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# INTRO PHYS MECH PHYS 1111

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This 21 page Class Notes was uploaded by Heath Hane III on Saturday September 12, 2015. The Class Notes belongs to PHYS 1111 at University of Georgia taught by Pan in Fall. Since its upload, it has received 38 views. For similar materials see /class/202487/phys-1111-university-of-georgia in Physics 2 at University of Georgia.

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N Chapter 8 Potential Energy and Conservation of Energy Answers to EvenNumbered Conceptual Questions As water vapor rises there is an increase in the gravitational potential energy of the system Part of this potential energy is released as snow falls onto the mountain If an avalanche occurs the snow on the mountain accelerates down slope converting more gravitational potential energy to kinetic energy The initial mechanical energy of the system is the gravitational potential energy of the massEarth system As the mass moves downward the gravitational potential energy of the system decreases At the same time the potential energy of the spring increases as it is compressed Initially the decrease in gravitational potential energy is greater than the increase in spring potential energy which means that the mass gains kinetic energy Eventually the increase in spring energy equals the decrease in gravitational energy and the mass comes to rest The object s kinetic energy is a maximum when it is released and a minimum when it reaches its greatest height The gravitational potential of the system is a minimum when the object is released and a maximum when the object reaches its greatest height The jumper s initial kinetic energy is largely converted to a compressional springlike potential energy as the pole bends The pole straightens out converting its potential energy into gravitational potential energy As the jumper falls the gravitational potential energy is converted into kinetic energy and finally the kinetic energy is converted to compressional potential energy as the cushioning pad on the ground is compresse When the toy frog is pressed downward work is done to compress the spring This work is stored in the spring as potential energy Later when the suction cup releases the spring the stored potential energy is converted into enough kinetic energy to lift the frog into the air The total mechanical energy decreases with time if air resistance is present The distance covered by the ball is the same on the way down as it is on the way up and hence the amount of time will be determined by the average speed of the ball on the two portions of its trip Note that air resistance does negative nonconservative work continuously on the ball as it moves Therefore its total mechanical energy is less on the way down than it is on the way up which means that its speed at any given elevation is less on the way down It follows that more time is required for the downward portion of the trip Answers to EvenNumbered Conceptual Exercises a The value you assign to the jumper s initial gravitational potential energy is less than the value assigned to that quantity by your friend b The change in gravitational potential energy in your calculation is equal to that in your friend s calculation The two balls have the same change in gravitational potential energy Thus ignoring air resistance they will also have the same change in kinetic energy a Your answers will disagree b Your answers will agree c Your answers will agree Note that the information given for part b shows that the kinetic energy at point B is K 0 I at rest and the potential energy is Ll 25 I It follows then that the total mechanical energy of the system is E U K 25 I 0 I 25 I Since the system is conservative it will have the same total mechanical energy at each pointin its motion a AtpointA Ll E K 25I 12I 13 I b AtpointCK E U25I 5I20I In the original situation all of the initial gravitational potential energy of ball 1 U mgh is converted to kinetic energy Therefore K mgh In the second case the initial gravitational energy of ball 2 is Ll 2mgh2 mgh It follows that the kinetic energy of ball 2 just before it lands is also K 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 215 Chapter 8 Potential Energy and Conservation of Energy 1 N 1 in 1 a James S Walker Physics 3rd Edition a The potential energy of the t u hi h is O potential lg down the hill b Your kinetic energy remains the same since your speed is constant c In order for your speed and kinetic energy to remain constant as you pedal down the hill a nonconservative force must have done negative work on you and your bicycle For example you may have applied the brakes to control your speed or the ground may be soft or muddy In any case the mechanical energy of this system has decreased as you move The mechanical energy of the shuttle when it lands is considerably less than when it is in orbit First its speed on landing is low compared to its speed in orbit giving a much smaller kinetic energy Second the source of energy responsible for heating the tiles is the gravitational potential energy that is released as the shuttle loses altitude Therefore both the kinetic and potential energy of the shuttle have much smaller values when the shuttle lands leading to a smaller value for the total mechanical energy Ignoring any type of frictional force the speed of the object is the same at points A and G Similarly the speed of the object at point B is the same as its speed at points D and F In addition the lower a point on the curve the greater its speed Combining these observations we arrive at the following ranking A G lt B D F lt E lt C Solutions to Problems Picture the Problem The three paths of the object are depicted at right Strategy Find the work done by gravity W mgy when the object is moved downward W 7mg when it is moved upward and zero when it is moved horizontally Sum the work done by gravity for each segment of each path Solution 1 Calculate the work for path 1 W1 mgliyt 0yZ 0y3 mg740 m10 m 10 m W 32 kg98l msz 720 m