Answer
The increase in thermal energy is $~~240~J$
Work Step by Step
We can find the work done by the frictional force:
$W = -mg~cos~\theta~\mu_k~d$
$W = -(40~kg)(9.8~m/s^2)(0.40)(2.0~m)~cos~40^{\circ}$
$W = -240~J$
The increase in thermal energy is $~~240~J$
