Question

In the arrangement shown in the figure a block of mass $m=2kg$ lies on a wedge of mass $M=8kg$. The initial acceleration of the wedge (if the surfaces are smooth) given by $\frac{3\sqrt{3}g}{x}m{s}^{-2}$, then $x$ is.

Answer 1

Let $a_1:$ acceleration of wedge w.r.t. ground.
$a_2:$ acceleration of block w.r.t. wedge

The block has two accelerations, one w.r.t. the wedge and our along the wedge.
Using total work done by tension is zero.
$W_{net}^T=0$
$-Ty+Txcos{60}^{\circ }\times +Tx=0$
$y=\frac{3x}{2}$
Differentiating twice w.r.t time.
$a_2=\frac{3a_1}{2}$
Let $a_1=a\Rightarrow a_2=\frac{3a}{2}$

FBD of $m$ w.r.t wedge



(Circled ones are pseudo forces)
Using Newtons second law
Along the plane:
$\frac{\sqrt{3}mg}{2}+\frac{ma}{2}-T=m{a}_2=\frac{3ma}{2}$
$T=\frac{\sqrt{3}mg}{2}-ma......\left(1\right)$
Perpendicular to the plane
$N+\frac{\sqrt{3}ma}{2}=\frac{mg}{2}...\left(2\right)$

FBD of wedge M in ground frame



$T+\frac{\sqrt{3}N}{2}=Ma...\left(3\right)$
Substituting $\left(1\right)and\left(2\right)in\left(3\right)$
$\frac{\sqrt{3}mg}{2}-ma+\frac{\sqrt{3}}{2}(\frac{mg}{2}-\frac{\sqrt{3}ma}{2})=Ma$
$\Rightarrow \frac{3\sqrt{3}mg}{4}=(M+\frac{7}{4}m)a$
$\Rightarrow a=\frac{3\sqrt{3}mg}{4(M+\frac{7}{4}m)}(m=2,M=8)$
$\Rightarrow a=\frac{3\sqrt{3}g}{23}$