A man pushing a crate of mass m = 92.0 kg at a speed of v = 0.865 m/s encounters

Question

A man pushing a crate of mass m = 92.0 kg at a speed of v = 0.865 m/s encounters a rough horizontal surface of length = 0.65 m as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.357 and he exerts a constant horizontal force of 293 N on the crate, find the following. (a) Find the magnitude and direction of the net force on the crate while it is on the rough surface. magnitude N direction (b) Find the net work done on the crate while it is on the rough surface. J (c) Find the speed of the crate when it reaches the end of the rough surface. m/s

Explanation / Answer

a) The net force will be (assuming to the right is positive):

(F_{net} = F_{push} - F_{friction} = F_{push} - mu_kmg = 293N - (0.357)(92kg)(9.81m/s^2))

--> (F_{net} = -29.2 N)

So, the net force is 29.2N to the LEFT.

b) The net work done will be the work due to the net force:

(W_{net} = F_{net}l = (-29.2N)(0.65m) = -18.98J)

c) Using the work energy theorem:

(W_{net} = K_{f} -K_i = (1/2)mv_f^2 - (1/2)mv_i^2)

--> (v_f^2 = 2(W_{net} + (1/2)mv_i^2)/m)

--> (v_f = sqrt{2*(-18.98J + (1/2)(92kg)(0.865m/s)^2)/(92kg)} = 0.579 m/s)

Hope this helps

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