A spring of spring constant K is fixed at one end and the other end is attached to the mass ‘m’. The co-efficient of friction between block and the inclined plane is $\mu$ . The block is released when the spring is in its natural length. Assuming that $\tan \theta >\mu$,find the maximum speed of the block during the motion.
 
 

V = Vmaxwhen a = 0
Let dispt. of block is $x_0$ when its speed is max
 $mg\sin \theta =\mu mg\cos \theta +k{x}_0$
$mg\sin \theta -\mu mg\cos \theta =k{x}_0$ ____(1)
By W.E.T,  $K.E_f=K.E_i+mg{x}_0\sin \theta -\frac{1}{2}k{x}_0^2-\mu mg{x}_0\cos \theta$
 $\frac{1}{2}m{V}_{\max }^2=0+mg{x}_0\sin \theta -\frac{1}{2}k{x}_0^2-\mu mg{x}_0\cos \theta$
$\frac{1}{2}m{V}_{\max }^2=(mg{x}_0\sin \theta -\mu mg\cos \theta )x_0-\frac{1}{2}k{x}_0^2$ ____(2)
(1) or (2) $\Rightarrow \frac{1}{2}m{V}_{\max }^2=k{x}_0^2-\frac{1}{2}k{x}_0^2=\frac{1}{2}k{x}_0^2$   
 $V_{\max }=\sqrt{\frac{K}{m}}{x}_0$ substituting eq (1), we get 
 $V_{\max }=\sqrt{\frac{K}{m}}\left[\frac{mg\sin \theta -\mu mg\cos \theta }{K}\right]$
$V_{\max }=(\sin \theta -\mu \cos \theta )g\sqrt{\frac{m}{k}}$