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A block of mass $m$ initially at rest is dropped from a height $h$ on to a spring of force constant $k$. the maximum compression in the spring is $x$ then
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Verified Answer
The correct answer is:
$m g(h+x)=\frac{1}{2} k x^2$
From Energy conservation, It is known that
Change in gravitational potential energy \(=\) Elastic potential energy stored in compressed spring
\(\begin{aligned}
& \mathrm{E}_{\mathrm{g}}=\mathrm{mg}(\mathrm{h}+\mathrm{x}) \\
& \mathrm{E}_{\mathrm{s}}=\frac{1}{2} \mathrm{kx}^2 \\
& \Rightarrow \mathrm{mg}(\mathrm{h}+\mathrm{x})=\frac{1}{2} \mathrm{kx}^2
\end{aligned}\)
Change in gravitational potential energy \(=\) Elastic potential energy stored in compressed spring
\(\begin{aligned}
& \mathrm{E}_{\mathrm{g}}=\mathrm{mg}(\mathrm{h}+\mathrm{x}) \\
& \mathrm{E}_{\mathrm{s}}=\frac{1}{2} \mathrm{kx}^2 \\
& \Rightarrow \mathrm{mg}(\mathrm{h}+\mathrm{x})=\frac{1}{2} \mathrm{kx}^2
\end{aligned}\)
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