Question

When a body slides down from rest along a smooth inclined plane making an angle of ${45}^0$ with the horizontal, it takes time T. When the same body slides down from rest along inclined plane making the same angle and through the same distance, it is seen to take time pT, where p is some number greater than 1. The co-efficient of friction between of friction between and the body and the rough plane is

• A. 1/p
• B. $\mu =(1-1/p^2)$
• C. $1/p^2$
• D. 2 - p

Answer 1

The correct option is B $\mu =(1-1/p^2)$

Given Situations are shown in figure,
For smooth inclined plane,
Here, $Mgsin\theta =Ma$
$\Rightarrow a=gsin\theta$
Let s=length of inclined plane
Using, $s=ut+\frac{1}{2}a{t}^2\Rightarrow s=0\times T+\frac{1}{2}\left(gsin\theta \right)T^2(\because t=T)$
$\Rightarrow s=\frac{1}{2}gsin\theta T^2\Rightarrow s=\frac{1}{2\sqrt{2}}g{T}^2(\because \theta ={45}^0)$ ...(i)
For rough inlined plane.
$f=\mu N=\mu Mgcos\theta$
$Mgsin\theta -f=M{a}^{\prime }$
$\Rightarrow a^{\prime }=(sin\theta -\mu cos\theta )g$
Using $S=ut+\frac{1}{2}{a}^{\prime }{t}^2$
$s=0\times \left(pT\right)+\frac{1}{2}(sin\theta -\mu cos\theta )g\times p^2{T}^2$(\because t=pT)
$s=\frac{1}{2\sqrt{2}}(1-\mu )g{p}^2{T}^2$ ..............(ii)
From equation (i) and (ii)
$\frac{1}{2\sqrt{2}}g{T}^2=\frac{1}{2\sqrt{2}}(1-\mu )g{p}^2{T}^2$
$1=(1-\mu )p^2=\mu =(1-\frac{1}{p^2})$