Answer to Question #155274 in Physics for Nazrul Naim

Question

A 12 kg object is moving on a rough, horizontal surface. It has 24 J of kinetic energy. The friction force acting on it is 0.5 N. How far will it slide?

Answer 1

According to the work-energy theorem the net work done by the forces on an object equals the change in its kinetic energy. Thus in order to stop the box the friction force should perform the following amount of work:

$$W = |K_f - K_i|$$

where $K_f = 0J$ is the final kinetic energy (after box stops), and $K_i = 24J$ is the initial kinetic energy.

On the other hand, by definition, the work is:

$$W = Fs$$

where $F = 0.5N$ is the friction force and $s$ is the distance the box slides. Thus, obtain:

$$Fs = K_i\\ s = \dfrac{K_i}{F} = \dfrac{24J}{0.5N} = 48m$$

Answer. 48m.