Chapter 5. Answers to Even Numbered Problems m kj. 6. (a) 900 J (b) (a) 31.9 J (b) 0 (c) 0 (d) 31.9 J. 10.

Similar documents
Chapter Seven - Potential Energy and Conservation of Energy

Chapter 8 Potential Energy and Conservation of Energy Important Terms (For chapters 7 and 8)

Conservation of Energy

where v means the change in velocity, and t is the

Chapter 07: Kinetic Energy and Work

TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES. PHYS 2211, Exam 2 Section 1 Version 1 October 18, 2013 Total Weight: 100 points

Chapter 8: Potential Energy and The Conservation of Total Energy

Chapter 3. r r. Position, Velocity, and Acceleration Revisited

EMU Physics Department

PHYS 1441 Section 002 Lecture #16

Physics 2A Chapter 3 HW Solutions

Conservation of Energy

Chapter 3 and Chapter 4

Problem While being compressed, A) What is the work done on it by gravity? B) What is the work done on it by the spring force?

PHYS 1443 Section 004 Lecture #12 Thursday, Oct. 2, 2014

Spring Force and Power

You will analyze the motion of the block at different moments using the law of conservation of energy.

Dynamics 4600:203 Homework 08 Due: March 28, Solution: We identify the displacements of the blocks A and B with the coordinates x and y,

CHAPTER 8 Potential Energy and Conservation of Energy

Chapter 7: Conservation of Energy

Physics for Scientists and Engineers. Chapter 9 Impulse and Momentum

Name: PHYS 110 Dr. McGovern Spring 2018 Exam 1. Multiple Choice: Circle the answer that best evaluates the statement or completes the statement.

Work is the change in energy of a system (neglecting heat transfer). To examine what could

Week3, Chapter 4. Position and Displacement. Motion in Two Dimensions. Instantaneous Velocity. Average Velocity

Lecture 16. Chapter 11. Energy Dissipation Linear Momentum. Physics I. Department of Physics and Applied Physics

Chapter 7. Potential Energy and Conservation of Energy

Physics 207 Lecture 13. Lecture 13

Study Guide For Exam Two

PHYS 1441 Section 002 Lecture #15

Physics 40 HW #4 Chapter 4 Key NEATNESS COUNTS! Solve but do not turn in the following problems from Chapter 4 Knight

AP Physics Enosburg Falls High School Mr. Bushey. Week 6: Work, Energy, Power

Lesson 4: Relative motion, Forces, Newton s laws (sections )

For the three forces. find: (a) the resultant force R~ (a) (b) the magnitude of the resultant force. Three coplanar forces, A

Period & Frequency. Work and Energy. Methods of Energy Transfer: Energy. Work-KE Theorem 3/4/16. Ranking: Which has the greatest kinetic energy?

Angular Momentum and Fixed Axis Rotation. 8.01t Nov 10, 2004

Physics 207, Lecture 13, Oct. 15. Energy

Energy and Energy Transfer

PHYSICS 203-NYA-05 MECHANICS

A Tale of Friction Basic Rollercoaster Physics. Fahrenheit Rollercoaster, Hershey, PA max height = 121 ft max speed = 58 mph

= 1.23 m/s 2 [W] Required: t. Solution:!t = = 17 m/s [W]! m/s [W] (two extra digits carried) = 2.1 m/s [W]

PHYS 1101 Practice problem set 12, Chapter 32: 21, 22, 24, 57, 61, 83 Chapter 33: 7, 12, 32, 38, 44, 49, 76

Page 1. Clicker Question 9: Physics 131: Lecture 15. Today s Agenda. Clicker Question 9: Energy. Energy is Conserved.

Week 11: Chapter 11. The Vector Product. The Vector Product Defined. The Vector Product and Torque. More About the Vector Product

Physics 207: Lecture 20. Today s Agenda Homework for Monday

Slide. King Saud University College of Science Physics & Astronomy Dept. PHYS 103 (GENERAL PHYSICS) CHAPTER 5: MOTION IN 1-D (PART 2) LECTURE NO.

