Question

A given object takes $n$ times as much times to slide down a ${45}^o$ rough incline as it takes to slide down a perfectly smooth ${45}^o$ incline. What is the coefficient of kinetic friction between the object and the incline?

Answer 1

Let me take $F.B.D$ as there are two condition

Case (i)

When the surface is smooth

case (ii)

when the surface is rough

in this case, additional force is friction

Friction force $=$ normal force $\times$ coefficient of friction

According to newton's first law normal force should be equal to $mg\cos \theta$

Now using the equation of motion in case 1

$\begin{array}{c}d=\frac{1}{2}\frac{g}{\sqrt{2}}{t}^2\\ \Rightarrow t_1=\sqrt{2\sqrt{2}\frac{d}{g}}\end{array}$

In case 2

$\begin{array}{c}ma=mg\sin \theta -\mu mg\cos \theta \\ ma=mg\left(\sin \theta -\mu \cos \theta \right)\\ a=g\left(\sin {45}^{\circ }-\mu \cos {45}^{\circ }\right)\\ a=\frac{g}{\sqrt{2}}\left(1-\mu \right)\\ and\\ d=\frac{1}{2}\frac{g}{\sqrt{2}}\left(1-\mu \right){t_2}^2\\ t_2=\sqrt{\frac{2\sqrt{2}.d}{g\left(1-\mu \right)}}\\ given,\\ n{t}_1=t_2\\ n\sqrt{\frac{2\sqrt{2}d}{g}}=\sqrt{\frac{2\sqrt{2}d}{g\left(1-\mu \right)}}\end{array}$

squaring both side we get

$\begin{array}{c}{n}^2=\frac{2\sqrt{2}d}{g}=\frac{2\sqrt{2}d}{g\left(1-\mu \right)}\\ n=\frac{1}{\sqrt{1-\mu }}\\ 1-\mu =\frac{1}{n^2}\\ \mu =1-\frac{1}{n^2}\end{array}$