, so if m is the same and Δ y is the same, Δ U will be the same.

Transcription

1 Phscs 40 Homework Solutons - Walker, Chapter 8 Problems and Conceptual Exercses 7 (a) The change n the tatonal potental energ s the same or the two balls because the change n ertcal poston s the same Δ mgδ, so m s the same and Δ s the same, Δ wll be the same (b) Explanaton II s the best explanaton or wh the change n potental energ s the same 7 Pcture: m ( m) cos35 35 m A B Clearl, the tatonal potental energ s gong to decrease as the bob goes rom A to B the bob goes down Its ntal tatonal potental energ s: and the nal potental ener s mg mg So the change n the tatonal potental energ s: ( ) mg( ) Δ mg But s just the dstance ndcated n the pcture So we need to nd an expresson or ths dstance n terms o the length o the strng and the angle that the strng makes wth the ertcal To do ths, consder the rght trangle shown n the pcture aboe From ths pcture, t s clear that: So: ( m ) cos 35 m Δ Δ ( 033 kg)( 98 m/s )[ m ( m) cos35 ] 070 J

2 Phscs 40 Homework Solutons - Walker, Chapter 8 8 (a) Ignorng ar resstance, the onl orce actng on the tenns ball wll be t, whch s a conserate orce So the total energ wll be consered Take the tatonal potental energ to be zero at the surace o the court e, at ground leel When the ball s at ths leel, t has knetc energ: m I the ball has a speed m/s when t s caught b the an, then the ball s knetc energ dropped b the amount: m m Snce the ball s total energ must sta constant, ts tatonal potental energ must hae ncreased b an equal amount The change n the tatonal potental energ s: So we hae: Solng or Δ mgδ Δ n Eq ges: Pluggng n numbers, I get: mgδ m m () Δ g Δ 57 m Ths s the heght aboe the court at whch the an catches the ball (b) The answer to ths queston can be ound n Eq () You don t need to know the mass o the tenns ball because t cancels n Eq () m appears n the expresson or the potental energ ganed and n the expresson or the knetc energ lost 9 Ignorng ar resstance, the onl orce actng on the apple wll be the orce o t, whch s a conserate orce So the total energ wll be consered The ntal energ (beore the apple alls e, when ts heght aboe the ground s 40 m) s all tatonal potental energ: and the knetc energ s 0 J When the apple s at 30 m ( 4 0 m) 8404 J 8 J E mg (keepng sg gs),, ts total energ wll stll be 8 J But ts tatonal potental energ wll now be: ( 30 m) 6803 J 6 J 3 0 mg

3 Phscs 40 Homework Solutons - Walker, Chapter 8 The remander o the total energ wll be knetc energ, so 3 0 E J J 060 J J, keepng two sg gs (Note that I used all o the dgts rom the preous calculatons n calculatng, to mnmze errors that mght hae resulted rom repeated roundng) 0 m : E 8 J 0 m : E 8 J 0 m : E 8 J ( 0 m) 40 J 4J 0 mg J ( 0 m) 060 J J 0 mg J J 8 J 4J 6 J 30 (a) kx m (Energ conseraton When the sprng s at ts mum compresson and the block s momentarl at rest all o the knetc energ that the block had pror to httng the sprng has been stored n the sprng as potental energ) Now just sole or k: ( 9 kg)( 6 m/s) ( 0048 m) m 3 k 3 0 N/m x (b) What does need to be x s to be cm (00 m)? Well, rom the conseraton-o-energ equaton aboe, I gure: k x m ( 00 m) N/m 9 kg 040 m/s 76 (a) The onl orce dong work on the chld wll be t (neglectng ar resstance) So the total energ wll be consered When he reaches hs mum ertcal heght, hs knetc energ goes to zero, so all o the knetc energ he had at the bottom has been turned nto tatonal potental energ I we choose the reerence leel or the tatonal potental energ to be at hs lowest pont, then the potental energ there s zero So at hs mum ertcal heght, the potental energ wll be: mg And, as we sad beore, ths must be equal to the knetc energ he had at the bottom: 3

4 Phscs 40 Homework Solutons - Walker, Chapter 8 Solng or, I nd: mg m 008 m g (b) I the ntal speed s haled, then ths means replacng b n the expresson or mmedatel aboe So the new alue o So would be: g 4 g would be one-ourth o what t was beore 86 The author means to sa that the block starts rom rest, though he doesn t sa ths Because the ramp s rctonless, the total energ o the block wll be consered (There are onl two orces on t the normal orce and the weght and the normal orce, whle not conserate, neer does an work on the block) So when the block reaches the bottom o the ramp, t wll hae acqured an amount o knetc energ equal to the potental energ t lost Ths potental energ lost s: ( 5 m 05 m) mg( 5 m) lost mg So ths must equal the knetc energ that the block has at the bottom o the ramp: Solng or ges: ( 5 m) m mg ( 5 m) 495 m/s 50 m/s g (keepng sg gs) Once the block leaes the slde, t becomes a projectle So or ths second part o the moton we can use all o the acts that we know about projectles Imagne a coordnate sstem whose orgn s at the poston o the block when t just leaes the ramp Take the poste x drecton to be to the let n the gure and the poste drecton to be upward Then what we want to know s, What s x when 05 m? Well, when does 05 m? Recall one o the ormulas or the moton o a projectle: t gt Snce we are told that the block s mong horzontall when t leaes the ramp, 0 0 And 0 0 because o the choce o orgn I made So we hae: 4

5 Phscs 40 Homework Solutons - Walker, Chapter 8 gt Solng or the tme ges: t g And I want to know what t s when 05 m Ths tme s: t ( 05 m) 98 m/s 06 s Ths s the tme when the block hts the ground Now what s x at ths tme? Well, rom what we know about the horzontal moton o a projectle: x x t 0 + x0 But x 0 0 b choce and x0 495 m/s (the eloct at the bottom o the ramp s all horzontal) So when the block hts the ground, ( 4 95 m/s)( 06 s) 8 m m x (keepng sg gs) 5