A solid sphere rolls without slipping and&nbs | vedclass

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Question

By conservation of energy

$\frac{1}{2} \mathrm{mv}^{2}\left(1+\frac{2}{5}\right)=\frac{1}{2} \mathrm{kx}^{2}$

$\Rightarrow \operatorname{mv}^{2}

Vedclass · Accepted Answer

$v\sqrt {\frac{{7m}}{{5k}}} $

Vedclass · Answer

By conservation of energy

$\frac{1}{2} \mathrm{mv}^{2}\left(1+\frac{2}{5}\right)=\frac{1}{2} \mathrm{kx}^{2}$

$\Rightarrow \operatorname{mv}^{2}\left(\frac{7}{5}\right)=\mathrm{kx}^{2}$

$\Rightarrow x=v \sqrt{\frac{7 m}{5 k}}$

A solid sphere rolls without slipping and presses a spring of spring constant $'k'$ as shown in figure. Then, the compression in the spring will be :-

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