A man pushing a crate of mass m = 92.0 kg at a speed of v = 0.880 m/s encounters

Question

A man pushing a crate of mass m = 92.0 kg at a speed of v = 0.880 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 0.360 and he exerts a constant horizontal force of 277 N on the crate. (a) Find the magnitude and direction of the net force on the crate while it is on the rough surface. magnitude 48 direction opposite as the motion of the crate (b) Find the net work done on the crate while it is on the rough surface. 30.72 (c) Find the speed of the crate when it reaches the end of the rough surface. 0.32645 X Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s

Explanation / Answer

(a)

frictional force f = - uk*m*g

applied force F = 277 N

net force = F + f = 277 - (0.36*92*9.81) = -49.7 N = -48 N

magnitude = 48 N

direction opposite to the motion of crate

(b)

work done W = F_net*displacment = -48*0.65 = -31.2 J

(c)

work done = change in kinetic energy

W = Kf - Ki

W = (1/2)*m*(vf^2 - vi^2)

-31.2 = (1/2)*92*(vf^2 - 0.88^2)

vf = 0.31 m/s

any doubts, post in comment box.rate the answer if it helps.

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