Answer: 8.66 ms$^{-2}$ and 74 N

Here is the force diagram for the problem.

According to Newton's 2nd law, $F = ma_x = F_P \cos 30^\circ$

Given that $F_P = 50N,\;$ and $m = 5\; Kg \rightarrow a_x = \large\frac{50N \times \cos 30 ^\circ}{5 kg}$$ = 8.66 m/s^2$

From the diagram above, we see that $N - mg - F_P \sin30^\circ = 0$

$\Rightarrow N = mg + F_P \sin 30^\circ = 5 \times 9.8 + 50 \sin 30 = 74 \;N$