l 39 10 quot l 30m Side View 2 Calculate Wfor path 2 WZ mg07y4 0 32 kg98l msz720 m 3 Calculate Wfor path 3 W mgyS 07y 32 kg981mszl0 m730 m Insight The work is pathindependent because gravity is a conservative force Picture the Problem The three paths ofthe sliding box are depicted at right Strategy The work done by friction is W ak mgd where dis the distance the box is pushed irregardless of direction because the friction force always acts in a direction opposite the motion Sum the work done by friction for each segment of each path Solution 1 Calculate the W7 m dd d d d workforpath 1 1 uk gi 1 z 3 4 5 qukmg4040l0l0l0 m W 702132 kg98l mszll0 m 2 Calculate W forpath 2 W2 i kmgw d7 d8 702132 kg98l msz 20 m20 ml0 m 3 Calculate the work for m3 VVK kmgid9dlndlll pa 702132 kg98l msz l0 m 30 m 30 m Insight The amount of work done depends upon the path because friction is a nonconservative force 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing 39om the publisher 216 Chapter 8 Potential Energy and Conservation of Energy 3 James S Walker Physics 3rd Edition Picture the Problem The physical situation is depicted at right Strategy Use equation W ch equation 78 to nd the work done by the spring but caution is in order This work is positive when the force exerted by the spring is in the same direction that the block is A B traveling but it is negative when they point in opposite directions One 7mm 0 m cm 43 cm way to keep track of that sign convention is to say that l 7 I W 7 7kx z 7x That way the work Wlll always be negative 1fyou Q i start out at 9g 0 because the spring force will always be in the opposite direction from the stretch or compression Solution 1 a Sum the work done by Wl 7x22 x 7x32 the spring for each segment of path 1 Z550 Nm02 70040 m20040 mZ 70020 mZ Wl7044 J033 J 2 Sum the work done by the spring for WZ k xf 7x42 x 7x32 each segment ofpath 2 550 Nm02 470020 m2 1 70020 mZ 70020 mZ l 2 WZ 701 J0 J 3 b The work done by the spring will if you increase the mass because the results have no dependence on the mass ofthe block Insight The work done by the spring is negative whenever you displace the block away from x 0 but it is positive when the displacement vector points towards x 0 Picture the Problem The two paths of the object are shown at right Side View Strategy Find the work done by gravity W mgy when the object is moved downward W 7mg when it is moved upward and zero when it is moved horizontally Sum the work done by gravity for each a segment of each path Solution 1 a Calculate the work for path 1 W1 mg0 yl 52 kg981 msz 10 m 2 Calculate the work for path 2 WZ mgyl 0 26 kg98lmszl0 m 3 b If you increase the mass of the object the work done by gravity will because it depends linearly on m Insight The work is pathindependent because gravity is a conservative force 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion ofthis material may be reproduced in any form or by any means without permission in writing from the publisher 217 Chapter 8 Potential Energy and Conservation of Energy James S Walker Physics 3rd Edition 5 Picture the Problem The physical situation is depicted at right Strategy Use WSp 7kx z 7 9 for the work done by the spring That way the work will always be negative if you start out at Jq Obecause the spring force will always be in the opposite direction A from the stretch or compression The work done by kinetic friction is 39Z39O a 0 20m 4 cm Wamp qukmgd where dis the distance the box is pushed irregardless l O of direction because the friction force always acts in a direction 1 opposite the motion Solution 1 a Sum the work done bythe WSp kxlz 7x42 xf 7x32 spring for each segment ofpath 2 480 Nm02 70020 mz70020 my 70020 mZ W 70096 J0 J Sp 2 Sum the work done by friction for Wamp qukmg ll dz eaCh segment ofpath 2 70l627 kg981 msZ 00200040 m 3 b Sum the work done by the W p k xi 7 x spring for the direct path from A to B 480 Nm0Z 70020 m2 l 4 Sum the work done by friction for Wfr 7tkmgd 70l627 kg98l msz0020 m the direct path from A to B Insight The work done by friction is always negative and increases in magnitude with the distance traveled 6 Picture the Problem The cliff diver plunges straight downward due to the force of gravity Strategy Solve equation 83 for the weight of the diver Let y 0 correspond to the surface of the water Solution Solve equation 83 for mg U mgy Q mg E w 540 N y 46 m Insight lfyou set U 0 at the top ofthe cliff then U 25 kl and y 746 m when the diver enters the water 7 Picture the Problem The climber stands at the top of Mt Everest Strategy Find the gravitational potential energy by using equation 83 Solution Calculate U mgy U mgy 83 kg981 msZ 8848 m 72x105 J Insight You are free to declare that the climber s potential energyis zero at the top ofMt Everest and 72 M at sea leve l 8 Picture the Problem The Jeopardy button is a spring that must be compressed a certain amount to activate the switch Strategy Use equation 85 to nd the work required to compress the spring an estimated 10 cm in order to activate the switch While the work the spring force does is WE iAU the work a contestant must do is W AU Solution Calculate the work 7 7 Z 1 Z ittakesby ndmg AU WciAU7ka 07Z120Nm0010m 70006OJ7 Insight The contestant does positive work on the spring because the force she exerts is in the same direction her nger is pushing The energy she expends is brie y stored as potential energy in the spring 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing 39om the publisher 218 Chapter 8 Potential Energy and Conservation of Energy Iames S Walker Physics 3 d Edition 9 Picture the Problem The Hawkmoth wing behaves as an ideal spring Strategy Use Hooke s Law equation 64 to find the force constant of the wing then use equation 85 to find the energy stored Combine both equations to find the force required to store twice the energy Solution 1 a Solve equation 64 for k k 75 M 0625 Nm 063 Nm x 700048 In 2 b Use equation 85 to find U U 5108 0625 Nm 00048 m2 72gtlt10 6 J 3 c Solve equation 85 for x taking the 2U 2U negative root because the wing is depressed in i x 3 x i 7 the negative