Laws of Motion - Result Question 41
41. 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 4.0s. The coefficients of static and kinetic friction between the box and the plank will be, respectively:
[2015 RS]
(a) 0.6 and 0.5
(b) 0.5 and 0.6
(c) 0.4 and 0.3
(d) 0.6 and 0.6
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Answer:
Correct Answer: 41. (a)
Solution:
- (a) Coefficient of static friction, $\mu_s=\tan \theta$
$\Rightarrow \mu_s=\tan 30^{\circ}=\frac{1}{\sqrt{3}}=0.577 \cong 0.6$
So, distance covered by plank,
$S=u t+\frac{1}{2} at^{2}$
where, $u=0$ and $a=g \sin \theta-\mu_k(g) \cos \theta$
$4=\frac{1}{2} a(4)^{2} \Rightarrow a=\frac{1}{2}=0.5$ $[\because \theta=30^{\circ}, s=4 m.$ and $t=4 s$ given $]$
$\Rightarrow \mu_k=\frac{0.9}{\sqrt{3}}=0.5$