Chapter 8. Potential Energy and Conservation of Energy

Physics 101 Lecture 9 Linear Momentum and Collisions

EN40: Dynamics and Vibrations. Homework 4: Work, Energy and Linear Momentum Due Friday March 1 st

Physics 141. Lecture 14. Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 14, Page 1

ˆ (0.10 m) E ( N m /C ) 36 ˆj ( j C m)

Physics 2A Chapters 6 - Work & Energy Fall 2017

PHYS 705: Classical Mechanics. Newtonian Mechanics

K = 100 J. [kg (m/s) ] K = mv = (0.15)(36.5) !!! Lethal energies. m [kg ] J s (Joule) Kinetic Energy (energy of motion) E or KE.

Mechanics Cycle 3 Chapter 9++ Chapter 9++

Supplemental Instruction sessions next week

Chapter 11 Angular Momentum

10/24/2013. PHY 113 C General Physics I 11 AM 12:15 PM TR Olin 101. Plan for Lecture 17: Review of Chapters 9-13, 15-16

GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME PHYSICAL SCIENCES GRADE 12 SESSION 1 (LEARNER NOTES)

11. Dynamics in Rotating Frames of Reference

EMU Physics Department.

Angular momentum. Instructor: Dr. Hoi Lam TAM ( 譚海嵐 )

Prof. Dr. I. Nasser T /16/2017

Recitation: Energy, Phys Energies. 1.2 Three stones. 1. Energy. 1. An acorn falling from an oak tree onto the sidewalk.

Conservation of Energy

ONE-DIMENSIONAL COLLISIONS

Week 6, Chapter 7 Sect 1-5

1 cos. where v v sin. Range Equations: for an object that lands at the same height at which it starts. v sin 2 i. t g. and. sin g

Physics 207 Lecture 6

Rotational and Translational Comparison. Conservation of Angular Momentum. Angular Momentum for a System of Particles

in state i at t i, Initial State E = E i

Physics 106 Lecture 6 Conservation of Angular Momentum SJ 7 th Ed.: Chap 11.4

Linear Momentum and Collisions

Chapter 2. Pythagorean Theorem. Right Hand Rule. Position. Distance Formula

Physics 105: Mechanics Lecture 13

Physics 114 Exam 2 Spring Name:

Module 11 Design of Joints for Special Loading. Version 2 ME, IIT Kharagpur

Physics 131: Lecture 16. Today s Agenda

Part C Dynamics and Statics of Rigid Body. Chapter 5 Rotation of a Rigid Body About a Fixed Axis

PHYSICS 231 Review problems for midterm 2

ENGN 40 Dynamics and Vibrations Homework # 7 Due: Friday, April 15

Physics 111: Mechanics Lecture 11

Page 1. SPH4U: Lecture 7. New Topic: Friction. Today s Agenda. Surface Friction... Surface Friction...

Physics 2A Chapter 9 HW Solutions

Physics 5153 Classical Mechanics. D Alembert s Principle and The Lagrangian-1

Lecture 22: Potential Energy

1. The number of significant figures in the number is a. 4 b. 5 c. 6 d. 7

10/9/2003 PHY Lecture 11 1

Physics 201 Lecture 9

10/23/2003 PHY Lecture 14R 1

Week 9 Chapter 10 Section 1-5

Ch04 Work, Energy and Power What is work in physics?

Chapter 11: Angular Momentum

Motion in One Dimension

Degrees of Freedom. Spherical (ball & socket) 3 (3 rotation) Two-Angle (universal) 2 (2 rotation)

Chapter 12 Equilibrium & Elasticity

Force = F Piston area = A

First Law: A body at rest remains at rest, a body in motion continues to move at constant velocity, unless acted upon by an external force.

measurement and the charge the electric field will be calculated using E =. The direction of the field will be the . But, so

Please initial the statement below to show that you have read it

Section 8.1 Exercises

Transcription:

Answers to Even Numbered Problems Chapter 5. 3.6 m 4..6 J 6. (a) 9 J (b).383 8. (a) 3.9 J (b) (c) (d) 3.9 J. 6 m s. (a) 68 J (b) 84 J (c) 5 J (d) 48 J (e) 5.64 m s 4. 9. J 6. (a). J (b) 5. m s (c) 6.3 J 8.. m..5 m. (a).768 m (b) 5.68 J 4. 6.5 m s 6. 5. m 8. (a) 9.9 m s (b) 7.67 m s 3. (a) v 5.94 m s, v 7.67 m s (b) 47 J B 3. 5. m s 34. 6 m 36. (a) 9.9 m s (b).8 J 38. (a) No, µ ( m sn θ) (b) cos θ, µ ( sn θ) C x m x (c) normal orce, ravtatonal orce, and vertcal component o appled orce (d) 4.3 N, 47.9 J, 6.9 J 43

