Question

A body takes $n$ times the time to slide down a rough inclined plane as it takes to slide down the same inclined plane when it is perfectly frictionless. The coefficient of kinetic friction between the body and the plane for an angle of inclination of ${45}^{\circ }$ is given by $\mu$. Then find $\mu$=?

• A. $1-\frac{1}{n}$
• B. $\frac{1}{n}$
• C. $(1-\frac{1}{n^2})$
• D. $(\frac{1}{n^2}-1)$

Answer 1

Answer: (C)

$(1-\frac{1}{n^2})$

For smooth surface, time$=t$
from fbd, $ma=mg\sin \theta \Rightarrow a=g\sin \theta$
now, $d=\frac{1}{2}a{t}^2$ (using $S=ut+\frac{1}{2}a{t}^2$) here d= distance from top to bottom of inclined plane.
$\therefore d=\frac{1}{2}g\sin \theta t^2.....\left(1\right)$
for rough surface, time$=nt$
$ma=mg\sin \theta -\mu mg\cos \theta \Rightarrow a=g(\sin \theta -\mu \cos \theta )$
now, $d=\frac{1}{2}a(nt{)}^2$
$\therefore d=\frac{1}{2}g(\sin \theta -\mu \cos \theta )n^2{t}^2....\left(2\right)$
solving $\left(1\right)$ and $\left(2\right)$, $\sin \theta =n^2\sin \theta -\mu n^2\cos \theta$
put $\theta =45$, we will get,
$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}{n}^2-\frac{1}{\sqrt{2}}\mu n^2$
$1=n^2(1-\mu )$
$\mu =1-\frac{1}{n^2}$
Ans:(C)