A block of mass m, attached to a spring of spring constant $k$, oscillates on a smooth horizontal table. The other end of the spring is fixed to a wall. The block has a speed $v$ when the spring is at its natural length. Before coming to an instantaneous rest, if the block moves a distance $x$ from the mean position, then
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(c) By using conservation of mechanical energy
$\frac{1}{2}k{x^2} = \frac{1}{2}m{v^2} $
$\frac{1}{2}k{x^2} = \frac{1}{2}m{v^2} $
$\Rightarrow x = v\sqrt {m/k} $
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