Question

1: Calculate the work done on the crate by: - the 200 N force - the weight of the crate - the normal contact force N. 2: Calculate the rate of work done against the frictional force F. ?

Answer 1

As,the crate is moving with constant velocity,that means,the amount of kinetic frictional force is just being balanced by the Horizontal component of the applied force,i.e `200 cos 30 N= 173.20 N`

Now,as the crate is moving with constant velocity,So,work done by the horizontal component of the `200 N` force is `200 cos 30 *(0.5*16) J` i.e `1385.6 J`

Now, as the crate is motionless vertically,we can say, `N+ 200 sin 30 = W` (where, `N` is the normal reaction,and `W` is the weight)

So,displacement is occurring perpendicular to these 3 forces,hence work done is zero.

And,work done by the frictional force is the same as that of the work done by the horizontal component of the 200 N force

So,rate of work done `(W/t)` i.e power = `1385.6 /16 W=86.6 W`