44 CHAPTER 5 4. 77 m s 4. (a).9 m s (b) 5.6 J h 5 44. ( 4sn θ + ) 46..5 m (meased alon the nclne) 48. (a) J (b).9 hp 5..9 m s 5. 94 m 54. (a) 7.9 hp (b) 4.9 hp 56. (a) 7.5 J (b) 5. J (c) 7.5 J (d) 3. J 58..9 m s 6..6 m 6..4 m s 64. 3.9 J 66. (a) 58 trps (b) 9.5 (. hp) 68. 895 J 7. (a) 3. J (b).5 J (c) (d).5 J 7. (a) 3.8 J or javeln, (b).9 N on javeln, (c) Yes 74. 4.9 J 76..5 78. (b).6 m s 7.3 J or dscus, 3.6 N on dscus, 8. J or shot 4. N on shot 8. (a) 7 5.3 J (b).6 8. (a) 5.8 m (b).77 or 7. m s

Enery 45 84. (a) Choose the sprn constant so the weht o a tray stretches all o sprns a dstance equal to the thcness o a tray. (b) 36 N m, The lenth and wdth o a tray are not needed. 86. (a). m s (b) 6. m s 88. (a) 4. m s (b) 7.9 J (c) 8 N (d) 77 N m 5. 9.8 m s 49. N and s drected 5.5 (a) The orce o ravty s ven by downwards. The anle between the orce o ravty and the drecton o moton s θ 9. - 3. 6., and so the wor done by ravty s ven as ( θ) cos s 49. N cos6..5 m 6.3 J (b) The normal orce exerted on the bloc by the nclne s rcton orce s n mcos3., so the µ n.436 49. N cos3. 8.5 N Ths orce s drected opposte to the dsplacement (that s θ 8 ), and the wor t does s ( θ) cos s 8.5 N cos8.5 m 46.3 J (c) Snce the normal orce s perpendcular to the dsplacement; θ 9, cos θ, and the wor done by the normal orce s zero.

46 CHAPTER 5 5.7 (a) Σ snθ + n m y n m sn θ Σ cos θ µ n x cos θ n µ cos θ m sn θ µ s. m n m n 8. q. m (.5 )( 8. )( 9.8 m s ) µ m 79.4 N µ snθ+ cosθ.5 sn. + cos. cos θ s 79.4 N cos.. m.49 J.49 J (b) 3 (c) cos θ 74.6 N 3 ( ) cos θ s 74.6 N cos8. m.49 J.49 J 5.5 (a) The nal netc enery o the bullet s KE (b) e now that. 3 mv 3 m s 9 J KE, and also ( cos θ av ) s. Thus, av KE 9 J scos θ.5 m cos.8 N net road cos s+ resst cos s N cos s+ 95 N cos8 s 5.7 ( θ ) ( θ ) net 3 N 95 N m. J Also, KE KE mv net, so 3 ( ). J v net m. m s

Enery 47 5.3 hle the motorcycle s n the ar, only the conservatve ravtatonal orce acts on cycle and rder. Thus, mv + my mv + my, whch ves ( 35. m s) ( 33. m s) v v y y y 9.8 m s 6.94 m 5.4 Let m be the mass o the ball, R the radus o the crcle, and the 3. N orce. th KE+ PE KE+ PE yelds y at the bottom o the crcle, nc ( cos ) πr mv + mv + m( R) or Thus, ( πr) v + v + 4R m v π 3. N.6 m + +.5 ( 5. m s) 4( 9.8 m s )(.6 m ) vn v 6.5 m s 5.7 Snce no non-conservatve orces do wor, we use conservaton o mechancal enery, wth the zero o potental enery selected at the level o the base o the hll. Then, mv+ my mv + my wth y yelds v v 3. m s y 9.8 m s.459 m Note that ths result s ndependent o the mass o the chld and sled.

