CHAPTER 6 WORK AND ENERGY

Transcription

1 CHAPTER 6 WORK AND ENERGY CONCEPTUAL QUESTIONS 16. REASONING AND SOLUTION A trapeze artist, starting rm rest, swings dwnward n the bar, lets g at the bttm the swing, and alls reely t the net. An assistant, standing n the same platrm as the trapeze artist, jumps rm rest straight dwnward. a. The wrk dne by gravity, n either persn, is W = mgh, where m is the mass the persn, and h is the magnitude the vertical cmpnent the persn's displacement. The value h is the same r bth the trapeze artist and the assistant; hwever, the value m is, in general, dierent r the trapeze artist and the assistant. Therere, gravity des the mst wrk n the mre massive persn. b. Bth the trapeze artist and the assistant strike the net with the same speed. Here's why. They bth start ut at rest n the platrm abve the net. As the trapeze artist swings dwnward, bere letting g the bar, she mves alng the arc a circle. The wrk dne by the tensin in the trapeze crds is zer because it pints perpendicular t the circular path mtin. Thus, i air resistance is ignred, the wrk dne by the nncnservative rces is zer, Wnc = J. The ttal mechanical energy each persn is cnserved, regardless the path taken t the net. Since bth the trapeze artist and the assistant had the same initial ttal mechanical energy (all ptential), they must have the same ttal mechanical energy (all kinetic) when they reach the net. That is, r either persn, mgh = (1/ )mv. Slving r v gives v= gh, independent the mass the persn. The value h is the same r bth the trapeze artist and the assistant; therere, they strike the net with the same speed. CHAPTER 6 WORK AND ENERGY PROBLEMS 3. REASONING The nly tw rces that act n the gymnast are his weight and the rce exerted n his hands by the high bar. The latter is the (nn-cnservative) reactin rce t the rce exerted n the bar by the gymnast, as predicted by Newtn's third law. This rce, hwever, des n wrk because it pints perpendicular t the circular path mtin. Thus, W nc = J, and we can apply the principle cnservatin mechanical energy. SOLUTION The cnservatin principle gives 1 mv + mgh = 1 mv + mgh E Since the gymnast's speed is mmentarily zer at the tp the swing, v = m/s. I we take h = m at the bttm the swing, then h = r, where r is the radius the circular path llwed by the gymnast's waist. Making these substitutins in the abve expressin and slving r v, we btain v = gh = g(r) = (9.8 m/s )( 1.1 m) = 6.6 m/s E

2 33. REASONING Since air resistance is being neglected, the nly rce that acts n the alling water is the cnservative gravitatinal rce (its weight). Since the height the alls and the speed the water at the bttm are knwn, we may use the cnservatin mechanical energy t ind the speed the water at the tp the alls. SOLUTION The cnservatin mechanical energy, as expressed by Equatin 6.9b, states that 1 1 mv + mgh + mgh E The mass m can be eliminated algebraically rm this equatin, since it appears as a actr in every term. Slving r v gives ( ) = + v v g h h ( ) ( ) ( ) = 5.8m/s + 9.8m/s m 33.m = 3.9m/s 34. REASONING AND SOLUTION The cnservatin energy gives E Rearranging gives h 1 1 mv + mgh + mgh ( 14.m/s) ( 13.m/s) h = = 1.4m 9.8m/s ( ) 4. REASONING AND SOLUTION I air resistance is ignred, the nly nncnservative rce that acts n the skier is the nrmal rce exerted n the skier by the snw. Since this rce is always perpendicular t the directin the displacement, the wrk dne by the nrmal rce is zer. We can cnclude, therere, that mechanical energy is cnserved. 1 mv + mgh = 1 mv + mgh Since the skier starts rm rest v = m/s. Let h deine the zer level r heights, then the inal gravitatinal ptential energy is zer. This gives mgh = 1 mv (1)

