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Question

By COME

$\mathrm{PE}_{i}+\mathrm{KE}_{i}=\mathrm{PE}_{f}+\mathrm{KE}_{f}\left[\mathrm{k} \cdot \varepsilon_{1}=\mathrm{k} \varepsilon_{\mathrm{f}}=

Vedclass · Accepted Answer

$\frac{{2mg(h + x)}}{{{x^2}}}$

Vedclass · Answer

By COME

$\mathrm{PE}_{i}+\mathrm{KE}_{i}=\mathrm{PE}_{f}+\mathrm{KE}_{f}\left[\mathrm{k} \cdot \varepsilon_{1}=\mathrm{k} \varepsilon_{\mathrm{f}}=0\right]$

$\Rightarrow m g(h+x)=\frac{1}{2} k x^{2}$

$\mathrm{k}=\frac{2 \mathrm{mg}(\mathrm{h}+\mathrm{x})}{\mathrm{x}^{2}}$

A ball of mass $m$ is dropped from a heigh $h$ on a platform fixed at the top of a vertical spring, as shown in figure. The platform is depressed by a distance $x.$ Then the spring constant is

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