In the given arrangement of a doubly inclined plane two blocks of masses $$M$$ and $$m$$ are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is $$0.25$$. The value of $$m$$, for which $$M = 10$$ kg will move down with an acceleration of $$2 \text{ m s}^{-2}$$, is: (take $$g = 10 \text{ m s}^{-2}$$ and $$\tan 37° = \frac{3}{4}$$)

Now lets write the equations for both the block

So for the 10 Kg block 

$$Mg\sin\theta\ _1-\mu\ _sMg\cos\theta\ _1-T=Ma$$

$$Mg\sin\ 53^{\circ\ }-0.25Mg\cos53^{\circ\ }-T=Ma$$

$$100\times\ \frac{4}{5}-0.25\times\ 100\times\ \frac{3}{5}-T=20$$

$$80-T=15$$

$$T=45$$

Now for the block of mass m
$$T-mg\sin\theta\ _2-\mu\ _smg\cos\ \theta\ _2=ma$$
$$T-mg\sin37^{\circ\ }-0.25mg\cos37^{\circ\ }=m\left(2\right)$$
$$45-mg\times\ \frac{3}{5}-0.25mg\times\ \frac{4}{5}\ =m\times\ 2$$
$$mg\times\ \frac{3}{5}+0.25mg\times\ \frac{4}{5}\ +m\times\ 2=45$$
$$6m+2m+2m=45$$
$$10m=45$$
$$m=4.5\ Kg$$