Answer
$$
\begin{aligned} a =2.17 \mathrm{\ m} / \mathrm{s}^{2} \end{aligned}
$$
Work Step by Step
If $\mu_{s}=0.400$ and $\mu_{k}=0.300,$ then the magnitude of $\vec{f}$ has a maximum value of
$$
f_{s, \max }=\mu_{s} F_{N}=(0.400)\left(m g-0.500 m g \sin 20^{\circ}\right)=0.332 m g
$$
In this case, $$F \cos \theta=0.500 m g \cos 20^{\circ}=0.470 m g>f_{s, \max } $$ Therefore, the acceleration of the block is
$$
\begin{aligned} a &=\frac{F}{m}\left(\cos \theta+\mu_{k} \sin \theta\right)-\mu_{k} g \\ &=(0.500)\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right)\left[\cos 20^{\circ}+(0.300) \sin 20^{\circ}\right]-(0.300)\left(9.80 \mathrm{m} / \mathrm{s}^{2}\right) \\ &=2.17 \mathrm{\ m} / \mathrm{s}^{2} \end{aligned}
$$
