r/mt)−−−−→ v2 =GmE

r

√−−→ v =

√GmE

r

v =

[(6.674× 10−11 N m2/kg2)(5.97× 1024 kg)

1.48× 107 m

]1/2

= 5190 m/s

Problem 1373. An asteroid is in circular orbit around the sun, at a distance of 1.50×1011

m from the sun’s center. The mass of the sun is 1.99 × 1030 kg. What is the asteroid’sspeed? Round your answer to three significant figures.

(a) 2.41× 104 m/s (b) 2.68× 104 m/s*(c) 2.98× 104 m/s (d) 3.27× 104 m/s


Fg = Frad =Gmams

r2=mav

2

r

·(r/ma)−−−−→ v2 =Gms

r

√−−→ v =

√Gms

r

v =

[(6.674× 10−11 N m2/kg2)(1.99× 1030 kg)

1.50× 1011 m

]1/2

= 2.98× 104 m/s

Problem 1374. A small moon is in circular orbit around Planet X, at a distance of1.68 × 107 m from the planet’s center. The moon’s orbital speed is 1120 m/s. What isthe mass of Planet X? Round your answer to three significant figures.

(a) 2.84× 1023 kg *(b) 3.16× 1023 kg(c) 3.47× 1023 kg (d) 3.82× 1023 kg


Fg = Frad =GmmmX

r2=mmv

2

r=

·(r2/Gmm)−−−−−−→ mX =v2r

G

mX =(1120 m/s)2(1.68× 107 m)

(6.674× 10−11 N m2/kg2)= 3.16× 1023 kg

478

Problem 1375. A small planet is in circular orbit around a star, at a distance of 3.78×1011 m from the star’s center. The orbital speed of the planet is 3.44 × 104 m/s. Whatis the star’s mass? Round your answer to three significant figures.

(a) 5.43× 1030 kg (b) 6.03× 1030 kg*(c) 6.70× 1030 kg (d) 7.37× 1030 kg


Fg = Frad =Gmpms

r2=mpv

2

r=

·(r2/Gmp)−−−−−−→ ms =v2r

G

ms =(3.44× 104 m/s)2(3.78× 1011 m)

(6.674× 10−11 N m2/kg2)= 6.70× 1030 kg

479

Problem 1376. A satellite has a mass of 6 kg. It is in circular orbit around a planetwith a mass of 6 × 1024 kg. The radius of the planet is 106 m; the radius of the orbitis 108 m. What is the satellite’s orbital period? For ease of calculation, assume thatG = 2

3× 10−10 N m2/kg2.

(a) π × 102 s *(b) π × 105 s(c) π × 1014 s (d) π × 1017 s(e) None of these


T =2πr3/2

√GM

where T is the period, r is the radius of the orbit, G is the gravitational constant, andM is the mass of the primary (in this case, the planet). The mass of the satellite andthe radius of the planet don’t enter into the formula at all, and have nothing to do withthis problem. Hence:

T =2πr3/2

√GM

=2π(108 m)3/2[

(23× 10−10 N m2/kg2)(6× 1024 kg)

]1/2We’ll look at the numbers first, then check the units and make sure that they work.

T =2π × 1012

[4× 1014]1/2=

2π × 1012

2× 107= π × 105

Now, we need to check the units. In the numerator, the units are m3/2. In the denomi-nator, remember that 1 N = 1 kg m/s2. Hence the units of the denominator are:[

N m2

kg2 ·kg

1

]1/2

=

[kg2m3

kg2s2

]1/2

=m3/2

s

Thus the units of the whole formula (numerator over denominator) are:

m3/2

m3/2/s= s

Our solution is: T = π × 105 s.

480

Problem 1377. A satellite with a mass of 150 kg is in circular orbit around a planetwith a mass of 1.5 × 1026 kg. The radius of the planet is 108 m; the radius of the orbitis 1010 m. What is the satellite’s orbital period? For ease of calculation, assume thatG = 2

3× 10−10 N m2/kg2.

(a) 2π × 104 s (b) 2π × 105 s(c) 2π × 106 s *(d) 2π × 107 s(e) None of these


T =2πr3/2

√GM

where T is the period, r is the radius of the orbit, G is the gravitational constant, andM is the mass of the primary (in this case, the planet). The mass of the satellite andthe radius of the planet don’t enter into the formula at all, and have nothing to do withthis problem. Hence:

T =2πr3/2

√GM

=2π(1010 m)3/2[

(23× 10−10 N m2/kg2)(1.5× 1026 kg)


T =2π × 1015

[1× 1016]1/2=

2π × 1015

108= 2π × 107


N m2

kg2 ·kg

1

]1/2

=

[kg2m3

kg2s2

]1/2

=m3/2

s


m3/2

m3/2/s= s

Our solution is: T = 2π × 107 s.

481

Problem 1378. A star has a mass of 2.4× 1031 kg and a radius of 109 m. A planet incircular orbit around the star has a mass of 2.4× 1025 kg and a radius of 8× 106 m. Theradius of the planet’s orbit is 4× 1012 m. What is the planet’s orbital period? For easeof calculation, assume that G = 2

3× 10−10 N m2/kg2.

*(a) 4π × 108 s (b) 4π × 109 s(c) 4π × 1010 s (d) 4π × 1011 s(e) None of these


T =2πr3/2

√GM

where T is the period, r is the radius of the orbit, G is the gravitational constant, andM is the mass of the primary (in this case, the star). The mass of the planet and theradii of the planet and of the star don’t enter into the formula at all, and have nothingto do with this problem. Hence:

T =2πr3/2

√GM

=2π(4× 1012 m)3/2[

(23× 10−10 N m2/kg2)(2.4× 1031 kg)


T =2π(8× 1018)

[1.6× 1021]1/2=

16π × 1018

[16× 1020]1/2=

16π × 1018

4× 1010= 4π × 108


N m2

kg2 ·kg

1

]1/2

=

[kg2m3

kg2s2

]1/2

=m3/2

s


m3/2

m3/2/s= s

Our solution is: T = 4π × 108 s.

482

Problem 1379. (Similarity Problem) A satellite with mass m1 is in circular orbit ata distance of r0 from the center of a planet with mass Mp. A second satellite with massm2 = 2m1 is in circular orbit around the same planet, at the same distance from thecenter. If the period of the first satellite is T1, what is the period of the second satellite?

(a) T1/4 (b) T1/2

*(c) T1 (d)√

2 T1

(e) None of these

Solution: The orbital period only depends on the orbital radius r0 and the mass ofthe primary (in this case, the planet) Mp. The mass of the satellite does not enter intothe calculation. Since both satellites orbit the same planet at the same distance, theyhave the same period.

Problem 1380. (Similarity Problem) A satellite with mass m1 is in circular orbitaround the Earth at a distance of r1 from the center of the planet; it has a period ofT1. Scientists would like to launch a second satellite into circular orbit with a period ofT2 = 2T1. What should they do to give the second satellite this period?

(a) Orbit the second satellite at a distance of r2 = 64r1

*(b) Orbit the second satellite at a distance of r2 = 41/3 · r1

(c) Give the second satellite a mass of m2 = 4m1

(d) Give the second satellite a mass of m2 = m1/4(e) None of these


T =2πr3/2

√GM

where T is the period, r is the radius of the orbit, G is the gravitational constant, and Mis the mass of the primary (in this case, the Earth). The mass of the satellite does notappear in the formula, so we can rule out answers (c) and (d). M and G are the samefrom one experiment to the other; so the only things that change are r and T . Separatingthe changing from the unchanging variables, we get:

T =2πr3/2

√GM

( )2−−→ T 2 =4π2r3

GM

÷r3−−→ T 2

r3=

4π2

GM

⇒ T 2

r3=T 2

1

r31

=T 2

2

r32

=(2T1)2

r32

=4T 2

1

r32

·r31r32/T 21−−−−−→ r3

2 = 4r31

( )1/3

−−−→ r2 = 41/3r1

483

Problem 1381. (Similarity Problem) Planet X has mass MX . A satellite with massmX is in circular orbit around it at a distance of rX from the center of the planet. PlanetY has mass MY ; a satellite with mass mY is in circular orbit around it at a distance ofrY = 2rX from the planet’s center. If the two satellites have the same period, which ofthe following is true?

(a) MY = MX/8 *(b) MY = 8MX

(c) mY = mX/8 (d) mY = 8mx

(e) None of these


T =2πr3/2

√GM

where T is the period, r is the radius of the orbit, G is the gravitational constant, andM is the mass of the primary (in this case, Planet X). The mass of the satellite does notappear in the formula, and has no bearing on the problem; thus we can rule out answers(c) and (d).In the two experiments, M and r are different; G and T are the same. Separating thechanging from the unchanging variables, we get:

T =2πr3/2

√GM

( )2−−→ T 2 =4π2r3

GM

·G/4π2

−−−−→ T 2G

4π2=r3

M

⇒ r3

M=

r3X

MX

=r3Y

MY

=(2rX)3

MY

=8r3

X

MY

·MXMY /r3X−−−−−−−→ MY = 8MX

484

13 Applications of Newton’s Laws to Rotating Bod-

ies

13.1 Torque

Problem 1382. A diving board extends 2 m horizontally from the side of a pool. Aswimmer weighing 600 N stands on the very end of the board. What torque does theswimmer exert about the base of the board? Round your answer to two significant figures.

(a) 300 N·m (b) 600 N·m*(c) 1200 N·m (d) 2400 N·m(e) None of these

Solution: Since the board is horizontal and gravity is vertical, the angle θ betweenthe board and the force on it is 90◦. Hence

τ = Fr sin θ = Fr sin 90◦ = Fr = (600 N)(2 m) = 1200 N·m

Problem 1383. A fisherman is holding a fishing rod horizontally over the edge of abridge. The rod is 150 cm long; from the end of it hangs a catfish with a mass of 6.0 kg.What torque does the fish exert about the base of the road? Round your answer to twosignificant figures.

(a) 0.92 N·m (b) 9.0 N·m*(c) 88 N·m (d) 900 N·m(e) None of these

Solution: Don’t forget to convert centimeters to meters, and to convert the catfish’smass to its weight. The rod is horizontal and the fish’s weight pulls vertically, so theangle θ between the rod and the force is 90◦. Hence

τ = Fr sin θ = mgr sin 90◦ = mgr = (6.0 kg)(9.8 m/s2)(1.50 m) = 88 N·m

Problem 1384. A cord wraps around a pulley with a radius of 8 cm. The pulley isrusty, and requires a torque of 120 N·m to make it turn. How much force do you haveto pull on the cord with in order to make the pulley turn? Round your answer to twosignificant figures.


Solution: When we’re applying force to a cord or a belt wrapped around a wheel,we can assume that the force is perpendicular to the radius; so τ = Fr sin 90◦ = Fr. Weknow τ and r; we want to know F . Don’t forget to convert centimeters to meters.

τ = Fr ⇒ F =τ

r=

120 N·m0.08 m

= 1500 N

485

Problem 1385. On a construction site, a plank is fastened at one end and projects 2.0m horizontally from the building. The fastening will break if subjected to a torque ofmore than 1500 N·m. What is the mass of the heaviest piece of equipment that can safelybe placed on the end of the plank? Round your answer to two significant figures.


Solution: The plank is horizontal and the force of weight is vertical; so τ =Fr sin 90◦ = Fr. We know r and τ ; we want to know m, for which we have to usethe fact that F = mg.

τ = Fr = mgr ⇒ m =τ

gr=

1500 Nm

(9.8 m/s2)(2.0 m)= 77 kg

Problem 1386. You are out in the middle of the desert with a flat tire. Your lug nutsare rusty, and require a torque of 300 N·m to loosen them. You are strong enough toexert a force of 500 N on the end of a wrench. How long does the wrench have to be inorder for you to get the lug nuts off? Round your answer to two significant figures.

(a) 15 cm (b) 17 cm*(c) 60 cm (d) 67 cm(e) None of these

Solution: Since no angle is mentioned, assume that you can apply the force at 90◦

to the wrench handle; so τ = Fr. We know τ and F ; we want to know r.

τ = Fr ⇒ r =τ

F=

300 N·m500 N

= 0.6 m = 60 cm

Problem 1387. An action-movie hero who weighs 70 kg is hanging from a flagpole thatprojects horizontally from the side of a building. The flagpole will break off if subjectedto a torque of more than 1200 N·m. How far out on the flagpole can the hero go? Roundyour answer to two significant figures.

(a) 0.84 m (b) 1.1 m(c) 1.4 m *(d) 1.7 m(e) None of these

Solution: Since the flagpole is horizontal and the hero’s weight is vertical, θ = 90◦

and τ = Fr. We know τ and m (which we can convert to F using F = mg); we want toknow r.

τ = Fr = mgr ⇒ r =τ

mg=

1200 N·m(70 kg)(9.8 m/s2)

= 1.7 m

486

Problem 1388. You are trying to loosen a rusty nut with a wrench 30 cm long. Becausethe nut is in an awkward place, you can only pull at an angle of 70◦ to the wrench handle.If you can pull with a force of 600 N, how much torque do you exert on the nut? Roundyour answer to two significant figures.


Solution: Don’t forget to convert centimeters to meters.

τ = Fr sin θ = (600 N)(0.30 m) sin 70◦ = 170 N·m

Problem 1389. A fisherman has a bass on the end of his line. His rod is 1.2 m long; theline makes an angle of 50◦ with the rod; and the bass is pulling on the line with a forceof 40 N. How much torque does the bass produce about the handle of the rod? Roundyour answer to two significant figures.

(a) 31 N·m (b) 33 N·m(c) 35 N·m *(d) 37 N·m(e) None of these

Solution: τ = Fr sin θ = (40 N)(1.2 m) sin 50◦ = 37 Nm

Problem 1390. A telephone pole is supported by a guy wire thatis attached 3.0 m up on the pole and makes an angle of 60◦ withthe ground. If the tension in the wire is 6000 N, how much torquedoes it exert about the base of the pole? Round your answer to twosignificant figures.

*(a) 9000 N·m (b) 11,000 N·m(c) 13,000 N·m (d) 16,000 N·m(e) None of these

JJJJJJ60◦

��)30◦

r F

Solution: Careful! The θ in the formula τ = Fr sin θ is the angle between the forceand the radius. In this case, the radius is vertical (from the base of the pole to the pointwhere the wire joins the pole), so θ = 90◦ − 60◦ = 30◦.

τ = Fr sin θ = (6000 N)(3.0 m) sin 30◦ = 9000 N·m

487

Problem 1391. A fisherman is holding a rod 120 cm long at an angleof 55◦ above the horizontal. A carp weighing 60 N is hanging from theend of the rod. How much torque does the fish exert about the handleof the rod? Round your answer to two significant figures.

(a) 29 N·m *(b) 41 N·m(c) 59 N·m (d) 72 N·m(e) None of these

��

r?F

55◦

θ = 35◦

Solution: Careful! The rod’s angle is 55◦ above the horizontal; the force of the carp’sweight is vertical. Thus the angle θ between the rod and the force is 90◦−55◦ = 35◦. It’sa good idea to sketch a diagram, as at right. Also, don’t forget to convert centimeters tometers.

τ = Fr sin θ = (60 N)(1.2 m) sin 35◦ = 41 N·m

Problem 1392. A flagpole with a length of 40 m has been installedcarelessly, and makes an angle of 12◦ to the vertical. The wind exertsa horizontal force of 300 N on the flag at the top of the pole. Howmuch torque does this produce about the base of the pole? Roundyour answer to two significant figures.

(a) 2500 N·m (b) 4200 N·m(c) 7000 N·m *(d) 12,000 N·m(e) None of these

��

r

12◦ -

�F

��3

θ = 78◦

Solution: Careful! The pole is at 12◦ to the vertical; the force of the wind is horizontal.Thus the angle θ between the force and the flagpole is 90◦− 12◦ = 78◦. It never hurts tosketch a diagram, as at right.

τ = Fr sin θ = (300 N)(40 m) sin 78◦ = 12, 000 N·m

Problem 1393. A bird with a mass of 4.2 kg lands on top of a pole 3.5 m long thatmakes an angle of 15◦ to the vertical. How much torque does the bird exert about thebase of the pole? Round your answer to two significant figures.


Solution: Don’t forget to calculate the bird’s weight from its mass.

τ = Fr sin θ = mgr sin θ = (4.2 kg)(9.8 m/s2)(3.5 m) sin 15◦ = 37 N·m

488

Problem 1394. A fisherman has a marlin on the end of his line. The line makes anangle of 70◦ with the rod, which is 2.2 m long. The fisherman will lose his grip on therod if it is subjected to a torque of more than 1200 N·m. How much force must themarlin exert on the line in order to break the fisherman’s grip? Round your answer totwo significant figures.


Solution: τ = Fr sin θ ⇒ F =τ

r sin θ=

1200 N·m(2.2 m) sin 70◦

= 580 N

Problem 1395. A flagpole 2.5 m long projects from the side of a building, at an angleof 35◦ to the vertical. A bird lands on the end of the flagpole, producing a torque of80 N·m about the flagpole’s base. What is the bird’s mass? Round your answer to twosignificant figures.

(a) 4.2 kg (b) 4.6 kg(c) 5.1 kg *(d) 5.7 kg(e) None of these

Solution: Use torque to calculate weight, then use that to calculate mass.

τ = Fr sin θ = mgr sin θ ⇒ m =τ

gr sin θ=

80 N·m(9.8 m/s2)(2.5 m) sin 35◦

= 5.7 kg

Problem 1396. Rebels have overthrown the President-for-life of Tyrannia, and are nowattempting to pull down a large statue of him. They have tied a cable around the neckof the statue, 12 m above the ground, and attached the other end to a pickup. The ropemakes an angle of 40◦ to the vertical. In order to topple the statue, it is necessary toproduce a torque of 80,000 N·m about its base. How much force does the pickup need toapply to the rope? Round your answer to two significant figures.

(a) 8700 N *(b) 10,000 N(c) 12,000 N (d) 15,000 N(e) None of these

Solution: τ = Fr sin θ ⇒ F =τ

r sin θ=

80, 000 N·m(12 m) sin 40◦

= 10, 000 N

489

Problem 1397. A flagpole is supported by a guy wire that is attached to the pole 5.0 mabove the ground, and that makes an angle of 65◦ to the horizontal. To keep the flagpolefrom blowing over in high winds, the wire must be able to produce a torque of 6000N·m about the base of the pole. How much tension must the wire be able to withstand?Round your answer to two significant figures.

(a) 1300 N *(b) 2800 N(c) 6100 N (d) 13,000 N(e) None of these

Solution: Be careful with your angle. The wire makes an angle of 65◦ with thehorizontal, so it makes an angle of θ = 90◦− 65◦ = 25◦ with the vertical flagpole. Sketcha diagram if you’re not sure about this.

τ = Fr sin θ ⇒ F =τ

r sin θ=

6000 N·m(5.0 m) sin 25◦

= 2800 N

Problem 1398. A branch projects diagonally upward from the trunk of a tree, makingan angle of 70◦ with the vertical trunk. A monkey weighing 240 N climbs out along thebranch. The branch will break off if it is subjected to a torque of more than 300 N·mabout its base. How far along the branch can the monkey go without breaking it? Roundyour answer to two significant figures.


Solution: τ = Fr sin θ ⇒ r =τ

F sin θ=

300 N·m(240 N) sin 70◦

= 1.3 m

Problem 1399. A partly open drawbridge makes an angle of 15◦ to the horizontal. Atruck weighing 1.2 × 105 N drives slowly up the slope. The drawbridge will collapse ifsubjected to a torque of more than 2.0 × 106 N. How far along the drawbridge can thetruck go? Round your answer to two significant figures.


Solution: Be careful with your angle. The bridge makes an angle of 15◦ to thehorizontal; but the force of the truck’s weight is vertical, so it makes an angle of 90◦−15◦ =75◦ with the bridge. Sketch a diagram if you’re unclear on this.

τ = Fr sin θ ⇒ r =τ

F sin θ=

2.0× 106 N·m(1.2× 105 N) sin 75◦

= 17 m

490

Problem 1400. The boom of a crane is 20 m long. A cement truck weighing 300,000N is hanging from the end of the crane. The crane will topple over if it is subjected toa torque about its base greater than 5.0 × 106 N·m. What is the largest angle with thevertical that the boom can safely make? Round your answer to the nearest degree.

(a) 34◦ (b) 41◦

(c) 49◦ *(d) 56◦

(e) None of these

Solution: τ = Fr sin θ ⇒ θ = sin−1[ τFr

]= sin−1

[5.0× 106 N·m

(300, 000 N)(20 m)

]= 56◦

Problem 1401. Rebels have overthrown the Dear Leader of Annexia, and are attemptingto pull down his statue. They have attached a cable around the statue’s neck, 11 m aboveground level, and have attached the other end to a truck. If the truck can pull with aforce of 50,000 N, and if it will require a torque of 400,000 N·m about the statue’s baseto pull it over, what angle must the cable make with the horizontal? Round your answerto the nearest degree.

*(a) 43◦ (b) 47◦

(c) 50◦ (d) 53◦

(e) None of these

Solution: Careful! The problem asks for the cable’s angle with the horizontal; butthe radius (from the statue’s base to the neck) is vertical. Thus the problem is asking forφ = 90◦ − θ, where θ is the angle between the cable and the vertical. Sketch a diagramif you’re not clear on this.

τ = Fr sin θ ⇒ θ = sin−1[ τFr

]⇒ φ = 90◦ − sin−1

[400, 000 N·m

(50, 000 N)(11 m)

]= 43◦

Problem 1402. Two people are standing on a horizontal diving board. One personweighs 400 N, and is standing 3 m from the base of the board. The other weighs 800 N,and is standing 2 m from the base. What is the magnitude of the net torque that thetwo people produce about the base of the board? Round your answer to two significantfigures.


Solution: Both people are standing on the same side of the base; so their torques pointin the same direction, and therefore have the same sign. Since the board is horizontaland the forces (weight) are vertical, θ = 90◦ and Fr sin θ = Fr.

τ = τ1 + τ2 = F1r1 + F2r2 = (400 N)(3 m) + (800 N)(2 m) = 2800 N·m

491

Problem 1403. A teeter-totter consists of a masslessboard pivoting on a fulcrum. With the board held hor-izontal, a physics instructor weighing w1 = 800 N sits1 m from the fulcrum. A physics student weighingw2 = 500 N sits on the opposite side, 2 m from thefulcrum. What is the net torque about the fulcrum?Assume that the instructor produces a positive torque.Round your answer to two significant figures.

��

TTT

w1 w2

(a) 200 N·m *(b) −200 N·m(c) 300 N·m (d) −300 N·m(e) None of these

Solution: We will use ri = 1 m and rs = −2 m, so that the instructor’s torque τi willbe positive and the student’s torque τs will be negative. Since the board is horizontaland the forces are vertical, θ = 90◦ and Fr sin θ = Fr.

τ = τi + τs = Firi + Fsrs = (800 N)(1 m) + (500 N)(−2 m) = −200 N·m

Problem 1404. Two pulleys are fixed on a shaft, asshown at right. One pulley has a radius of 15 cm; a weightof w1 = 6 N hangs from that pulley on the left side of theshaft. The second pulley has a radius of 10 cm; a weightof w2 = 4 N hangs from it, on the right side of the shaft.What is the net torque on the shaft? Assume that w1

produces a positive torque. Round your answer to twosignificant figures.

*(a) 0.50 N·m (b) −0.50 N·m(c) 1.3 N·m (d) −1.3 N·m(e) None of these

u

w1 w2

Solution: Don’t forget to convert centimeters to meters. We will use r1 = 0.15 mand r2 = −0.10 m, so that the directions of the torques will be right.

τ = τ1 + τ2 = w1r1 + w2r2 = (6 N)(0.15 m) + (4 N)(−0.10 m) = 0.50 N·m

492

Problem 1405. You are trying to close a door; your dog is on the opposite side, tryingto push it open. You apply a force of 200 N at a distance of 80 cm from the hinges.The dog applies a force of 400 N at a distance of 60 cm from the hinges. What is thenet torque on the door? Assume that you are applying a positive torque. Round youranswer to two significant figures.

(a) −160 N·m *(b) −80 N·m(c) 120 N·m (d) 400 N·m(e) None of these

Solution: We will use Fy = 200 N and Fd = −400 N so that the torques will havethe right signs. Since no angles are mentioned, we’ll assume that you and the dog areboth pushing at right angles to the door, so sin θ = 1 for both. Don’t forget to convertcentimeters to meters.

