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

The Correct Option is 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)
$=mg \sin\theta-\mu_kN$ (as $ f_k=\mu_kN$)
$=mg \sin\theta-\mu_k mg \cos \theta$ (as $N=mg \cos \theta$)
$a=g(\sin\theta-\mu_k \cos \theta)$
$A\, sg = 10\, ms^2$ and $ \theta = 30^{\circ}$
$\therefore a=(10\, ms^2)(\sin 30^{\circ} - \mu_k \cos 30^{\circ})$ ...(i)
If s is the distance travelled by the block in time t, then
$s=\frac{1}{2} at^2$ (as u=0)
or $a=\frac{2_s}{t^2}$
But s - 4.0 m and t = 4.0 s (given)
$\therefore a=\frac{2(4.0m)}{(4.0s)^2}=\frac{1}{2}\, ms^{-2}$
Substituting this value of a in eqn. (i), we get
$\frac{1}{2} \,ms^{-2}=(10 \,ms^{-2})\Bigg(\frac{1}{2}-\mu_k\frac{\sqrt3}{2}\Bigg)$
$\frac{1}{10}=1-\sqrt3\mu_k$ or $\sqrt3\mu_k=1-\frac{1}{10}=\frac{9}{10}=0.9$
$\mu_k=\frac{0.9}{\sqrt3}=0.5$