Answer
$$
F=39 \mathrm{N}
$$
Work Step by Step
Finally, we are dealing with kinetic friction (pointing downhill, so that
$$
\begin{array}{l}{0=F-m g \sin \theta-f_{k}=m a} \\ {0=F_{N}-m g \cos \theta}\end{array}
$$
$\left.\text { along with } f_{k}=\mu_{k} F_{N} \text { (where } \mu_{k}=0.15\right)$ brings us to
$$
F=m g\left(\sin \theta+\mu_{k} \cos \theta\right)=39 \mathrm{N}
$$
