A block of mass m is hung vertically from an | vedclass

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Question

$m g=k x$

$m g=\frac {m g}{a} x$

$x=a$

$V_{m} \Rightarrow \sqrt{\frac{k}{m}} a$

$K E=\frac{1}{2} m v^{2}=\frac{1}{2} m \frac{K}{m} a^{2}$

Vedclass · Accepted Answer

$mga/2$

Vedclass · Answer

$m g=k x$

$m g=\frac {m g}{a} x$

$x=a$

$V_{m} \Rightarrow \sqrt{\frac{k}{m}} a$

$K E=\frac{1}{2} m v^{2}=\frac{1}{2} m \frac{K}{m} a^{2}$

$\Rightarrow \frac{m g}{2 a} a^{2}$

$\Rightarrow \operatorname{armg}_{2} / 2$

$mga/2$

$A$ block of mass $m$ is hung vertically from an elastic thread of force constant $mg/a$. Initially the thread was at its natural length and the block is allowed to fall freely. The kinetic energy of the block when it passes through the equilibrium position will be :

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