Question

A 0.75 kg block slides with a constant velocity down a 20° inclined plane.a. How much work is done by the force of friction on the block as it slides the total length of the plane?b. What is the net work done on the block?c. Discuss the net work done if the angle of incline is adjusted so that the block accelerates down the plane.d. In part (c) is it possible for the frictional work to be greater in magnitude than the gravitational work? What would this mean in terms of the block’s speed?

Answer 1

As the block slides down with constant velocity, its acceleration is zero, therefore the net force applied on the block in the direction of the plane (I will denote this direction as 'x') is zero.

This means that the x-component of the gravitational force is equal in magnitude and opposite in directions with the friction force.

a)

We can demonstrate the physics in this question by starting with Newton's Second Law in the x-direction:

`F_x = ma_x  = 0\\F_x = mg\sin(20^\circ) - F_f = 0\\F_f = mg\sin(20^\circ) = (0.75)(9.8)(0.34) = 2.51 N`

Since we don't know the length of the plane, we can denote it as 'd'.

The work done by the force of friction is

`W_f = F_fd = 2.51d`

b)

The net work done on the block is zero.

This can be derived from the work-energy theorem. The total work done on an object is equal to the change in the kinetic energy of the object. In this question, the block slides down with constant velocity, so its kinetic energy is constant as well. This means total work done on the block is zero.

c)

If the block would accelerate down the plane, this would mean that the x-component of the gravitational force is bigger than the friction force. In that case the work done on the block would be positive, since the net force and the direction of the motion are in the same direction.

Another approach to this case would again be using work-energy theorem. If the block accelerates down the plane, its final velocity would be higher than its initial velocity, so its final kinetic energy would be higher than its initial.

`W_(total) = K_2 - K_1`

Total work done on the block would be positive.

d)

No. The frictional work cannot be greater than the gravitational work. The maximum value that the frictional work can have is the same value that the gravitational work has. By definition, the frictional force cannot be greater than the applied force, otherwise the object would slide up the plane without any external force.