Question

A 0.40-kg block initially at rest on a frictionless horizontal surface is acted upon by a force of 7.0 N for a distance of 3.5 m. How much kinetic energy does the block gain?

Answer 1

24.5 J

Step-by-step explanation:

The work done on the block by the force is:

W = Fdcos(theta)

where F is the applied force, d is the distance over which the force is applied, and theta is the angle between the force and the displacement. In this case, the force is in the same direction as the displacement, so cos(theta) = 1.

W = (7.0 N)(3.5 m)(1) = 24.5 J

Since the surface is frictionless, all the work done on the block goes into increasing its kinetic energy. The kinetic energy gained by the block is therefore:

K = W = 24.5 J

So the block gains 24.5 J of kinetic energy.