A block of mass m and wedge M is arranged as shown in the figure.

Initially the system is kept stationary, certain time system is released from rest. If acceleration of M is found to be $A=\sqrt{5}\mathrm{m}/{\mathrm{s}}^2$ towards right, then find net acceleration of m with respect to ground

If system is released, the wedge will slide towards right. Let wedge slide towards right by distance .r. The string of length 3x will be adjusted on the inclined face.

$\text{ Let }{d}_{\text{wedge }}=x\text{ (towards right) }$

$\text{ Hence }{d}_{\text{block, wedge }}=3x$

$\begin{array}{l}{\overrightarrow{d}}_{\text{block }}={\overrightarrow{d}}_{\text{block, wedge }}+{\overrightarrow{d}}_{\text{wedge }}\\ =\sqrt{(3x{)}^2+x^2+2(3x\left)\right(x)\cos \left(180-{37}^{\circ }\right)}\\ =\sqrt{10x^2-6x^2\cos {37}^{\circ }}=x\sqrt{10-6\cos {37}^{\circ }}\\ =x\sqrt{10-6\times \frac{4}{5}}=x\sqrt{\frac{26}{5}}\end{array}$

Hence, acceleration of block

$\left|{\overrightarrow{a}}_m\right|=A\sqrt{\frac{26}{5}}=\sqrt{26}\mathrm{m}/{\mathrm{s}}^2$