τ = τy + τd = Fyry + Fdrd = (200 N)(0.80 m) + (−400 N)(0.60 m) = −80 N·m

Problem 1406. Two civil engineering students are attempting to loosen a nut on abridge, using a very large wrench. Both students are pulling in the same direction. Onestudent is applying a force of 400 N at an angle of 90◦ to the handle of the wrench, 1.2m from the nut; the other student is applying a force of 500 N at an angle of 80◦ to thehandle, 1.6 m from the nut. What is the net torque applied to the nut? Round youranswer to two significant figures.


Solution: Since both students are producing torque in the same direction, we don’thave to worry about signs.

τ = τ1 + τ2

= F1r1 sin θ1 + F2r2 sin θ2

= (400 N)(1.2 m) sin 90◦ + (500 N)(1.6 m) sin 80◦

= 1300 N·m

493

Problem 1407. Rebels have overthrown the Tsar of Fanatistan, and are now trying topull down his statue. However, the two rebel organizations are in disagreement on howto do it. The People’s Front has attached a rope 6.0 m up on the statue, and is pullingnorthward with a force of 20,000 N at an angle of 40◦ to the horizontal. The PopularFront has attached their rope 7.5 m up on the statue, and is pulling southward with aforce of 15,000 N at an angle of 35◦ to the horizontal. What is the net torque about thebase of the statue? Assume that the People’s Front is applying positive torque. Roundyour answer to the nearest 1000 N·m.

*(a) 0 N·m (b) −13, 000 N·m(c) 38,000 N·m (d) 180,000 N·m(e) None of these

Solution: Careful! Notice that we are given angles to the horizontal, but thatto calculate the torque we need angles measured to the vertical. Also, to make thesigns of the torques work out correctly, we’ll give the Popular Front a negative force:F2 = −15, 000 N.

τ = τ1 + τ2

= F1r1 sin θ1 + F2r2 sin θ2

= (20, 000 N)(6.0 m) sin 50◦ + (−15, 000 N)(7.5 m) sin 55◦

= 0 N·m

Problem 1408. A massless plank is lying on the ground and extending partly out overthe edge of the Grand Canyon. A tourist weighing 800 N wants to walk to the end of theplank, 2.5 m beyond the cliff edge, and have his wife take his picture. If the wife weighs500 N, how far back from the cliff edge must she stand on the plank to keep her husbandfrom falling into the canyon? Round your answer to two significant figures.

(a) 1.6 m (b) 2.5 m*(c) 4.0 m (d) 6.4 m(e) None of these

Solution: If the wife really doesn’t want her husband to fall in, then she must producea torque about the cliff edge that matches his. We assume that the plank is horizontal,so Fr sin θ = Fr. Then

Fwrw = Fhrh ⇒ rw =FhrhFw

=(800 N)(2.5 m)

500 N= 4.0 m

494

Problem 1409. A teeter-totter consists of a masslessboard pivoting on a fulcrum. A physics instructor weigh-ing w1 = 800 N sits 1.2 m from the fulcrum. A physicsstudent weighing w2 = 500 N sits on the opposite side.How far from the fulcrum must the student sit to bal-ance the instructor’s torque? Round your answer to twosignificant figures.

��

TTT

w1 w2

(a) 0.75 m (b) 1.4 m(c) 1.6 m *(d) 1.9 m(e) None of these

Solution: Since no angle is mentioned, we can assume that the board is horizontal; soFr sin θ = Fr. To balance, the torques produced by instructor and student must match.

w1r1 = w2r2 ⇒ r2 =w1r1

w2

=(800 N)(1.2 m)

500 N= 1.9 m

Problem 1410. A vertical pulley has a radius of 5 cm. A thin string is wound aroundthe outside of the pulley, and a weight of 8 N is suspended from it. The pulley has abrake, located 3 cm from the pulley’s axle. How much force must the brake exert to keepthe pulley from turning? Round your answer to two significant figures.

(a) 1.9 N (b) 4.8 N*(c) 13 N (d) 120 N(e) None of these

Solution: To keep the pulley from turning, the torques produced by the weight andby the brake must match.

F1r1 = F2r2 ⇒ F2 =F1r1

r2

=(8 N)(0.05 m)

0.03 m= 13 N

Problem 1411. A massless horizontal pole projects 1.6 m from theside of a building. It is held up by a cable running from the side ofthe building to the end of the pole, making an angle of 35◦ to thehorizontal. A sign weighing w = 30 N hangs from the pole 1.5 m outfrom the building. What is the tension in the cable? Round youranswer to two significant figures.


w

HHH

HHHH

Solution: Since the system is not moving, the torques produced by the weight andby the tension must match. Let T be the tension.

TrT sin θT = wrw ⇒ T =wrw

rT sin θ=

(30 N)(1.5 m)

(1.6 m) sin 35◦= 49 N

495

Problem 1412. A flagpole 2.4 m long projects from the side of abuilding, making an angle of 40◦ to the vertical. The flagpole has amass of 80 kg; its center of mass is at its midpoint. It is supportedby a horizontal cable running from the side of the building to the endof the pole. What is the tension in the cable? Round your answerto two significant figures.


��

r40◦

r?w = mg

T50◦

Solution: We need to balance two torques about the base of the pole. The first isproduced by the weight of the flagpole: F1 = w = mg, r1 = 1.2 m (the midpoint ofthe pole), and θ1 = 40◦. The second is produced by the tension in the cable: F2 = T ,r2 = 2.4 m (the end of the pole), and θ2 = 50◦.

mgr1 sin θ1 = Tr2 sin θ2

⇒ T =mgr1 sin θ1

r2 sin θ2

=(80 kg)(9.8 m/s2)(1.2 m) sin 40◦

(2.4 m) sin 50◦= 330 N

Problem 1413. A flagpole 7 m high is supported by a guy wireattached 5 m high on the north side of the pole and meeting theground at an angle of θ = 50◦. The wind blows southward, exerting aforce of 200 N on the flag at the top of the pole. What is the tensionin the wire? Round your answer to two significant figures.


JJJJJJθ

Solution: Since the pole doesn’t move, the torque produced by the wire must balancethe torque produced by the wind on the flag. Be careful with the angles; the anglebetween the wire and the flagpole is φ = 90◦ − 50◦ = 40◦. Let T be the tension; then

F1r1 = Tr2 sinφ ⇒ T =F1r1

r2 sinφ=

(200 N)(7 m)

(5 m) sin 40◦= 440 N

496

Problem 1414. A pendulum consists of a lead sinker with weight Wattached to the end of a thin string with length L; the other end ofthe string is attached to a hook in the ceiling. The pendulum is pulledback and released. When the string makes an angle of θ with thevertical, it has a tension of T . At that point, what is the net torqueabout the hook in the ceiling?

(a) 0 (b) (T +W ) sin θ

*(c) WL sin θ (d) (T −W )L cos θ

(e) None of these

AAAAAAAA

L

uθ

?Wθ

Solution: The tension is pulling straight away from the hook, so it makes no contri-bution to the torque: r = 0. The weight W is pulling straight downward, at an angle of180◦ − θ to the string and at a distance of L from the hook. Hence

τ = WL sin(180◦ − θ) = WL sin θ

13.2 Angular acceleration

Problem 1415. A windmill has a moment of inertia of 5.0 kg·m2. When the wind beginsto blow, it accelerates at 0.40 rad/s2. What is the torque on the windmill? Round youranswer to two significant figures.

(a) 0.80 N·m *(b) 2.0 N·m(c) 5.0 N·m (d) 13 N·m(e) None of these

Solution: τ = Iα = (5.0 kg·m2)(0.40 rad/s2) = 2.0 N·m

Problem 1416. A shaft’s moment of inertia is 24 kg·m2. How much torque does it taketo accelerate the shaft at 3.0 rad/s2? Round your answer to two significant figures.

(a) 8.0 N·m (b) 17 N·m(c) 35 N·m *(d) 72 N·m(e) None of these

Solution: τ = Iα = (24 kg·m2)(3.0 rad/s2) = 72 N·m

497

Problem 1417. A flywheel has a mass of 1200 kg, a radius of 0.75 m, and a moment ofinertia of 480 kg·m2. When a motor connected to the flywheel is switched on, the flywheelaccelerates at 0.80 rad/s2. How much torque does the motor apply to the flywheel?Round your answer to two significant figures.


Solution: We don’t need the mass or the radius for this problem.

τ = Iα = (480 kg·m2)(0.80 rad/s2) = 380 N·m

Problem 1418. A bicycle wheel’s moment of inertia is 0.30 kg·m2. If you apply a torqueof 20 N·m to it, what is its angular acceleration? Round your answer to two significantfigures.

(a) 1.8 rad/s2 (b) 6.0 rad/s2

(c) 20 rad/s2 *(d) 67 rad/s2

(e) None of these

Solution: τ = Iα ⇒ α =τ

I=

20 N·m0.30 kg·m2 = 67 rad/s2

Problem 1419. A revolving door has moment of inertia 120 kg·m2. You apply a torqueof 50 N·m to it. What is its angular acceleration? Round your answer to two significantfigures.

(a) 0.12 rad/s2 *(b) 0.42 rad/s2

(c) 1.4 rad/s2 (d) 2.4 rad/s2

(e) None of these

Solution: τ = Iα ⇒ α =τ

I=

50 N·m120 kg·m2 = 0.42 rad/s2

Problem 1420. You are trying to determine the moment of inertia of a roast turkeyon a massless turntable. Unfortunately, your physics book does not include a formulafor the moment of inertia of a turkey, roasted or otherwise. You apply a torque of 0.060N·m to the turntable, and it accelerates at 1.4 rad/s2. What is the turkey’s moment ofinertia? Round your answer to two significant figures.

*(a) 0.043 kg·m2 (b) 0.054 kg·m2

(c) 0.067 kg·m2 (d) 0.084 kg·m2

(e) None of these

Solution: τ = Iα ⇒ I =τ

α=

0.60 N·m1.4 rad/s2 = 0.043 kg·m2

498

Problem 1421. A motor supplies a torque of 160,000 N·m to a merry-go-round, whichaccelerates at 0.3 rad/s2. What is the merry-go-round’s moment of inertia? Round youranswer to two significant figures.

(a) 48,000 kg·m2 (b) 110,000 kg·m2

(c) 240,000 kg·m2 *(d) 530,000 kg·m2

(e) None of these

Solution: τ = Iα ⇒ I =τ

α=

160, 000 N·m0.3 rad/s2 = 530, 000 kg·m2

Problem 1422. A flywheel has a radius of 60 cm and a moment of inertia of 150 kg·m2.A belt around the rim of the wheel imparts a force of 80 N. What is the flywheel’s angularacceleration? Round your answer to two significant figures.

(a) 0.29 rad/s2 *(b) 0.32 rad/s2

(c) 0.35 rad/s2 (d) 0.39 rad/s2

(e) None of these

Solution: τ = Fr = Iα ⇒ α =Fr

I=

(80 N)(0.60 m)

150 kg·m2 = 0.32 rad/s2

Problem 1423. A gate is 2 m long, and has a moment of inertia of 200 kg·m2. A cowpushes on the gate 1.5 m from the hinges, with a force of 400 N. What is the gate’sangular acceleration? Round your answer to two significant figures.

(a) 1.5 rad/s2 (b) 2.3 rad/s2

*(c) 3.0 rad/s2 (d) 4.0 rad/s2

(e) None of these

Solution: Since no angle is mentioned, we assume that the cow pushes at a rightangle; so τ = Fr.

τ = Fr = Iα ⇒ α =Fr

I=

(400 N)(1.5 m)

200 kg·m2 = 3.0 rad/s2

Notice that we don’t need the length of the gate.

Problem 1424. A shaft is 10 m long, with a radius of 10 cm and a moment of inertia of12 kg·m2. A belt around the shaft imparts a force of 50 N. What is the shaft’s angularacceleration? Round your answer to two significant figures.

(a) 0.30 rad/s2 (b) 0.34 rad/s2

(c) 0.38 rad/s2 *(d) 0.42 rad/s2

(e) None of these

Solution: The length of the shaft is irrelevant to this problem.

τ = Fr = Iα ⇒ α =FR

I=

(50 N)(0.10 m)

12 kg·m2 = 0.42 rad/s2

499

Problem 1425. A large gear has a radius of 1.2 m, a mass of 4100 kg, and a moment ofinertia of 4000 kg·m2. A second gear engages the teeth on the outer edge of the first oneand imparts a force of 200 N. What is the angular acceleration of the first gear? Roundyour answer to two significant figures.

(a) 0.050 rad/s2 *(b) 0.060 rad/s2

(c) 0.067 rad/s2 (d) 0.075 rad/s2

(e) None of these

Solution: The mass of the gear doesn’t matter.


I=

(200 N)(1.2 m)

4000 kg·m2 = 0.060 rad/s2

Problem 1426. A merry-go-round has a radius of 2.0 m. A physics professor pusheswith a force of 120 N tangent to the outer edge of the merry-go-round, giving it an angularacceleration of 1.8 rad/s2. What is the merry-go-round’s moment of inertia? Round youranswer to two significant figures.

(a) 110 kg·m2 (b) 120 kg·m2

*(c) 130 kg·m2 (d) 150 kg·m2

(e) None of these

Solution: τ = Fr = Iα ⇒ I =Fr

α=

(120 N)(2.0 m)

1.8 rad/s2 = 130 kg·m2

Problem 1427. A belt runs over a pulley with a radius of 12 cm. When the belt ispulled with a force of 60 N, the pulley experiences a radial acceleration of 2.5 m/s2. Whatis the pulley’s moment of inertia? Round your answer to two significant figures.

(a) 2.1 kg·m2 (b) 2.3 kg·m2

(c) 2.6 kg·m2 *(d) 2.9 kg·m2

(e) None of these

Solution: τ = Fr = Iα ⇒ I =Fr

α=

(60 N)(0.12 m)

2.5 m/s2 = 2.9 kg·m2

Problem 1428. A fountain features a stone sphere with a radius of 2.8 m, suspendedfrictionlessly on a film of water in a hemispherical basin. The sphere has a moment ofinertia of 2800 kg·m2. How much force do you have to apply along the sphere’s equatorin order to produce an acceleration of 0.04 m/s2? Round your answer to two significantfigures.


Solution: τ = Fr = Iα ⇒ F =Iα

r=

(2800 kg·m2)(0.04 m/s2)

2.8 m= 40 N

500

Problem 1429. A gate is 2.1 m long, and has a moment of inertia of 260 kg·m2. Howhard do you have to push on the outer edge of the gate to induce an acceleration of1.5 rad/s2? Round your answer to two significant figures.


Solution: Since we don’t mention an angle, assume that you push at right angles tothe gate.

τ = Fr = Iα ⇒ F =Iα

r=

(260 kg·m2)(1.5 rad/s2)

2.1 m= 190 N

Problem 1430. A windmill’s moment of inertia is 20 kg·m2. It is initially at rest; whenthe wind starts to blow, it begins turning, reaching a speed of 4.5 rad/s in 8 seconds.What is the magnitude of the torque that the wind applied to the windmill? Round youranswer to two significant figures.


Solution: We can’t use τ = Iα directly, since we don’t know α. However, we knowω0 = 0, ω, and t; so we can calculate α.

ω = ω0 + αt = αt ⇒ α =ω

t

τ = Iα =Iω

t=

(20 kg·m2)(4.5 rad/s)

8 s= 11 N·m

Problem 1431. A shaft has a radius of 8 cm and a moment of inertia of 15 kg·m2. It isdriven by a belt wrapped around it. If the shaft is initially at rest and the belt is pulledwith a force of 20 N, how long does it take for the shaft to turn 10 radians? Round youranswer to two significant figures.

(a) 12 s *(b) 14 s(c) 15 s (d) 17 s(e) None of these

Solution: Since we know r and F , we can calculate τ ; since we know I, we cancalculate α. Then we know α, ω0 = 0, θ0 = 0, and θ; we want to know t.


I

θ = θ0 + ω0t+1

2αt2 =

1

2αt2 =

Fr

2It2

⇒ t =

√2Iθ

Fr=

[2(15 kg·m2)(10 rad)

(20 N)(0.08 m)

]1/2

= 14 s

501

Problem 1432. A wheel’s moment of inertia is 30 kg·m2. It is spinning at 25 rad/swhen a brake is applied at a distance of 30 cm from the wheel’s center. The wheel turnsthrough 15 radians before coming to a halt. How much force was produced by the brake?Round your answer to two significant figures.


Solution: We know r and we want to know F . If we knew τ , we could calculate itdirectly. We know I; if we knew α, we could calculate τ . We know ω0, ∆θ, and ω = 0,from which we can calculate α.

tau = Iα = Fr ⇒ F =Iα

r

2α∆θ = ω2 − ω20 = −ω2

0 ⇒ α = − ω20

2∆θ

Since we only care about the magnitude of the force applied by the brake, we’ll ignorethe negative sign.

F =Iω2

0

2r∆θ=

(30 kg·m2)(25 rad/s)2

2(0.30 m)(15 rad)= 2100 N

Problem 1433. An evil physics professor decides to stop the rotation of the Earth. Heembeds a large rocket in a mountain on the equator, pointing westward. The Earthhas a radius of 6.4 × 106 m and a moment of inertia of 9.8 × 1037 kg·m2; it is initiallyturning at 7.3 × 10−5 rad/s. If the professor wants to stop the rotation in 3.2× 107 s(approximately one year), how much thrust must his rocket produce? Round your answerto two significant figures.

(a) 3.2× 1018 N *(b) 3.5× 1019 N(c) 3.9× 1020 N (d) 4.2× 1021 N(e) None of these

Solution: We know r. If we knew τ , we could calculate F . We know I; if we knewα, we could calculate τ . We know ω0, ω = 0, and t; from these, we can calculate α.


r

ω = 0 = ω0 + αt ⇒ α = −ω0

t

Since we want the magnitude of the force, we will ignore the negative sign.

F =Iω0

rt=

(9.8× 1037 kg·m2)(7.3× 10−5 rad/s)

(6.4× 106 m)(3.2× 107 s)= 3.5× 1019 N

502

Problem 1434. A large gear’s radius is 30 cm, and its moment of inertia is 40 kg·m2.A second gear engages the teeth on the outer edge of the first one and applies a force of20 N. The gears are initially at rest. After the first gear has turned through 5 rad, howfast will it be turning? Round your answer to two significant figures.

(a) 1.0 rad/s (b) 1.1 rad/s*(c) 1.2 rad/s (d) 1.3 rad/s(e) None of these

Solution: We know ω0 = 0 and ∆θ; we want to know ω. If we knew α, we couldcalculate ω. We know r and F , so we can calculate τ ; and we know I, so we can calculateα.


I

2α∆θ = ω2 − ω20 = ω2

⇒ ω =√

2α∆θ =

[2Fr∆θ

I

]1/2

=

[2(20 N)(0.30 m)(5 rad)

40 kg·m2

]1/2

= 1.2 rad/s

Problem 1435. A flywheel with a radius of 0.50 m and a moment of inertia of 120 kg·m2

is driven by a belt around the rim. The flywheel is initially at rest. You would like it toturn through 20 rad in 6 seconds. What force do you need to apply to the belt? Roundyour answer to two significant figures.


Solution: We know r and I; we want to know F . If we knew α, we could calculateτ , and from that we could calculate F . We know θ0 = 0, ω0 = 0, θ, and t. From these,we can calculate α.

θ = θ0 + ω0t+1

2αt2 =

1

2αt2 ⇒ α =

2θ

t2


r=

2Iθ

rt2=

2(120 kg·m2)(20 rad)

(0.50 m)(6 s)2= 270 N

503

13.3 Moment of inertia

Problem 1436. A physics instructor has a mass of 90 kg. His wife has a mass of 60 kg.The two are riding together on a merry-go-round with two concentric circles of woodenhorses. In which of the following situations is the system’s moment of inertia the smallest?

(a) Both sit together one one of the outer horses(b) Both sit on outer horses, on opposite sides of the merry-go-round*(c) She sits on one of the outer horses; he sits on the inner horse next to her(d) He sits on one of the outer horses; she sits on an inner horse on the opposite side

Solution: In all of the possible solutions, the mass is the same. We can reduce themoment of inertia by moving mass closer to the axis of rotation. We don’t have theoption of putting both people on inner horses (making both rp and rw small). Sincemp > mw, our best option is to put the professor on an inner horse (making rp small).

Problem 1437. A flywheel consists of a solid iron disc with two holes drilled throughit, turning on a shaft through its center. In which of the following arrangements of holesis the moment of inertia the largest?

*(a)

uk k(b)

uk k(c)

ukk(d)

u kk

Solution: To make the moment of inertia large, we want as much mass as possible tobe far away from the axis of rotation. In situation (a), we are drilling our holes near theaxis, where the mass removed contributed little to I. In the other situations, the massremoved contributed more to I, so drilling holes there makes greater reductions in I.

Problem 1438. A lead block with mass mb is attached to a massless turntable, at adistance of r1 from the center. The system’s moment of inertia is I1. The block is thenmoved to a distance of 2r1 from the center. What is the new moment of inertia?

(a) I1 (b)√

2 I1

(c) 2I1 *(d) 4I1

(e) None of these

We assume that the block is small enough to be treated as a point mass. Then

I1 = mbr21 and I2 = mb(2r1)2 = 4mbr

21 = 4I1

504

Problem 1439. A lead block with mass mb is attached to a massless turntable, at adistance of r1 from the center. The system’s moment of inertia is I1. A second block withthe same mass is then attached at the same distance from the center, on the oppositeside. What is the new moment of inertia?

(a) I1 (b)√

2 I1

*(c) 2I1 (d) 4I1

(e) None of these

Solution: We assume that the block is small enough to be treated as a point mass.Then

I1 = mbr21 and I2 = mbr

21 +mbr

21 = 2mbr

21 = 2I1

Problem 1440. A system consists of a massless turntable with a lead block with massm1 attached at a distance of r1 from the center. A second system has the same momentof inertia, but uses a block with mass m2 = 2m1. At what distance r2 is this blockattached?

(a) r1/4 (b) r1/2

*(c) r1/√

2 (d) r1

(e) None of these

Solution: We assume that the blocks can be treated as point masses.

I1 = m1r21 and I2 = I1 = m2r

22 = (2m1)r2

2 ⇒ r21 = 2r2

2 ⇒ r2 =r1√

2

Problem 1441. A gate can be regarded as a thin rectangular plate hinged along oneside, with a mass of 80 kg and a length of 3.0 m. The gate is pushed with a force of 50N, 2.0 m from its hinges. What angular acceleration does it experience? Round youranswer to two significant figures.

(a) 0.30 rad/s2 (b) 0.34 rad/s2

(c) 0.38 rad/s2 *(d) 0.42 rad/s2

(e) None of these

Solution: We’ll use L = 3.0 m to denote the length of the gate, and r = 2.0 m todenote the distance at which force is applied. We know r and F , so we can calculate τ .To calculate α, we need to know I. We can calculate I from the mass and dimensions ofthe gate.

I =mL2

3


I=

3Fr

mL2=

3(50 N)(2.0 m)

(80 kg)(3.0 m)2= 0.42 rad/s2

505

Problem 1442. A wind turbine has three blades. Each blade has a mass of 1200 kgand a length of 20 m. Each blade can be regarded as a thin rod connected at one endto the axis. How much torque must be applied to the turbine to make it accelerate at0.10 rad/s2? Round your answer to two significant figures.

(a) 35,000 N·m (b) 39,000 N·m(c) 43,000 N·m *(d) 48,000 N·m(e) None of these

Solution: We know α. We want to know τ . We can calculate I from the descriptionof the turbine. We’ll use L = 20 m to denote the length of a blade.

Iturbine = 3Iblade = 3

(ML2

3

)= ML2

τ = Iα = ML2α = (1200 kg)(20 m)2(0.10 rad/s2) = 48, 000 N·m

Problem 1443. A flywheel consists of a solid cylinder with a radius of 70 cm and athickness of z = 8 cm; it is made of steel with a density of ρ = 7.5 g/cm3. The flywheelis driven by a belt that wraps around its rim. If you want it to go from rest to a speedof 12 rad/s in 5 seconds, how much force do you need to apply to the belt? Round youranswer to two significant figures.


