## University Physics with Modern Physics (14th Edition)

Let $W_f$ be the work done by friction. Let $W_g$ be the work done by gravity. $W_{total} = K_2 - K_1$ $W_f+W_g = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$ $W_f - mgh = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$ $W_f = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 + 2mgR$ $W_f = \frac{1}{2}(120~kg)(8.0~m/s)^2 - \frac{1}{2}(120~kg)(25.0~m/s)^2 + (2)(120~kg)(9.80~m/s^2)(12.0~m)$ $W_f = -5440~J$ The work done by friction is -5440 J.