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A crate is pushed horizontally with 100N across a 5m floor. If the frictional force between the crate and the floor is 40N, then the kinetic energy gained by the crate is . 200j 240j 250j 300j 500j
Contant horizontal force of magnitude F = 100 N is acting on the crate. It displaces the crate by s = 5 m.Thus, work done by this force is F.s = F.s = 100 . 5 = 500 J. the direction of force is same as that of displacement.Frictional force between crate and floor is f = 40 N. It is too acting for the displaceement s = 5 m.Its work done is f.s = – f.s = – 40 . 5 = – 200 J. the direction of force is opposite to that of displacement.Total work done by the forces acting on the crate is 500 J + (– 200 J) = 300 J.As we know that net work done by all the forces acting on a body is equal to change in its kinetic energy,300 J = KE = KEf – KEi . Assuming the crate to be at rest initially, KEi = 0. KEf = 300 J .
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