direction as in step 1 4 Use Hooke s Law to nd the force F ikx 7k 7 izUk wlek 22gtlt72gtlt10 6 I0625 Nm 00042 N Insight We bent the rules for significant figures for k a bit in steps 2 and 4 in order to avoid rounding error Notice that in step 4 the force required to double the stored energy is J2 bigger than the original force 0 Picture the Problem The mass is suspended from a vertical spring As the spring is stretched it stores potential energy Strategy Doubling the mass doubles the force exerted on the spring and therefore doubles the stretch distance due to Hooke s Law F ikx Use a ratio to find the increase in spring potential energy when the stretch distance is doubled Solution 1 a If the mass attached to the spring is doubled the stored potential energy in the spring will increase by a factor of four because the stretch distance will doubleA doubling of the mass will double the extension of the spring Doubling the extension of the spring will increase the potential energy by 2 b Find an expression for the stretch U1 mgx1x12 distance as a function of Uand m 7 1 2 mgxl 7amp7 20962 J 20065m mg 30kg98l ms x1 3 Doubling the mass doubles the force and x1 quotlgk therefore doubles the stretch distance x2 2mgk 2x1 4 Calculate the ratio U2 U1 U 7 U kx22 013m2 we 0065 m2 U2 4U1 40962 J Insight Note that the change in gravitational potential energy also quadruples as the new mass is hung on the spring It doubles because there is twice as much mass and it doubles again because the spring stretches twice as far i i Picture the Problem The spring in the soap dispenser is compressed by the applied force Strategy Use equation 85 to find the spring constant using the given energy and compression distance data Solve the same equation for x in order to answer part 2 00025 1 Solution 1 a Solve equation 85 for k k g 2 00 Nm 020 kNm x 00050 In I 2 00084 I 2 b Solve equation 85 for x x Q 200 Nm Insight To compress the spring of this dispenser 050 cm requires 10 N or about 4 lb of force 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 219 Cha ter 8 Potential Ener and Conservation ofEner James S Walker Ph n39cx 3rd Edition 12 Picture the Problem The spring is stretched by the applied force and stores potential energy Strategy Use equation 64 F 40c to find the spring constant andthen equation 85 U gkxz to find the stretch distance Solution 1 a Solve equation 64 for k k77 74397 N 350 Nm x 0013 rn 2 0020 I 2 Solve equation 85 for x x E quotg 105 cm k 360 Nm 2 0080 I 3 b Repeat step Zwith the new U x g k 360 Nm Insight Notice that in part b the stretch distance doubled but the stored potential energy quadrupled because Uis proportional to 3 Picture the Problem A graph of the potential energy vs stretch distance is depicted at right Strategy The work that you must do to stretch a spring is equal to minus the work done by the spring because the force you exert is in the opposite U direction from the force the spring exerts Use equations 81 and 8 together to find the spring constant and the required work to stretch the spring the specified distance Solution 1 a Because the stored potential energy in a spring is proportional to the stretch distance squared the work required to stretch the spring from 500 cm to 600 cm will be than the work required to stretch it from 400 cm to 500 cm 2 b Use equations 81 and Wm1 7W5 g 77AU U5 7U4 85 tofindk kx 77104 7kx52 7x1 ZWM1 7 2305 J i if 678gtlt104 Nm X 0 00500 m 700400m 3 Use kand equations 81 and 85 to find the new W 39 req W PM 7x 578x10A Nm00600 m2 700500 my Insight Using the same procedure we discover that it would take 441 Ito stretch the spring from 600 cm to 700 cm 14 Picture the Problem The pendulum bob swings from point A to point B and loses altitude and thus gravitational potential energy See the figure at right l Strategy Use the geometry ofthe problem to find the change in altitude Ay of the i go pendulum bob and then use equation 83 to find its change in gravitational potential 3 1 2 m energy 39 Solution 1 Find the height change Ay ofthe pendulum bob AyLC sg LLC 59 1 7777 7701 757 2 Use Ay to findAU AU mgAy mgLcos971 033 kg981ms212 mcos35 71 Insight Note that Ay is negative because the pendulum swings from A to B Likewise Ay is positive and the pendulum gains potential energy if it swings from B to A 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion ofthis material may be reproduced in any form or by any means without permission in writing from the publisher 220 Cha ter 8 Potential Ener and Conservation ofEner James S Walker Ph n39cx 3rd Edition 15 Picture the Problem The swimmer descends through a vertical height of 261 In as she slides without friction Strategy As the sw c loss in gravitational potential energy equal to the ga in kinetic energy by setting her change in mechanica equal to zero so that AE Ef 7 0 or Ef E Let y 0 at the bottom ofthe slide v 0 at the top immer des ends the slide her gravitational potential energy is converted into kinetic energy Set the in 39 39 1 energy Solution Set Ebm Emp and solve for mm 1 E K balm Emp Kmp Ump bottom U mm 1 2 7 7mm 0 a 0 my Vbam xllgymp 2981ms2261m Insight If she has amass of40 kg the swimmer loses 102 k ofpotential energy and gains 102 k of kinetic energy ox Picture the Problem The swimmer descends through a vertical height of 261 rn as she slides without friction Strategy As the swimmer descends the slide her gravitational potential energy is converted into kinetic energy Set the loss in gravitational potential energy equal to the gain in kinetic energy by setting her change in mechanical energy equal to zero so that AE Ef 7 0 or Ef E Let y 0 at the bottom ofthe slide v 0840 ms at the top Solution Set EMAm Emp and solve for vwm Emum Emp Kbmm U mm Kmp U 09 l 2 7 i 2 imam 0 a 7mm mgymp mm W Zgymp 1 ms2261m Insight Note that she is not going 0840 ms faster than the 716 ms she would be traveling if she started from rest That s because the 141 J of kinetic energy she has at the start if she has a mass of 40 kg is tiny compared with the 1020 J of kinetic energy she gains on the way down 17 Picture the Problem As the ball ies