A man pushing a crate of mass m = 92.0 kg at a speed of v = 0.875 m/s encounters

Question

A man pushing a crate of mass m = 92.0 kg at a speed of v = 0.875 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 280 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.

(b) Find the net work done on the crate while it is on the rough surface.
3 J

(c) Find the speed of the crate when it reaches the end of the rough surface.
4 m/s

Explanation / Answer

a ) net force = 280 - N

= 280 - 0.36 *92 *9.8 = - 44.576 N,

magnitude = 44.576 N,

direction : opposte to the direction of applied force

b ) net work done = net force * displacement

now we have to find displacement

deceleration = net force/mass = 44.576/92 = 0.48452 m/sec2

V^2 = u^2 +2as

0 = (0.875)^2 - 2 * 0.48452 * s

s = 0.7900 m

since this is greater than 0.65 so the block will not stop on rough surface

so displacement = 0.65 m

work done = 44.576 * 0.65 = 28.97 J

c ) speed, v^2 = u^2 - 2as

v^2 = (0.875)^2 - 2 * 0.48452 *0.65

= 0.135749

v = 0.3684 m/sec

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