Solution: We know the dimensions and density of the flywheel. From these, we cancalculate m, and then I. We know ω0 = 0, ω, and t; from these, we can calculate α.With that information, we can calculate τ and then F .

m = ρV = πr2zρ

I =mr2

2=πr4zρ

2

ω = ω0 + αt = αt ⇒ α =ω

tτ = Fr = Iα

⇒ F =Iα

r=πr3zρω

2t=π(0.70 m)3(0.08 m)(7500 kg/m3)(12 rad/s)

2(5 s)= 780 N

506

Problem 1444. A flywheel can be regarded as a cylindrical shell with an inside radiusof 50 cm, an outside radius of 60 cm, a thickness (length) of z = 7.5 cm, and a densityof ρ = 7.5 g/cm3. It is mounted on a shaft with radius 1.5 cm. (The moment of inertiaof the shaft, and of the wheel’s spokes, can be ignored.) A belt around the shaft drivesthe flywheel. How much force do you need to apply to the belt if you want the wheel,starting from rest, to reach a speed of 1.2 rad/s by the time it has made one completerevolution? Round your answer to two significant figures.


Solution: We know the dimensions and density of the flywheel; from these we cancalculate m and then I. We know ω0 = 0, ω, and ∆θ; from these, we can calculate α.That will allow us to calculate τ ; and since we know the radius r of the shaft, we cancalculate F . We’ll use Ri and Ro for the inner and outer radii of the flywheel.

m = ρV = πρz(R2o −R2

i )

I =m(R2

o +R2i )

2=πρz(R4

o −R4i )

2

2α∆θ = ω2 − ω20 = ω2 ⇒ α =

ω2

2∆θ


r=πρzω2(R4

o −R4i )

4r∆θ

=π(7500 kg/m3)(0.075 m)(1.2 rad/s)2[(0.60 m)4 − (0.50 m)4]

4(0.015 m)(2π rad)= 450 N

507

Problem 1445. (Similarity Problem) A solid sphere, solid cylinder, thin-walled hol-low cylinder, and thin-walled hollow sphere all having the same mass M and radius Rare released from rest from the same height H on a ramp. Assuming that the ramp issmooth and each of the four objects rolls without slipping, which of these objects will bethe first one to reach the bottom of the hill?

*(a) solid sphere (b) solid cylinder(c) thin-walled hollow cylinder (d) thin-walled hollow sphere(e) None of these

Solution: Take the y-axis pointing upward and the origin y = 0 to be at the bottomof the ramp. Since each object has the same mass M and initial height y = H, theyhave the same initial potential energy Ui = MgH. Since they also start from rest, theyall have the same initial kinetic energy Ki = 0. Thus, the initial mechanical energy isMEi = MgH.

At each point along the ramp they all have the same potential energy. When the objectis y units above ground level its potential energy is Uf (y) = Mgy. Notice that when theyreach ground level that the potential energy is zero. The final kinetic energy at heighty above the ground is the sum of the kinetic energy due to translation and rotation:Kf (y) = 1

2Mv2

cm + 12Icmω

2. When an object with radius R has rotated through onecomplete revolution (2π radians), it has rolled a distance equal to its circumference(2πR). Thus the distance travelled during any time interval is R times the angularvelocity. It follows that vcm = Rω. Substituting ω = vcm/R in the expression for kineticenergy gives

Kf (y) =1

2Mv2

cm +1

2Icm

v2cm

R2=

1

2Mv2

cm

(1 +

IcmMR2

).

We now make an observation about the moment of inertia about the center of mass ofeach of the four objects. They all have the structure: Icm = βMR2, where 0 < β ≤ 1.Thus the final mechanical energy can be expressed as

MEf = Mgy +1

2Mv2

cm (1 + β) .

Ignoring any loss of energy due to rolling friction we can apply the conservation ofmechanical energy to arrive at the equation

MgH = MEi = MEf = Mgy +1

2Mv2

cm (1 + β)

−Mgy−−−−−−−→ Mg(H − y) =1

2Mv2

cm (1 + β)

× 2M

(1+β)−1

−−−−−−−−−−−→ 2g(H − y) (1 + β)−1 = v2cm

√−−−−−−→ vcm =

√2g(H − y)√

1 + β(13.1)

Notice that the final velocity depends on the parameters H, g, y, and β. This is anamazing result; it says that the final velocity is independent of the mass M and theradius R!

508

If we compare any two objects at a time, then we can treat them like two similar ex-periments with common parameters: H, g, y. We wish to find the similarity betweenthe two different velocities. Treating the above equation as the governing equation andput all of the common parameters on the right-hand side of the equation and all of theexperiment-dependent parameters on the left we arrive at

×√

1+β−−−−−−−−→ vcm√

1 + β =√

2g(H − y)

Comparing two similar experiments (objects) with β1 < β2 and using the above equationwith the common parameters as our link we have

vcm,1√

1 + β1 =√

2g(H − y) = vcm,2√

1 + β2

⇒ vcm,1vcm,2

=

√1 + β2√1 + β1

(13.2)

Since β1 < β2 ⇒√

1 + β1 <√

1 + β2, it follows from equation (13.2) that vcm,1 > vcm,2.

Define the following subscripts:ss = solid spheresc = solid cylinderhc = thin-walled hollow cylinderhs = hollow sphere.Then βss = 2

5< βsc = 1

2< βhs = 2

3< βhc = 1 ⇒ vcm,ss > vcm,sc > vcm,hs > vcm,hc.

Thus, the solid sphere reaches the bottom of the hill first.

509

Part V

Work, Energy, and Momentum

510

14 Work and mechanical energy

14.1 Work

Problem 1446. A rocket is fired upward. As it rises, the work done by gravity on therocket is:


Solution: The rocket is moving upward; gravity is pulling downward. Since the forceis in the opposite direction of the motion, the work done by the force is negative.

Problem 1447. A piano is dropped from a high window. As it falls, the work done bygravity on the piano is:


Solution: The piano moves downward; gravity is pulling downward. Since the forceand the motion are in the same direction, the work done by the force is positive.

Problem 1448. An ice skater is gliding across the frictionless surface of a frozen pond.As he crosses the pond, the work done by gravity on the skater is:


Solution: The force of gravity is pulling straight down. The skater is moving hori-zontally. Since the force is perpendicular to the motion, the work done by the force iszero.

Problem 1449. A pickup is towing a car along a level road. The two are connected bya horizontal tow-strap, which has a tension of 4000 N. How much work does the pickupdo in towing the car a distance of 100 m in a time of 10 s?

(a) 2× 104 J (b) 2× 105 J*(c) 4× 105 J (d) Not enough information: need car’s mass(e) None of these

Solution: The force and the motion are in exactly the same direction, so the angle θbetween force and motion is zero. Hence

W = Fs cos θ = Fs = (4000 N)(100 m) = 4× 105 J

The time it takes to move the car is irrelevant to the problem.

511

Problem 1450. You are pushing a shopping cart across a level parking lot. The carthas a mass of 10 kg. You push it with a horizontal force of 100 N to move it at a constantspeed of 4 m/s. How much work do you do on the cart?

(a) 400 J (b) 800 J(c) 1000 J *(d) Not enough information: need the distance(e) None of these

Solution: The force and the motion are in exactly the same direction, so the angle θbetween force and motion is zero. Hence W = Fs cos θ = Fs. We know F = 100 N, butwe don’t know the distance s, and we have no way of finding it out from the informationwe’re given in the problem. Hence: (d)

Problem 1451. A filing cabinet has a mass of 69 kg. You push on it horizontally inorder to move it at a constant velocity. The coefficient of kinetic friction between thecabinet and the floor is 0.77. How much work do you do in moving the cabinet 0.85 m?Round your answer to two significant figures.

(a) 45 J (b) 59 J*(c) 440 J (d) 520 J(e) None of these

Solution: To move the cabinet at a constant velocity, the force with which you pushit has to equal the force of kinetic friction. The work done is this force, times the distancethat you push the cabinet:

W = fk · s = mgµks = (69 kg)(9.8m/s2)(0.77)(0.85 m) = 440 J

Problem 1452. A crate full of pineapples has a mass of 170 kg. A worker pusheshorizontally on the crate to move it 12 m across the floor at a constant speed. Thecoefficient of kinetic friction between the sliding crate and the floor is 1.12. How muchwork does the worker do on the crate? Round your answer to the nearest 100 J.

(a) 900 J (b) 2300 J(c) 20,000 J *(d) 22,400 J(e) None of these

Solution: To overcome friction, the worker has to push with a horizontal force ofF = mgµk. Exerting this force over a horizontal distance of s does work:

W = Fs = mgµks = (170 kg)(9.8 m/s2)(1.12)(12 m) = 22, 400 J

512

Problem 1453. A sidewalk runs parallel to a street. A skateboarder is being toweddown the sidewalk by a rope attached to the back of a car on the street. The rope makesan angle of 23◦ to the direction in which the car and the skateboarder are moving. Thetension in the rope is 185 N. How much work does the car do in towing the skateboarder78 m down the sidewalk? Round your answer to the nearest 100 J.

(a) 5600 J *(b) 13,300 J(c) 14,400 J (d) 15,700 J(e) None of these

Solution: The component of the tension in the direction of the sidewalk is T cos 23◦.The work is this parallel component of force, times the distance through which it acts:

W = (T cos 23◦)s = (185 N)(78 m) cos 23◦ = 13, 300 J

Problem 1454. A road runs beside and parallel to a straight stretch of railroad track.A train robber attaches his pickup truck to a boxcar, using a horizontal rope that makesan angle of 17◦ to the tracks, and tows the boxcar 1200 m down the tracks. If thetension in the rope was 7400 N, how much work was done on the boxcar? Give youranswer in scientific notation, rounded to two significant figures.

(a) 2.4× 106 J (b) 2.6× 106 J*(c) 8.5× 106 J (d) 9.3× 106 J(e) None of these

��

��

17◦

Solution: The component of force parallel to the track is T cos θ = (7400 N) cos 17◦.The work done is this component of force times the distance that the boxcar was moved:

W = Tx cos θ = (7400 N)(1200 m) cos 17◦ = 8.5× 106 N·m = 8.5× 106 J

Problem 1455. You are taking your German shepherd for a walk in Reid Park whenshe decides to chase a Black-crowned Night-Heron by the pond. You pull back with aforce of T on her leash, which makes an angle of θ with the horizontal. However, the dogis stronger than you are, and pulls you forward a distance of x. How much work do youdo on the dog?

(a) Tx cos θ *(b) −Tx cos θ(c) Tx sin θ (d) −Tx sin θ(e) None of these

Solution: The dog is moving horizontally forward; the force is directed backwardat an angle of θ to the horizontal. Hence the component of force parallel to the dog’sdisplacement is −T cos θ. Since the dog moves a distance of x against this force, the workdone is −Tx cos θ.

513

Problem 1456. A case of ball bearings has a mass of 145 kg. A worker pushes horizon-tally on the case to move it 4.2 m across the floor at a constant speed. The coefficient offriction between the sliding case and the floor is 0.58. How much work does the workerdo on the case? Round your answer to two significant figures.

(a) 180 J (b) 350 J(c) 1700 J *(d) 3500 J(e) None of these

Solution: To overcome friction, the worker has to push with a horizontal force ofF = mgµ. Exerting this force over a horizontal distance of s does work:

W = Fs = mgµs = (145 kg)(9.8 m/s2)(0.58)(4.2 m) = 3500 J

14.2 Power

Problem 1457. You are pushing a shopping cart with a mass of 23 kg across a parkinglot. The forces of resistance total 52 N. How much power is necessary to push the cartat 1.8 m/s? Round your answer to two significant figures.

(a) 47 W *(b) 94 W(c) 460 W (d) 920 W(e) None of these

Solution: Use the equation for power as a product of force times velocity: P = Fv.Then P = Fpush ·vcart = (52N)(1.8m/s) = 94 W. To find Fpush, draw a free-body diagramand apply Newton’s second law, along with the fact that the cart is moving at a constantspeed (so it’s not accelerating). That will give you: Fpush = fk.

Problem 1458. (Power) How much power, in kilowatts, must be developed by theelectric motor of a 1600-kg car moving at 26 m/s on a level road if the forces of resistancetotal 720 N? Round your answer to the nearest tenth of a kilowatt.

(a) 95.2 kW (b) 36.3 kW(c) 133.2 kW *(d) 18.7 kW(e) None of these

Solution: Use the equation for power as a product of force times velocity: P = Fv.Then P = Fmotor · vcar = (720N)(26m/s) = 18.7 kW. To find Fmotor, draw a free-bodydiagram and apply Newton’s second law together with the fact that the car is moving ata constant speed (so it’s not accelerating) to arrive at Fmotor = fk. See my solutions onthe web for the details.

514

14.3 Kinetic energy (K.E.)

Problem 1459. A model airplane with a mass of 6 kg is flying at a speed of 10 m/s.What is the airplane’s kinetic energy?


Solution: K = 12mv2 = 1

2(6 kg)(10m/s)2 = 300 J

Problem 1460. A skier with a mass of 80 kg is sliding down a slope at 20 m/s. Whatis the skier’s kinetic energy?

(a) 3.2× 103 J (b) 6.4× 103 J*(c) 1.6× 104 J (d) 3.2× 104 J(e) None of these


2(80 kg)(20 m/s)2 = 1.6× 104 J

Problem 1461. A suitcase with a mass of 20 kg falls out of an airplane flying at a heightof 500 m. Just before it hits the ground, the suitcase is falling with a speed of 80 m/s.What is the suitcase’s kinetic energy at this point?

(a) 1.6× 103 J (b) 5× 104 J*(c) 6.4× 104 J (d) 1× 105 J(e) None of these


2(20 kg)(80 m/s)2 = 6.4× 104 J

The height from which the suitcase falls is irrelevant to the problem. If you’re feelingadvanced, you can calculate the potential energy of the suitcase at that height. It’s lessthan the kinetic energy of the suitcase hitting the ground, so the suitcase lost mechanicalenergy (probably to air resistance) as it fell.

Problem 1462. (Similarity Problem) A car with a mass of M and a truck with amass of 9M are moving at the same speed V . If K1 is the kinetic energy of the car andK2 is the kinetic energy of the truck, which of the following equations is true?

(a) K2 = K1 (b) K2 = 3K1

*(c) K2 = 9K1 (d) K2 = 81K1

(e) None of these

Solution: Since K = 12mv2:

K1 =MV 2

2K2 =

(9M)V 2

2= 9K1

515

Problem 1463. (Similarity Problem) When a car is moving at speed v1, its kineticenergy is K1. When it is moving at speed v2, its kinetic energy is K2 = 2K1. What isthe relationship between v2 and v1?

*(a) v2 =√

2 v1 (b) v2 = 2v1

(c) v2 = 2√

2 v1 (d) v2 = 4v1

(e) None of these

Solution: Use K = 12mv2. In this problem, the mass doesn’t change; the speed and

kinetic energy do. We separate the variables that don’t change from the ones that do:

m

2=K

v2=K1

v21

=K2

v22

⇒ v22 =

K2

K1

v21 = 2v2

1 ⇒ v2 =√

2 v1

Problem 1464. (Similarity Problem) When a crate is pushed at speed vc across asmooth floor with a low coefficient of friction, it has kinetic energy K1. When the samecrate is pushed at the same speed across a rough floor with a higher coefficient of friction,its kinetic energy is K2. What is the relationship between K1 and K2?

(a) K1 > K2 *(b) K1 = K2

(c) K1 < K2 (d) Not enough information: need mass of crate(e) None of these

Solution: The kinetic energy depends only on the mass of the crate and its speed.Since these are the same in both situations, the kinetic energy is the same.

Problem 1465. You are testing your new potato gun at the ballistics lab. A potatofired at a speed of v has kinetic energy of KE. What is the mass of the potato?

(a)

√2KE

m(b)

√KE

2m

(c)

√m

2KE(d)

√KE

2m(e) None of these

Solution: Use the formula for kinetic energy:

K =1

2mv2 ⇒ v2 =

2K

m⇒ v =

√2K

m=

[2 · 60 J

0.37 kg

]1/2

= 18 m/s

516

Problem 1466. You have just acquired a Civil War cannon, and you want to know itsmuzzle velocity. You fire a cannonball with a mass of 7.3 kg into a ballistic pendulum,and discover that the kinetic energy of the cannonball was 370,000 J. What was thespeed of the cannonball? Round your answer to the nearest 10 m/s.


Solution: Since K = 12mv2, solving for v gives:

v =

√2K

m=

√2 · 370, 000 J

7.3 kg= 320 m/s

Problem 1467. (Impulse and K.E.) Two railroad cars are at rest on two horizontalfrictionless tracks. One car is three times as heavy as the other. Each car is pushed withthe same horizontal force of F for the same time T . At the end of this time, the kineticenergy of the lighter car is K. What is the kinetic energy of the heavier car?

(a) 3K (b)√

3K

(c) K/√

3 *(d) K/3(e) None of these

Solution: Let the mass of the light car be M . We’ll construct a table with thekinematic properties of the two cars:

Light car Heavy carMass M 3MAcceleration = force/mass F/M F/3MSpeed = acceleration × time FT/M FT/3M

Kinetic energy = 12×mass× speed2 K =

F 2T 2

2M

F 2T 2

6M=K

3

517

14.4 Potential energy (P.E.)

Problem 1468. Which of the following statements is true?

(a) If a force is directed downward, it does negative work.(b) The work done by a force on an object equals the increase in the object’s potential

energy.(c) If a force has moved an object around a circle and back to its starting point,

then the force has done no work on the object.*(d) If an object is moving in the positive direction, and if a force is applied to it in

the negative direction, then the force does negative work on the object.

Solution: We can rule out (a): what matters is not whether the force is directedupward or downward or sideways, but whether it has a positive or negative componentin the direction of the object’s motion. We can rule out (b), since the work done onan object might have the effect of increasing its kinetic energy without changing thepotential energy (e.g. accelerating a glider along a horizontal air track). We can ruleout (c) for much the same reason: if the object is moving faster when it returns to itsstarting point than it was when it started, it has greater kinetic energy, so work hasbeen done on it. Only (d) is true.

Problem 1469. You move 56 kg of potatoes from the floor to a shelf 1.9 m high. Byhow much did you increase the gravitational potential energy of the potatoes? Roundyour answer to the nearest joule.


Solution: The gravitational potential of the potatoes is

Ugrav = mgy = (56 kg)(9.8 m/s2)(1.9 m) = 1043 J

Problem 1470. A case of rat food has a mass of 10.3 kg. To keep rats from eating it,you move it from the floor to a shelf 2.2 m high. By how much did you increase thepotential energy of the case? Round your answer to the nearest 10 J.


Solution: Let y = 0 be the level of the floor. Then the initial potential energy of thecase is zero. The potential energy of the case on the shelf is

Ugrav = mgy = (10.3 kg)(9.8m/s2)(2.2 m) = 220 J

518

Problem 1471. A cat has a mass of 4.3 kg. It jumps up from the floor onto a countertop1.2 m high. How much did the cat’s gravitational potential energy increase in going fromthe floor to the countertop? Round your answer to the nearest joule.


Solution: The gravitational potential energy of an object with mass m at a height habove the floor is

Ugrav = mgh = (4.3 kg)(9.8 m/s2)(1.2 m) = 51 J

Problem 1472. A spring can store 36 J of energy when compressed by 0.12 m. Whatis the spring constant of the spring?


Solution: Since the energy stored in a spring is U = 12kx2,

k =2U

x2=

2 · 36 J

(0.12 m)2= 5000 N/m

We were given x in meters, so we didn’t have to convert it.

Problem 1473. You move 41 kg of physics books from the floor to a shelf 2.1 m high.By how much did you increase the gravitational potential energy of the books? Roundyour answer to the nearest joule.


Solution: The gravitational potential of the books is

Ugrav = mgy = (41 kg)(9.8 m/s2)(2.1 m) = 844 J

519

14.5 Work-Mechanical-Energy Theorem

Work-Mechanical-Energy Theorem: Wext = ∆ME.

14.5.1 Work-Kinetic-Energy Theorem

Special case 1: Work-Kinetic-Energy Theorem: Wext = ∆KE.

Problem 1474. A bartender slides a glass of beer with a mass of 480 g along the top ofa level bar. The coefficient of kinetic friction between the glass and the bar is µk = 0.13.When the glass leaves the bartender’s hand, it is moving at a speed of v0 = 2.1 m/s. Howfar does the glass travel before coming to a stop? Round your answer to the nearest 10cm.(a) 140 cm (b) 160 cm*(c) 170 cm (d) 190 cm(e) None of these

Solution: We know the mass m, the coefficient of friction µk, the initial velocity vi,and the final velocity vf = 0. We want the stopping distance ∆x.

Apply the work-energy principle: Wnet = ∆K = Kf −Ki. The net force after the glassis released is friction, so

Wnet = Fnet · (distance) = −µkN∆x

Since the top of the bar is level, N = mg. Hence Wnet = −µkmg∆x.

Since vf = 0, Kf = 12mv2

f = 0. Hence Wnet = −Ki = −12mv2

0. Hence

− µkmg∆x = −1

2mv2

0

÷(−m)−−−−−−−→ µkg∆x =v2

0

2÷µkg−−−−−−→ ∆x =

v20

2µkg=

(2.1 m/s)2

2(0.13)(9.8 m/s2)= 1.7 m = 170 cm

520

Problem 1475. A skier slides down a slope and onto a frozen lake, where he slowsdown and comes to a halt. When he first reaches the lake, he is moving at 24 m/s. Thecoefficient of kinetic friction between his skis and the ice is 0.13. How far does he travelacross the lake before stopping? Round your answer to the nearest 10 m.


Solution: Use the work-energy principle. If the skier’s mass is m and his initial speedis vi, then his initial kinetic energy is Ki = 1

2mv2

i . He will slide until the negative workdone by friction equals Ki. Since he is on a level surface, N = mg; so the force of kineticfriction is Fk = µkmg. If he slides a total distance of s, then the negative work done byfriction is Fks = µkmgs. Hence:

µkmgs = Ki =1

2mv2

i ⇒ s =mv2

i

2µkmg=

v2i

2µkg=

(24 m/s)2

2(0.13)(9.8 m/s2)= 230 m

Problem 1476. A skier slides down a small slope and onto a frozen lake, where heslows down and comes to a halt. When he first reaches the lake, he is moving at 10 m/s.He travels 50 meters across the lake before coming to a stop. What is the coefficientof kinetic friction between his skis and the ice? Round your answer to the nearesthundredth and approximate g by 10 m/s2.

(a) .15 *(b) .10(c) .29 (d) .19(e) None of these

Solution: Use the work-energy principle. If the skier’s mass is m and his initial speedis vi, then his initial kinetic energy is Ki = 1

2mv2

i . He will slide until the negative workdone by friction equals Ki. Since he is on a level surface, the normal force is equal toweight N = mg; so the force of kinetic friction is fk = µkmg. If he slides a total distanceof s, then the negative work done by friction is fks = µkmgs. Hence:

µkmgs = Wfriction = Ki =1

2mv2

i

÷mgs−−−−−−−→ µk =mv2

i

2smg=

v2i

2sg

evaluate−−−−−−−−→ µk =(10 m/s)2

2(50 m)(10 m/s2)= .10

521

Problem 1477. A car has a weight of 13,500 N. The coefficient of kinetic frictionbetween its wheels and the wet highway is 0.47. The car is travelling at 18.4 m/s whenthe driver locks up the brakes. How far does the car travel before it comes to a completestop? Round your answer to the nearest meter.


Solution:Step 1: Draw a picture and a free-body diagram (not shown here).

Step 2: Write down what you’re given and what you want.

given:

w = mg = 13, 500N (weight = mass × gravity)

µk = 0.47 (coefficient of kinetic friction-car is moving)

vi = 18.4m/s (initial velocity)

vf = 0m/s (final velocity)

want: ∆x (the stopping distance)

Note: Time is not given, nor is it needed in this problem.

Step 3: Use the appropriate fundamental kinematic equation and solve for the unknown.Alternatively, we can use the work-kinetic-energy theorem, which is what we’ll do.We can determine the mass m of the car from its weight w = mg, where g = 9.8 m/s2.We can determine the kinetic energy K from the mass and the speed v. We can determinethe force of kinetic friction fk from the weight and the coefficient of friction µk. Thenthe force of friction times the stopping distance x equals the kinetic energy.

m =w

g=

13, 500 N

9.8 m/s2

Ki =1

2mv2

i =1

2

w

gv2i =

wv2i

2gand Kf = 0 (final K.E. - no motion)

fk = µw = (0.47)(13, 500 N)

∆K = Kf−Ki = −fk·∆x ⇒ ∆x =Ki

fk=wv2/2g

µw=

v2

2µg=

(18.4 m/s)2

2(4.7)(9.8 m/s2)= 37 m

Notice that the weight of the car cancels out. A heavy car would require more force tostop it than a light one; but it would have a greater frictional force.

522

Problem 1478. (Similarity Problem) You are driving at a speed of V on a straightand level road when you see a deer in front of you and lock up the brakes. You travel adistance D before coming to a complete stop. If your original speed had been 2V whenyou hit the brakes, what distance would you have travelled before stopping?