through the air and gains altitude some of its initial kinetic energy is converted into gravitational potential F energy C Strategy Set the mechanical energy at the start ofthe throw equal to the mechanical energy at its highest point Let the height be y 0 at the start of the throw and nd yf at the highest point Solution 1 a Set E Ef and solve for yr E Ef K A Kr Ur mv2 0 7 2 mv mgyf 1 830 ms 2 7 710 ms 2 wiw g gt lt I gt 2g 2981ms 2 b The height change is independent ofthe mass so oubling the ball s mass would cause no change to a Insi ht A more massive ball would have more kinetic energy at the start but would require more energy to change its height by 0942 rn so the mass cancels out 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion ofthis material may be reproduced in any form or by any means without permission in writing from the publisher 221 Chapter 8 Potential Energy and Conservation of Energy Iames S Walker Physics 3 d Edition 18 Picture the Problem As the ball ies through the air and gains altitude some of its initial kinetic energy is converted into gravitational potential energy Strategy Set the mechanical energy just after the bounce equal to the mechanical energy when it is caught Let the height be y 0 at the bounce and find yf at the catch Solution 1 a Set E Ef and solve for yf E Ef K U Kf Uf mv 2 0 gmva mgyf 7i 7 2 7 16 my 712 ms2 2g f 2981ms2 yr 2 b The height change is independent of the mass so it is not necessary to know the mass of the tennis ball Insight A more massive ball would have more kinetic energy at the start but would require more energy to change in height by 0942 In so the mass cancels out 19 Picture the Problem The apple falls straight down from rest accelerating at a rate of 981 msz Strategy As the apple falls its gravitational potential energy is converted into kinetic energy The sum of the gravitational and kinetic energies equals the mechanical energy which remains constant throughout the fall Use equations 76 and 83 to find the kinetic and gravitational potential energies respectively Solution 1 a Find Ua when ya 40 m U mgya 021kg981ms240 m 824 I 82 I 2 The apple falls from rest so va 0 ms Ka 9 3 The total energy is the sum ofK and U Ea Ka Ua E 0824 I 82 I 4 b Find U1 when yb 30 In I Ub mgyb 021kg981ms2 30 m 618 I 2 I 5 The total energyremains 82 I always so find Kb Kb EiUb 8247618 I 1 I 6 c d e Repeat steps 4 and 5 y m 40 30 20 10 UI 82 62 41 21 0 K I 0 21 41 62 82 EU 82 82 82 82 82 Insight We bent the rules a bit on significant figures in step 5 to avoid rounding errors This is especially a problem when subtracting two large numbers in order to obtain a small number as in step 5 Notice the progression as the mechanical energy begins as entirely gravitational potential energy and ends as entirely kinetic energy 20 Picture the Problem The block slides on a frictionless horizontal surface encounters a spring compresses it and brie y comes to rest Strategy Set the mechanical energy when sliding freely equal to the mechanical energy when the spring is fully compressed and the block is at rest Solve the resulting equation for the spring constant k then repeat the procedure to find the initial speed required to compress the spring only 12 cm before coming to rest Solution 1 a Set E Ef where the initial K U Kf Uf state is when it is sliding freely and the final gmv Z 0 0 Jrgeriax state is when 1t is at rest hav1ng compressed the 2 27 k 14 2 spnng k quot2 M 2300 Nm 23 kNm xma 0048 m 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 222 Chapter 8 Potential Energy and Conservation of Energy Iames S Walker Physics 3 d Edition N N N 2 b Solve the equation from step 1 for v 2 2300 Nm0012 m 27 kg 2 kx max v 7 m Insight The kinetic energy of the sliding block is stored as potential energy in the spring Moments later the spring will have released all is potential energy the block would have gained is kinetic energy again and would then be sliding at the same speed but in the opposite direction Picture the Problem The trajectory of the rock is depicted at right Strategy The rock starts at height h rises to ymax comes brie y to rest then falls down to the base of the cliff at y 0 Set the mechanical energy at the point of release equal to the mechanical energy at the base of the cliff and at the maximum height ymax inorder to find v and ymax Solution 1 a Set E Ef and solve for v K 4amp1 Kf Uf mv 2 mgh mvf2 0 v 1 vf272gh 2 29 ms 72981 mSZ l T l 43m 32 m 2 b Now set Eymax Ef K and solve for max 2 29 m 2 y 0mgymax mvf20 2 ymaxvif4 S 2g 2981 ms2 Insight In part a the initial energy is a combination of potential and kinetic but becomes all kinetic just before impact with the ground In part b the rock at the peak of is ight has zero kinetic energy39 all of is energy is potential energy Picture the Problem The block slides on a frictionless horizontal surface encounters a spring compresses it and brie y comes to rest when the spring compression is 415 cm Strategy As the block compresses the spring is kinetic energy is converted into spring potential energy The sum of the spring potential and kinetic energies equals the mechanical energy which remains constant throughout Use equations 76 and 85 to find the kinetic and spring potential energies respectively Solution 1 3 Find Ka when va 0950 ms K mv l40 kg0950 ms2 0632 I 2 The spring is not compressed so xa 0 cm Ua 7 1lac 9 3 3 The total energy is the sum ofK and U Ea Ka Ua E 0632 J 0 0632 I 4 b Find Ubwhen xb l00 cm Ub kx 734 Nm00100 m2 00367 I 5 The total energy remains 0632 I always so find Kb Kb EiUb 0632700367 I 0595 I 639 L d 9 Repeat Steps 4 and 5 y cm 000 100 200 300 400 UJ 000 0037 0147 0330 0587 K I 0632 0595 0485 0302 0045 E I 0632 0632 0632 0632 0632 Insight The initial kinetic energy of the sliding block is stored as potential energy in the spring when it comes to rest Moments later the spring will have released all its potential energy the block would have gained its kinetic energy again and would then be sliding at the same speed but in the opposite direction 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 223 vhanmx v t M 