\( \Rightarrow \,x = \frac{F}{K}\)

Apply work energy theorem 

\(\,{W_{sp}}\) + \(w_f\) = \(\Delta K.E\)

\(\begin{array}{l}
 - \frac{1}{2}K{x^2} + F.x = \Delta K.E\,\,\,;\,\, - \frac{1}{2}K\frac{{{F^2}}}{{{K^2}}} + \frac{{{F^2}}}{K}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}mu_{\max }^2\\
\frac{{{F^2}}}{{2K}} = \frac{1}{2}mu_{\max }^{2\,}\,\,\,;\,\,\,\frac{F}{{\sqrt {mK} }} = {V_{\max }}
\end{array}\)