Question

In the given figure, a ball of mass $m$ is connected to an elastic spring of force constant $K=m g / L$ through an inextensible string. Now the ball is released from rest. The maximum potential energy stored in the spring during the fall is $m g L[p+\sqrt{q}]$. Find $(p+q)$.

Answer 1

Using energy conservation between initial moment and equilibrium moment when $T=m g$.
$\frac{1}{2} \frac{m g}{L} \times L^{2}+\frac{1}{2} m v^{2}=m g \times 2 L $
$v=\sqrt{3 g L}$
If ball falls a maximum distance of $L+x$, then
$m g(L+x)=\frac{1}{2} \frac{m g}{L} x^{2} $
$x=L(1+\sqrt{3}) $
$\therefore P E_{\max }=m g L(2+\sqrt{3})$