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 $4.0 \,m$ down the plank in $4.0\, s$

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 $m a=m g \sin \theta-f_{k}$
(where $f_{k}$ is force of kinetic friction)
$=m g \sin \theta-\mu_{k} N$ (as $\left.f_{k}=\mu_{k} N\right)$
$=m g \sin \theta-\mu_{k} m g \cos \theta($ as $N=m g \cos \theta)$
$a=g\left(\sin \theta-\mu_{k} \cos \theta\right)$
A s $g=10 m s^{2}$ and $\theta=30^{\circ}$
$\therefore a=\left(10 m s^{2}\right)\left(\sin 30^{\circ}-\mu_{k} \cos 30^{\circ}\right) \ldots$ (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 $s-4.0 m$ and t $=4.0 s ($ given)
$\therefore a=\frac{2(4.0 m)}{(4.0 s)^{2}}=\frac{1}{2} m s^{-2}$
Substituting this value of a in eqn. (i), we get
$\frac{1}{2} m s^{-2}=\left(10 m s^{-2}\right)\left(\frac{1}{2}-\mu_{k} \frac{\sqrt{3}}{2}\right)$
$\frac{1}{10}=1-\sqrt{3} \mu_{k}$
or $\sqrt{3} \mu_{k}=1-\frac{1}{10}=\frac{9}{10}=0.9$
$\mu_{k}=\frac{0.9}{\sqrt{3}}=0.5$