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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 :
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Verified Answer
The correct answer is:
$2 \mathrm{Mg} / \mathrm{k}$
Loss in grav. $\mathrm{PE}=$ gain in spring $\mathrm{PE}$
At maximum elongation
$$
\begin{aligned}
\mathrm{Mgx} & =\frac{1}{2} \mathrm{kx}^2 \\
\mathrm{x} & =\frac{2 \mathrm{Mg}}{\mathrm{k}}
\end{aligned}
$$
At maximum elongation
$$
\begin{aligned}
\mathrm{Mgx} & =\frac{1}{2} \mathrm{kx}^2 \\
\mathrm{x} & =\frac{2 \mathrm{Mg}}{\mathrm{k}}
\end{aligned}
$$
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