(a) D (b)√

2D(c) 2D *(d) 4D(e) None of these

Solution: We will use the work-energy principle. Your car will move until the negativework done by friction equals its initial kinetic energy. If m is the mass of the car, v is itsinitial speed, Fk is the force of kinetic friction, and d is the stopping distance, then

1

2mv2 = Fkd ⇒ d =

mv2

2Fk

The mass of the car and the force of friction don’t change between trials. In the firsttrial, your initial speed is V , so

D =mV 2

2Fk

In the second trial, your initial speed is 2V , so

d =m(2V )2

2Fk=

4mV 2

2Fk= 4D

14.5.2 Work-Potential-Energy Theorem

Special case 2: Work-Potential-Energy Theorem: Wext = ∆PE.

Problem 1479. (The potential energy in a spring) A giant spring with springconstant 50 N/m is 10 m long in its unstretched position. Suppose a force acts on thespring and stretches it to 12 m long. How much energy is stored in the spring?

(a) 50 J (b) 100 J(c) 150 J (d) 200 J(e) None of these

Solution: Step 1: Write down what you’re given and what you want.

Given: Lu = 10 m, s = 12 m, and k = 50 N/m.Want: Uspring

The equation for the potential energy stored in a spring is:

Uspring(x) =1

2kx2 , where x = Ls−Lu is the displacement. Here x = 12 m− 10 m = 2 m.

Uspring(x = 2) =1

250 22 J = 100 J

523

Problem 1480. A spring is hanging from the ceiling in the physics lab. When thereis no weight hanging from it, the spring is 42 cm long; it has a spring constant of 8300N/m. You hang a lead block with a mass of 87 kg from the spring, stretching it. Whatis the length of the spring with the weight hanging from it? Round your answer to thenearest centimeter.(a) 10 cm (b) 21 cm(c) 32 cm *(d) 52 cm(e) None of these

Solution: We know the unstretched length of the spring Lu = 42 cm = 0.42 m; thespring constant k = 8300 N/m; and the mass m = 87 kg.

If y is the change in length of the spring, we want the stretched length: Ls = Lu + y.

Use the equation |Fspring| = ky to find y. We assume that the system is static, so thereis no acceleration; so the two forces on the weight match: |Fspring| = mg = ky. Divideby k:

y =mg

k⇒ Ls = Lu +

mg

k= 0.42 m +

(87 kg)(9.8 m/s2)

8300 N/m= 0.52 m = 52 cm

Problem 1481. An unstretched spring is 15.0 cm long. If you hang an object with amass of 8.8 kg from it, it stretches to a length of 19.2 cm. What is the spring constantof the spring? Round your answer to two significant figures.

(a) 4.5 N/m (b) 22 N/m(c) 210 N/m *(d) 2100 N/m(e) None of these

Solution: From Hooke’s law, F = kx. The force acting on the spring is F = mg, wherem = 8.8 kg. The value of x is the difference between the stretched and the unstretchedlength of the spring: x = 0.192 m− 0.150 m = 0.042 m. (The answers are given in N/m,so we have to convert centimeters to meters.) Hence

k =F

x=mg

x=

(8.8 kg)(9.8 m/s2)

0.042 m= 2100 N/m

524

Problem 1482. An unstretched spring has a length of 22.8 cm. When an object with amass of 12.0 kg is hung from it, it stretches to a length of 35.5 cm. What is the springconstant of the spring? Round your answer to the nearest N/m.




x=

(12.0 kg)(9.8 m/s2)

0.127 m= 926 N/m

Problem 1483. A box has a mass of 22.2 kg. It is sitting on an icy sidewalk, whichconstitutes a horizontal frictionless surface. A spring with a spring constant of 103 N/mis attached to the box and pulled horizontally, so that the box accelerates at 1.31 m/s2.How much does the spring stretch? Round your answer to the nearest 0.01 m.


Solution: The spring is pulled horizontally and there is no friction, so the forceexerted by the spring is the mass of the box times its acceleration: F = ma. By Hooke’slaw, F = kx. Hence

ma = kx ⇒ x =ma

k=

(22.2 kg)(1.31 m/s2)

103 N/m= 0.28 m

Problem 1484. A spring has a spring constant of 1700 N/m. You pull on it with a forcethat increases to 120 N until it stops stretching. How much work have you done on thespring? Round your answer to two significant digits.

*(a) 4.2 J (b) 8.5 J(c) 14 J (d) 60 J(e) None of these

Solution: When you pull on a spring with a force of magnitude F , the amount bywhich it stretches is given by: F = kx. Solving this for x gives: x = F/k. The work thatit takes to stretch a spring by x is:

W =1

2kx2 =

k

2

(F

k

)2

=F 2

2k=

(120 N)2

2 · 1700N/m= 4.2 J

525

Problem 1485. A spring has a spring constant of 680 N/m. You pull on it with a forcethat increases to 170 N until it stops stretching. How much work have you done on thespring? Round your answer to two significant digits.

*(a) 21 J (b) 43 J(c) 5400 J (d) 11000 J(e) None of these

Solution: When you pull on a spring with a force of magnitude F , the amount bywhich it stretches is given by: F = kx. Solving this for x gives: x = F/k. The work thatit takes to stretch a spring by x is:

W =1

2kx2 =

k

2

(F

k

)2

=F 2

2k=

(170 N)2

2 · 680N/m= 21 J

Problem 1486. With no force applied to it, a spring is 24 cm long. You pull on it witha force of 330 N, stretching it to a length of 30 cm. How much work have you done instretching the spring?

*(a) 9.9 J (b) 49.5 J(c) 990 J (d) 4950 J(e) None of these

Solution: The work that you have done is equal to the potential energy stored in thespring. We can find that using the formula: U = 1

2kx2. Here x is the change in length

from equilibrium: x = 30 cm − 24 cm = 6 cm = 0.06 m. We can calculate k using thespring formula: F = kx ⇒ k = F/x. Then the potential energy is:

U =kx2

2=Fx2

2x=Fx

2=

(330 N)(0.06 m)

2= 9.9 J

Problem 1487. An unstretched spring is 0.24 m long. When a force that increases to190 N is applied, it stretches until its total length is 0.30 m. How much work does it taketo stretch the spring to this length?

*(a) 5.7 J (b) 11.4 J(c) 28.5 J (d) 57 J(e) None of these

Solution: The change in the spring’s length is x = 0.30 m − 0.24 m = 0.06 m. ByHooke’s law, F = kx, so k = F/x. The work that it takes to stretch the spring is

W =kx2

2=Fx

2=

(190 N)(0.06 m)

2= 5.7 J

The spring’s lengths were given in meters, so we didn’t have to convert.

526

Problem 1488. An unstretched spring has a length of 23.8 cm. When an object with amass of 18.2 kg is hung from it, it stretches to a length of 32.2 cm. What is the springconstant of the spring? Round your answer to the nearest N/m.

(a) 217 N/m (b) 554 N/m*(c) 2123 N/m (d) 9435 N/m(e) None of these



x=

(18.2 kg)(9.8 m/s2)

0.084 m= 2123 N/m

Problem 1489. With no force applied to it, a spring hanging from the ceiling is 24 cmlong. You hang a weight of 330 N from it, stretching it to a length of 30 cm. How muchwork was done by gravity in stretching the spring?

(a) 9.9 J (b) 49.5 J(c) 990 J (d) 4950 J(e) None of these

Solution: By applying a force (the weight of the object) through a distance x (thedisplacement) gravity has done work on the system. The work done on the spring-mass system is equal to the potential energy stored in the spring. We can find thatusing the formula: U = 1

2kx2. Here x is the change in length from equilibrium: x =

Ls−Lu = 30 cm−24 cm = 6 cm = 0.06 m. We can calculate k using the spring formula:F = kx ⇒ k = F/x. Then the potential energy is:

U =kx2

2=Fx2

2x=Fx

2=

(330 N)(0.06 m)

2= 9.9 J

527

Problem 1490. (Derive Problem-Requires Calculus)[Approximating the strength of gravity as a function of height] The magnitudeof the gravitational attraction between two particles of mass m1 and m2 is given by

|Fg(r)| = Gm1m2

r2,

where r is the distance between the two particles and the force that each particle exertson the other is directed along the radial direction between the two particles, and G isthe universal gravitational constant.

(a) What is the potential energy function U = Ug(r), if we take U0 = limr→∞

Ug(r) = 0?

That is, we take our reference point out at r =∞.

(b) How much work Wg must be done against the conservative gravitational force toincrease the separation distance from r = R to r = R + h?

(c) Let m1 = Me the mass of the earth and m2 = m the mass of a small point particle onthe earth, where we assume that m�Me. Let R = Re be the mean radius of the earthand h be the height that the particle is above the ground, where h� Re. Compute thework done against gravity in moving the object from the earth’s surface at r = Re to aheight h above the ground r = Re + h. Denote this work by Wg. Express your answer

in terms of the parameter g =GMe

R2e

, and the variableh

Re

. Since Re, Me, m, and G are

fixed, the work will be a function of h. However, it will turn out to be advantageous to

think of the work as a function ofh

Re

.

(d) We are now going to repeat the calculation in part (c) except this time we aregoing to approximate the force of gravity as a constant force Fgrav = −mg, where g =GMe

R2e

is the constant magnitude of gravity that is assumed to be independent of theseparation distance between the two attracting masses. This is nothing new, when weapply Newton’s second law to trajectory problems involving gravity on the earth’s surface,such as computing the path of a cannon ball launched from the ground, we take the forceof gravity as a constant.

Compute the work Wgrav done in moving the same object of mass m in part (c) from theground at y = 0 to height y = h against the constant force of gravity Fgrav = −mg.

(e) We now want to compare the exact result for the work found in part (c) with theapproximate value of work found in part (d). What is the size of the relative errorbetween the work found in part (c) and (d)? Hint: W.L.O.G. we can take Wgrav as ourreference measure. Show that the relative error is

Wg −Wgrav

Wgrav

= O(h

Re

).

(f) The mean radius of the earth is approximately 6370 km. Use the relative error thatyou find to predict how high you can go above the earth’s surface and safely ignore thefact that gravity is changing with height when you compute the work done in lifting anobject from the earth’s surface.

528

Figure 7: Earth-mass system.

Solution: See hand-written solutions.

529

14.6 Conservation of mechanical energy (M.E.)

Problem 1491. Planet X has no air, hence no air resistance. An astronaut standingatop a cliff on Planet X drops a rock down into the lava lake below. If the gravitationalpotential energy of the rock just before being dropped is Ugrav, and its kinetic energy justas it hits the lava is K, what is the relationship between Ugrav and K? Use the level ofthe lava lake as y = 0.

(a) Ugrav < K*(b) Ugrav = K(c) Ugrav > K(d) Not enough information: it depends on Planet X’s gravity gX .

Solution: Since there is no air resistance, we assume that mechanical energy isconserved: Ui +Ki = Uf +Kf . Initially, just as the astronaut drops the rock, it has zerovelocity; so Ki = 0. Just as the rock hits the lava, it is at y = 0, so Uf = 0. Hence

Ugrav = Ui = Kf = K

Problem 1492. Two identical stone blocks are transported from the bottom to the topof a cliff. The first block is fastened to a rope and pulled straight up through the air tothe top. The second block is pulled up a frictionless ramp to the top of the cliff. If thework done on the first block is W1 and the work done on the second block is W2, whatis the relationship between W1 and W2?

(a) W1 < W2 *(b) W1 = W2

(c) W1 > W2 (d) Not enough information: any of these could be true

Solution: Since there is no friction, conservation of energy applies; so the work doneon each block is equal to the change in its potential energy. Since the blocks are identical,they have the same mass; and they are raised by the same height. Hence the potentialenergy change, and therefore the work, is the same for both blocks.

Problem 1493. Two frictionless inclined planes have the same vertical height Y . Plane 1makes an angle of 60◦ to the horizontal; plane 2 makes an angle of 30◦ to the horizontal.Two objects with the same mass M are released from the tops of the planes to slidedown. The object that slides down plane 1 has speed v1 when it reaches the bottom; theobject that slides down plane 2 has speed v2 when it reaches the bottom. What is therelationship between v1 and v2?

*(a) v1 = v2 (b) v1 = 2v2

(c) v1 sin 30◦ = v2 sin 60◦ (d) v1 cos 60◦ = v2 cos 30◦

(e) None of these

Solution: The key word is “frictionless”. Since there are no nonconservative forces,mechanical energy must be conserved. Initially, both objects have zero kinetic energyand the same potential energy (since they have the same mass M and they’re at thesame height Y ). When they reach the bottom of their planes, they still have the samemechanical energy; since they have height zero, their potential energy is zero; so theirmechanical energy is all kinetic, and it’s the same for both. Since they have the samekinetic energy and the same mass, their speeds must be the same.

530

Problem 1494. Ricky, the 5 kg snowboarding rac-coon, is on his snowboard atop a frictionless ice-coveredquarter-pipe with radius 3 meters (see figure at right). Ifhe starts from rest at the top of the quarter-pipe, whatis his speed at the bottom of the quarter-pipe? For easeof calculation, assume that g = 10 m/s2. Round youranswer to the nearest 0.1 m/s.


x� Raccoon: 5 kg

3 m

3 m

Solution:We are given: m = 5 Kg, R = 3 m, vi = 0 m/s, and g = 10 m/s2.We want: vf =?We assume: no friction, hence the mechanical energy of the system is conserved.

Since the system, Ricky and the earth, has no external forces on it, such as friction, wecan apply the conservation of mechanical energy: MEi = MEf . Now, the initial potentialdue to gravity is Ugrav = mgh = mgR and the initial kinetic energy is Ki = 0 since Rickyis at rest. The final P.E. is zero, since Ricky is now at ground level h = 0. The final K.E.is Kf = 1

2mv2

f . Using the definition of mechanical energy: ME = PE +KE gives

mgR = PEi +KEi = MEi = MEf = PEf +KEf =1

2mv2

f

÷ 12m

−−−−−−→ v2f = 2Rg

√−−−−−−→ vf =

√2Rg =

[(2)(3 m)(10 m/s2)

]1/2= 7.7 m/s

Notice that since R = h, this is the same speed that Ricky would have if he had jumpedstraight down (go look back at our 1-D kinematic problems). The only difference is thedirection of his speed would have been different. If he had jumped straight down, thenhis velocity would also have been straight down, and into the ground would go Ricky; butthe quarter pipe redirected his motion to being tangent to the ground, while preservinghis speed. So instead of busting his head, and winding up dead; Ricky sped off to work,where he was a clerk. Were it not for conservation of mechanical energy he would havebeen late for his job, and replaced by his competitor Rob.

531

Problem 1495. The surface gravity on Planet X is gX . An astronaut standing on thesurface of the planet fires a bullet with mass m straight upward with kinetic energy K.What is the maximum height hmax reached by the bullet?

(a)

√2K

mgX*(b)

K

mgX

(c)

√2K

m(d)

K2

mgX

(e) None of these

Solution: Apply the conservation of mechanical energy. Take the initial position tobe the ground at the moment the potato is launched, and the final position to be thehighest point of the trajectory. Let y = 0 correspond to the initial position. At the initialand final points:{Ui = mgy = 0 (y = 0)

Ki = K

and{Uf = mgXhmax

Kf = 0 (ball is momentarily at rest at the top of the trajectory)

Notice that at its highest point, the potato will have zero kinetic energy, so all of itsenergy will be potential. At this height hmax,

K = mgXhmax ⇒ hmax =K

mgX

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Problem 1496. Planet X has a surface gravity of 13.6 m/s2 and no atmosphere. Anastronaut on Planet X fires a potato gun straight upward; the potato has a mass of 0.73kg, and leaves the gun with a speed of 34.3 m/s. What is the maximum height that thepotato reaches? Round your answer to the nearest meter.


Solution: Apply the conservation of mechanical energy. Take the initial position tobe the ground at the moment the potato is launched, and with the final position to bethe highest point of the trajectory. Let y = 0 correspond to the initial position. At theinitial and final points:{Ui = mgy = 0 (y = 0)

Ki = 12mv2

i = 430J

and{Uf = mgXhmax

Kf = 0 (ball is momentarily at rest at the top of the trajectory)

Notice that at its highest point, the potato will have zero kinetic energy, so all of itsenergy will be potential. At this height hmax,

Ki = mgXhmax ⇒ hmax =Ki

mgX=

430 J

(0.73 kg)(13.6 m/s2)= 43 m

533

Problem 1497. (Lab Problem) A spring cannon shoots a ball with a mass of 28 gvertically into the air from ground level. In this situation, air resistance can be ignored.If the spring is compressed by a distance of 1.2 cm, the ball reaches a maximum heightof 2.4 m. What is the spring constant of the spring in the cannon? Round your answerto the nearest 100 N/m.


Solution: We know the mass of the ball m, the compression distance y of the spring,the maximum height hmax of the ball, and the value of g. We want to know the springconstant k.

Since there’s no air resistance and no friction, we can assume that mechanical energy isconserved: Ki + Ui = Kf + Uf . We will take the initial position (subscript i) as thatof the cannon with its spring compressed, just before it is fired; and the final position(subscript f) as the system when the ball has reached its maximum height. Since theball isn’t moving in either case, Ki = Kf = 0. Hence Ui = Uf .

The initial potential energy is all in the spring: Ui = 12ky2. The final potential energy is

all gravitational: Uf = mghmax. Hence

1

2ky2 = mghmax

· (2/y2)−−−−−→ k =2mghmax

y2=

2(0.028 kg)(9.8 m/s2)(2.4 m)

(0.012 m)2= 9100 N/m

Problem 1498. (Lab Problem) A spring cannon shoots a ball with a mass of 9.6 gvertically into the air from ground level. In this situation, air resistance can be ignored.The spring is compressed by a distance of 3.4 cm below the mouth of the cannon. If theball reaches a maximum height of 96 cm above the mouth of the cannon, then what isthe spring constant of the spring in the cannon? Round your answer to the nearest 10N/m. Take y = 0 to be at the mouth of the cannon.

(a) 140 N/m (b) 150 N/m*(c) 160 N/m (d) 170 N/m(e) None of these

Solution: We know the mass of the ball m, the compression distance ∆y of thespring, the maximum height ymax of the ball, and the value of g. We want to know thespring constant k.

Since there’s no air resistance and no friction, we can assume that mechanical energy isconserved: Ki + Ui = Kf + Uf . We will take the initial position (subscript i) as thatof the cannon with its spring compressed, just before it is fired; and the final position(subscript f) as the system when the ball has reached its maximum height ymax. Sincethe ball isn’t moving in either case, Ki = Kf = 0. Hence Ui = Uf .

The initial potential energy is the energy stored in the spring plus the negative gravi-tational potential energy (because the ball is ∆y below the zero gravitational reference

534

potential line at y = 0): Ui = Ugrav + Uspring = −mg∆y + 12k(∆y)2. The final potential

energy is all gravitational: Uf = mgymax. Hence Ui = Uf becomes

−mg∆y +1

2k(∆y)2 = mgymax

+mg∆y−−−−−−→ 1

2k(∆y)2 = mgymax +mg∆y

· 2

(∆y)2

−−−−−−→ k =2mg

(∆y)2(ymax + ∆y)

evaluate−−−−−−−−→ k =

(2(0.0096 kg)(9.8 m/s2)

(0.034 m)2

)(0.96 + 0.034) m = 162 N/m ≈ 160 N/m

Note: We could have taken the zero reference line to be the top of the ball when it isin its initial position sitting on the compressed spring. In this case the maximum heightwould be htextmax = ymax + ∆y.

Problem 1499. (Lab Similarity Problem) A spring cannon is used to launch twoballs straight upward: first, a ball with mass M ; then a ball with mass 2M . The springin the cannon is compressed by the same amount in both launchings. Which of thefollowing statements is true?

(a) The two balls reach the same maximum height.(b) The larger ball will leave the spring cannon at a higher speed than the smaller

ball.*(c) The two balls leave the spring cannon with the same kinetic energy.(d) At their maximum heights, the smaller ball has greater potential energy than

the larger ball.

Solution: In both trials, the spring is compressed by the same amount; so in bothtrials, it has the same potential energy. When the balls are launched, the potentialenergy of the spring is entirely converted into kinetic energy of the ball. Thus both ballsleave the spring cannon with the same kinetic energy.

Since the balls have different masses, they leave the spring cannon with different speeds:the small ball at a greater speed than the large ball. That lets us rule out (b). At theirmaximum heights, each ball stops moving; so its initial kinetic energy is completelyconverted into gravitational potential energy. Since they started with the same kineticenergy, they must have the same potential energy at their maximum heights; so wecan rule out (d). Since they have different masses, they must reach different maximumheights in order to have the same potential energy there; that rules out (a).

535

Problem 1500. A baseball with mass M and a cannonball with mass 9M are thrownvertically upward from ground level at the same initial speed V , with no air resistance.The initial kinetic energy of the baseball is Kb; it reaches a maximum height of Yb; andat this height, its gravitational potential energy is Ub. The corresponding values for thecannonball are Kc, Yc, and Uc. Which of the following is true?

(a) Yb = 3Yc *(b) Uc = 9Ub(c) Kb = Kc (d) Kc = 81Kb

(e) None of these

Solution: Since there is no air resistance, we assume that mechanical energy isconserved. Initially at ground level, the potential energy is zero and the mechanicalenergy consists entirely of the ball’s kinetic energy: Kb = MV 2/2 and Kc = (9M)V 2/2 =9Kb. At the high point, the ball is motionless, so its kinetic energy is zero. At this point,the potential energy of each ball equals its initial kinetic energy; so Uc = Kc = 9Kb = 9Ub.

Problem 1501. (Lab Problem) A frictionless horizontal air track has a horizontalspring at one end, with a spring constant of 800 N/m. A glider with a mass of 500 gis pressed against the spring, compressing it by 4 cm. When the glider is released, thespring pushes it down the track. How fast is the glider moving when it leaves the spring?

(a) 20 cm/s (b) 40 cm/s(c) 80 cm/s *(d) 160 cm/s(e) None of these

Solution: We assume that energy is conserved; so the potential energy of the springbefore the glider is released equals the kinetic energy of the glider after it leaves thespring. Don’t forget to convert units to kilograms and meters.

Uel =1

2kx2 = K =

1

2mv2

⇒ kx2 = mv2

⇒ v2 =kx2

m

⇒ v =

√k

mx =

[800 N/m

0.5 kg

]1/2

(0.04 m) = 1.6 m/s = 160 cm/s

536

Problem 1502. (Lab Problem) A horizontal spring with spring constant k1 is at oneend of a frictionless horizontal air track. A glider is pushed against the spring, compress-ing it by a distance of x1, so that the energy stored in the spring is 100 J. After the glideris released, it reaches a speed of v1 and a kinetic energy of K1. The procedure is thenrepeated, with the spring replaced by one with spring constant k2 > k1; the new springis compressed by a distance of x2, so that the energy stored in it is once again 100 J.When the glider is released this time, it reaches a speed of v2 and a kinetic energy of K2.Which of the following is true?

(a) x1 = x2 (b) K1 < K2

*(c) v1 = v2 (d) k1x21 < k2x

22

(e) None of these

Solution: The air track is frictionless, so we assume that there are no nonconser-vative forces, which means that mechanical energy is conserved. Initially, the spring iscompressed and the glider is not moving; so the kinetic energy is zero and the mechan-ical energy equals the potential energy of the spring. After the glider has been releasedand has reached its maximum speed, there is no potential energy left in the spring, andthe mechanical energy equals the kinetic energy of the glider. In both trials, the initialpotential energy is U1 = U2 = 100 J. Hence in both trials, the final kinetic energy isK1 = K2 = 100 J. Since the mass of the glider doesn’t change from the first trial to thesecond, and since its kinetic energy is the same in both trials, its velocity must be thesame: v1 = v2.

537

Problem 1503. (Lab Problem) A frictionless horizontal air track has a spring ateither end. The spring on the left has a spring constant of kL; the one on the righthas a spring constant of kR. A glider with a mass of m is pressed against the left-handspring, compressing it by a distance of xL from its equilibrium position. The glider isthen released, so that the spring propels it rightward. It slides along the track and intothe right-hand spring. Find a formula for the maximum compression of the right-handspring xR as a function of the given parameters: xL, kL, kR, m. (The diagram followingthe answers shows the situation before the glider is released, when the left-hand springis compressed and the glider is not moving.)

