A wedge of mass M=1.0 kg with an inclined face at 60^° to the horizontal rests on a smooth horizontal table and is free to slide. A block of mass…

Let the wedge accelerate horizontally with $A$ and let $N$ be the contact force. Horizontal motion of the wedge: $MA=N\sin\theta$. Perpendicular to the face the block has no motion relative to the wedge, so its acceleration component along the normal equals that of the wedge, giving $mg\cos\theta-N=mA\sin\theta$. Eliminating $A$, $N=\dfrac{Mmg\cos\theta}{M+m\sin^2\theta}$. With $M=1$, $m=4$, $\theta=60^\circ$: $N=\dfrac{1\times4\times10\times0.5}{1+4\times0.75}=\dfrac{20}{4}=5\ \mathrm{N}$. (Check: $A=N\sin60^\circ/M=4.33\ \mathrm{m\,s^{-2}}$, and $4\times10\times0.5-5=15=4\times4.33\times0.866$.) Note that $N$ is much smaller than $mg\cos\theta=20\ \mathrm{N}$ because the wedge recoils.