48 CHAPTER 5 5.3 Realze that all three masses have dentcal speeds at each pont n the moton and that v. Then, conservaton o mechancal enery ves KE PE PE, or m m m v m y y m y y m y y ( + + 3) ( ) + ( ) + 3( 3 3) 3. 5. 4. m. 5. 4. m 9.8 m s Thus, v ( ) + + ( + ) yeldn v 5. m s 5.33 (a) Use conservaton o mechancal enery rom when the projectle s at rest wthn the un untl t reaches maxmum heht. Then, ( KE PE PEs) ( KE PE PEs) + + + + becomes + mymax + + + x or max 3 ( ) my. 9.8 m s. m x (. m ) 544 N m (b) Ths tme, we use conservaton o mechancal enery rom when the projectle s at rest wthn the un untl t reaches the equlbrum poston o the sprn. Ths ves ( ) KE PE + PE PE + PE m x + x + s s v x x m 544 N m 3. (. m ) ( 9.8 m s )(. m ) yeldn v 9.7 m s

Enery 49 5.39 e shall tae PE at the lowest level reached by the dver under the water. The dver alls a total o 5 m, but the non-conservatve orce due to water resstance acts only dn the last 5. m o all. The wor-enery theorem then ves KE+ PE KE+ PE nc av cos8 5. m 7 9.8 m s 5 m or ( ) ( + ) + Ths ves the averae resstance orce as 3 av. N. N 5.44 Choose PE at water level and use ( KE PE) ( KE PE) + + or the trp down the cved slde. Ths ves h mv m + + mh, so the 5 speed o the chld as she leaves v 4h 5 the end o the slde s h q h/5 y The vertcal component o ths launch velocty s 4h v y vsnθ snθ 5 At the top o the arc, y vy v y + ay y ves the maxmum heht the chld reaches dn the arborne trp as v. Thus, 4h h θ + ( ) 5 5 sn ymax Ths may be solved or max h 5 y to yeld ymax ( 4sn θ + )

5 CHAPTER 5 5.45 Choose PE at the level o the base o the hll and let x represent the dstance the ser moves alon the horzontal porton beore comn to rest. The normal orce exerted on the ser by the snow whle on the hll s n mcos.5 and, whle on the horzontal porton, n m. Consder the entre trp, startn rom rest at the top o the hll untl the ser comes to rest on the horzontal porton. The wor done by rcton orces s nc cos8 m cos8 + x ( mcos.5 )( m ) µ µ Applyn nc ( KE PE ) ( KE PE ) m x + + to ths complete trp ves ( m ) ( m) x [ ] m µ cos.5 m µ + + m sn.5 sn.5 µ or x cos.5 ( m ). I µ.75, then x 89 m v v 8. m s 5.53 (a) The acceleraton o the car s.5 m s a. Thus, the t. s constant orward orce due to the enne s ound rom Σ enne ar ma as enne ar + ma + 3 3 4 N.5.5 m s.65 N The averae velocty o the car dn ths nterval s v averae power nput rom the enne dn ths tme s P av v+ v 3 hp ennevav (.65 N )( 9. m s) 3. hp 746 av 9. m s, so the (b) At t. s, the nstantaneous velocty o the car s v 8. m s and the nstantaneous power nput rom the enne s 3 hp P ennev (.65 N )( 8. m s) 63.9 hp 746

Enery 5 5.55 The wor done on the partcle by the orce as the partcle moves rom x x to x x s the area under the cve rom x to x. (a) or x to x 8. m, area otranle ABC AC alttude x (N) B 6 4 A C E 4 6 8 4 D x (m) (b) or 8 ( 8. m )( 6. N ) 4. J x 8. m to x. m, 8 area otranle CDE CE alttude (. m )( 3. N ) 3. J (c) 8 + 8 4. J+ 3. J. J 5.6 (a) rom v v a y ( y) +, we nd the speed just beore touchn the round as v + 9.8 m s. m 4.4 m s (b) Choose PE at the level where the eet come to rest. Then KE+ PE KE+ PE becomes nc s + mv + ms ( av cos8 ) ( ) or ( 75 )( 4.4 m s) ( ) mv + m + s 5. m av 3 5 75 9.8 m s.5 N

