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Unformatted text preview: olve the yequation to give n = mgcosq. Use this in the friction model to get f k = m k mg cos q . Now substitute this result for fk into the xequation:
a0 = mg sin q  m k mg cos q m =  g (sinq + m k cos q ) =  9.8 m / s ( 2 )(sin 30 + 0.20 cos 30 ) = 6.60 m /s
2 2 Kinematics now gives
v1 = v0 + 2a 0 ( x1  x0 ) fi x1 =
2 2 v12  v 2 0 2a 0 = 0 m 2 /s 2  (10 m / s) 2 6.60 m / s 2 ( ) = 7.58 m The block's height is then h = x1sinq = (7.58 m)sin30 = 3.79 m. r (b) For the return trip, f k points up the slope, so the xcomponent of the second law is a x = a1 = (F )
net x m = w sin q + f k m =  mg sin q + f k m
2 Note the sign change. The yequation and the frict...
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This note was uploaded on 05/20/2008 for the course PHYS 161 taught by Professor Hammer during the Spring '07 term at Maryland.
 Spring '07
 Hammer
 mechanics, Work

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