3 At the crest the secnd hill, the tw rces that act n the skier are the nrmal rce and the weight the skier. The resultant these tw rces prvides the necessary centripetal rce t keep the skier mving alng the circular arc the hill. When the skier just lses cntact with the snw, the nrmal rce is zer and the weight the skier must prvide the necessary centripetal rce. F N mg mv mg = s that v =gr r () Substituting this expressin r v int Equatin (1) gives Slving r h gives mgh = 1 mgr r 36 m h = = = 18 m CHAPTER 7 IMPULSE AND MOMENTUM CONCEPTUAL QUESTIONS 3. REASONING AND SOLUTION a. Yes. Mmentum is a vectr, and the tw bjects have the same mmentum. This means that the directin each bject s mmentum is the same. Mmentum is mass times velcity, and the directin the mmentum is the same as the directin the velcity. Thus, the velcity directins must be the same. b. N. Mmentum is mass times velcity. The act that the bjects have the same mmentum means that the prduct the mass and the magnitude the velcity is the same r each. Thus, the magnitude the velcity ne bject can be smaller, r example, as lng as the mass that bject is prprtinally greater t keep the prduct mass and velcity unchanged. 8. REASONING AND SOLUTION The impulse-mmentum therem, Equatin 7.4, states that F t mv. Assuming that the gl ball is at rest when it is struck with the club, F t. During a gd "llw-thrugh" when driving a gl ball, the club is in cntact with the ball r the lngest pssible time. Frm the impulse-mmentum therem, it is clear that when the cntact time t is a maximum, the inal linear mmentum mv the ball is a maximum. In ther wrds, during a gd "llw thrugh" the maximum amunt mmentum is transerred t the ball. Therere, the ball will travel thrugh a larger hrizntal distance. CHAPTER 7 IMPULSE AND MOMENTUM

4 PROBLEMS 6. REASONING AND SOLUTION a. Accrding t Equatin 7.4, the impulse-mmentum therem, F t mv. Since the galie catches the puck, v =. Slving r the average rce exerted n the puck, we have m( v v ).17kg[ 65m / s] 3 F = = =. 1 N 3 t 5. 1 s By Newtn s third law, the rce exerted n the galie by the puck is equal in magnitude, but ppsite in directin, t the rce exerted n the puck by the galie. Thus, the average rce exerted n the galie is N. b. I, instead catching the puck, the galie slaps it with his stick and returns the puck straight back t the player with a velcity 65 m/s, then the average rce exerted n the puck by the galie is m( v v ).17kg[( 65m / s) ( + 65m / s)] 3 F = = = 4. 1 N 3 t 5. 1 s The average rce exerted n the galie by the puck is thus N. The answer in part (b) is twice that in part (a). This is cnsistent with the cnclusin Cnceptual Example 3. The change in the mmentum the puck is greater when the puck rebunds rm the stick. Thus, the puck exerts a greater impulse, and hence a greater rce, n the galie. 1. REASONING This is a prblem in vectr additin, and we will use the cmpnent methd r vectr additin. Using this methd, we will add the cmpnents the individual mmenta in the directin due nrth t btain the cmpnent the vectr sum in the directin due nrth. We will btain the cmpnent the vectr sum in the directin due east in a similar ashin rm the individual cmpnents in that directin. Fr each jgger the mmentum is the mass times the velcity. SOLUTION Assuming that the directins nrth and east are psitive, the cmpnents the jggers mmenta are as shwn in the llwing table: Directin due east Directin due nrth 85 kg jgger 85kg.m / s = 17kg kg m/s 55 kg jgger 55kg (3.m / s cs3 ) = 14kg 55kg (3.m / s sin 3 ) = 87kg m/ s Ttal 31 kg m/s 87 kg m/s Using the Pythagrean therem, we ind that the magnitude the ttal mmentum is (31kg ) + (87kg ) = 3kg

5 The ttal mmentum vectr pints nrth east by an angle θ, which is given by θ 87kg = tan 1 = 16 31kg m / s 16. REASONING AND SOLUTION The mmentum the spaceship is transerred t the rcket, s (m s + m r )v s = m r v r, and the rcket's velcity is v r = (m 6 s + m r )v s (4. 1 kg + 13 kg)(3 m/s) = m r 13 kg = m/s