Answer
$$
W=432 \mathrm{J}
$$
Work Step by Step
Let the distance (length) the crate walk relative to the conveyor belt before it stops slipping be $d .$ Then from Eq. $2-16\left(v^{2}=2 a d=2(f / m) d\right)$ we find
$$
\Delta E_{\text {th }}=f d=\frac{1}{2} m v^{2}=K
$$
Thus, the total energy that must be supplied by the motor is
$$
W=K+\Delta E_{\mathrm{th}}=2 K=(2)(216 \mathrm{J})=432 \mathrm{J}
$$
