Homework 4Sol

# First solve the y equation to give n mgcosq use this

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Unformatted text preview: olve the y-equation 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 x-equation: 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 x-component 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 y-equation 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.

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