Answer
The increase in thermal energy of block and plane is $~~25~J$
Work Step by Step
In part (b), we found that the block travels $~~d = 1.5~m~~$ along the plane.
We can find the work done by the frictional force:
$W = -mg~cos~\theta~\mu_k~d$
$W = -(5.0~kg)(9.8~m/s^2)(0.40)(1.5~m)~(cos~30^{\circ})$
$W = -25~J$
The increase in thermal energy of block and plane is $~~25~J$