*(a) xR =√

kL

kRxL (b) xR =

√kR

kLxL

(c) xR = kL

kRxL (d) xR = xL

(e) None of these

��EEEE C

CCCCC ��

m

Solution: We know the mass m of the glider, the spring constants KL and KR of theleft- and right-hand springs, and the compression xL of the left-hand spring. We want toknow the compression xR of the right-hand spring.

There is no friction, so we can assume that mechanical energy is conserved: Ui + Ki =Uf + Kf . We will take the initial situation to be that in which the left-hand spring isfully compressed, before the glider is released to move; and the final situation to be thatin which the glider has pushed the right-hand spring to its maximum compression. Sincethe glider isn’t moving in either case, Ki = Kf = 0; so Ui = Uf .

In both situations, the potential energy is all in a spring: Ui = 12kLx

2L = 1

2kRx

2R = Uf .

We can solve this for xR:

· 2−−−−→ kRx2R = kLx

2L

÷kR−−−−−−→ x2R =

kLx2L

kR√

−−−−−→ xR =

√kLx2

L

kR=

√kLkR

xL

Notice that we never needed the mass of the glider. Its only function in the problem wasto carry all the energy from one spring to the other.

538

Problem 1504. (Lab Problem) A frictionless horizontal air track has a spring ateither end. The spring on the left has a spring constant of 1800 N/m; the one on theright has a spring constant of 1300 N/m. A glider with a mass of 4.30 kg is pressedagainst the left-hand spring, compressing it by 3.9 cm. The glider is then released, sothat the spring propels it rightward. It slides along the track and into the right-handspring. What is the maximum compression of the right-hand spring? Round youranswer to the nearest 0.1 cm. (The diagram following the answers shows the situationbefore the glider is released, when the left-hand spring is compressed and the glider isnot moving.)

(a) 2.0 cm (b) 2.8 cm*(c) 4.6 cm (d) 5.4 cm(e) None of these

��EEEE C

CCCCC ��

4.30 kg

Solution: We know the mass m of the glider, the spring constants KL and KR of theleft- and right-hand springs, and the compression xL of the left-hand spring. We want toknow the compression xR of the right-hand spring.

There is no friction, so we can assume that mechanical energy is conserved: Ui + Ki =Uf + Kf . We will take the initial situation to be that in which the left-hand spring isfully compressed, before the glider is released to move; and the final situation to be thatin which the glider has pushed the right-hand spring to its maximum compression. Sincethe glider isn’t moving in either case, Ki = Kf = 0; so Ui = Uf .

In both situations, the potential energy is all in a spring: Ui = 12kLx

2L = 1

2kRx

2R = Uf .

We can solve this for xR:

· 2−−−−→ kRx2R = kLx

2L

÷kR−−−−−−→ x2R =

kLx2L

kR√

−−−−−→ xR =

√kLx2

L

kR=

√kLkR

xL =

√1800 N/m

1300 N/m(0.039 m) = 0.046 m = 4.6 cm

Notice that we never needed the mass of the glider. Its only function in the problem wasto carry all the energy from one spring to the other.

539

Problem 1505. You drop a rock off the Navajo Bridge and let it fall 467 feet to theColorado River. This is a situation in which air resistance is too large to be ignored. Ifthe gravitational potential energy of the rock just before being dropped is Ugrav, and itskinetic energy just as it hits the water is K, what is the relationship between Ugrav andK? Use the level of the river as y = 0.

(a) Ugrav < K (b) Ugrav = K*(c) Ugrav > K (d) Not enough information: any of these could be true

Solution: If there were no air resistance, then the mechanical energy (M.E.) of therock-earth system would be conserved: Ui+Ki = Uf +Kf . However, air resistance meansthat the rock losses some of its kinetic energy (K.E.) to the molecules in the atmosphere(the atmosphere is not considered part of the rock earth system). The atmosphere actslike a K.E. sink, it converts K.E. into heat! Thus, the final M.E. will be less than theinitial M.E. In particular, Ui+Ki = Uf +Kf +Heat > Uf +Kf . Initially, the rock has noK.E. since it starts from rest (zero velocity); so Ki = 0. Just as the rock hits the river,it is at y = 0, so Uf = 0. Hence

Ugrav = Ui > Kf = K

Problem 1506. A frozen pond acts as a frictionless horizontal surface. A spring with aspring constant of 32,000 N/m is attached to a wall at the edge of the pond, and a pianowith a mass of 450 kg is pushed against the spring, compressing it by 0.18 m. When thepiano is released, how fast does it slide across the ice? Round your answer to the nearest0.1 m/s.


Solution: Before the piano is released, it is stationary; so the initial kinetic energy ofthe spring-piano system is Ki = 0. The potential energy of the system is that stored inthe spring: Ui = 1

2kx2. Hence the total mechanical energy of the system is Ui+Ki = 1

2kx2.

After the piano is released, this energy is completely transferred from the spring to thepiano. The final kinetic energy of the system is Kf = 1

2mv2, and the final potential

energy is Uf = 0. Thus the total mechanical energy of the system is Uf +Kf = 12mv2.

By conservation of energy, the initial and final mechanical energies of the system mustbe the same. Hence:

kx2

2=mv2

2⇒ v =

√kx2

m=

√(32, 000 N/m)(0.18 m)2

450 kg= 1.5 m/s

540

Problem 1507. (Derive Problem [Ball in the bucket]) You want to shoot a ballof mass m out of a spring cannon into a bucket. The spring cannon is aimed to fire theball in the horizontal direction. The bucket’s radius is just a little larger than the ball’sradius. The bucket is designed to capture the ball so long as the ball’s center of mass hitsthe center of the top of the bucket. The cannon’s manufacturer gives the spring constantas k. You measure the following distances: H, the height from the top of the bucket tothe center of mass of the ball as it sits in the cannon; and R, the distance from the endof the muzzle of the cannon (where the ball exits the cannon) to the center of the topof the bucket. Neglecting friction and air drag, and treating the ball as a point particle,derive a formula for the distance d that the spring must be compressed in order to shootthe ball into the bucket (see figure below). Your final answer should be an expression ford as a function of m, k, g, H, and R. Justify your answer.

(a)

√mg

2Hk*(b) R

√mg

2Hk

(c) R

√Hk

mg(d) R

√mg

Hk

(e) None of these

Figure 8: Ball in the bucket.

Solution: See hand-written solutions

541

Problem 1508. (Derive Problem) A small block of mass m slides along a frictionlessloop-the-loop track. The radius of the loop is R. At what height H above the bottom ofthe track should the block be released from so that it just makes it through the loop-the-loop without losing contact with the track? Hint: We want the maximum height suchthat the normal force exerted by the track on the block at the top of the loop is zero.That is, if we release the block from a height of H + ε, where ε is arbitrarily small, thenthe normal force at the top of the loop will not be zero. In order to receive any creditfor this problem, you must show all the work necessary to arrive at the answer. Simplywriting down the final formula will earn you a grade of zero.

(a)2

3R (b)

3

4R

*(c)5

2R (d)

3

2

(e) None of these

Figure 9: The loop-the-loop.


542

Problem 1509. (Derive Problem) A small block of mass m sits at rest at the topof a frictionless hemispherical mound of ice of radius R. The block is given a slight tapby a very small bug and begins to slide down the side of the mound. Find the height Habove the ground where the block leaves the ice. Express this height as a function of theradius R.Hint: Look for the position where the normal force first vanishes. This will be the pointwhere the block leaves the ice.

*(a)2

3R (b)

3

4R

(c)5

2R (d)

3

2

(e) None of these

Figure 10: Block sliding off a hemispherical piece of ice.


543

15 Conservation of Linear Momentum

15.1 Computing Linear Momentum p = mv

Problem 1510. A skateboarder with a mass of 60 kg is moving at a speed of 3 m/s.What is his momentum?

*(a) 180 kg m/s (b) 270 kg m/s(c) 360 kg m/s (d) 540 kg m/s(e) None of these

Solution: p = mv = (60 kg)(3 m/s) = 180 kg m/s.

Problem 1511. A shopping cart with a mass of 13.4 kg is rolling across a parking lotat 4.0 m/s. What is the cart’s momentum?

(a) 26.8 kg m/s *(b) 53.6 kg m/s(c) 107.2 kg m/s (d) 214.4 kg m/s(e) None of these

Solution: p = mv = (13.4 kg)(4.0 m/s) = 53.6 kg m/s.

Problem 1512. Your potato gun launches a potato with a mass of 370 g at a speed of21 m/s. What is the magnitude of the potato’s momentum? Round your answer to twosignificant digits.

(a) 3.9 kg m/s *(b) 7.8 kg m/s(c) 82 kg m/s (d) 160 kg m/s(e) None of these

Solution: p = mv = (0.37 kg)(21 m/s) = 7.8 kg m/s. Don’t forget to convert gramsto kilograms.

Problem 1513. A flying duck with a mass of 2 kg has a momentum of 40 kg m/s. Whatis the duck’s speed?


Solution: p = mv ⇒ v =p

m=

40 kg m/s

2 kg= 20 m/s

544

Problem 1514. An artillery shell has a mass of 5 kg and a momentum of 4000 kg m/s.What is the shell’s speed?

(a) 40 m/s (b) 80 m/s(c) 160 m/s *(d) 800 m/s(e) None of these

Solution: p = mv ⇒ v =p

m=

4000 kg m/s

5 kg= 800 m/s

15.2 Applying conservation of momentum to isolated systems

Problem 1515. You drop an object from rest on a planet with no atmosphere. As theobject falls, which of the quantities of the object listed below is conserved?

(a) Momentum

*(b) Mechanical energy

(c) Potential energy

(d) Kinetic energy

Solution: The object is accelerating in a straight line due to the force of gravity. Itsspeed increases as it falls; so its momentum and its kinetic energy are both increasing.Thus they are not conserved. The potential energy decreases as the object falls, so itis not conserved. Since the planet has no atmosphere, there is no air drag or otherdissipative force. Thus the object-planet system is an isolated system and conservesmechanical energy.

Problem 1516. You drop a wrench from the top of a bridge. Which property of thewrench is conserved as it falls? Assume that there is no air resistance.

(a) Momentum (b) Kinetic energy(c) Potential energy *(d) Mechanical energy(e) None of these

Problem 1517. A cannon has a mass of 1100 kg. It fires a cannonball with a mass of5.3 kg at a muzzle velocity of 390 m/s. How fast does the cannon move backward whenthe ball is fired? Round your answer to two significant digits.

*(a) 1.9 m/s (b) 2.8 m/s(c) 18 m/s (d) 27 m/s(e) None of these

Solution: Assuming that the cannon-cannonball system is initially at rest, the totalmomentum is zero. Hence the backward momentum of the cannon must equal the forwardmomentum of the cannonball:

mcvc = mbvb ⇒ vc =mbvbmc

=(5.3 kg)(390 m/s)

1100 kg= 1.9 m/s

545

Problem 1518. A cannon has a mass of 1500 kg. It fires a cannonball with a mass of5.5 kg at a muzzle velocity of 370 m/s. How fast does the cannon move backward whenthe ball is fired? Round your answer to two significant digits.

*(a) -1.4 m/s (b) -2.7 m/s(c) -11 m/s (d) -22 m/s(e) None of these

Solution: Assuming that the cannon-cannonball system is an isolated system that isinitially at rest, the total momentum is zero. Hence by conservation of momentum, themagnitude of the backward momentum of the cannon must equal the forward momentumof the cannonball:

Psys,f = mcvc +mbvb = 0 ⇒ vc = −mbvbmc

= −(5.5 kg)(370 m/s)

1500 kg= −1.4 m/s

Problem 1519. An astronaut with a mass of 80 kg is adrift in space outside of his spacestation. Fortunately, he is holding a space wrench with a mass of 2 kg. How fast musthe throw the wrench to give himself a recoil speed of 20 cm/s?


Solution: The initial momentum of the system is zero; so the backward momentum ofthe astronaut must equal the forward momentum of the wrench. Don’t forget to convertcentimeters to meters. Starting from ∆Pastronaunt = −∆Pwrench and ignoring the negativesign, since we know the two objects move in opposite directions, we have

mava = mwvw ⇒ vw =mavamw

=(80 kg)(0.2 m/s)

2 kg= 8 m/s

546

Problem 1520. A cannon with a mass of mc is sitting on a frictionless frozen lake. Itfires a cannonball with a mass of mb horizontally at a speed of vb. After it has fired theball, what is the cannon’s backward recoil speed?

(a)

√mbv2

b

mb +mc

(b)mbvb

mb +mc

(c)

√mbv2

b

mc

*(d)mbvbmc

(e) None of these

Solution: We know the masses and initial velocities (zero) of the two bodies, and thefinal velocity of the cannonball. We want to know the final velocity vc of the cannon. Wecan use conservation of momentum. The initial momentum is zero, so after the cannonis fired, its backward momentum must equal the forward momentum of the ball:

mcvc = mbvb÷mc−−−→ vc =

mbvbmc

Problem 1521. (Lab Problem: The spring-cannon glider - a discrete rocket)A spring cannon is fastened to a glider at rest on a horizontal frictionless air track. Thespring-cannon-glider assembly has a mass of mc. It has been previously determined thatthe cannon can launch a ball with a mass of mb horizontally at a speed of vb. Use theconservation of momentum to determine the recoil speed of the cannon after it launchesthe ball. Take the positive x-axis in the direction of the ball.

(a) −√mb

mg

vb (b) −(

mb

mb +mg

)vb

*(c) −mbvbmg

(d) −vb

(e) None of these

Solution: We take our system to be the ball-spring-cannon-glider system. Since weare only interested in the motion of the spring-cannon glider, we can ignore any effects ofgravity on the ball. We can then treat the system as isolated and apply the conservationof momentum to the system, which gives psys,i = psys,f . Since the system is initially atrest, psys,i = 0. It follows that

psys,f = mgvg +mbvb = 0 ⇒ vg = −mbvbmg

.

Note: This problem simulates, in a simple discrete way, the idea behind a rocket. Themain difference is that a rocket expels gas continuously, where as the spring cannon onlydoes it once. This type of problem is known as an explosion problem and it acts like aperfectly inelastic collision in reverse (if we ran the film backwards).

547

Problem 1522. (Lab Problem) A spring cannon is fastened to a glider at rest on ahorizontal frictionless air track. The combined mass of the cannon and the glider is 400g. The cannon fires a ball with a mass of 20 g horizontally at a speed of 8 m/s. How fastdoes the cannon-glider assembly move backward after the cannon is fired?

(a) 20 cm/s (b) 25 cm/s*(c) 40 cm/s (d) 50 cm/s(e) None of these

Solution: Use conservation of momentum: psys,i = psys,f . Since the system is initiallyat rest, its momentum is zero. After the cannon is fired, the forward momentum of theball must equal the backward momentum of the cannon. Don’t forget to convert units.

mbvb = mcvc ⇒ vc =mbvbmc

=(0.02 kg)(8 m/s)

0.4 kg= 0.4 m/s = 40 cm/s

548

15.3 Impulse

Problem 1523. An irate constituent throws a pie at a congressman. The pie has a massof 1.2 kg and hits the congressman with a horizontal speed of 14 m/s. How large is theimpulse that the pie gives to the congressman? Round your answer to two significantfigures.

(a) 8.4 kg m/s *(b) 17 kg m/s(c) 120 kg m/s (d) 140 kg m/s(e) None of these

Solution: The impulse given by the pie is equal to the change ∆p in its momentum.We assume that the politician is large and heavy compared to the pie, so that the piecomes to a complete stop when it hits him. Then the final momentum of the pie is zero;so the impulse equals the original momentum of the pie:

J = ppie = mv = (1.2 kg)(14 m/s) = 17 kg m/s

Problem 1524. An irate constituent throws a pie at a congressman. The pie has a massof 1.6 kg and hits the congressman with a horizontal speed of 14 m/s. How large is theimpulse that the pie gives to the congressman? Round your answer to two significantfigures.

*(a) 22 kg m/s (b) 45 kg m/s(c) 160 kg m/s (d) 310 kg m/s(e) None of these

Solution: The impulse given by the pie is equal to the change ∆p in its momentum.We assume that the politician is large and heavy compared to the pie, so that the piecomes to a complete stop when it hits him. Then the final momentum of the pie is zero;so the impulse equals the original momentum of the pie:

J = ppie = mv = (1.6 kg)(14 m/s) = 22 kg m/s

Problem 1525. A golf ball has a mass of 0.071 kg. It is placed on a tee and hit, givingit a speed of 42 m/s. Golf scientists determine that the club is in contact with the ballfor 0.046 s. Assuming constant acceleration, what is the average force exerted on the ballby the club? Round your answer to the nearest newton.


Solution: The golf ball starts at rest, so the impulse given it by the club is equal toits final momentum. This is J = p = mv. Since we’re assuming constant acceleration,the force is the impulse divided by the time:

F =mv

t=

(0.071 kg)(42 m/s)

0.046 s= 65 N

549

Problem 1526. A ball weighing 120 g and moving horizontally at 9.2 m/s strikes a walland rebounds horizontally at 8.4 m/s. What is the magnitude of the impulse that thewall gives to the ball? Round your answer to two significant digits.

(a) 0.096 kg m/s (b) 0.94 kg m/s*(c) 2.1 kg m/s (d) 21 kg m/s(e) None of these

Solution: The impulse is the ball’s change in momentum. If we take the direction awayfrom the wall as positive, then the ball’s initial momentum is pi = (0.12 kg)(−9.2 m/s);its final momentum is pf = (0.12 kg)(8.4 m/s). The change in momentum is

∆p = pf − pi = (0.12 kg)(8.4 m/s)− (0.12 kg)(−9.2 m/s)

= (0.12 kg)(17.6 m/s) = 2.1 kg m/s

Problem 1527. A ball weighing 290 g and moving horizontally at 14 m/s strikes a walland rebounds horizontally at 10 m/s. What is the magnitude of the impulse that thewall gives to the ball? Round your answer to two significant digits.

(a) 1.2 kg m/s *(b) 7.0 kg m/s(c) 14 kg m/s (d) 43 kg m/s(e) None of these

Solution: The impulse is the ball’s change in momentum. If we take the direction awayfrom the wall as positive, then the ball’s initial momentum is pi = (0.29 kg)(−14 m/s);its final momentum is pf = (0.29 kg)(10 m/s). The change in momentum is

∆p = pf − pi = (0.29 kg)(10 m/s)− (0.29 kg)(−14 m/s)

= (0.29 kg)(24 m/s) = 7.0 kg m/s

Problem 1528. A large truck travelling down the highway collides with a small station-ary deer. Which of the following statements is true?

(a) The average force acting on the deer during the collision is greater than theaverage force acting on the car.

(b) After the collision, the magnitude of the deer’s velocity relative to the car isthe same as it was before the collision.

(c) The change in the car’s momentum is greater than the change in the deer’smomentum.

*(d) The magnitude of the impulse given to the car by the deer is the same as themagnitude of the impulse given to the deer by the car.

Solution: We are not told whether this collision is elastic or inelastic; and judging byexperience and common sense, it’s probably not perfectly elastic. Thus we can rule out(b). We know that momentum is conserved, so we can rule out (a) and (c). That leavesus with (d), which is true: the changes in momentum of the car and the deer must beequal and opposite, which means that the impulses acting on them must be equal andopposite.

550

Problem 1529. A large freight train traveling down the tracks collides with a small,cute, cuddly stationary squirrel burying a nut. Which of the following statements is true?

(a) The average force acting on the squirrel during the collision is greater than theaverage force acting on the train.

(b) After the collision, the magnitude of the squirrel’s velocity relative to the trainis the same as it was before the collision.

(c) The change in the train’s momentum is greater than the change in the squirrel’smomentum.

*(d) The magnitude of the impulse given to the train by the squirrel is the same asthe magnitude of the impulse given to the squirrel by the train.

Solution: We are not told whether this collision is elastic or inelastic; and judging byexperience and common sense, it’s probably not perfectly elastic. Thus we can rule out(b). We know that momentum is conserved, so we can rule out (a) and (c). That leavesus with (d), which is true: the changes in momentum of the train and the squirrel mustbe equal and opposite, which means that the impulses acting on them must be equal andopposite.

Problem 1530. (Impulse over long-time period) A crate with a mass of 30 kg issliding down a ramp. Starting from rest, it reaches a speed of 3 m/s after 10 s. Whatis the magnitude of the net force on the crate? Assume the net force on the crate isconstant.


Solution: The crate begins at rest, so the impulse given to it is equal to its finalmomentum. Thus

J = ∆p = F∆t = mv ⇒ F =mv

∆t=

(30 kg)(3 m/s)

10 s= 9 N

Problem 1531. A golf ball has a mass of 0.071 kg. It is placed on a tee and hit, givingit a speed of 42 m/s. Golf scientists determine that the club is in contact with the ballfor 0.046 s. Assuming constant acceleration, what is the average force exerted on the ballby the club? Round your answer to the nearest newton.


Solution: The golf ball starts at rest, so the impulse given it by the club is equal toits final momentum. This is J = p = mv. Since we’re assuming constant acceleration,the force is the impulse divided by the time:

F =mv

t=

(0.071 kg)(42 m/s)

0.046 s= 65 N

551

Problem 1532. (Similarity Problem) Two railroad cars are standing on two hori-zontal frictionless tracks. The first car has mass M ; the second car has mass 3M . Eachcar is pulled along its track with a force of F for a time of T . At the end of this time,the momentum of the first car is P . What is the momentum of the second car?

*(a) P (b) P/3

(c) 3P (d)√

3P(e) None of these

Solution: Since both cars are subjected to the same force F for the same time T ,they have the same final momentum P . The second car only experiences one-third theacceleration of the first, so it only ends up going at one-third the velocity; but since ithas three times the mass, its momentum is the same.

Problem 1533. (Similarity Lab Problem) Two gliders are at rest on parallelhorizontal frictionless air-tracks. Glider 1 has mass m1; glider 2 has mass m2. Eachglider is pushed with a horizontal force of Fpush for the same total time T . At the end ofthis time, glider 1 has kinetic energy K1. What is the relationship between the kineticenergy of glider 2 in terms of the kinetic energy of glider 1?

*(a) K/2 (b) K

(c)√

2K (d) 4K(e) None of these

Solution: This is a similarity problem. We have two experiments with certainparameters that are in common to both. Since the same force acts on both gliders forthe same amount of time, both gliders experience the same impulse. This is our commonparameter. Thus p1,f − p1,i = ∆p1 = J = ∆p2 = p2,f − p2,i. Since the gliders both startfrom rest we have v1,i = v2,i = 0. Substituting this result into the previous expressiongives

m1v1 = p1,f = p2,f = m2v2÷m2−−−−−−→ v2 =

m1

m2

v1 .

The kinetic energy of glider 1 is K1 = 12m1v

21; the kinetic energy of glider 2 is

K2 =1

2m2v

22

=1

2m2

(m1

m2

v1

)2

(substitute for v2 for the above equation)

=1

2m1v

21

(m1

m2

)=

(m1

m2

)K1

Comments:

552

1. Notice that the ratio of the kinetic energies is

K2

K1

=M1

M2

This follows directly from the fact that the momentums are the same (p1 = p2), so

K2

K1

=p2 v2

p1 v1

=v2

v1

=m1

m2

Problem 1534. (Similarity Problem) Two railroad cars are at rest on two horizontalfrictionless tracks. One car is three times as heavy as the other. Each car is pushed withthe same horizontal force of F for the same time T . At the end of this time, the kineticenergy of the lighter car is K. What is the kinetic energy of the heavier car?

(a) 3K (b)√

3K

(c) K/√

3 *(d) K/3(e) None of these

Solution: Let the mass of the light car be M . We’ll construct a table with thekinematic properties of the two cars:

Light car Heavy carMass M 3MAcceleration = force/mass F/M F/3MSpeed = acceleration × time FT/M FT/3M

Kinetic energy = 12×mass× speed2 K =

F 2T 2

2M

F 2T 2

6M=K

3

Problem 1535. (Similarity Problem) Two carts are at rest on parallel horizontalfrictionless tracks. One cart has mass M ; the second has mass 2M . Each cart is pushedwith a horizontal force of F for a time T . At the end of this time, the magnitude of thefirst cart’s momentum is P . What is the magnitude of the second cart’s momentum?

(a) P/2 *(b) P(c) 2P (d) 4P(e) None of these

Solution: Newton’s second law allows us to calculate the accelerations of bothcarts. For the first, A = F/M ; for the second, A2 = F/2M = A/2. Both carts arepushed for time T ; so the speed of the first is V = AT and the speed of the secondis V2 = A2T = AT/2 = V/2. Then the momentum of the first cart is P = MV ; themomentum of the second is P2 = M2V2 = (2M)(V/2) = MV = P .

