A 7 kg crate is sliding at a constant speed of 14 m/s along a frictionless horiz

Question

A 7 kg crate is sliding at a constant speed of 14 m/s along a frictionless horizontal table, when it encounters a friction patch (s=0.6, k=0.2).

a.) Find the work done by each force acting on the crate, after it has slid 5 m:
work done by normal force: ?J
work done by gravity: ?J
work done by friction: ?J

b.) Find the speed of the crate after sliding this distance.
?m/s

c.) Some time later, the crate comes to rest. How much total work is done on the crate, from start to finish?* (from when it enters the friction patch to when it stops)
?J

Explanation / Answer

The mass of the crate m = 7.0kg the coefficient of kinetic friction k = 0.2 the distance traveled by crate d = 5m (a) The work done by normal force W = 0       The workdone by gravity W = 0 Since there is no motion along vertically The workdone by frictional force           W = f*s = -k mg*d                         = -(0.2)(7)(9.8)(5)                        = -68.6 J (b) From work energy theorem the initial speed of the crate vi = 14m/s        W = (1/2)mvf^2 - (1/2)mvi^2         -68.6 = (0.5)(7) [ vf^2 - (14)^2]        -19.6 = vf^2 - 196 then      vf^2 = 176.4 therefore the speed        vf = 13.28 m/s (c) Finally the crate comes to rest so again from work energy theorem      W = - (0.5)(7)(14)^2 =- 686 J

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