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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 b g a f = + v f v i f = 1 33 . ± m s (b) F y = 0: n = 60 0 9 80 0 . . kg m s 2 b g f n k k = = = µ 0 400 588 235 . N N a f f i k = 235 N ± FIG. P9.55 (c) For the person, p I p i f + = mv Ft mv i f + = 60 0 4 00 235 60 0 1 33 0 680 . . . . . kg m s N kg m s s b g a f b g = = t t (d) person: m m f i v v i = = 60 0 1 33 4 00 160 . . . ± kg m s N s a f cart: 120 1 33 0 160 kg m s N s . ± b g = + i (e) x x v v t f i i f = + = + = 1 2 1 2 4 00 1 33 0 680 1 81 d i a f . . . . m s s m (f) x x v v t f i i f = + = + = 1 2 1 2 0 1 33 0 680 0 454 d i b g . . . m s s m (g) 1 2 1 2 1 2 60 0 1 33 1 2 60 0 4 00 427 2 2 2 2 mv mv f i = = . . . . kg m s kg m s J b g b g (h) 1 2 1 2 1 2 120 0 1 33 0 107 2 2 2 mv mv f i = = . . kg m s J b g (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
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