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A block of mass $m$, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant $k$. The other end of the spring is fixed, as shown in the figure. The block is initially at rest in a equilibrium position. If now the block is pulled with a constant force $F$, the maximum speed of the block is
$\frac{{2F}}{{\sqrt {mk} }}$
$\frac{F}{{\pi \sqrt {mk} }}$
$\frac{{\pi F}}{{\sqrt {mk} }}$
$\frac{F}{{\sqrt {mk} }}$
Solution
When $\,{v_{\max }}$ $\Rightarrow $ acceleration $= 0$
$ \Rightarrow \,x = \frac{F}{K}$
Apply work energy theorem
$\,{W_{sp}}$ + $w_f$ = $\Delta K.E$
$\begin{array}{l}
– \frac{1}{2}K{x^2} + F.x = \Delta K.E\,\,\,;\,\, – \frac{1}{2}K\frac{{{F^2}}}{{{K^2}}} + \frac{{{F^2}}}{K}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}mu_{\max }^2\\
\frac{{{F^2}}}{{2K}} = \frac{1}{2}mu_{\max }^{2\,}\,\,\,;\,\,\,\frac{F}{{\sqrt {mK} }} = {V_{\max }}
\end{array}$