1 v James s WalkerPhysxcs 3 Edrtron 2 23 Strategy that y 0 and y0 The posmve ya1ue sothat the sum ofthe two W111 remam zero The mechamcal energy remams zero througnout Snlntinn 1 a The rnrtra1 lunatic energy K 15 zero beeause the rockfalls from rest 39 Set E E atter the rock falls to yr K U K mgy 75 00 kg9 81rns e2 morn 3Ca1eu1ateA1lt Mag K ggHjI 4 h The mmal kmeuc energy 15 9811 K 5 Set E 5 a ertherockfallsto y 74 00m KmUh KerZh Kx7 mgyx7 1ltfm mgym Kn Kn m yrh yn 98115 00 kg9 81 m522 00 m4 00 rn 5Ca1eu1ateAK Magek mzeggn Insight lt lt to lt lt Fonnstance H lt y 12 earn y 1000mandym800m nttt partaeular ya1ue that rnatter m physresw 24 39 A ahatude and thus grayrtataona1 potentra1 energy See thefxgure at nght Strategy Use the geometry of the problem to ndthe ehange m altde Ay of the energy App1y conservauon of energy between pomts B and A to fmd the speed at A n 13 1 Snlntinn 1 a Frndthe herght Ay EL we 5 L 1 we a ehange Ayofthependulum bob W quotO A A 2 Use Ayto ndAU AUmgAymgL1eeos5 W 033kg9 81rns 12rn1eeos35 B 3 h Set EH EA and solve for VA KE UE 1g UA mv lzmvi AU 2 0701 VA y eLAU 2 4 rns2 77 33 j m kg L r t 1 1 y energy depends hnear1y on the rnass 1 energy W111 rnerease the ehange m lunatic energy W111 also mcrease si t lt That rneans the formulafor yprn step 3 15 rndependent ofmass J nu Uppev Sadd e Rwev NJ AH HgHs veserved an p ngm M as they Hquot m 1 V 224 Chapter 8 Potential F nergv and Con ervatinn ofEnergy 2 James S Walker Physics 3rd Edition Picture the Problem The pendulum bob swings from point B to point A and gains altitude and thus gravitational potential energy See the gure at right Strategy Use equation 76 to nd the kinetic energy ofthe bob at point B Use the geometry of the problem to nd the maximum change in altitude AynEX of the pendulum bob and then use equation 83 to nd its maximum change in gravitational potential energy Apply conservation of energy between points B and the endpoint of its travel to e maximum ang e am the string makes with the vertical Solution 1 a Use equation 76 to KB mv a0 kg24 15 KE B 2 b Since there is no iction mechanical energy is conserved If we take the potential energy at point B to be zero we can say that all ofthe bob s kinetic energy will become potential energy when the bob reaches is maximum height and comes momentme to rest Therefore the change in potential energy between point B and the point where the bob comes to rest is 3 c Find the height change AynEX of the pendulum bob AynEX LiLcosQquotEX L17cost9m 4 Use equation 83 and the result of AU mgAyquotEX mgL 17 cos 6quot part b to solve for rm 1 AU 095 J Hum cos 1 cos 1 mgL 033 kg981msz12 m Insight The pendulum bob cannot swing any further than 41 because there is not enough energy available to raise the mass to a higher elevation Picture the Problem The motions ofthe masses in the Atwood s machine are depicted in the gure at righ Strategy Mechanical energy is conserved since there is no iction Set 13quot Ef and solve for vf The speeds ofeach mass must always be the same because they are connected by a rope Solution 1 a Set E Er and solve for vf K U va idf 00izmlvf2 mzvgJrrlrl gy rngy2 0i2m1m2vmlghngih m27m1ghizmlmzvrz 2 b Use the expression E from part a to nd vf Insight The mass m2 loses more gravitational potential energy than ml gains so there is extra energy available to give the system kinetic energy We bent the rules for signi cant gures a bit in step 2 because by the rules ofsubtraction 4137 kg 04 kg only one signi cant gure 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion ofthis material may be reproduced in any rm or y 225 anymeans without permission In writing 39om the publisher Chapter 8 Potential Energy and Conservation of Energy 27 N 00 N 0 James S Walker Physics 3rd Edition Picture the Problem The motions ofthe masses in the Atwood s machine are depicted in the gure at right Strategy Mechanical energy is conserved since there is no friction Set E Ef and solve for h when vf 0 The speeds of each mass must always be the same because they are connected by a rope Solution 1 Set E Ef and then set vf 0 KU waf immz mvf 0 may my quot11 HEN 0MIghMZgh 2 Solve for h h im quot2 jvf mjo2o msZ 004 m 2g mpmZ 981 msz 41737kg Insight The kinetic energy of the twomass system is converted to a difference in potential energy between mass 2 and mass 1 By the rules for subtraction of signi cant gures 41 i 37 kg 04 kg only one signi cant gure Picture the Problem The surfer changes altitude and speed losing mechanical energy along the way due to friction Strategy The nonconservative work equals the difference in mechanical energy between the beginning and the end of t erun Solution Use equation 89 to nd Wm Wm AB Ef 7E MV mgyfmvf Huey ml vfrvfhgwrx 72 kg 82 msZ 713 msZ981 msZ 07175 m Wm 1100 Insight We usually expect the nonconservative work to be negative because friction steals mechanical energy and converts it into heat But in this case the force from the wave propels the surfer and he ends up with more mechanical energy 1200 J than he had at the start 61 J Picture the Problem The child slides down the slide changing altitude and speed and loses mechanical energy along the way due to friction Strategy The nonconservative work equals the difference in mechanical energy between the top and the bottom of the slide Use equation 89 to nd the mechanical energy at the bottom of the slide and equations 76 and 83 to nd the speed of the child Sety 0 at the bottom ofthe slide Wm AE Ef iE imvf mgyf 7mvf my imvi Wu mgyf timvf mgy Solution 1 Use equations 89 76 and 83 to solve for Kf 2 Multiply both sides by 2 m and take the square root to nd vf Vi ZWmm2gy yfvf J27361J19 kg2981msz2170 m0Z Vf Insight Ifthe slide had been frictionless the child s speed would have been 64 ms and her mechanical energy would have remained 391 J throughout the trip down the slide 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing 39om the publisher 226 Chapter 8 Potential Energy and Conservation of Energy Iames S Walker Physics 3 d Edition 30 Picture the Problem The athlete accelerates horizontally through the water from rest to 120 ms while