Question

In an ideal pulley particle system, mass $m_2(>m_1)$ is connected with a vertical spring of stiffness $k$. If mass $m_2$ is released from rest, when the spring is undeformed, find the maximum compression of the spring.

• A. $x=\frac{2(m_2-m_1)g}{k}$
• B. $x=\frac{2\left(m_2\right)g}{k}$
• C. $x=\frac{2(m_2+m_1)g}{k}$
• D. $x=\frac{(m_2-m_1)g}{k}$

Answer 1

The correct option is A $x=\frac{2(m_2-m_1)g}{k}$

Work done by tension and normal reaction at the pulley are zero (since they are internal constraint forces)
So, the total mechanical energy of the system will be conserved
$\Delta M.E=0$
or $\Delta K+\Delta U=0$


Let $x$ be the maximum compression of the spring.
Then, $\Delta U_{sp}=\frac{k}{2}{x}^2......\left(i\right)$

$m_2$ will descend by the same distance $x$.
i.e $\Delta U_{g{r}_2}=-m_{2}gx.....\left(ii\right)$
and $m_1$ will ascend by $x$
i.e $\Delta U_{g{r}_1}=m_{1}gx....\left(iii\right)$

Since both $m_1$ and $m_2$ comes to rest simultaneously.
$\Delta K=0.....\left(iv\right)$

From conservation of energy :
$\Delta K+\Delta U=0$
i.e $[\Delta U_{sp}+\Delta U_{g{r}_1}+\Delta U_{g{r}_2}]+\Delta K=0$
$\Rightarrow (\frac{1}{2}k{x}^2+m_{1}gx-m_{2}gx)+0=0$
$\Rightarrow x=\frac{2(m_2-m_1)g}{k}$