5 CHAPTER 5 5.64 The normal orce the nclne exerts on bloc A s n ( m )cos37, and the rcton orce s µ n µ m cos37. The vertcal dstance bloc A rses s A A m sn37 m, whle the vertcal dsplacement o bloc B s m. y A e nd the common nal speed o the two blocs by use o KE+ PE KE+ PE KE+ PE nc Ths ves ( µ ) m Acos37 s m A + m B v + m A( ya) + m B( yb) or A A y B v ( µ ) m B yb m A ya m A cos37 s m + m A B 9.8 m s m 5 m.5 5 m cos37 5 whch yelds v 57 m s The chane n the netc enery o bloc A s then 3 KEA m Av ( 5 )( 57 m s ) 3.9 J 3.9 J 5.7 rst, determne the mantude o the appled orce by consdern a ree-body daram o the bloc. Snce the bloc moves wth constant velocty, Σ Σ rom Σ x, we see that n cos3 x y 3 n 5 m Thus, µ n µ cos3 and Σ y becomes sn 3 m+ µ cos3, or ( 5. )( 9.8 m s ) m sn 3 µ cos3 sn3.3 cos3. N

Enery 53 (a) The appled orce maes a 6 anle wth the dsplacement up the wall. Thereore, cos6. N cos6 3. m 3. J s mcos8 s 49 N. 3. m.5 J (b) ncos9 s (c) n PE m y 49 N 3. m.5 J (d) 5.73 The potental enery assocated wth the wnd s PEw x, where x s meased horzontally rom drectly below the pvot o the swn and postve when movn nto the wnd, neatve when movn wth the wnd. e choose PE at the level o the pvot as shown n the e. Also, note that D Lsnφ + Lsn θ so D φ sn sn θ L, or L cos wnd drecton q L x L sn L snq v L cosq 5. m φ sn sn5. 8.94. 4. m D (a) Use conservaton o mechancal enery, ncludn the potental enery assocated wth the wnd. The nal netc enery s zero Jane barely maes t to the other sde, and ( KE PE PEw ) ( KE PE PEw ) + + + + becomes + m Lcos + + Lsn mv + m Lcos + Lsn ( φ) ( φ) ( θ) ( θ) L m or v L( cosθ cosφ) + ( snθ+ sn φ) where m s the mass o Jane alone. Ths yelds v 6.5 m s

54 CHAPTER 5 (b) Aan, usn conservaton o mechancal enery wth KE ( KE PE PEw ) ( KE PE PEw ) + + + + ves + M Lcos + Lsn Mv + M Lcos + + Lsn where M 3 s the combned mass o Tarzan and Jane. Thus, ( θ) ( θ) ( φ) ( φ) L v L( cosφ cosθ) ( snθ+ sn φ) whch ves v 9.87 m s M 5.76 hen the bloc moves dstance x down the nclne, the wor done by the rcton orce s cos8 x µ nx µ mcos θ x. rom the wor-enery theorem, KE+ PE + PE KE+ PE + PE, we nd nc s s ( cos ) nc µ m θ x KE+ PE + PEs. Snce the bloc s at rest at both the start and the end, ths ves µ ( 9.6 N cos37. )(. m ) + ( 9.6 N )(. m sn37. ) + ( N m )(. m ) or µ.5 5.83 Careul examnaton o e P5.83 reveals that, dn some tme nterval, bloc B moves upward. cm, bloc A wll move downward. cm and the dstance separatn the two blocs ncreases by 3. cm. Generalzn, we conclude that when the vertcal separaton between the blocs ncreases by h, bloc B moves upward dstance h 3 and bloc A moves downward dstance h 3. Also, at any nstant dn the moton bloc A has speed v, the speed o bloc B wll be v. Choosn y at the level where both blocs start rom rest and man use o the above observatons, conservaton o mechancal enery ves ( A) + ( B) + ( A) + ( B) ( A) + ( B) + ( A) + ( B) m v m v m y m y m v m v m y m y or mv m( v ) m( h ) m( h ) + + 3 + + 3 + + + Ths reduces to 5 h 8 3 v, or h 8 8h v 3 5 5