Answer
(a) 30 N
(b) $-4.60\,m/s^{2}$, the negative sign indicates that the direction is downwards.
Work Step by Step
(a) The normal force is equal in magnitude and opposite in direction to the horizontal component of the force F.
Therefore, N$= F\cos \theta=60\,N\times\cos60^{\circ}=30\,N$
(b) $F_{vertical}-W=ma$
Or $a=\frac{F_{vertical}-W}{m}=\frac{F\sin\theta-mg}{m}=\frac{60\,N\sin60^{\circ}-10.0\,kg\times9.8\,m/s^{2}}{10.0\,kg}=-4.60\,m/s^{2}$
The negative sign indicates that the direction is downwards.
