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

Transcription

1 Physcs 00A Homework 7 Chapter 8 Where s the Energy? In ths problem, we wll consder the ollowng stuaton as depcted n the dagram: A block o mass m sldes at a speed v along a horzontal smooth table. It next sldes down a smooth ramp, descendng a heght h, and then sldes along a horzontal rough loor, stoppng eventually. Assume that the block sldes slowly enough so that t does not lose contact wth the supportng suraces (table, ramp, or loor. You wll analyze the moton o the block at derent moments usng the law o conservaton o energy. A Whch word n the statement o ths problem allows you to assume that the table s rctonless? Smooth I the problem had rcton t would say rough. B Expresson o conservaton o energy. Wthout rcton there s no non-conservatve work. Energy s conserved: mv + mgh = mv + mgh C As the block sldes down the nclne K ncreases, U decreases, E stays the same D Speed at the bottom. The answer assumes v = v = v h = h = h v = v h = h = 0 top top bottom bottom From the conservaton o energy equaton n part B vb = ( v + gh E From the bottom o the ramp untl the block stops. The potental energy s zero, snce n ths case the loor has been chosen or U=0. The ntal knetc energy s dmnshed due to the acton o rcton. Copyrght 00 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst. No porton o ths materal may be reproduced, n any orm or by any means, wthout permsson n wrtng rom the publsher. 8

2 Chapter 8: Potental Energy and Conservaton o Energy Work Energy Theorem Wnc = K + U K U Wnc = K -K K + Wnc = K mv + Wnc = mv F As the block sldes across the loor K decreases, U stays the same, E decreases G Frcton s responsble or the decrease n mechancal energy. H Snce the block comes to a stop at the level where U=0, the energy lost to rcton s E = mv + mgh 8. Calculate the work done by gravty as a 3. kg object s moved rom pont A to pont B n the gure along paths,, and Pcture the Problem: The three paths o the object are depcted at rght. Strategy: Fnd the work done by gravty W = mgy when the object s moved downward, W = mgy when t s moved upward, and zero when t s moved horzontally. Sum the work done by gravty or each segment o each path. Soluton:. Calculate the work or path : [ 0 0 ] mg ( 4.0 m (.0 m + (.0 m ( ( ( W = mg y+ + y + + y3 = + W = 3. kg 9.8 m/s.0 m = 63 J [ ] [ ]. Calculate W or path : W mg y4 ( ( = = 3. kg 9.8 m/s.0 m = 63 J [ ] ( ( ( 3. Calculate W or path 3: W3 mg y5 y6 ( = + 0 = 3. kg 9.8 m/s.0 m 3.0 m = 63 J Insght: The work s path-ndependent because gravty s a conservatve orce. 8.4 A 4. kg block s attached to a sprng wth a orce constant o 550 N/m, as shown n the gure. Fnd the work done by the sprng on the block as the block moves rom A to B along paths and. 4. Pcture the Problem: The physcal stuaton s depcted at rght. Strategy: Use equaton W = kx (equaton 7-8 to nd the work done by the sprng, but cauton s n order: Ths work s postve when the orce exerted by the sprng s n the same drecton that the block s travelng, but t s negatve when they pont n opposte drectons. One way to keep track o that sgn conventon s to say that W = k( x x. That way the work wll always be negatve you Copyrght 00 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst. No porton o ths materal may be reproduced, n any orm or by any means, wthout permsson n wrtng rom the publsher. 8

