Free-body diagrams, equilibrium and motion, talked through.

Down the slope: $mg\sin\theta = 5 \times 9.81 \times \sin 30^\circ = 24.5$ N

Normal force: $N = mg\cos\theta = 42.5$ N, so static friction can hold at most $\mu_s N = 0.4 \times 42.5 = 17.0$ N.

$24.5 > 17.0$, so the block slides, and kinetic friction acts up the slope:

$a = g(\sin\theta - \mu_k \cos\theta) = 9.81(0.5 - 0.3 \times 0.866) = 2.36\ \text{m/s}^2$