doing nonconservative work against the drag from the water Strategy The total nonconservative work done on the athlete changes his mechanical energy according to equation 89 This nonconservative work includes the positive work Wm1 done by the athlete s muscles and the negative work Wm2 done by the water Use this relationship and the known change in kinetic energy to find Wm2 Solution Set the nonconservative work equal to the Wm Wm1 Wn52 AE Ef iE Kf 70 change in mechanical energy and solve for Wm2 The W 2 Kf 7W 1 gmva 7W 1 initial mechanical energy is zero 720 kg 120 mS2 7061 I Insight The drag force from the water reduced the swimmer s mechanical energy but his muscles increased it by a greater amount resulting in a net gain in mechanical energy L l Picture the Problem The airplane travels in a straight line slowing down and coming to rest after landing at high spee Strategy The total nonconservative work done on the airplane changes its mechanical energy according to equation 89 The nonconservative work equals the change in mechanical energy which is known from the initial and final speeds of the airplane there is no change to its gravitational potential energy Solution Apply equation 89 directly W AE Ef 7E va 7mv2 07l7000kg82ms2 W 7 57gtlt107 J M E Insight The aircraft brakes and the recovery cables remove the airplane s kinetic energy converting it into heat and sound Heat management in large braking systems like this one is an important engineering issue L N Picture the Problem The car travels in a straight line slowing down and coming to rest after applying the brakes Strategy The total nonconservative work done on the car changes its mechanical energy according to equation 89 The nonconservative work equals the change inmechanical energy which is known from the initial and final speeds of the car there is no change to its gravitational potential energy In this case AE AK Solution 1 a Apply equation 89 directly Wm AE Ef E mvf2 i mvf AK 2 AK 1100 kg12 ms2 7 17 ms 780X10 I 2 b The mechanical energy of the car is converted into heat by friction in the brakes Insight Lots of hard braking in a car can cause heat management problems because the heat created by friction can melt or warp parts of the brake system 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 227 Chapter 8 Potential Energy and Conservation ofEnergy James S Walker Physics 3rd Edition 33 Picture the Problem The initial and nal states ofthe system L 5 are depicted at right The values given in the examp e are m1240 kg m2 180 kg d 0500 m and uk 0450 Strategy The nonconservative Work done by friction changes the mechanical energy of the system Use equation 89 to nd AE and set it equal to the Work done by iction Solve the resulting expression for the nal velocity of the system W Solution 1 Write equation m 89 to nd Wm Er E KrlUr KxUi 17m1m2Vmighngyzr imimz lz mighngyzr Wm im1 mz 7quot ngy2r Tyzi 2 The nonconservative Work is done by friction W ifkd i km gd 3 Substitute the expression from step 2 into step solve for v 7v vng07d V3 2 2981 msz 0500m180 kg70450240 kg V 240180 kg kmigdimrmzvi land gdmziukmlmlmzv 7 2gdmz 7 v2 7 7 f m1m2 l68m2s2 vf lv hrl68mZs2 13 ms2l68mZs2 Insight lfthe blocks are not given the initial speed ofl3 ms their nal speed is 130 ms as in example 810 4 Finally solve for vf Picture the Problem The seal slides down the ramp changing altitude and speed and loses mechanical energy along the Way due to friction Strategy The nonconservative Work done by friction changes the mechanical energy of the seal Use equation 89 and the given information to find the nonconservative Work Then use Newton s Second Law to find the normal force on the seal and use the normal force to nd the force of kinetic friction Solve the resulting expression for k Let y 0 at the Water s surface Solution 1 a Use equations 89 76 and 83 to nd Wm WmAEErE izmvf2 mgyf7izmv 2 mgy mv vmgyry 420 kg 440 ms2 7011981 msz 07175 m W n iggjilijrcwteon s Second Law to ndthe Nimgcosg 0 3 N mgcos 6 3 Use the right triangle formed by the ramp h h to nd the distance the seal slides 5quot 6 7 3 d d s1n 6 4 b Set Wm equal to the Work done by iction and solve for k Wm 74d 7 kNd ink mgcos hsm 7 Wm sing 7314 Jtan350 7 mgcos h 420kg981ms2l75m 305 pk Insight Verify for yourselfthat ifthe slide Were frictionless the seal would enter the Water at 586 ms 2007 Pearson Prentice Hall Upper Saddle River NJ currently exist No portion ofthis material may be reproduced in any rm or 228 All rights reserved This material is protected under all copyright laws as they y any means without permission in writing 39om the publisher Chapter 8 Potential Energy and Conservation of Energy 35 L 0 Iames S Walker Physics 3 d Edition Picture the Problem The rock falls straight downward under the influence of both gravity and water resistance Strategy The nonconservative work done by water resistance equals minus the force times the distance This work reduces the mechanical energy of the rock Use equation 83 to find the potential energy of the rock and then equation 89 to determine the kinetic energy of the rock Let y 0 at the bottom of the pond Solution 1 a Calculate the work done by water resistance Wm Fd 746 N 0 m I 0 2 Find the potential energy ofthe rock at y l8700 m Ua mgya 19 kg 981 ms2l8 m 3 I 3 Use equation 89 to find the kinetic energy of the rock Wm AE Kf Uf 7 K U KKUILW0700Q 4SumUanthof1ndE EaUaKa34JOJ3 I 5 b c Repeat steps 1 through 4 W U K 0 34 0 24 70 15 14 E 34 31 29 723 746 Insight If there were no water resistance the rock would be in freefall and the mechanical energy would remain constant at 34 1 Note that the nonconservative work only removes kinetic energy the potential energy remains mgy Picture the Problem The car drives up the hill changing its kinetic and gravitational potential energies while both the engine force and friction do nonconservative work on the car Strategy The total nonconservative work done on the car changes its mechanical energy according to equation 89 This nonconservative work includes the positive work W 2 done by the uni done by the engine and the negative work Wn