Answer
The work done by friction is -5440 J.
Work Step by Step
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.