3 Chapter 8: Potental Energy and Conservaton o Energy start out at x = 0 because the sprng orce wll always be n the opposte drecton rom the stretch or compresson. Soluton:. (a Sum the work done by the sprng or each segment o path :. Sum the work done by the sprng or each segment o path : ( ( W = k x x + x x 3 { } ( 550 N/m 0 ( m ( m ( 0.00 m ( ( = + W = 0.44 J J = 0. J ( ( W = k x x + x x ( 550 N/m 0 ( 0.00 m ( 0.00 m ( 0.00 m ( ( W = 0. J + 0 J = 0. J { } = + 3. (b The work done by the sprng wll stay the same you ncrease the mass because the results do not depend on the mass o the block. Insght: The work done by the sprng s negatve whenever you dsplace the block away rom x = 0, but t s postve when the dsplacement vector ponts toward x = 0. Introducton to Potental Energy The work energy theorem: Wtotal = K K A A orce actng on a partcle over a dstance changes the knetc energy o the partcle. B To calculate the change energy, you must know the orce as a uncton o dstance. x to C To llustrate the work-energy concept, consder the case o a stone allng rom x under the nluence gravty. Usng the work-energy concept, we say that work s done by the gravtatonal orce, resultng n an ncrease o the knetc energy o the stone. The work o conservatve orces can be wrtten n terms o the change n potental energy: Wc = Δ U = ( U U Wtotal = Wnc + W c = ΔK Wnc Δ U = ΔK W =Δ K + Δ U =Δ E nc D Rather than ascrbng the ncreased knetc energy o the stone to the work o gravty, we now (when usng potental energy rather than work-energy say that the ncreased knetc energy comes rom the change o the potental energy. E Ths process happens n such a way that total mechancal energy, equal to the sum o the knetc and potental energes, s conserved. Copyrght 00 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst. No porton o ths materal may be reproduced, n any orm or by any means, wthout permsson n wrtng rom the publsher. 8 3

4 Chapter 8: Potental Energy and Conservaton o Energy Stretchng a Sprng As llustrated n the gure, a sprng wth sprng constant k s stretched rom (x=0 to (x=3d, where (x=0 s the equlbrum poston o the sprng. (Intro gure ARemember that the amount o work done by the sprng s equal to the negatve o the change n potental energy. The work done by us s opposte to the one o the sprng. $W_{hand}=-W_{sprng}=U_{}-U_{}$ From x=0 to x=d: From x=d to x=d: Whand = k ( (0 = d k kd 3 Whand = k ( ( = d k d kd 5 From x=d to x=3d: Whand = k (3 d k( d = kd B The same energy s needed to compress or stretch the sprng. From x=0 to x=d: Whand = k ( d k(0 = kd From x=0 to x=-d: Whand = k ( d k(0 = kd C Now consder two sprngs A and B that are attached to a wall. Sprng A has a sprng constant that s our tmes that o the sprng constant o sprng B. I the same amount o energy s requred to stretch both sprngs, what can be sad about the dstance each sprng s stretched? k A = 4k Gven condton B k x 4k x = k x A A = kbxb B A B B x A = x B The energes are equal Sprng A must stretch hal the dstance sprng B stretches. Copyrght 00 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst. No porton o ths materal may be reproduced, n any orm or by any means, wthout permsson n wrtng rom the publsher. 8 4

5 Chapter 8: Potental Energy and Conservaton o Energy D Two dentcal sprngs are attached to two derent masses, M A and M B, where M A s greater than The masses le on a rctonless surace. Both sprngs are compressed the same dstance, d, as shown n the M B. gure. Whch o the ollowng statements descrbes the energy requred to compress sprng A and sprng B? (Part D gure The energy assocated wth a sprng s ndependent o mass. Sprng A requres the same amount o energy as sprng B. 8.0 Fnd the gravtatonal potental energy o an 88 kg person standng atop Mt. Everest at an alttude o 8848 m. Use sea level as the locaton or y=0. 0. Pcture the Problem: The clmber stands at the top o Mt. Everest. Strategy: Fnd the gravtatonal potental energy by usng equaton 8-3. Soluton: Calculate U m gy : = U mgy ( ( ( = = = = 6 88 kg 9.8 m/s 8848 m J 7.6 MJ Insght: You are ree to declare that the clmber s potental energy s zero at the top o Mt. Everest and 7. MJ at sea level! 8.7 A 0.33 kg pendulum bob s attached to a strng. m long. What s the change n the gravtatonal potental energy o the system as the bob swngs rom pont A to pont B n the gure 7. Pcture the Problem: The pendulum bob swngs rom pont A to pont B and loses alttude and thus gravtatonal potental energy. See the gure at rght. Strategy: Use the geometry o the problem to nd the change n alttude Δy o the pendulum bob, and then use equaton 8-3 to nd ts change n gravtatonal potental energy. Soluton:. Take the apex o the y = ya = Lcosθ Pendulum as the place or y=0. Let L be the length o the pendulum. y = yb = L. Fnd ΔU : Δ U = U U Δ U = mgy mgy Δ U = mg( L -( Lcos θ Copyrght 00 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst. No porton o ths materal may be reproduced, n any orm or by any means, wthout permsson n wrtng rom the publsher. 8 5

