A wedge of mass M= 10 kg. height h=3m and angle of inclination  $\alpha =37°$ is at rest on a horizontal surface.  There is a small point–like object (mass $m=0.5kg$ ) next to the slope as shown in the figure.  At what acceleration must the wedge be moved in  order that the point-like object reaches its top in time $t=5s$ (Neglect the friction between point-like object and wedge)

Let Wedge is moving rightward with acceleration a and mass m has an acceleration A with respect to wedge along the surface of the wedge in upward direction, so
 $\frac{h}{\sin \alpha }=\frac{1}{2}A{t}^2\Rightarrow A=\frac{2h}{t^2\sin \alpha }\mathrm{.....}\left(1\right)$
With the help of FBD of mass m in the frame of wedge, we can write

$$\begin{gathered} A=a\cos \alpha -g\sin \alpha \\ \Rightarrow a=g\tan \alpha +\frac{2h}{t^2\sin \alpha \cos \alpha }=10\times \frac{3}{4}+2\times 3\times \frac{5}{3}\times \frac{5}{4}\times \frac{1}{5\times 5}=8m/s^2 \end{gathered}$$

(OR)

Acceleration of the object along the inclination is 

$a_O=a\cos \alpha -g\sin \alpha$

Time taken is $t=5s$

The distance covered is 5 m. 

By kinematic equations of motion

$5=\frac{1}{2}{a}_O{t}^2$

$\Rightarrow a_O=\frac{2}{5}$

$\Rightarrow a=8 m/s^2$