Samyak Jain
Last Activity: 5 Years ago
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 = KE
f – KE
i . Assuming the crate to be at rest initially, KE
i = 0.
KEf = 300 J .