6 Chapter 8: Potental Energy and Conservaton o Energy mgl( cosθ ( 0.33 kg( 9.8 m/s (. m( cos 35 Δ U = = Δ U = 0.70 J Insght: Note that the change n heght Δy s negatve because the pendulum swngs rom A to B. Lkewse, the change n heght s postve and the pendulum gans potental energy t swngs rom B to A. Conservaton o Energy Rankng Task A Rank each pendulum on the bass o ts ntal gravtatonal potental energy (beore beng released. Possble combnatons or potental energy a m=8 kg; h=5 cm; U = (8(9.8(0.5 =.76 J b m= kg; h=60cm; U = ((9.8(0.6 = 5.88 J c m= kg; h=60 cm; U = ((9.8(0.6 =.76 J d m=3 kg; h=45 cm; U = (3(9.8(0.45 = 3.3 J e m=4 kg; h=30 cm; U = (4(9.8(0.3 =.76 J m= kg; h=30 cm; U = ((9.8(0.3 = 5.88J From largest to smallest: d, (a,c,d, (b, B Rank each pendulum on the bass o the maxmum knetc energy t attans ater release. Energy s conserved. All the potental energy converts nto knetc, when the maxmum knetc s reached. So the order o the pendulums accordng to possble knetc energes s the same as above. C Rank each pendulum on the bass o ts maxmum speed. Convertng the potental nto knetc energy v mv = mgh = gh The larger the heght the larger the speed, ndependent o the mass. Copyrght 00 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst. No porton o ths materal may be reproduced, n any orm or by any means, wthout permsson n wrtng rom the publsher. 8 6