friction Use this relationship and the known change in potential energy to findAK Solution Set the nonconservative W W sz AE Ef iE work equal tothe change anechanical Wm1 sz Kf UfK U AK mgyf y energy and solve for AK Wncl Wm2 7mgyf yl 644gtlt105 J 7311gtlt105 J71250 kg981 ms2162 m A1lt134gtlt105 1134k1 Insight The friction force reduces the car s mechanical energy but the engine increased it by a greater amount resulting in a net gain in both kinetic and potential energy The car gained speed while traveling uphill Picture the Problem The skater travels up a hill we know this for reasons given below changing his kinetic and gravitational potential energies while both his muscles and friction do nonconservative work on him Strategy The total nonconservative work done on the skater changes his mechanical energy according to equation 89 This nonconservative work includes the positive work Wm1 done by his muscles and the negative work sz done by the friction Use this relationship and the known change in potential energy to find Ay Solution 1 a The skater has gone because the work done by the skater is larger than that done by friction so the skater has gained mechanical energy However the final speed of the skater is less than the initial speed so he has lost kinetic energy Therefore he must have gained potential energy and has gone uphill 2 b Set the nonconservative Wm Wncl Wm2 AE Ef iE workequalto the change in Wm Wm2 Kf Uf 7K U mvf2 7K2mgAy mechanical energy and solve fOl Ay AyWnclmc27mv 7vi2mg 3420 7715 J7810 kgl22 ms27250 ms2 365 810 kg981 ms2 m Insight Verify for yourself that ifthe skates had been frictionless but the skater s muscles did the same amount of work the skater s final speed would have been 437 ms He would have sped up if it weren t for friction 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 229 C 38 w 5 ha ter 8 Potential Ener and Conservation of Ener James S Walker Ph rics 3rd Edition Picture the Problem The initial and final states ofthe system are depicted at right The masses given in the example are m 240 kg m2 180 kg but bk 0350 and dis unknown in this problem Strategy The nonconservative work done by friction changes the mechanical energy ofthe system Use equation 89 to nd AE and set it equal to the work done by friction Solve the resulting expression for the travel distance d Solution 1 Write equation 89 to obtain an expression for Wm Wm E E K UfKquot U Wm WM Wighm yu 3m mm Wighwegyhi W mi quot00 v3MZgyzryz 2 The nonconservative work is done by friction W fkd kmigd kWigdmr mzv 02ng0nd Agom2 kmiimi m2Vr2 d quotA mzV 2gmrkmi 3 Substitute the expression from step 2 into step 1 and solve for d 240 180 kg205 ms2 2981ms2180 kg 0350240 kg 4 b The conservative work changes the potential energy W AU ngAy 180 kg981ms20937 m 5 c The nonconservative work 7 7 2 is done by imon Wm kmigd 0350240 kg981ms 0937 m 6 d Verify Wmm AK A1ltm1 m2v 4 240 180 kg205 ms2 02883 J WW W Wm 165772 J 8 JAK veri ed 7 Verify Wc AU 65 JAU was verified in step 4 8 Verify Wm AE AKAU 883 J165 J 7 J Wm verified Insight The blocks travel almost twice as far in this problem when compared with Example 810 and are traveling almost twice as fast when W17 hits the floor but the similarity of these comparisons is simply coincidental Picture the Problem The truck travels down a hill changing its kinetic and gravitational potential energies Strategy Find the change in gravitational potential energy by using equation 83 and then the change in kinetic energy by using equation 76 Sum the two changes to find the change in mechanical energy Solution 1 a Use equation 83 to find AU AUmgyf y 15 800 kg981ms21440 m 1630 m 294gtlt107 J 2 b Use equation 76 to find AK AK mv vf 0 5800 kg290 ms2 120 msf 551gtlt106 J 3 c The total mechanical energy changes by AE AK AU 239 M Insight The loss in mechanical energy means a quot In this case some the truck driver either applied the brakes or relied on frictional forces to slow the truck 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion ofthis material may be reproduced in any form or by any means without permission in writing from the publisher 230 Chapter 8 Potential Energy and Conservation of Energy James S Walker Physics 3rd Edition 40 Picture the Problem The block slides horizontally on a rough surface encounters a spring and compresses it a distance of 1 10 cm before coming to rest Strategy The nonconservative Work done by friction changes the mechanical energy of the system Use equation 89 to nd AE and set it equal to the Work done by friction Solve the resulting expression for the spring constant k Solution 1 Write equation89 to obtain an Wm Ef E Kf Uf K U expression for Wm 0izkxZ 7mv 2 0 2 The nonconservative Work is done by Wm ifkd ipkmgx friction as the block travels a distance x 3 Substitute the expression from step 2 into irtkmgx ilk2 filmy step 1 and solve for k 2 2 k7 mv 72ukmgx 7 quotYD z kgx 7 2 7 7 x x 7usog aomm izmijgmms miwm QHOmY k 415 Nm Insight The force exerted by the spring at x 0110 m is 7 45 7 N Verify for yourselfthat ifthe coef cient of static friction with the rough surface is less than 5 259 then the spring is strong enough to accelerate the block again in the opposite direction and the block will not remain at rest 4 Picture the Problem The Uvs x plot is depicted at right Strategy Describe the motion of the object keeping in mind that objects tend to move to the minimum potential energy and When they do their kinetic energy is maximum The turning points on the potential energy plot are poinw A and E Solution At point A the object is at rest As the object travels from point A to point B some of its potential energy is converted into kinetic energy and the obj ect s speed increases As the object travels from point B to point C some of its kinetic energy is converted back into potential energy and its speed decreases From point C to point D the speed increases again and from point D to point E the speed decreases 10m 20m 30m 10m 50m Insight The object momentarily comes to rest at point E but then turns aron and accelerates back towards D and retraces its path all the Way