Answer
The increase in thermal energy is $~~509~J$
Work Step by Step
We can find the work done by the frictional force:
$W = -mg~cos~\theta~\mu_k~d$
$W = -(50~kg)(9.8~m/s^2)(0.20)(6.0~m)~cos~30^{\circ}$
$W = -509~J$
The increase in thermal energy is $~~509~J$