7 Chapter 8: Potental Energy and Conservaton o Energy 8.4 Startng at rest at the edge o a swmmng pool, a 7.0 kg athlete swms along the surace o the water and reaches a speed o.0 m/s by dong the work W = 6 J. Fnd the nonconservatve work, W, done by the water on the athlete. nc nc 4. Pcture the Problem: The athlete accelerates horzontally through the water rom rest to.0 m/s whle dong nonconservatve work aganst the drag rom the water. Strategy: The total nonconservatve work done on the athlete changes hs mechancal energy accordng to equaton 8-9. Ths nonconservatve work ncludes the postve work W done by the athlete s muscles and the negatve work W done by the water. Use ths relatonshp and the known change n knetc energy to nd. Soluton: Set the nonconservatve work equal to the change n mechancal energy and solve or W. The ntal mechancal energy s zero: ( ( ( nc nc W nc Wnc = Wnc + Wnc =Δ E = E E = K 0 Wnc = K Wnc = mv Wnc = 7.0 kg.0 m/s 6 J = 09 J Insght: The drag orce rom the water reduced the swmmer s mechancal energy, but hs muscles ncreased t by a greater amount, resultng n a net gan n mechancal energy. nc 8.50 An 8.0 kg n-lne skater does 340 J o nonconservatve work by pushng aganst the ground wth hs skates. In addton, rcton does -75 J o nonconservatve work on the skater. The skater's ntal and nal speeds are.50 m/s and. m/s, respectvely. 50. Pcture the Problem: The skater travels up a hll (we know ths or reasons gven below, changng hs knetc and gravtatonal potental energes, whle both hs muscles and rcton do nonconservatve work on hm. Strategy: The total nonconservatve work done on the skater changes hs mechancal energy accordng to equaton 8-9. Ths nonconservatve work ncludes the postve work Wnc done by hs muscles and the negatve work Wnc done by the rcton. Use ths relatonshp and the known change n potental energy to nd Δ y. Soluton:. (a The skater has gone uphll because the work done by the skater s larger than that done by rcton, so the skater has ganed mechancal energy. However, the nal speed o the skater s less than the ntal speed, so he has lost knetc energy. Thereore he must have ganed potental energy, and has gone uphll.. (b Set the nonconser- Wnc = Wnc + Wnc =Δ E = E E vatve work equal to the Wnc + Wnc = ( K + U ( K + U = m( v v + mgδy change n mechancal energy and solve or Δ y : Δ y = Wnc Wnc m( v v + mg {( 340 J + ( 75 J ( 8.0 kg (. m/s (.50 m/s } = = 3.65 m 8.0 kg 9.8 m/s ( ( Insght: Very or yoursel that the skates had been rctonless but the skater s muscles dd the same amount o work, the skater s nal speed would have been 4.37 m/s. He would have sped up t weren t or rcton! 8.30 A.9 kg block sldes wth a speed o.6 m/s on a rctonless horzontal surace untl t encounters a sprng. A I the block compresses the sprng 4.8 cm beore comng to rest, what s the orce constant o the sprng? B What ntal speed should the block have to compress the sprng by.cm? 30. Pcture the Problem: A block sldes on a rctonless, horzontal surace and encounters a horzontal sprng. It compresses the sprng and brely comes to rest. Strategy: Set the mechancal energy when sldng reely equal to the mechancal energy when the sprng s ully compressed and the block s at rest. Solve the resultng equaton or the sprng constant k, then repeat the procedure to nd the ntal speed requred to compress the sprng only. cm beore comng to rest. Copyrght 00 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst. No porton o ths materal may be reproduced, n any orm or by any means, wthout permsson n wrtng rom the publsher. 8 7

8 Chapter 8: Potental Energy and Conservaton o Energy Soluton:. (a Set E = E where the ntal state s when t s sldng reely and the nal state s when t s at rest, havng compressed the sprng. K + U = K + U mv + 0= 0+ kx max (.9 kg(.6 m/s ( m mv k = = = 300 N/m = 3. kn/m x max. (b Solve the equaton rom step or : ( ( v kx 300 N/m 0.0 m v max = = = m.9 kg 0.40 m/s Insght: The knetc energy o the sldng block s stored as potental energy n the sprng. Moments later the sprng wll have released all ts potental energy, the block would have ganed ts knetc energy agan, and would then be sldng at the same speed but n the opposte drecton A 50 kg car drves up a hll that s 6. m hgh. Durng the drve, two nonconservatve orces do work on the car: ( the orce o rcton, and ( the orce generated by the car's engne. The work done by rcton s J ; the work done by the engne s J. 49. Pcture the Problem: The car drves up the hll, changng ts knetc and gravtatonal potental energes, whle both the engne orce and rcton do nonconservatve work on the car. Strategy: The total nonconservatve work done on the car changes ts mechancal energy accordng to equaton 8-9. Ths nonconservatve work ncludes the postve work Wnc done by the engne and the negatve work Wnc done by the rcton. Use ths relatonshp and the known change n potental energy to nd Δ K. Soluton: Set the nonconservatve Wnc = Wnc + Wnc = Δ E = E E work equal to the change n me- Wnc + Wnc = ( K + U ( K + U = Δ K + mg( y y chancal energy and solve or Δ K : Δ K = Wnc + Wnc mg( y y 5 5 = J J 50 kg 9.8 m/s 6. m ( ( ( ( ( J 34 kj Δ K = = Insght: The rcton orce reduces the car s mechancal energy, but the engne ncreased t by a greater amount, resultng n a net gan n both knetc and potental energy. The car ganed speed whle travelng uphll. Copyrght 00 Pearson Educaton, Inc. All rghts reserved. Ths materal s protected under all copyrght laws as they currently exst. No porton o ths materal may be reproduced, n any orm or by any means, wthout permsson n wrtng rom the publsher. 8 8