Question

All surfaces shown in the figure are smooth. The system is released with the spring unstretched. In equilibrium, compression in the spring will be -

• A. $\frac{m g}{\sqrt{2} k}$
• B. $\frac{2 m g}{k}$
• C. $\frac{\left(M + m\right) g}{\sqrt{2} k}$
• D. $\frac{m g}{k}$

Answer 1

Answer: (D)

Let $N$ be the normal reaction between $m$ and $M$ ,
Equilibrium of $M$
$N \, sin 45^{^\circ }=kx$ ... (i)
Equilibrium of $m$ in vertical direction gives
$N \, cos \, 45^{^\circ }=mg$ ... (ii)
From Eqs. (i) and (ii), we get
$x=\frac{m g}{k}$