553

15.4 Collisions

Problem 1536. (Lab Problem) Two gliders collide on a horizontal frictionless airtrack. Which of the following statements must be true regardless of whether the collisionis elastic or inelastic?

(a) Neither momentum nor kinetic energy is necessarily conserved.

(b) Kinetic energy is conserved; momentum is not necessarily conserved.

*(c) Momentum is conserved; kinetic energy is not necessarily conserved.

(d) Momentum and kinetic energy are both conserved.

Solution: In all collisions in an isolated system (no external forces), momentum isconserved. In elastic collisions, kinetic energy is conserved; in inelastic collisions, it is not.

Problem 1537. (Lab Problem) Two gliders collide on a frictionless horizontal airtrack. If the collision is elastic, which of the following statements must be true?

*(a) Momentum and kinetic energy are both conserved.

(b) Momentum is conserved; kinetic energy is not conserved.

(c) Kinetic energy is conserved; momentum is not conserved

(d) Neither momentum nor kinetic energy is conserved.

Solution: In all collisions in an isolated system (no external forces), momentum isconserved. In elastic collisions, kinetic energy is conserved (by definition).

Problem 1538. (Lab Problem) Two gliders collide on a horizontal frictionless airtrack. If the collision is inelastic, which of the following statements must be true?

(a) Momentum and kinetic energy are both conserved.

*(b) Momentum is conserved; kinetic energy is not conserved.

(c) Kinetic energy is conserved; momentum is not conserved

(d) Neither momentum nor kinetic energy is conserved.

Solution: In all collisions in an isolated system (no external forces), momentum isconserved. In inelastic collisions, kinetic energy is not conserved.

554

Problem 1539. Two railroad cars are rolling toward one another on a frictionless hori-zontal track. The two cars have the same mass and are moving at the same speed. Whenthey collide, their couplings engage so that they remain attached together. Which of thefollowing statements is true?

(a) The kinetic energy of the system is conserved during the collision.

(b) The momentum of each car is the same before and after the collision.

*(c) The two cars lose all of their kinetic energy during the collision.

(d) The momentum of the system decreases during the collision.

Solution: Since the two cars have the same mass and the same initial speed in op-posite directions, the initial momentum of the system is zero. Thus when the cars arecoupled together, their speed must be zero, which means their total kinetic energy iszero. Thus (c) is true. We can rule out (a) because kinetic energy is lost; (b), becauseeach car has nonzero momentum before the collision and zero momentum afterward; and(d), because the total momentum of the system is conserved.

Problem 1540. (Lab Problem) Aglider with a mass of 500 g is at rest on africtionless horizontal air track. A secondglider with a mass of 250 g is launched to-ward it at a speed of V . The second gliderhas a spring attached, which is compressedas it collides with the first glider. Whenthe spring is at its maximum compression,which of the following statements is true?

500 g250 g

��

��

��C

CC

CCC

-V

(a) The second glider is at rest relative to the ground.

*(b) Both gliders have the same velocity.

(c) Both gliders have the same momentum.

(d) Both gliders have the same kinetic energy.

555

Problem 1541. (Lab Problem) Aglider with a mass of M is at rest on ahorizontal frictionless air track. A secondglider with a mass of m, where m < M , islaunched toward it at a speed of V . Thesecond glider has a spring attached, whichis compressed as it collides with the firstglider. When the spring is at its maxi-mum compression, which of the followingstatements is true?

Mm

��

��

��C

CC

CCC

-V

(a) The second glider is at rest relative to the ground.

*(b) Both gliders have the same velocity.

(c) Both gliders have the same momentum.

(d) Both gliders have the same kinetic energy.

Problem 1542. (Lab Problem) Two gliders collide on a frictionless horizontal airtrack. The first glider has a mass of 2.0 kg; before the collision, it is moving rightward at5.0 m/s. The second glider has a mass of 3.0 kg; before the collision, it is at rest. Afterthe collision, the second glider is moving rightward at 4.0 m/s. What is the velocity ofthe first glider after the collision?

*(a) 1.0 m/s leftward (b) 1.0 m/s rightward(c) 3.0 m/s leftward (d) 3.0 m/s rightward(e) None of these

Solution: We know the mass and initial velocity of each glider; and we know the finalvelocity of the second glider. We want to know the final velocity of the first glider. Wecan use conservation of momentum. Taking rightward as the positive direction, we get:

P = m1v1i +m2v2i = m1v1f +m2v2f

v2i=0−−−−→ m1v1i = m1v1f +m2v2f

−m2v2f−−−−−→ m1v1f = m1v1i −m2v2f

÷m1−−−→ v1f =m1v1i −m2v2f

m1

=(2.0 kg)(5.0 m/s)− (3.0 kg)(4.0 m/s)

2.0 kg= −1.0 m/s

Since the sign is negative, the glider is moving leftward.

556

Problem 1543. (Lab Problem) A glider with a mass of 2.0 kg is moving rightwardat a speed of 3.0 m/s on a frictionless horizontal air track. It collides with a stationaryglider whose mass is 4.0 kg. If the collision is elastic, what is the speed of the heavierglider after the collision? Round your answer to the nearest 0.1 m/s.


Solution: Since the collision is elastic, both momentum and kinetic energy areconserved. Since the heavier glider is stationary, the initial momentum and kinetic energyare only those of the first glider. Taking the positive direction as rightward, and usingthe subscript 1 for the lighter glider and 2 for the heavier, we find:

pi = m1v1i = (2.0 kg)(3.0 m/s) = 6.0 kg m/s

Ki =1

2m1v

21i =

1

2(2.0 kg)(3.0 m/s)2 = 9.0 J

By conservation of momentum and kinetic energy, pf = pi and Kf = Ki. For brevity’ssake, we will drop the ”f” subscript and refer to the post-collision velocities as v1 and v2.

m1v1 +m2v2 = pi and1

2m1v

21 +

1

2m2v

22 = Ki

Substituting numbers, we get

2v1 + 4v2 = 6 and 1v21 + 2v2

2 = 9

We want to know the speed of the heavier glider after the collision: v2. We’ll solve thefirst equation for v1, then substitute that into the second equation; that will give us anequation in v2 alone, which we will solve for v2.

Solving the first equation for v1:

2v1 + 4v2 = 6 ⇒ v1 =6− 4v2

2= 3− 2v2

Substitute this expression for v1 into the kinetic-energy equation:

v21 + 2v2

2 = 9 ⇒ (3− v2)2 + 2v22 = 9

Expand to eliminate the parentheses:

9− 6v2 + v22 + 2v2

2 = 9 ⇒ 3v22 − 6v2 = 0

Factor:3v2(v2 − 2) = 0 ⇒ v2 = 0 or v2 = 2

The solution v2 = 0 refers to the velocity of the glider before the collision; so the velocityafter the collision is 2.0 m/s.

557

Problem 1544. (Lab Problem) Two gliders collide on a frictionless horizontal airtrack. Glider 1 has a mass of 0.88 kg; glider 2 has a mass of 1.13 kg. Before the collision,glider 1 is moving rightward at 3.3 m/s; glider 2 is at rest. After the collision, glider 1is moving rightward at 1.1 m/s. How fast is glider 2 moving? Round your answer to thenearest 0.1 m/s.


Solution: We don’t know if this collision is elastic or inelastic. We know the massesof both gliders, and the velocities of both before the collision, so we can calculate theinitial momentum. We know the velocity of glider 1 after the collision, so we can useconservation of momentum to calculate the post-collision velocity of glider 2.

Since the initial velocity of glider 2 is zero, the initial momentum of the system is:

pi = p1,i + p2,i = m1v1,i +m2v2,i = m1v1,i

After the collision, the momentum of the system is

pf = p1,f + p2,f = m1v1,f +m2v2,f

Since momentum is conserved, pf = pi; so

m1v1,f +m2v2,f = m1v1,i

Solve for v2,f :

v2,f =m1(v1,i − v1,f )

m2

=(0.88 kg)(3.3 m/s− 1.1 m/s)

1.13 kg= 1.7 m/s

Problem 1545. A block of wood with a mass of 2300 g is sitting on a frictionlesshorizontal tabletop. A rifle bullet with a mass of 44 g is fired horizontally into the blockat a speed of 270 m/s and stops inside the block. After the collision, what is the speedof the combined block and bullet? Round your answer to the nearest 0.1 m/s.


Solution: Since the bullet stops inside the block, we have an inelastic collision. Weuse conservation of momentum. If mb and vbi are the mass and initial speed of the bullet,and mw and vwi = 0 are the mass and initial speed of the wood block, then P = mbvbi.If vf is the speed of block plus bullet after the collision, then

P = (mb +mw)vf

⇒ vf =P

mb +mw

=mbvbi

mb +mw

=(0.044 kg)(270 m/s)

0.044 kg + 2.3 kg= 5.1 m/s

558

Problem 1546. A block of wood with a mass of 1300 g is sitting on a horizontal tabletop.The coefficient of kinetic friction between the block and the table is µk = 0.74. A riflebullet with a mass of 39 g is fired horizontally into the block at a speed of 290 m/s andstops inside the block. How far does the block with the bullet embedded in it slide beforecoming to a stop? Round your answer to the nearest 0.1 m.


Solution: There are two processes going on here: an inelastic collision between thebullet and the block, followed by the block’s sliding to a halt as friction dissipates itskinetic energy.

Since the bullet-wood collision is inelastic, we can’t use conservation of energy there. Wehave to use conservation of momentum. We know the masses and initial velocities ofthe bullet and the block: mb, vbi, mw, and vwi = 0. (We’ll use the subscript “w” for“wood”, since “block” and “bullet” both begin with the same letter.) We want to knowthe kinetic energy Ks of the block-bullet system immediately after the collision, and wecan calculate that if we know the velocity vs at that time. By conservation of momentum,

mbvbi +mwvwi = mbvbi = (mb +mw)vs÷(mb+mw)−−−−−−−→ vs =

mbvbimb +mw

Now we can calculate the energy of the system immediately after the collision. There isno potential energy involved.

Ks =1

2msv

2x =

(mb +mw)(mbvbi)2

2(mb +mw)2=

m2bv

2bi

2(mb +mw)

As the block-bullet system slides to a halt, the only force working on it is friction. It willmove until the work done against the frictional force equals its initial energy. Since thetabletop is horizontal, the normal force equals the weight; so the force of kinetic frictionis

Fk = µkN = µk(mb +mw)g

The work done against friction is the force times the distance x that the block slides:W = Fkx. We want to know x; and we can find it by setting W = Ks and then solvingfor x.

W = Fkx = µk(mb +mw)gx =m2bv

2bi

2(mb +mw)= Ks

÷µk(mb+mw)g−−−−−−−−−→ x =m2bv

2bi

2µkg(mb +mw)2=

(0.039 g)2(290 m/s)2

2(0.74)(9.8 m/s2)(0.039 g + 1300 g)2= 4.9 m

559

Problem 1547. (Derive Problem) A bullet of mass mb with an initial velocity of vb,istrikes a wooden block of mass mw at rest on a frictionless horizontal surface. The bulletemerges with its speed reduced to vb,f . Derive a formula for the resulting velocity vw,f ofthe block. Assume there is no loss of mass in the wooden block when the bullet passesthrough the block.

Solution: This is a collision problem between the bullet and the wooden block. Ifwe take the system to be the wooden block plus the bullet, then we would like to ap-ply the conservation of momentum to the system before and after the collision. Thisrequires demonstrating that we can ignore all of the external forces on the system inthe direction tangent to the surface. Take this to be the x-direction. Since we havea horizontal frictionless surface, there is no forces due to gravity and friction on thesystem in the x-direction, and ignoring any effects due to air drag over this short timeperiod, we are justified in applying conservation of momentum. We now proceed in steps.

Step 1: Choose the coordinate system.

Choose a fixed coordinate system with the x-axis running along the frictionless surfaceand the y-direction normal to the surface opposite the direction of gravity.

Step 2: Draw a before and after picture.

Figure 11: Bullet-wooden-block collision.

560

Step 3: Write down what you are given and what you want. Also include any assump-tions you are making to simplify the problem.

given:

mb = mass of the bullet

mw = mass of the wooden block

vb,i = initial speed of the bullet (before impact)

vb,i = final speed of the bullet (after impact)

vw,i = 0 (initial speed of the wooden block (after impact))

want: vb,i = final speed of the wooden block (after impact)

assumptions: This is a 1-D motion problem; no external forces on the system

Step 4: Apply conservation of momentum to the system: psys,i = psys,f.

mbvb,i +mwvw,i = psys,i = psys,f = mbvb,f +mwvw,fset vw,i=0−−−−−→ mbvb,i + 0 = mbvb,f +mwvw,f−mbvb,f−−−−−→ mwvw,f = mbvb,i −mbvb,f÷mw−−−→ vw,f = −mb

mw

∆vb

A faster way would be to use ∆pw = −∆pb with pw,i = 0.

Problem 1548. (Lab Problem: Explosion problem) Two gliders, each with massM , are at rest on a frictionless horizontal air track. The gliders are tied together with ahorizontal spring between them; the spring has been compressed so that it stores a po-tential energy of U . When the string connecting the two gliders is cut, the spring pushesthem apart. After the spring has fully expanded, what is the speed of each glider?

(a)U

2M(b)

U

M

(c)

√2U

M*(d)

√U

M

(e) None of these

Solution: There is no friction in the system, so we will use conservation of energy.Since the gliders are initially at rest, the momentum of the system is zero. Therefore,since the gliders have the same mass M , they will move at the same speed v in oppo-site directions after the string is cut. Hence the kinetic energy of each glider will beKg = 1

2Mv2; so the kinetic energy of the whole system will be Ks = 2Kg = Mv2. By

conservation of energy,

U = Ks = Mv2 ⇒ v2 =U

M⇒ v =

√U

M

561

Problem 1549. Two identical gliders have an elastic collision on a horizontal frictionlessair track. Assume there are no external forces on the two-glider system, and the motionis one-dimensional. Before the collision, the initial velocity of first glider is v1,i 6= 0, andthe initial velocity of the second glider is v2,i = 0. Suppose that after the collision thefirst glider is observed to be at rest. What is the velocity of the second glider after thecollision?

*(a) v1,i (b) −v1,i

(c) 0 (d) −v1,i/2(e) None of these

Solution: This problem gives you extra information. It follows immediately fromconservation of momentum that v2,f = v1,i. We could have come to the same conclusionwithout knowing the final velocity of the first glider.

In general, in an elastic collision between two objects of the same mass with one of theobjects initially at rest, the two objects will switch velocities after the collision. To seethis, apply conservation of momentum and K.E.:{

mv1,i + 0 = mv1,f +mv2,f÷m−−→ v1,i + 0 = v1,f + v2,f

12mv2

1,i + 0 = 12mv2

1,f + 12mv2

2,f

÷(m/2)−−−−→ v21,i + 0 = v2

1,f + v22,f

Solving the first equation for v2,f = v1,i − v1,f and substituting this expression into thesecond equation leads to

v21,i − v2

1,f = (v1,i − v1,f )2 factor the LHS and ÷ (v1,i−v1,f )−−−−−−−−−−−−−−−−−−→ v1,i + v1,f = v1,i − v1,f

−v1,f +v1,f−−−−−−→ 2v1,f = 0

⇒ v2,f = v1,i .

562

Problem 1550. (Derive Problems) At either end of a horizontal frictionless air trackis a spring. The spring on the left has spring constant kL. The spring on the right hasspring constant kR. A glider with a mass of mL is pressed against the left-hand spring,compressing it a distance xL from equilibrium. The glider is initially at rest and heldin place by a trigger mechanism. The glider is then released, so that the spring propelsit rightward. In the middle of the track is a second glider with a mass of mR. The twogliders have small magnets attached, so when the first glider strikes the second, they sticktogether and slide into the right-hand spring. Let xR be the maximum compression of theright-hand spring. Derive a formula for xR. (The diagram shows the situation before theleft-hand glider is released, when the left-hand spring is compressed and neither glider ismoving.)

��EEEE C

CCCCC ��

mLmR

Solution: This problem requires several steps.

Step 1: Apply the conservation of mechanical energy to the left-hand-glider-spring sys-tem.

We will use the compression of the left-hand spring to calculate its potential energy.When it is released, the potential energy stored in the spring will be converted to kineticenergy of the left-hand glider; we will use that to determine the initial velocity of theglider, and hence the glider’s initial momentum.

Step 2: Apply the conservation of momentum to the collision of the two gliders.

When the gliders collide, they will not be in contact with either spring, so we can treatthe pair of gliders as an isolated system. The collision will be perfectly inelastic, soapplying the conservation of momentum will give us the speed of the combined glidersafter they’ve stuck together.

Step 3: Apply the conservation of mechanical energy to the right-hand-glider-pair-springsystem.

If we know the speed of the pair of gliders after the collision, then we know their kineticenergy. When the right-hand spring is fully compressed, all of that kinetic energy will

563

be stored in the spring; we’ll use that to determine the compression.

We begin with step 1. Let xL be the compression of the left-hand spring. Then theinitial potential energy in the spring, and hence the left-hand glider, is UL,i = 1

2kLx

2L.

The initial kinetic energy of the left-hand glider is KL,i = 0, since the glider is at rest.

After the left-hand spring is released, this energy is completely transferred to the left-hand glider. The final kinetic energy of the glider will be KL,f = 1

2mLv

2L, and its potential

is UL,f = 0, since the glider is no longer experiencing any forcing by the spring. Equatingthe initial and final mechanical energy of the left-hand-glider-spring system gives

1

2kLx

2L + 0 = UL,i +KL,i = UL,f +KL,f = 0 +

1

2mLv

2L

⇒ vL =

√kLx2

L

mL

⇒ pL,i = mLvL =√mLkLx2

L

(initial momentum of left-hand-glider before collision)

⇒ psys,i = pL,i + pR,i =√mLkLx2

L

(initial momentum of two-glider system before collision)

Notice that the initial momentum of the right-hand glider is zero since it is initially atrest. This completes step 1 and gives us the initial momentum for step 2.

After the perfectly inelastic collision, the combined gliders will move at a speed vLR,conserving momentum. We’ll use mLR = mL + mR for the combined mass of the twogliders. The subscript LR on the mass and velocity is used to remind us that the massesare stuck together. Applying conservation of momentum just before and after the collisionyields:

mLRvLR = psys,f = psys,i =√mLkLx2

L ⇒ vLR =

√mLkLx2

L

mLR

This completes step 2 and begins step 3.

In the last step of the analysis, we again apply the conservation of mechanical energy.The kinetic energy of the combined gliders will be

KR,i =1

2mLRv

2LR =

mLkLx2L

2mLR

.

Since the right-hand spring is not yet compressed, the initial potential of the mass-springsystem is zero. Thus the initial mechanical energy is

MER,i = KR,i + UR,i =mLkLx

2L

2mLR

+ 0

564

When the right-hand spring is fully compressed, all of the kinetic energy in the gliderswill be stored in the right-hand spring. The final kinetic energy of the mass (gliders)-spring system is KR,f = 0. The final potential energy of the mass-spring system isUR,f = Uspring = kRx

2R/2, where xR is to be determined. Thus the final mechanical

energy of the system is

MER,f = KR,f + UR,f = kRx2R/2 + 0

We now apply the conservation of mechanical energy to arrive at the expression for xR:

kRx2R

2= UR,f +KR,f = UR,i +KR,i =

mLkLx2L

2mLR

⇒ xR =

√kLkR

√mL

mLR

xL =

√kLkR

√mL

mL +mR

xL

Notice that if kL = kR and mR = 0 (no second glider), then xR = xL.

Comment: Just remember, if you see this problem on an exam, say to yourself: Iwant me-mom-me! Use conservation of M.E., then momentum (mom), then one moreapplication of M.E.

565

Problem 1551. (Dervive Problem [The Ballistic Pendulum])Below is a sketch of the ballistic pendulum from Pima community college’s physics lab.The apparatus is designed to measure the speed of the steel ball as it leaves the mouthof the spring cannon. The ball is fired horizontally into a heavy bob attached to a lightrod that is free to pivot at one end like a pendulum. The pendulum catches the ball as itleaves the muzzle of the cannon. We treat this as a perfectly-inelastic collision betweenthe ball and the pendulum. The pendulum arm then swings away from the cannon andupward; a device on the apparatus allows one to measure the angle θ = Θmax that therod swings through. When the arm swings, it lifts the ball to a height of hmax above theball’s initial height when it was in the spring cannon. Although it would be very hard tomeasure this height directly, we can use trigonometry to compute the maximum heightof the ball in terms of the measured angle Θmax. In the lab we can measure the valuesof the mass of the ball Mball; the mass of the pendulum Mpend; the distance between thepivot point of the pendulum and the center of mass of the ball-pendulum system, whichwe shall refer to as the length of the rod Lrod; and the angle Θmax of the swing after thecannon is fired.

(a) Find the initial velocity of the ball as it leaves the muzzle v0 as a function of the givendata: Mball,Mpend, Lrod,Θmax. Your answer should not explicitly contain hmax, since itcan be expressed as a function of these data variables. Note: I have used capital let-ters to denote known parameters that can be measured in the lab prior to the experiment.

(b) If the spring is compressed from its equilibrium position by a distance X, what is thespring constant of the cannon as a function of the measured quantities?

Hint: This is a 3-step problem. In step 1, use conservation of mechanical energy to findthe relationship between the spring constant and the velocity of the ball as it leaves thecannon. In step 2, use conservation of momentum over the inelastic collision. Approx-imate the collision as a one-dimension collision in the horizontal direction. In step 3,apply conservation of mechanical energy to the ball-pendulum system. Take the initialstate of the system just after the collision, and the final state of the system at the pointwhere the pendulum is momentarily at rest at the highest point of its upswing.

566

Figure 12: Ballistic Pendulum found in lab

Figure 13: Ballistic Pendulum: Stage 1


567


568

16 Rotational kinetic energy and angular momen-

tum

Problem 1552. Two physics students are on a merry-go-round that is rotating friction-lessly. Each student has a mass of 80 kg. Initially, one student is standing at the center,while the other is standing at the outer edge. The two change places, walking past oneanother at the same speed. How does the angular speed of the merry-go-round changeas the students move?

(a) The speed of the merry-go-round doesn’t change.

(b) The speed decreases until the students meet halfway between the edge and thecenter; then it increases.

*(c) The speed increases until the students meet halfway between the edge and thecenter; then it decreases.

(d) The speed increases during the whole time that the students move, becausethey are adding energy to the system.

Solution: There are no external torques, so angular momentum is conserved: L = Iω.Hence if I increases, ω decreases, and vice versa. We want to see if I changes as thestudents move. Let R be the merry-go-round’s radius, and m the mass of each student.When one is at the center and the other is at the edge, I = mR2. When both arehalfway from the center to the edge, I = (2m)(R/2)2 = mR2/2. Since I gets smaller asthe students walk toward the halfway point, then gets larger again, the merry-go-roundfirst speeds up, then slows down.

569

Problem 1553. A physics instructor has a mass of M ; his wife has a mass of M/2.The two are on a merry-go-round that is rotating frictionlessly. Initially, the instructoris at the center and his wife is at the outer edge. The two trade places, passing oneanother halfway from the center to the edge. At which of the following three points isthe merry-go-round’s angular speed the greatest?

(a) When the instructor is at the center and the wife is at the edge.

(b) When the wife is at the center and the instructor is at the edge.

*(c) When the two are both halfway from the center to the edge.

(d) The merry-go-round’s speed doesn’t change.

Solution: There are no external torques on the system, so angular momentum isconserved: L = Iω ⇒ ω = L/I. The angular speed ω is the greatest when themoment of inertia I is the smallest. The contribution of each person to the total momentof inertia is mr2, where m is their mass and r is their distance from the center. Let Rbe the radius of the merry-go-round, so that someone at the edge is at r = R. Look atthe three situations:

Instructor at center, wife at edge I = M(0)2 +M

2R2 =

MR2

2

Wife at center, instructor at edge I =M

2(0)2 +MR2 = MR2

Both halfway from center to edge I = M

(R

2

)2

+M

2

(R

2

)2

=3MR2

8

Since I changes, the speed of the merry-go-round changes; so we can rule out (d). SinceI is the smallest in situation (c), the merry-go-round’s speed is the greatest in thatsituation.

570

Problem 1554. A physics instructor has a mass of M ; his wife has a mass of M/2.The two are on a merry-go-round that is rotating frictionlessly. Initially, the instructoris at the center and his wife is at the outer edge. The two trade places, passing oneanother halfway from the center to the edge. At which of the following three points isthe merry-go-round’s angular speed the smallest?

(a) When the instructor is at the center and the wife is at the edge.

