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Unformatted text preview: Physics 120  Homework 5 2.RQ.16 B. 2.RQ.20 There is no contradiction here. When you pull a sled across a level eld at constant velocity, the force that you exert just balances the frictional force so that the net force is zero and therefore there is no change in velocity. 2.P.25 m = 2200 kg , v i = h 25 , , i m/s , v f = h , , i , x = 0 . 8 m F net =  p t Assuming that F net is constant during the collision, we can nd the time interval for the collision t from t = x v avgx = x 1 2 ( v ix + v fx ) = . 8 m 1 2 (25 m/s +0) = 0 . 064 s F net = m ( v f v i ) t = 2200 kg . 064 s [ h , , i  h 25 , , i m/s ] = h 8 . 6 10 5 , , i N F net = 8 . 6 10 5 N 2.P.28 m . 4 kg , v i x = 3 . 5 m/s , t = 0 . 002 s , = 20 (a) B (b) Note that F net x = 0 and therefore v f x = v i x = 3 . 5 m/s. Also, v f y = v f x tan = (3 . 5 m/s ) tan 20 = 1 . 3 m/s v i = h 3 . 5 , , i m/s and v f = h 3 . 5 , 1 . 3 , i m/s...
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 Winter '08
 Labrake
 Force, Friction, Mass, Work, 0.8 m, 1.3 M/s, Vfy

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