Question

When a body slides down from rest along a smooth inclined plane making an angle of $45^?$ with the horizontal, it takes time $T$. When the same body slides down from rest along a rough 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 the body and the rough plane is

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

Answer 1

Answer: (B)

$\mu$ $=\left(1-1 /p^{2}\right)$

Answer 2

Given situations are shown in figure, For smooth inclined plane, Here, $Mg \,sin \,\theta$ $=Ma$ $\Rightarrow$ $a=g \,sin \,\theta$ Let s = Length of inclined plane Using, $s=ut+\frac{1}{2}at^{2}$ $\Rightarrow$ $s=0\times T+\frac{1}{2}$ $\left(g\,sin\,\theta\right)T^{2}$ $\left(\because\,t=T\right)$ $\Rightarrow\quad$ $s=\frac{1}{2} g\, sin\, \theta\, T^{2}$ $\Rightarrow$ $s=\frac{1}{2\sqrt{2}}\,g \,T^{2}$ $\left(\because\,\theta=45^{\circ}\right)$ $\quad...\left(i\right)$ For rough inclined plane. $f=\mu N$ $=\mu Mg\, cos\, \theta$ $Mg \,sin \,\theta-f=Ma'$ $\Rightarrow$ $a'=\left(sin \,\theta-\mu cos \theta\right)g$ Using $s=ut+\frac{1}{2}a' t^{2}$ $\Rightarrow\quad$ $s=0\times\left(pT\right)+\frac{1}{2}\left(sin \theta-\mu cos \theta\right)g\times p^{2}T^{2}$ $\left(\because\,t=pT\right)$ $\Rightarrow \quad$ $s=\frac{1}{2\sqrt{2}}\left(1-\mu\right)gp^{2}T^{2}$ $\quad\ldots\left(ii\right)$ From equation (i) and (ii) $\frac{1}{2\sqrt{2}} gT^{2}$ $=\frac{1}{2\sqrt{2}} \left(1-\mu\right)$ $gp^{2}T^{2}$ $\Rightarrow \quad$ $1=\left(1-\mu\right)p^{2}$ $\Rightarrow$ $\mu=\left(1-\frac{1}{p^{2}}\right)$