*(b) When the wife is at the center and the instructor is at the edge.

(c) When the two are both halfway from the center to the edge.

(d) The merry-go-round’s speed doesn’t change.

Solution: There are no external torques on the system, so angular momentum isconserved: L = Iω ⇒ ω = L/I. The angular speed ω is the smallest when themoment of inertia I is the largest. The contribution of each person to the total momentof inertia is mr2, where m is their mass and r is their distance from the center. Let Rbe the radius of the merry-go-round, so that someone at the edge is at r = R. Look atthe three situations:

Instructor at center, wife at edge I = M(0)2 +M

2R2 =

MR2

2

Wife at center, instructor at edge I =M

2(0)2 +MR2 = MR2

Both halfway from center to edge I = M

(R

2

)2

+M

2

(R

2

)2

=3MR2

8

Since I changes, the speed of the merry-go-round changes; so we can rule out (d). Since Iis the largest in situation (b), the merry-go-round’s speed is the smallest in that situation.

Problem 1555. A flywheel’s moment of inertia is 80 kg·m2. Starting at rest, it issubjected to a torque of 15 N·m for 6 s. At the end of this time, what is its kineticenergy? Round your answer to two significant figures.


Solution: We know I, τ , and t. We want to know K. If we knew ω, we could useK = 1

2Iω2. If we knew α, we could calculate ω = αt. We know I and τ , so we can

calculate α using τ = Iα.

K =1

2Iω2 =

1

2I(αt)2 =

1

2I(τIt)2

=τ 2t2

2I=

(15 N·m)2(6 s)2

2(80 kg·m2)= 51 J

571

Problem 1556. A wind turbine has three blades. Each blade has a mass of 1000 kg anda length of 20 m. Each blade can be regarded as a thin rod connected at one end to theaxis. If the turbine is turning at a rate of one revolution every 3.0 s, what is its kineticenergy? Round your answer to two significant figures.

*(a) 8.8× 105 J (b) 9.7× 105 J(c) 1.1× 106 J (d) 1.2× 106 J(e) None of these

Solution: We can calculate the moment of inertia of an individual blade, then multiplyby 3 to get I for the whole turbine. We know the period T , so we can use ω = 2π/T tocalculate ω. Then we can use K = 1

2Iω2.

Iblade =mbR

2

3⇒ I = 3Iblade = mbR

2

K =1

2Iω2 =

1

2(mbR

2)

(2π

T

)2

=2π2mbR

2

T 2=

2π2(1000 kg)(20 m)2

(3.0 s)2= 8.8× 105 J

Problem 1557. A shaft can be regarded as a solid cylinder with radius 4 cm and mass120 kg. If the shaft is turning at 20 revolutions per second, what is its kinetic energy?Round your answer to two significant figures.

*(a) 760 J (b) 830 J(c) 890 J (d) 960 J(e) None of these

Solution: We know the mass m and radius R of the shaft, and its frequency f . Wewant to know its kinetic energy K. If we knew I and ω, we could use K = 1

2Iω2. We

can use I = mR2/2 and ω = 2πf .

K =1

2Iω2 =

1

2

(mR2

2

)(2πf)2 = π2mf 2R2

= π2(120 kg)(20 s−1)2(0.04 m)2 = 760 J

572

Problem 1558. A windmill whose moment of inertia is I is initially at rest. The windbegins to blow, producing a torque of τ about the windmill’s axis. After a time t, whatis the windmill’s kinetic energy?

*(a)τ 2t2

2I(b)

Iτ 2t2

2

(c)τt2

2I(d)

Iτt2

2

(e) None of these

Solution: We know I, τ , and t. We want to know K. If we knew ω, we could useK = 1

2Iω2. If we knew L, we could calculate ω from L = Iω. We know τ and t, and

ω0 = 0, so we can calculate L = τt.

K =1

2Iω2 =

1

2I

(L

I

)2

=1

2I

(τt

I

)2

=τ 2t2

2I

Problem 1559. Two flywheels, one with moment of inertia 240 kg·m2 and one withmoment of inertia 150 kg·m2, turn on the same shaft. The first flywheel is fixed on theshaft; the second one is connected by a clutch, so that it can be made to turn either withthe first or independently of it. Initially, the first wheel is turning at 30 rad/s; the secondis at rest. The clutch is then operated, so that both wheels turn together. What is theircommon angular speed? Round your answer to two significant figures.

(a) 17 rad/s *(b) 18 rad/s(c) 20 rad/s (d) 22 rad/s(e) None of these

Solution: Use conservation of angular momentum. The initial angular momentum isthat of the first flywheel, since the second is at rest. After the two are connected, theyturn at the same angular speed. We’ll use ω1 for the initial angular speed of the firstflywheel, and ωc (for “combined”) for the final angular speed.

I1ω1 = (I1 + I2)ωc ⇒ ωc =I1ω1

I1 + I2

=(240 kg·m2)(30 rad/s)

240 kg·m2 + 150 kg·m2 = 18 rad/s

573

Problem 1560. A door’s moment of inertia is 8 kg·m2. A bullet with a mass of 25g moving at 300 m/s strikes the door perpendicularly at a distance of 75 cm from thehinges. The bullet remains embedded in the door. What is the door’s angular speed afterthe collision? Assume that the embedded bullet does not change the door’s moment ofinertia. Round your answer to two significant figures.


Solution: Use conservation of angular momentum. The initial angular momentum isthat of the bullet. The final angular momentum is that of the swinging door.

mbvbR = Idωd ⇒ ωd =mbvbR

Id=

(0.025 kg)(300 m/s)(0.75 m)

8 kg·m2 = 0.70 rad/s

(In case you’re wondering about the validity of the assumption, the correct moment ofinertia for the door with the embedded bullet is

I = Id + Ib = Id +mbR2 = 8 kg·m2 + (0.025 kg)(0.75 m)2 = 8.014 kg·m2

The difference is less than 1%; it shouldn’t affect the results rounded to two significantfigures.)

Problem 1561. A door’s moment of inertia is 8 kg·m2. A bullet with a mass of 25 g,moving horizontally at 300 m/s, strikes the door at an angle of 60◦ to the door’s surface,at a distance of 75 cm from the hinges. The bullet remains embedded in the door. Whatis the door’s angular speed after the collision? Assume that the embedded bullet doesnot change the door’s moment of inertia. Round your answer to two significant figures.


Solution: Use conservation of angular momentum. The initial angular momentumis that of the bullet. The final angular momentum is that of the swinging door. Sincethe bullet doesn’t strike the door at a right angle, we need to calculate the angularmomentum using only the component of its velocity perpendicular to the door.

mbvbR cos θ = Idωd

⇒ ωd =mbvbR cos θ

Id=

(0.025 kg)(300 m/s)(0.75 m) cos 30◦

8 kg·m2 = 0.61 rad/s

574

Problem 1562. A door’s moment of inertia is 10 kg·m2. A ball with a mass of 40g moving at 30 m/s strikes the door perpendicularly at a distance of 70 cm from thehinges; the ball bounces backward at 25 m/s. What is the door’s angular speed after thecollision? Round your answer to two significant figures.


Solution: Use conservation of angular momentum. Before the collision, the totalangular momentum of the system is that of the ball. After the collision, it consists of theangular momentum of the swinging door plus that of the rebounding ball. We’ll use mfor the mass of the ball, vi for its initial velocity, and vf for its final velocity. We’ll takevi to be positive, so vf will be negative.

Li = Lf = mviR = mvfR + Iωf

⇒ ωf =mR(vi − vf )

I=

(0.040 kg)(0.70 m)[30 m/s− (−25 m/s)]

10 kg·m2 = 0.15 rad/s

Problem 1563. A merry-go-round’s moment of inertia is 900 kg·m2; its radius is 2 m.It is initially at rest. A physics student with a mass of 60 kg is standing on the outeredge. The student jumps off at 2.5 m/s, in a direction tangent to the outer edge. Afterthe student has jumped off, how long does it take for the merry-go-round to make onecomplete revolution? Round your answer to two significant figures.

*(a) 19 s (b) 21 s(c) 23 s (d) 25 s(e) None of these

Solution: Use conservation of angular momentum. Since the system is initially atrest, the final angular momentum of the merry-go-round must match the final angularmomentum of the student. We want to know the period T that it takes to make onecomplete revolution; that’s related to the angular velocity ω by ω = 2π/T .

Ls = Lm = mvR = Iω =2πI

T

⇒ T =2πI

mvR=

2π(900 kg·m2)

(60 kg)(2.5 m/s)(2 m)= 19 s

575

Problem 1564. A merry-go-round’s moment of inertia is 6000 kg·m2. It is initiallyturning frictionlessly at a rate of one revolution per 15 s. A physics instructor with amass of 100 kg is initially standing at the center of the merry-go-round. The instructorthen moves to the outer edge, 4 m from the center. When he has reached the outeredge, how long does it take for the merry-go-round to make one revolution? Round youranswer to two significant figures.


Solution: Since no external torques are applied to the system, angular momentum isconserved: L = Iω. I increases as the instructor moves outward, so ω must decrease tocompensate. We’ll use ωi and ωf to indicate the initial and final angular velocities; Im forthe moment of inertia of the merry-go-round by itself; and Ip for the physics instructor’scontribution to I. We want to know the period T , which is related to ω by ω = 2π/T .

Li = Lf = Imωi = (Im + Ip)ωf =2πImTi

=2π(Im + Ip)

Tf

⇒ Tf =(Im + Ip)Ti

Im=

(Im +mR2)TiIm

=[6000 kg·m2 + (100 kg)(4 m)2](15 s)

6000 kg·m2 = 19 s

Problem 1565. A small wind turbine has a moment of inertia of 400 kg·m2, and blades2 m long. The turbine is at rest, with one blade pointing upward at an angle of 30◦ to thevertical. A goose with a mass of 8 kg, flying horizontally in the plane of the turbine at 20m/s, strikes the end of that blade. Upon colliding with the blade, the goose momentarilycomes to a stop before falling to the ground. What is the turbine’s angular speed afterthe collision? Round your answer to two significant figures.


Solution: Use conservation of angular momentum. Initially, the total angular mo-mentum is that of the goose. After the collision, the goose is at rest, so the total angularmomentum is that of the turbine. The goose doesn’t hit the blade at a right angle, so incalculating its initial angular momentum, we need to use the component of its velocityperpendicular to the blade. Since the goose is flying horizontally and the blade is inclinedat 30◦ to the vertical, the goose hits the blade at θ = 60◦.

Li = Lf = mvR sin θ = Iω

⇒ ω =mvR sin θ

I=

(8 kg)(20 m/s)(2 m) sin 60◦

400 kg·m2 = 0.69 rad/s

576

Problem 1566. A hollow pipe has radius R. It is initially at rest on top of a hill withheight y. The pipe is given a nudge so that it rolls down the hill. How fast is it goingwhen it reaches the bottom?

(a) v =√

2g(y +R) (b) v =√g(y +R)

(c) v =

[2g

(y +

R

2

)]1/2

*(d) v =√gy

(e) None of these

Solution: Use conservation of mechanical energy. Initially, the pipe is at rest, so thetotal mechanical energy is the gravitational potential energy. At the bottom of the hill,the potential energy is zero, and the total energy is kinetic: partly the kinetic energy ofthe pipe’s center of mass moving in a line, and partly the kinetic energy of the pipe’srotation. To determine how to partition the kinetic energy among these, use the no-slipcondition: if a point where the pipe touches the ground is at rest relative to the ground,then relative to the pipe’s center, a point on the outer surface is rotating with a speedequal to the center’s linear speed relative to the ground. Thus ω = v/R.

U0 = Kf = mgy =1

2mv2 +

1

2Iω2 =

1

2mv2 +

1

2(mR2)

( vR

)2

= mv2

⇒ v =√gy

Problem 1567. A thin cylindrical pipe filled with concrete has radius R. It is initiallyat rest on top of a hill with height y. The pipe is given a nudge so that it rolls down thehill. How fast is it going when it reaches the bottom?

(a) v =√

2g(y +R) (b) v =

[2g

(y +

R

2

)]1/2

*(c) v =

√4gy

3(d) v =

√3gy

2

(e) None of these

Solution: Use conservation of mechanical energy. Initially, the pipe is at rest, so thetotal mechanical energy is the gravitational potential energy. At the bottom of the hill,the potential energy is zero, and the total energy is kinetic: partly the kinetic energy ofthe pipe’s center of mass moving in a line, and partly the kinetic energy of the pipe’srotation. To determine how to partition the kinetic energy among these, use the no-slipcondition: if a point where the pipe touches the ground is at rest relative to the ground,then relative to the pipe’s center, a point on the outer surface is rotating with a speedequal to the center’s linear speed relative to the ground. Thus ω = v/R.

U0 = Kf = mgy =1

2mv2 +

1

2Iω2 =

1

2mv2 +

1

2

(mR2

2

)( vR

)2

=3mv2

4

⇒ v =

√4gy

3

577

Problem 1568. (Requires calculus) A physics instructor has mass mi; his wife hasmass mw. The two are standing on a merry-go-round that is rotating frictionlessly.Initially, the instructor is standing at the outer edge, at a distance R from the center; thewife is standing at the center. The two change places, walking past one another at thesame speed in opposite directions. How far from the center will the instructor be whenthe merry-go-round is rotating at its maximum speed?

(a)miR

mi +mw

*(b)mwR

mi +mw

(c)

√miR2

mi +mw

(d)

√mwR2

mi +mw

(e) None of these

Solution: There are no external torques, so angular momentum is conserved: L = Iω.Hence ω will be at its maximum when I is at its minimum. If the instructor is at distancer from the center, his wife is at distance R − r; so their contribution to the moment ofinertia is

I(r) = mir2 +mw(R− r)2

To find the value of r at which I(r) is at a minimum, we want

dI

dr= 0 = 2mir − 2mw(R− r) = 2(mi +mw)r − 2mwR ⇒ r =

mwR

mi +mw

To check that this is really a minimum, note that d2I/dr2 = 2(mi +mw) > 0; so I(r) isconcave upward, and therefore has a minimum at our value of r.

578

Part VI

Applications of Mechanics

579

17 Applications of Mechanics

17.1 Harmonic Motion

Problem 1569. A pendulum goes through 20 complete cycles in 60 seconds. What isits frequency?

*(a)1

3Hz (b)

2π

3Hz

(c) 3 Hz (d)3

2πHz

(e) None of these

Solution: 1 Hz = 1 cycle/s; so

f =20 cycles

60 s=

1

3Hz

Problem 1570. An oscillator is made using a 10 kg block sliding on a frictionless hori-zontal surface, attached to a horizontal spring with a force constant of 4000 N/m. Thespring is initially stretched by 10 cm and held for a moment, then released. As the blockoscillates, what is its maximum speed?


Solution: Use conservation of energy. Initially, the system is not moving, so its totalenergy is the potential energy of the spring. When the block is at its maximum speed,the potential energy of the spring is zero, so the total energy of the system is the kineticenergy of the block. Hence:

U0 +K0 = U0 =1

2kx2

0

Uf +Kf = Kf =1

2mv2

f

⇒ kx20 = mv2

f

⇒ v2f =

kx20

m

⇒ vf =

√k

mx0 =

[4000 N/m

10 kg

]1/2

(0.1 m) = 2 m/s

580

Problem 1571. An oscillator is made using a 5 kg mass sliding on a frictionless hori-zontal surface, attached to a horizontal spring with a force constant of 20 N/m. What isthe period of the oscillator in seconds?

(a) 1 s *(b) π s(c) 2 s (d) 2π s(e) None of these

Solution: We know the mass m and the spring constant k. We want to know theperiod T . We don’t have a formula that gives us the period directly, but we have one forthe angular frequency ω, and we know that ω = 2π/T .

ω =

√k

m=

2π

T⇒ T = 2π

√m

k= 2π

[5 kg

20 N/m

]1/2

= π s

Problem 1572. An astronaut moves his furniture to Planet X. On Earth, the pendulumin his grandfather clock has a period of 1.0 s. On Planet X, the pendulum’s period is 0.5s. What is the gravitational acceleration gX of Planet X in terms of Earth’s gravity gE?

(a) gE/4 (b) gE/2(c) 2gE *(d) 4gE(e) None of these

Solution: We have a formula for the angular frequency ω of a pendulum given themass m and the gravitational acceleration g. We want to know the period T . We knowthat ω = 2π/T . We don’t know the pendulum length L, but it is the same on bothplanets.

ω =2π

T=

√g

L

⇒ 2π√L = T

√g = TE

√gE = TX

√gX

⇒ T 2EgE = T 2

XgX

⇒ gX =T 2EgET 2X

=

[1.0 s

0.5 s

]2

gE = 4gE

581

17.2 Mechanical waves and sound

Problem 1573. A tsunami consists of a series of waves travelling at a speed of 800km/hr, with a wavelength of 200 km. What is the period of the waves?

(a) 15 s (b) 4 min*(c) 15 min (d) 4 hr(e) None of these

Solution: Notice that the speed is given in km/hr, and that the wavelength is givenin km. This means that we don’t need to convert units. We would have had to do so if,for instance, the speed had been given in m/s and the wavelength in km.

v =λ

T⇒ T =

λ

v=

200 km

800 km/hr=

1

4hr = 15 min

Problem 1574. A cable with a linear density of 3 kg/m is stretched between two poles30 m apart. If the cable is plucked, the resulting transverse wave travels from one poleto the other in 1 second. What is the tension in the cable?


Solution: We know the linear density µ of the cable, and we can calculate the speedv from the length l and the time t. We want the tension FT .

v =

√FTµ

⇒ v2 =FTµ

⇒ FT = v2µ =l2µ

t2=

(30 m)2(3 kg/m)

(1 s)2= 2700 N

Problem 1575. A violin string is tuned so that its third harmonic has a frequency of1200 Hz. What is the fundamental frequency of the string?

(a) 150 Hz *(b) 400 Hz(c) 3600 Hz (d) 9600 Hz(e) None of these

Solution: fn = nf1 ⇒ f1 =fnn

⇒ f1 =f3

3=

1200 Hz

3= 400 Hz

582

Problem 1576. An obscure stringed instrument has two strings of identical length andcomposition, called the P and Q strings. The P string has a tension of TP , so that itsfundamental frequency is fP1. The Q string is tuned so that its fundamental frequencyfQ1 is the same as the second harmonic of the P string. What is the tension of the Qstring?

(a) TP/2 (b)√

2TP(c) 2TP *(d) 4TP(e) None of these

Solution: We will start by finding the relationship between fQ1 and fP1.

fQ1 = fP2 = 2fP1

Since f1 = v/2L, and since the strings have equal length (LP = LQ = L),

fQ1 =vQ2L

= 2fP1 =2vP2L

⇒ vQ = 2vP

Since the strings have the same composition, they have the same length density: µP =µQ = µ. Hence

vQ =

√TQµ

= 2vP = 2

√TPµ

⇒ TQµ

=4TPµ

⇒ TQ = 4TP

Problem 1577. A violinist is playing a note with a frequency of 480 Hz. The violinistnext to him is playing a note with a frequency of 500 Hz. What is the beat frequencyproduced by the two notes?

*(a) 20 Hz (b) 460 Hz(c) 490 Hz (d) 520 Hz(e) None of these

Solution: fbeat = f1 − f2 = 500 Hz− 480 Hz = 20 Hz

583

Problem 1578. Two wires of equal length L are stretched between two supports. Bothwires are subjected to the same tension FT . The first wire has a mass of MA; the second,of 3MA. If the fundamental frequency of the first wire is fA and that of the second wireis fB, which of the following equations is true?

(a) fB =fA3

*(b) fB =fA√

3

(c) fB =√

3 fA (d) fB = 3fA

(e) None of these

Solution: We have an equation for fundamental frequency, which requires the wavespeed v. We have an equation for wave speed, which requires the mass per unit lengthµ = M/L.

f1 =v

2Land v =

√FTµ

⇒ f1 =

√FT

2L√µ

=

√FTL

2L√M

=

√FT

2√LM

In this problem, L and FT are the same for both wires; M and f1 are different. Weseparate the constant part from the variable one:

f1

√M =

FT

2√L

= fA√MA = fB

√3MA

⇒ fB =fA√MA√

3MA

=fA√

3

Problem 1579. At a distance of 4 m from an omnidirectional speaker, the sound inten-sity is 9 W/m2. At what distance is the intensity 1 W/m2?


Solution: The intensity of sound varies inversely with the square of the distance. Letr1 = 4 m, I1 = 9 W/m2, and I2 = 1 W/m2. We want the distance r2.

I =I0

r2

⇒ I0 = Ir2 = I1r21 = I2r

22

⇒ r22 =

I1r21

I2

⇒ r2 =

√I1

I2

r2 =

[9 W/m2

1 W/m2

]1/2

(4 m) = 12 m

584

Problem 1580. A physics professor is in a car stalled on a railroad track. An oncomingtrain sounds its horn, and the professor notes that he hears it at a frequency of fp. Heknows that when the train is standing still, its horn has a frequency of ft; and that thelocal speed of sound is v0. How fast is the train approaching him?

(a)fp − ftft

v0 *(b)fp − ftfp

v0

(c)ft

fp − ftv0 (d)

fpfp − ft

v0

(e) None of these

Solution: We know the source frequency fS = ft and the listener frequency fL = fp,the listener speed vL = 0, and the local speed of sound v = v0. We have a formulafor Doppler shift. We want to know the train’s approach speed vt. We will modify thetextbook’s formula slightly, since in it the source speed is positive when it’s going awayfrom the listener; we want vt to be the speed toward the professor, so we’ll use vS = −vt.

fL =v + vLv + vS

fS

substitute−−−−−→ fp =v0

v0 − vtft

×(v0−vt)−−−−−→ v0fp − vtfp = v0ft+vtfp−v0ft−−−−−−→ vtfp = v0fp − v0ft = v0(fp − ft)

÷fp−−→ vt =fp − ftfp

v0

17.3 Elasticity

Problem 1581. A mop handle makes an angle of θ to the horizontal. A force of F isapplied along the mop handle; the mop contacts the floor over an area of A. What is thepressure exerted by the mop on the floor?

(a)F

A cos θ(b)

F

A sin θ

(c)F cos θ

A*(d)

F sin θ

A

(e) None of these

Solution: The component of F perpendicular to the floor is Fy = F sin θ. This forceis distributed over an area of A; so the pressure is Fy/A = F sin θ/A.

585

Problem 1582. Two wires hang from the ceiling. The wires are identical, except thatone is twice as long as the other. If a 20 kg mass is attached to the shorter wire, itstretches by 1 mm. If the same mass is attached to the longer wire, how much will itstretch?

(a) 1 mm (b)√

2 mm*(c) 2 mm (d) 4 mm(e) None of these

Solution: This is a situation of tensile stress. Since the two wires are identical incomposition, they have the same Young’s modulus Y ; and they have the same cross-sectional area A. Their lengths are different: l1 and l2 = 2l1. Each is subjected to thesame tensile force, F⊥ = 20 kg · g. The first wire stretches by ∆l1 = 1 mm. We want toknow the length change ∆l2 of the second wire.In this situation, Hooke’s law is

F⊥A

= Y∆l

l

Since only l and ∆l differ from one wire to the other, we can write:

F⊥Y A

=∆l

l=

∆l1l1

=∆l2l2

⇒ ∆l2 =l2∆l1l1

=2l1∆l1l1

= 2∆l1 = 2(1 mm) = 2 mm

586

Problem 1583. Two balls made of different materials are dropped from the deck of aship and sink to the bottom of an ocean trench. On the surface, the first ball has avolume of 60 cm3 and the second ball a volume of 120 cm3. At the bottom of the trench,the first has a volume of 50 cm3 and the second a volume of 110 cm3. If the bulk modulusof the first ball is B1 and that of the second is B2, which of the following is true?

*(a) B1 < B2 (b) B1 = B2

(c) B1 > B2 (d) Not enough information(e) None of these

Solution: We know the initial and final volumes for the two balls. We can calculatethe volume changes:

∆V1 = 60 cm3 − 50 cm3 = 10 cm3

∆V2 = 120 cm3 − 110 cm3 = 10 cm3

We don’t know the pressure at the bottom of the trench, but we know that it’s the samefor both balls. We want to know about the bulk moduli B1 and B2. The relationshipbetween volume, pressure, and bulk modulus is

P = B∆V

V

Since the pressure is the same for both balls, we can write

B1∆V1

V1

=B2∆V2

V2

⇒ B1

B2

=V1

∆V1

· ∆V2

V2

=60 cm3

10 cm3· 10 cm3

120 cm3=

1

2

⇒ 2B1 = B2

Since B is positive, B1 < B2.

