A block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from a ceiling and…
A block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value $\mathrm{k}$. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be :
$\mathrm{Mg} / 2 \mathrm{k}$
$\mathrm{Mg} / \mathrm{k}$
$2 \mathrm{Mg} / \mathrm{k}$
$4 \mathrm{Mg} / \mathrm{k}$
Solution
Loss in grav. $\mathrm{PE}=$ gain in spring $\mathrm{PE}$
At maximum elongation
$\begin{aligned}
\mathrm{Mgx} & =\frac{1}{2} \mathrm{kx}^2 \\
\mathrm{x} & =\frac{2 \mathrm{Mg}}{\mathrm{k}}
\end{aligned}$