Answer
$5.1\times 10^5~J~~$ of energy is transferred to thermal energy.
Work Step by Step
We can find the angle of the hill above the horizontal:
$sin~\theta = \frac{300~m}{500~m}$
$\theta = sin^{-1}~(\frac{300~m}{500~m})$
$\theta = 36.87^{\circ}$
We can find the work done on the rock by the force of friction
$W = -(mg)~(cos~\theta)~(d)~(\mu_k)$
$W = -(520~kg)(9.8~m/s^2)~(cos~36.87^{\circ})~(500~m)~(0.25)$
$W = -5.1\times 10^5~J$
$5.1\times 10^5~J~~$ of energy is transferred to thermal energy.
