A given object takes n times as much time to slide down a 45° rough incline as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between the object and the incline is given by

On smooth inclined plane, the acceleration of the body,

$\mathrm{a}=\mathrm{gsin}\mathrm{\theta }$

If s be the distance traveled, then

$\mathrm{s}=\frac{1}{2}\mathrm{gsin}{\mathrm{\theta t}}_1^2$   …(1)

On rough inclined plane, the acceleration is given by

$\begin{array}{ll}\mathrm{a}& =\mathrm{gsin}\mathrm{\theta }-\mathrm{\mu gcos}\mathrm{\theta }\\ \therefore \mathrm{s}& =\frac{1}{2}(\mathrm{gsin}\mathrm{\theta }-\mathrm{\mu gcos}\mathrm{\theta }){\mathrm{t}}_2^2 ...\left(2\right)\end{array}$

From eqs. (1) and (2),

$\frac{{\mathrm{t}}_2^2}{{\mathrm{t}}_1^2}=\frac{\sin \mathrm{\theta }}{\sin \mathrm{\theta }-\mathrm{\mu cos}\mathrm{\theta }}$

$\text{But}{\mathrm{t}}_2={\mathrm{nt}}_1\text{or}{\mathrm{t}}_2^2/{\mathrm{t}}_1^2={\mathrm{n}}^2$

$\therefore {\mathrm{n}}^2=\frac{\sin \mathrm{\theta }}{\sin \mathrm{\theta }-\mathrm{\mu cos}\mathrm{\theta }}$

Solving, we get

$\begin{array}{rl}\mathrm{\mu }& =\frac{{\mathrm{n}}^2-1}{{\mathrm{n}}^2}\times \frac{\sin \mathrm{\theta }}{\cos \mathrm{\theta }}=\frac{{\mathrm{n}}^2-1}{{\mathrm{n}}^2}\times \frac{\sin {45}^{\circ }}{\cos {45}^{\circ }}\\ & =\frac{({\mathrm{n}}^2-1)}{{\mathrm{n}}^2}\times \frac{(1/\sqrt{2})}{(1/\sqrt{2})}=\frac{{\mathrm{n}}^2-1}{{\mathrm{n}}^2}=(1-\frac{1}{{\mathrm{n}}^2}).\end{array}$