Question

A smooth block is released at rest from ${45}^{\circ }$ incline slides a distance $d$. The time taken to slide is $n$ time as much to slide on a rough incline plane than on a smooth incline. Find the coefficient of friction.

• A. ${\mu }_k=1-\frac{1}{n^2}$
• B. ${\mu }_k=\sqrt{1-\frac{1}{n^2}}$
• C. ${\mu }_x=1-\frac{1}{n^2}$
• D. ${\mu }_x=\sqrt{1-\frac{1}{n^2}}$

Answer 1

Answer: (A)

${\mu }_k=1-\frac{1}{n^2}$

For Smooth surface, $ma=mg\sin \theta \Rightarrow a=g\sin \theta$
here, time $=t$ and distance$=d$, so using formula $S=ut+\frac{1}{2}a{t}^2$, we get
$d=0+\frac{1}{2}\left(g\sin \theta \right)t^2$ ( initially block is at rest so $u=0$)
$\therefore d=\frac{1}{2}\left(g\sin \theta \right)t^2......\left(1\right)$
For rough surface,
$ma=mg\sin \theta -f=mg\sin \theta -\mu mg\cos \theta \Rightarrow a=g\sin \theta -\mu g\cos \theta$
here, time$=nt$ so,
$d=\frac{1}{2}(g\sin \theta -\mu g\cos \theta )(nt{)}^2.....(2)$
$\left(2\right)/\left(1\right),1=\frac{(g\sin \theta -\mu g\cos \theta )(nt{)}^2}{\left(g\sin \theta \right)t^2}$
put $\theta =45$,
$1=\frac{(\frac{1}{\sqrt{2}}-\mu \frac{1}{\sqrt{2}})n^2}{\frac{1}{\sqrt{2}}}$
or,$(1-\mu )n^2=1$
$\therefore \mu =1-\frac{1}{n^2}$