A 1 kg block situated on a rough incline is connected to a spring of negligible mass having spring constant $100{\mathrm{Nm}}^{-1}$ as shown in the figure. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. The coefficient of friction between the block and the incline is

(Take $\mathrm{g}=10{\mathrm{ms}}^{-2}$ and assume that the pulley is frictionless)

Here, $\mathrm{m}=1\mathrm{kg},\mathrm{\theta }={45}^{\circ },\mathrm{k}=100{\mathrm{Nm}}^{-1}$

From figure, $\mathrm{N}=\mathrm{mgcos}\mathrm{\theta }$

$\mathrm{f}=\mathrm{\mu N}=\mathrm{\mu mgcos}\mathrm{\theta }$

where $\mathrm{\mu }$ is the coefficient of friction between the block and the incline.

Net force on the block down the incline,

$=\mathrm{mgsin}\mathrm{\theta }-\mathrm{f}$

$=\mathrm{mgsin}\mathrm{\theta }-\mathrm{\mu mgcos}\mathrm{\theta }=\mathrm{mg}(\sin \mathrm{\theta }-\mathrm{\mu cos}\mathrm{\theta })$

Distance moved, $\mathrm{x}=10\mathrm{cm}=10\times {10}^{-2}\mathrm{m}$

Work done = Potential energy of stretched spring 

$\mathrm{mg}(\sin \mathrm{\theta }-\mathrm{\mu cos}\mathrm{\theta })\mathrm{x}=\frac{1}{2}{\mathrm{kx}}^2$

$2\mathrm{mg}(\sin \mathrm{\theta }-\mathrm{\mu cos}\mathrm{\theta })=\mathrm{kx}$

$2\times 1\times 10\times \left(\sin {45}^{\circ }-\mathrm{\mu cos}{45}^{\circ }\right)=100\times 10\times {10}^{-2}$

$\sin {45}^{\circ }-\mathrm{\mu cos}{45}^{\circ }=\frac{1}{2}\Rightarrow \frac{1}{\sqrt{2}}-\frac{\mathrm{\mu }}{\sqrt{2}}=\frac{1}{2}$

$1-\mathrm{\mu }=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\Rightarrow \mathrm{\mu }=1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}$

$\therefore \mathrm{\mu }=0.3$