Answer to Question #268543 in Physics for A 15.2 kg block is

Question

A 15.2 kg block is dragged over a rough, horizontal surface by a constant force of 133 N acting at an angle of 34.2 ◦ above the horizontal. The block is displaced 50.6 m, and the coefficient of kinetic friction is 0.176. Find the magnitude of the work done by the force of friction. 

Answer 1

By the definition of the work done, we have:

$$W_{fr}=F_{fr}d.$$

The friction force can be found as follows:

$$F_{fr}=\mu_kN=\mu_k(mg-F_{appl}sin\theta).$$

Substituting the friction force into the previous equation, we get:

$$W_{fr}=\mu_kd(mg-F_{appl}sin\theta),$$ $$W_{fr}=0.176\times50.6\ m\times(15.2\ kg\times9.8\ \dfrac{m}{s^2}-133\ N\times sin34.2^{\circ}),$$ $$W_{fr}=661\ J.$$