to point A at Which time the cycle begins again 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion ofthis material may be reproduced in any form or by any means without permission in writing 39om the publisher 231 Chapter 8 Potential Energy and Con ervati on of Energy James S Walker Physics 3rd Edition 42 Picture the Problem The UVs xplotis depicted at right Strategy The loss ofpotential energy from point A to point B becomes a ain in kinetic energy Use the magnitude of AU to determine the Velocity at points B C and D by using the conservation ofmechanical energy equation 89 Solution 1 a Set E Ef and solVe for vf K U Kf Uf 0100 Ji2mvfz 14 V 2100 J Ur 10m 20m 30m 40m m m f m 2100 1720 J 2 Use the expression in step 1 to nd VB VB 11 kg 2 100 1760 J 3 b Use the expression in step 1 to nd 110 vc g 11 kg 2100175OJ 4 c Use the expression in step 1 to nd VD VD T g 5 d If the object starts at rest at point A the potential energy at point A is the maximum energy of the object Therefore poinw are the turning points of its motion Insight Note that the lower the object is on the potential energy curve the higher its speed Picture the Problem The UVs xplotis depicted at right Strategy The object will be at its turning point when is total mechanical w m A energy equals its potential energy Use the provided information to nd the quot E total mechanical energy of the particle then draw a horizontal line at the corresponding Value on the U axis to nd the turning points 20 c Solution 1 Determine E from the E gmv UC 5139 D 39d d 39 f ti Pro 6 m on on 144 kg125 ms2 60 I 201 B E 71 J r 111m 20 m 31lm 411m 511m 2 Draw a horizontal line on the plot at U 71 J and determine where the line crosses the potential energy curve In this case it crosses at approximately x 05 m and x 46 m These are the locations ofthe turning points Insight In order for the turning points to be at A and C the object needs a speed of236 ms at point C 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion ofthis material may be reproduced in any form or y any means without permission in writing 39om the publisher 232 r h James S Walker Physcs 3 Edltron a amm X P V uh on a swmgls shown at nghttogether wrth the plot of Uversus 9 me an expresslon forthepotentlal energy ofapendulum as a 9 5 and eonstruetthe l L Strategy Determ funetaon of 9 Then ndthe energy for 9 plot Lety 0 eonespondto the lowestpornt ofthe chlld39s rnotron Snlntinn 1Use the nght tnanglern the dlagram to nd the herghty ofthe ehrld 2UseEq 8r3 o ndUE bula memes UmgymgLleeoss U90 mgLleeosgo 23kg 981m5220m10 U90 e 501 VFW 1 S90 3 Fmd the rnaxrrnurn U 4 Theplot of Uversus 9 ls shown at nght othe potentral energy e ehrld Insi gle nds to the mlmmum andls the equlllbnum posmon forthe ehrld on the swrng onee th turnrng pmnts Ifthere ls faction the rneehanreal energy slowly deereases and Joe the turnrng pornts occur at smaller andsmaller angles untrl the chlld comes to rest atE Picturethe Prnhlem The plot of Uversus Sfor the ehrldrs shown at nght Strategy From the prevlous problem we know the herght of the chlld ls glven memes Set the potentaal energy at the turnrng pornts equal to the energy when the ropes are yertreal and when U 0 beeausethatrs the eehanreal energy ofthe ehrld by y kmetl total rn U mgy mguleeoss 5 2 Set U KM and solve forthe angle 5M Snllltinn 1Use equataon 83 to ndU 2 712 1239 0 89msjz 29 81 rns 2 0 rn 3 Deterrnrnethe two values of em pumping Tnci ht UppEv Saddle Men M AH thls veserved e nu Hquot ml 233 Chapter 8 Potential Energy and nn ervati on of Energy James S Walker Physics 3rd Edition 46 Picture the Problem The Uvs xplotis depicted at right Strategy Use the provided information to nd the total mechanical energy ofthe particle then draw a horizontal line at the corresponding value on the 8 E Uaxis to nd the turning points J a Solution 1 a DetermineE from E KU 3650 J 231 C the provided information From the 839 J 7 D graphwe seethat U10 m250J 2m 39 B 10m 26 m 3llm 46 m lm quot 2 Draw a horizontal line on the plot at U 86 J and determine where the line crosses the potential energy curve In this x 48 m These are the locations ofthe turning points Therefore the m and the largest value ofx the particle can reach is c A 8 m case it crosses at approximately x 02 m and smallest value ofx the particle can reach is b Insight In order for the turning points to be at A and C the object needs a kinetic energy of50 J at x 10 m Picture the Problem The plot of Uversus xis shown at right UI 4 gt1 Strategy Use equation 85 to nd an expression for the potential energy as a function ofx Use the provided information to nd the total mechanical energy of the block Set the potential energy at the turning points equal to the kinetic energy at x 0 because that is the total mechanical energy ofthe block Solution 1 a Determine an U gkxz 775 Nmx2 expression for Ux and draw the 388 Nmx2 plot The plot is shown at right MS00 cm 388 Nm00500 m 0 9 J cm 2 b Set U Kmax and solve izkx 2 7L 2 max zmvmax max f or x x i l vmi l13msioo46m k 775 Nm Insight In order to make the turning points equal to i500 cm the block needs a speed ofI43 ms atx 0 4 00 Picture the Problem The plot of Uas a function ofy is shown at right U0 Strategy Use equation 83 to nd an expression for the potential energy as a function ofy Use the mass and velocity to nd the kinetic energy ofthe ball when it 37 is thrown The kinetic energy at that point is also its total mechanical energy ofthe 30 block Set the potential energy equal to the total mechanical energy to nd the turning point Solution 1 a Determine an U mgy 075 kg981mSly expression for Uy and draw the plot 7 4 Ny y In I 40 50 The plot is shown at right 2 b Set U Kmax and solve for ym mgymax izmvrznax v2 89 ms2 ymaxi 2g 2981ms2 Insight The ball is thrown with 30 J ofkinetic energy and rises 40 m until its gravitational potential energy is 30 J 2007 Pearson Prentice Hall Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion ofthis material may be reproduced in any form or by any means without permission in writing 39om the publisher 234

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