In the figure shown, a spring of spring constant k is fixed at one end and the other end is attached to the mass 'm' . The coefficient of friction between block and the inclined plane is '$\mathrm{\mu }$'. The block is released when the spring is in its natural length. Assuming that tan $\mathrm{\theta }>\mathrm{\mu }$, find the maximum speed of the block during the motion.

 

The speed is maximum when acceleration is zero. 

Let displacement of block is x₀ when the speed of block is maximum. 

At equilibrium, applying Newton's law to the block along the incline

$\begin{array}{l}\mathrm{mgsin}\mathrm{\theta }=\mathrm{\mu mgcos}\mathrm{\theta }+{\mathrm{kx}}_0\\ (\mathrm{mgsin}\mathrm{\theta }-\mathrm{\mu mgcos}\mathrm{\theta })={\mathrm{kx}}_0 ......\left(1\right)\end{array}$

Applying work-energy theorem to block between initial and final position, we get
$\begin{array}{l}(\mathrm{KE}{)}_{\mathrm{f}}=(\mathrm{KE}{)}_{\mathrm{i}}+{\mathrm{mgx}}_0\sin \mathrm{\theta }-\frac{1}{2}{\mathrm{kx}}_0^2-{\mathrm{\mu mgx}}_0\cos \mathrm{\theta }\\ \frac{1}{2}{\mathrm{mv}}_{max}^2=0+{\mathrm{mgx}}_0\sin \mathrm{\theta }-\frac{1}{2}{\mathrm{kx}}_0^2-{\mathrm{\mu mgx}}_0\cos \mathrm{\theta } ........\left(2\right)\end{array}$

$\frac{1}{2}{\mathrm{mv}}_{max}^2=(\mathrm{mgsin}\mathrm{\theta }-\mathrm{\mu mgcos}\mathrm{\theta }){\mathrm{x}}_0-\frac{1}{2}{\mathrm{kx}}_0^2$

Solving (1) and (2), we get 

       ${\mathrm{v}}_{max}=(\sin \mathrm{\theta }-\mathrm{\mu cos}\mathrm{\theta })\mathrm{g}\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$