Question

A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches $30^{\circ }$, the box starts to slip and slides $4.0m$ down the plank in $4.0s$

The coefficients of static and kinetic friction between the box and the plank will be, respectively

• A. 0.5 and 0.6
• B. 0.4 and 0.3
• C. 0.6 and 0.6
• D. 0.6 and 0.5

Answer 1

Answer: (D)

Let ${\mu }_s$ and ${\mu }_k$ be the coefficients of static and kinetic friction between the box and the plank respectively.
When the angle of inclination $\theta$ reaches $30^{\circ }$, the block just slides,
$\therefore {\mu }_s=\tan \theta =\tan 30^{\circ }=\frac{1}{\sqrt{3}}=0.6$

If a is the acceleration produced in the block, then $ma=mg\sin \theta -f_k$
(where $f_k$ is force of kinetic friction)
$=mg\sin \theta -{\mu }_{k}N$ (as $f_k={\mu }_{k}N)$
$=mg\sin \theta -{\mu }_{k}mg\cos \theta ($ as $N=mg\cos \theta )$
$a=g\left(\sin \theta -{\mu }_k\cos \theta \right)$
A s $g=10m{s}^2$ and $\theta =30^{\circ }$
$\therefore a=\left(10m{s}^2\right)\left(\sin 30^{\circ }-{\mu }_k\cos 30^{\circ }\right)\dots$ (i)
If s is the distance travelled by the block in time $t$,
then $s=\frac{1}{2}a{t}^2($ as us $=0)$
or $a=\frac{2_s}{t^2}$
But $s-4.0m$ and t $=4.0s($ given)
$\therefore a=\frac{2(4.0m)}{(4.0s{)}^2}=\frac{1}{2}m{s}^{-2}$
Substituting this value of a in eqn. (i), we get
$\frac{1}{2}m{s}^{-2}=\left(10m{s}^{-2}\right)\left(\frac{1}{2}-{\mu }_k\frac{\sqrt{3}}{2}\right)$
$\frac{1}{10}=1-\sqrt{3}{\mu }_k$
or $\sqrt{3}{\mu }_k=1-\frac{1}{10}=\frac{9}{10}=0.9$
${\mu }_k=\frac{0.9}{\sqrt{3}}=0.5$