Ok i haven\'t done physics this simple in some time and cannotremember how to do

Question

Ok i haven't done physics this simple in some time and cannotremember how to do this type of problem

An initially stationary 15.0 kg crate is pulled by cable a distanceL = 5.7 m up a frictionless ramp to height of 2.5m where itstops.
a) How much work is done on the crate by gravitational force?
b) How much work is doen on the crate by tension in the cable?

Explanation / Answer

The angle made by the ramp is           sin = 2.5 / 5.7           ==> = 26o a) The work is done on the crate by gravitational force           W = Fg.L =-mgsin * L = - 15 * 9.8 * sin26 * 5.7 = -367.5J b) T - mgsin = 0    ==> T = mgsin Therefore work is done on the crate by tension in thecable         WT = T.L =367.5J

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