Problem 1584. A cable has a cross-sectional area of 4 cm2. It will break if subjectedto a tensile stress greater than 2× 108 Pa. How much force must be applied to the cablein order to break it?

*(a) 8× 104 N (b) 5× 107 N(c) 8× 108 N (d) 5× 1012 N(e) None of these

Solution: Don’t forget to convert centimeters to meters. Remember that 1 Pa = 1N/m2.

P =F

A⇒ F = PA = (2.0× 108 N/m2)(4× 10−4 m2) = 8× 104 N

587

17.4 Static Fluids

17.4.1 Pascal’s Law and Archimedes’s principle

Problem 1585. Which of the following is a statement of Pascal’s law?

(a) If a gas is maintained at a constant temperature, then its volume will beinversely proportional to its pressure.

*(b) Pressure applied to an enclosed fluid is transmitted undiminished to everyportion of the fluid and to the walls of the container.

(c) A body immersed in a fluid is buoyed up with a force equal to the weight ofthe fluid displaced by the body.

(d) One mole of any ideal gas at standard temperature and pressure will have avolume of 22.4 liters.

Problem 1586. Which of the following is a statement of Archimedes’s principle?

(a) If a gas is maintained at a constant pressure, then its volume will be propor-tional to its absolute temperature.

(b) Pressure applied to an enclosed fluid is transmitted undiminished to everyportion of the fluid and to the walls of the container.

*(c) A body immersed in a fluid is buoyed up with a force equal to the weight ofthe fluid displaced by the body.

(d) Don’t be rude to armed Romans.

17.4.2 Density, specific volume, specific weight

Problem 1587. The density of air is 1.3 kg/m3. A room is 5.6 m long by 4.9 m deep by2.4 m high. What is the mass of the air in the room? Round your answer to the nearestkm.

(a) 62 kg (b) 69 kg(c) 77 kg *(d) 86 kg(e) None of these

Solution: The mass equals the density times the volume. We know the density ρ,the length l, the width w, and the height h. Hence

m = ρV = ρlwh = (1.3 kg/m3)(5.6 m)(4.9 m)(2.4 m) = 86 kg

588

Problem 1588. The density of air at sea level is 1.20 kg/m3. The living room in yourseaside cottage is 5.0 m wide, 4.0 m deep, and 3.0 m high. What is the weight of the airin the room? Give your answer to the nearest pound. Recall: 1 lb = 4.448 N.


Solution: The weight of a volume of gas is the mass times gravity = density timesvolume times gravity, and the the volume of a rectangular box is Volume = length ×width × height:

w = mg = (ρV )g = gρlwh = (1.20 kg/m3)(5.0 m)(4.0 m)(3.0 m) = 710 N.

Lastly, we must convert Newtons to pounds:

710 N = 710 N

(1 lb

4.448 N

)= 160 lb.

Problem 1589. Your new roommate has left the water running in the sink of yourrectangular water-tight dorm room, and the room is now full of water. If the dimensionsof the room are: length = 5.0 m, width = 4.0 m, and height = 3.0 m, and the density ofwater is 1.0× 103 kg/m3, then what is the weight of the water in the room? Round youranswer to two significant figures.

(a) 6.5× 106 N *(b) 5.9× 105 N(c) 6.0× 104 N (d) 5.4× 103 N(e) None of these

Solution: wwater = mwaterg = ρwaterV g, where V is the volume of the container. Wehave used the definition of density ρ = m/V to find the mass of the water. The volumeof a rectangular box is length × width × height = LWH. Thus

w = ρwaterV g = ρwaterLWHg = (103 kg

m3)(5m)(4m)(3m)gm/s2 = 64(9.8)N ≈ 65 .

Problem 1590. Your new roommate has filled his room with nitrous oxide gas, whichhas a density of 1.96 kg/m3. The room is 4.0 m long by 5.0 m wide by 3.0 m high. Whatis the mass of the N2O in the room? Round your answer to the nearest kilogram.

(a) 115 kg (b) 57 kg*(c) 118 kg (d) 164 kg(e) None of these

Solution: From the definition of density ρ = m/V we have mN2O = ρN2OV , wherethe volume of the rectangular room is V = LWH. Thus

mN2O = ρwaterLWH = (1.96kg

m3)(4m)(5m)(3m)m/s2 ≈ 120kg .

589

Problem 1591. Your new roommate has filled his room with helium, which has a densityof 0.18 kg/m3. The room is 4.0 m long by 5.0 m wide by 3.0 m high. What is the massof the helium in the room? Round your answer to two significant figures.

(a) 0.71 kg *(b) 11 kg(c) 1.0 kg (d) 28 kg(e) None of these

Solution: From the definition of density and the volume of a rectangular room:

mhelium = ρheliumLWH = (0.18kg

m3)(4m)(5m)(3m)m/s2 = 10.8 ≈ 11kg ,

where L = length, W = width, and H = height.

Problem 1592. The mass of an unknown gas mixture in a room that is 3 m × 4 m ×5 m is known to be 500 kg. What is the density of the gas (ρ = m

V)?

Solution: From the definition of density and the volume of a rectangular room:

ρ =m

LWH=

500 kg

(5 m)(4 m)(3 m)=

25

3

kg

m3= 8.33

kg

m3,

where m = mass, L = length, W = width, and H = height.

17.4.3 Static Pressure

Problem 1593. A solar-water heating system uses solar panels on the roof of a largebuild, 40.0 m above the storage tank. The pressure at the panels is 1 atmosphere. Whatis the absolute pressure at the top of the tank?

Solution: Use the formula for static pressure: Pabs = Patm + ρgh.

Problem 1594. A solar-water heating system uses solar panels on the roof of a largebuild 10 m above the top of a large storage tank. The pressure at the panels is 1atmosphere. What is the gage pressure at the top of the tank? Take the density of waterto be ρ = 1000 kg

m3 and g = 10 ms2

.

(a) 1 kPa (b) 10 kPa*(c) 102 kPa (d) 103 kPa(e) None of these

Solution: Use the equation

Pgage = ∆P = Pabs − Patm = ρgh = (103 kg

m3)(10

m

s2)(10m) = 105 N

m2= 105 Pa = 102 kPa

590

Problem 1595. A town in the middle-of-nowhere is built up around a large 100 meterhill. Since the hill is in the center of a town, it is proposed at a town-hall meeting that awater tower be placed at the top of the hill. However, the town barber points out thatwhen water flows through a pipe there is a great amount of resistance owing to viscosityand turbulent mixing of the fluid. He estimates that the gauge pressure in the pipe at thebottom of the hill must be at least 100 kPa in order for water to make it to the outskirtsof town. Find what the gauge pressure would be at the bottom of the hill.

Solution: Use Pgage = Pabs − Patm = ρgh.

Problem 1596. A submarine has a circular top hatch with a 1 m diameter. What isthe force exerted on the hatch from the pressure due to the water above the hatch whenthe sub dives down to a depth of 200 m? Take the approximate density of sea water tobe ρ = 1000 kg

m3 and g = 10 ms2

.Note: we are ignoring atmospheric pressure Patm in this problem and we are assumingthe submarine is a rigid body (i.e., we can ignore the pressure from inside the sub).(a) 2π × 105 N (b) π × 106 N(c) 2π × 106 N *(d) π

2× 106 N

(e) None of these

Solution: First we’ll find the pressure on the hatch, then we’ll use the definition ofpressure to find the force Fhatch = PhatchAhatch, where Fhatch is the force on the hatchfrom the pressure of the sea water above the hatch, Phatch is the pressure on the hatch,and Ahatch is the area of the hatch.

Phatch = ρgh = (103 kg

m3)(10

m

s2)(2× 102 m) = 2× 106 Pa = 2× 106 N

m2.

Using the definition of pressure we find:

Fhatch = PhatchAhatch = (πr2hatch)Phatch = π

(1

2

)2

m2(2× 106

) N

m2=π

2× 106 N

Problem 1597. A vertical, frictionless piston-cylinder device contains a gas that is inequilibrium with the weight of the piston balanced by the internal pressure for the gas.Suppose the mass of the piston is mpiston, the area of the piston cylinder is Acyl, and theatmospheric pressure outside the cylinder is Patm. Derive a formula for the gas containedin the cylinder Pgas in terms of the given information.

Solution: Using a free-body diagram it can be seen that

Fgas = Fatm +mpistongAcyl−−−−−−→ Pgas = Patm +

mpistong

Acyl.

591

Problem 1598. The basic barometer can used to measure the height of a building. Todo this a measurement is made at the bottom of the building at ground level and a secondmeasurement is taken at the top of the building on the roof. One can then relate thesereadings to pressure using the equation for a mercury-filled barometer: Patm(hHg) =ρHgghHg, where ρHg = 13, 600 kg/m3 is the density of mercury, g is the acceleration dueto gravity, and hHg is the height of the mercury column in the barometer measured inmillimeters. The approximate height of the building can then be found by making a fewsimplifying assumptions. For moderate sized buildings a reasonable assumption for theaverage density of air under fair weather conditions is approximately ρair ≈ 1.18 Kg/m3.

If the barometric readings at the top and bottom of a certain building are 730 mmHgand 755 mmHg respectively, what is the height of the building?

Solution: See handwritten solutions

Problem 1599. A gas is contained in a vertical, frictionless piston-cylinder device. Thepiston has a mass of 4 kg and cross-sectional area of 35 cm2. A compressed spring abovethe piston exerts a force of 60 N on the piston. If the atmospheric pressure is 95 kPa,determine the pressure inside the cylinder.


Problem 1600. High-altitude balloons are often filled with helium gas because it weighsonly about one-seventh of what air weighs under identical conditions. The buoyancyforce, which can be expressed as Fb = ρairgVballoon, will push the balloon upward. If weapproximate the shape of the balloon as a sphere with a radius of 5 m and a total payloadof 140kg (including the ropes and cage), determine the acceleration of the balloon whenit is first released. Assume the density of air is ρair = 1.16 kg/m3.


Problem 1601. A glass tube is attached to a water pipe as shown in the figure below.If the water pressure at the bottom of the tube is 115 kPa and the local atmosphericpressure is Patm = 92 kPa, determine how high the water will rise in the tube, in m.Assume g = 9.8 m/s2 at that location and take the density of water to be 103 kg/m3.

592

direction of fluid flow

h = ?

Patm

Figure 16: Vertical glass tube attached to horizontal water pipe.


Problem 1602. A pressure cooker cooks a lot faster than an ordinary pan by maintaininga higher pressure and temperature inside the cooker. The lid of a pressure cooker is wellsealed, and the steam can escape only through an opening in the middle of the lid. Aseparate piece of metal, the petcock, sits on top of this opening and prevents steam fromescaping until the pressure inside the cooker provides enough force on the petcock to overcome the weight of the petcock allowing the hot gas to escape. This mechanism providesa safety device that limits the maximum pressure that can form in the pressure cookerby providing a periodic release of the high-pressure gas in the container. This in turnprevents extremely high pressures from building up in the cooker that could lead to apotentially deadly explosion of hot gas and metal shards.

Assuming that once the cooker reaches the maximum pressure that the gas remainsconstant in the pressure cooker, determine the mass of the petcock needed for a pressurecooker that is designed to have an operation gage pressure of 100 kPa and has an openingcross-sectional area of 4 mm2. Assume an atmospheric pressure of 101 kPa. Be sure anddraw a free-body diagram of the petcock to accompany your solution.


593

18 Introduction to Thermodynamics

18.1 Introductory Concepts and Definitions

18.1.1 What is the study of thermodynamics?

Problem 1603. Which of the following items does not describe the main focus of thestudy of thermodynamics.(a) energy storage(b) the transfer of energy through heat and work(c) how energy transforms from one form of energy, such as kinetic, into another form*(d) how energy is bought and sold on the open market

Solution: While holding down cost and making equipment energy efficient is alwaysa concern for engineerings, the particulars of how energy is bought and sold is typicallynot a concern. Answer (d).

18.1.2 Defining Systems

Problem 1604. Which statement best describes the concept of a system?*(a) the subject of the analysis(b) an object with fixed set of molecules(c) a fluid together with some sort of solid(d) an object that radiates heat into its environment

Solution: (a)

Problem 1605. Consider a refrigerator in a kitchen. Take the refrigerator and every-thing in it to be our system. Which best describes the system’s surroundings?(a) All of the air in the kitchen.(b) Any one standing in the kitchen.(c) The air inside the refrigerator.*(d) Everything in the universe external to the system.

Solution: (d)

Problem 1606. Consider a refrigerator in a kitchen. Take the refrigerator and every-thing in it to be our system. Which best describes the system’s boundary?(a) All of the air in the kitchen.(b) The thin region separating the system from everything else.(c) The air inside the refrigerator.(d) The walls of the refrigerator.

Solution: Only (b) is correct. The walls are part of the system, but out could thinkof the thin metal layer as the boundary in practice.

594

18.1.3 Closed systems

Problem 1607. Identify all of the true statements:(a) A closed system is necessarily an isolated system*(b) An isolated system is necessarily a closed system(c) A system cannot be both closed and isolated.(d) The concepts of closed and isolated in regards to a system are independent concepts.

Solution: Only (b) is true. In an isolated system no energy can escape. If mass wereto escape, then by e = mc2, energy would also escape.

18.1.4 Control volume

Problem 1608. Identify all of the true statements:*(a) closed system = control mass(b) closed system = control volume(c) open system = control mass*(d) open system = control volume

Solution: Only (a) and (d) are true.

Problem 1609. Which of the following situations would be well suited for using a controlvolume in the thermodynamic analysis of the system? Do not worry about the details ofthe analysis.(a) compression of air in a cylinder(b) expansion of gases in a cylinder after a combustion(c) the air in a balloon*(d) filling a bike tire with air from a compressor

Solution: The answer is (d). The bike tire is not a closed system, but the volume isclearly defined as the inside of the tube in the tire. Notice that the control volume canchange shape.

Problem 1610. Which of the following situations would be well suited for using a controlmass in the thermodynamic analysis of the system? Do not worry about the details ofthe analysis.*(a) compression of air in a sealed cylinder(b) a window unit air conditioner(c) a jet engine(d) a fire hose spurting out water

Solution: Only (a) the system in (a) is a closed system.

595

18.1.5 Property, state, and process

Problem 1611. Consider the following two statements:(i) A system is in steady state if its properties are independent of space.(ii) A system is in steady state if its properties are independent of time.Which of the following statements is true:(a) Statement (i) is true and statement (ii) is true(b) Statement (i) is true and statement (ii) is false*(c) Statement (i) is false and statement (ii) is true(d) Statement (i) is false and statement (ii) is false

Solution: Steady-state ⇒ time-independent

18.1.6 Extensive and intensive properties

Problem 1612. Which of the following is not an extensive property?(a) Kinetic Energy (b) Momentum(c) Mass *(d) Density(e) None of these

Problem 1613. Which of the following is not an intensive property?(a) Velocity *(b) Volume(c) Pressure (d) Temperature(e) None of these

18.1.7 Equilibrium, quasi-equilibrium, and processes

Problem 1614. Two metal blocks, one at 50◦, and the other at 0◦ are set next to eachother in a perfectly insulted box at time t = 0. At this instant are the blocks in thermalequilibrium? Now suppose you wait a long time. Are they in equilibrium now?

Solution: At time t = 0 the blocks are not in thermal equilibrium. As time increases(i.e., t→∞) the system approaches equilibrium.

Problem 1615. In a quasi-equilibrium process, the pressure in a system(a) is held constant throughout the entire process*(b) is approximately spatially uniform throughout the system at each moment in time(c) increases if volume increases(d) always varies with temperature

Problem 1616. Which of the following process is a quasi-equilibrium process?(a) the stirring and mixing of cold creamer in hot coffee(b) a balloon bursting(c) combustion*(d) the slow and steady compression of air in a cylinder

596

18.1.8 Base SI units for mass, length, time, Force, energy, and pressure

Problem 1617. Express kinetic energy in terms of the base SI units: the kilogram,meter, and second. Write your answer in terms of the abbreviations ({kg, m, s}), andthen identify the resulting derived unit.

Solution: Apply the uninator:

[KE] = [1

2mv2] = [m] · [v]2 = kg

(m

s

)2

=(

kgm

s2

)·m = N ·m = J

Problem 1618. Express work (force × distance) in terms of the base SI units: thekilogram, meter, and second. Express your answer in terms of the abbreviations (kg, m,s).


[W ] = [F · d] = [F ] · [d] = N ·m = J

Problem 1619. Express PV , where P is pressure and V is volume, in terms of thebase SI units: the kilogram, meter, and second. Express your answer in terms of theabbreviations (kg, m, s). Compare this answer to the previous one. What do you notice?


[PV−] =

[F

A· V−]

=

(N

m2

)·m3 = J

Thus, preesure times volume has the units of work. In fact, it will be the most commonform of work that we will encounter. It is the work done by an expanding gas in movinga boundary.

Problem 1620. Express power in terms of the base SI units: the kilogram, meter, andsecond. Write your answer in terms of the abbreviations ({kg, m, s}), and then identifythe resulting derived unit.


[P ] =

[W

t

]=

J

s= watt

Problem 1621. Express specific weight in terms of the base SI units: the kilogram,meter, and second. Express your answer in terms of the abbreviations ({kg, m, s}), andthen identify the resulting derived unit.


[ρg] =

[m

V−g

]=

kg

m3· N

kg=

Pa

m⇒ [ρgh] = Pa,

as expected, since ρgh is known to be the pressure owing to the weight the fluid h unitsabove the point where the pressure is measured.

597

Problem 1622. Express specific volume in terms of the base SI units: the kilogram,meter, and second. Express your answer in terms of the abbreviations ({kg, m, s}).Solution: The specific volume is just the reciprocal of density. Work it out!

Problem 1623. Which one of the following expressions can be converted to the unit ofa Joule?(a) Pa ·m2 *(b) Pa ·m3

(c) Pa/m2 (d) N/kg(e) None of these

Solution: Since Pa = N/m2 and J = N·m, it follows that (b) is the answer.

Problem 1624. Which of the following is not an acceptable “extended” SI unit? Re-call: The SI system is the MKS system (Meter, Kilogram, Second), but we’ll allow theextended SI system to include the cgs system (centimeter, gram, second), but you can’tmix these two systems.(a) distance measured in centimeters(b) pressure measured in newtons per square meter(c) volume measured in cubic meters*(d) density measured in grams per cubic meter

Solution: Can’t mix grams and meters.

598

18.2 Temperature Scales

Use the equations TC = TK − 273 and TF = 95TC + 32 to answer the following questions.

Round your answers to the nearest integer.

Problem 1625. If heat is added to a system and the temperature of a system increases,without knowing anything else, which form of energy will be definitely increase?(a) The kinetic energy of the system(b) The potential energy of the system(c) The work done by the system*(d) The internal energy (i.e., the molecular energy) of the system

Solution: If the temperature increases, then the internal energy, which dependson temperature, will certainly rise. But adding heat to a system need not result in anincrease in bulk energy of the system. For example, consider a pot of water sitting ona hot-plate. If we take the water as our system, then clearly heat is being added to thesystem, but since the pot is not moving, its potential and kinetic energy are not changing.Since the boundary of the system is not moving, there is no work being done to, or bythe system.

Problem 1626. Convert 98◦F to ◦C.

(a) 32◦C (b) 208◦C(c) 20◦C *(d) 37◦C(e) None of these

Solution: TC =5

9(TF−32) =

5

9(98−32)◦C =

5

9·66◦C =

5

3·22◦C = (35+

5

3)◦C ≈ 37◦C


(a) 12◦C (b) 17◦C*(c) 20◦C (d) 37◦C(e) None of these

Solution: TC =5

9(TF − 32) = 20◦C


(a) 50◦C *(b) 43◦C(c) 20◦C (d) 37◦C(e) None of these

Solution: TC =5

9(TF − 32) = 43.3◦C ≈ 43◦C

Problem 1629. Convert 20◦C to ◦F.

*(a) 68◦F (b) 98◦F(c) 120◦F (d) 212◦F(e) None of these

Solution: TF =9

5TC + 32 = 68◦F

599

Problem 1630. Convert 100◦C to ◦F.

(a) 68◦F (b) 98◦F(c) 120◦F *(d) 212◦F(e) None of these

Solution: TF =9

5TC + 32 = 212◦F

Problem 1631. Convert 20◦C to K.

(a) −253 K *(b) 293 K(c) 68 K (d) 0 K(e) None of these

Solution: TK = TC + 273 K = (20 + 273) K = 293 K.

Problem 1632. Convert 10 K to ◦C.

*(a) −263 K (b) 293 K(c) 68 K (d) 0 K(e) None of these

Solution: TC = TK − 273◦C = (10− 273)◦C = −263◦C.

Problem 1633. Convert 98◦F to ◦R (degrees rankine).

(a) 460◦R (b) −558◦R*(c) 558◦R (d) 200◦R(e) None of these

Solution: TR = TF + 459.67 ≈ TF + 460 = 558◦R.

Note: Some books say do not use the degree symbol with the Rankine scale, since it’san absolute scale like the Kelvin scale.

600

Problem 1634. (Measuring wind chill) It is well-known that cold air feels muchcolder in windy weather than what the thermometer reading indicates because of the“chilling effect” of the wind. This effect is due to the increase in the convection heattransfer coefficient with increasing air velocities. The equivalent wind chill temperaturein ◦F is given by (all variables with an asterisk superscript (∗) are dimensional quantities)

T ∗equiv,F (◦F) = T ∗adj,F (◦F) + (T ∗ambient,F (◦F)− T ∗adj,F (◦F)) f(vb)

where f is the dimensionless function (which must have dimensionless arguments in orderto be dimensionally consistent! - for any smooth function this can be proved via a Taylorseries expansion)

f(vb) = 0.475− 0.0203vb + 0.304√vb ,

and T ∗adj,F = 91.4◦F is a temperature adjustment term, vb is a dimensionless wind speedthat has been non-dimensionalized by miles per hour (i.e. v∗b = vb , (miles/hour andT ∗ambient is the ambient air temperature in ◦F in calm air, which is taken to be lightwinds at speeds of no more than 4 mph. The constant 91.4◦F in the above equation isthe mean temperature of a resting person in a comfortable environment. Windy air attemperature Tambient and velocity v will feel as cold as the calm air at temperature Tequiv.Notice that the units of the coefficients in the second factor of the second term are, inorder: dimensionless, hours per mile, and the square root of hours per mile so that theresulting second factor is dimensionless after the dimensional variables are substitutedinto the equation.

Using proper conversion factors, obtain an equivalent relation in SI units where vSI isthe dimensionless wind speed based on m/s and Tambient is ambient air temperature in◦C.

Hint: When working with physical equations all terms must be dimensionally consistent.To preserve the dimensional consistency whatever we do to one side of the equation, wemust do to the other, just as you learned in algebra.

Solution: Starting from the original equation we first transform the dimensionalcomponent (i.e., the temperature component has physical dimensions, but the factorf(vb) is dimensionless).

T ∗equiv,F (◦F) = T ∗adj,F (◦F) + (T ∗ambient,F (◦F)− T ∗adj,F (◦F)) f(vb)

−32◦F−−−−−−−→ T ∗equiv,F (◦F)− 32◦F

= (T ∗adj,F (◦F)− 32◦F) + (T ∗ambient,F (◦F)− T ∗adj,F (◦F)) f(vb)

5◦C9◦F−−−−−−→

(5◦C

9◦F

)(T ∗equiv,F (◦F)− 32◦F

)=

(5◦C

9◦F

)(T ∗adj,F (◦F)− 32◦F)

+

(5◦C

9◦F

)(T ∗ambient,F (◦F)− T ∗adj,F (◦F)

)f(vb) .

601

Using the formula T ∗C(◦C) =

(5◦C

9◦F

)(T ∗F (◦F)− 32◦F) we can rewrite the equation as

T ∗equiv,C (◦C) = 33◦C +

(5◦C

9◦F

)(T ∗ambient,F (◦F)− T ∗adj,F (◦F)

)f(vb)

rewrite−−−−−−−→ T ∗equiv,C (◦C)

= 33◦C +

(5◦C

9◦F

)((T ∗ambient,F (◦F)− 32◦F)− (T ∗adj,F (◦F)− 32◦F)

)f(vb)

= 33◦C

+((5◦C

9◦F

)(T ∗ambient,F (◦F)− 32◦F)−

(5◦C

9◦F

)(T ∗adj,F (◦F)− 32◦F)

)f(vb)

= 33◦C +(T ∗ambient,C (◦C)− 33◦C

)f(vb)

The last step is to convert the correction factor f(vb).

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