271_Physics ProblemsTechnical Physics

271_Physics ProblemsTechnical Physics - P9.55 (a) b60.0 kg...

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Chapter 9 273 P9.55 (a) 60 0 4 00 120 60 0 .. . kg m s kg bg af =+ v f vi f = 133 . ± ms (b) F y = 0: n −= 60 0 9 80 0 kg m s 2 fn kk == = µ 0 400 588 235 . N N fi k =− 235 N ± FIG. P9.55 (c) For the person, pIp if += mv Ft mv 60 0 4 00 235 60 0 1 33 0680 . kg m s N kg m s s = t t (d) person: mm vv i = 60 0 1 33 4 00 160 . ± kg m s N s cart: 120 1 33 0 160 kg m s N s . ± −=+ i (e) xx v v t + = + = 1 2 1 2 400 133 181 di . . s m (f) v v t + = + = 1 2 1 2 0 1 33 0 680 0 454 . s m (g) 1 2 1 2 1 2 60 0 1 33 1 2 60 0 4 00 427 22 mv mv = kg m s kg m s J (h) 1 2 1 2 1 2 120 0 1 33 0 107 2 mv mv = kg m s J (i) The force exerted by the person on the cart must equal in magnitude and opposite in direction to the force exerted by the cart on the person. The changes in momentum of the two objects must be equal in magnitude and must add to zero. Their changes in kinetic energy are different in magnitude and do not add to zero. The following represent two ways of thinking about ’why.’ The distance the cart moves is different from the distance moved by the point of application of the friction force to the cart.
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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