Question

A block is moving up an inclined plane of inclination $60^{\circ }$ with velocity of $20m/s$ and stops after $2s$. If $g=10m/s^2$ , then the approximate value of coefficient of friction is

• A. 3
• B. 3.3
• C. 0.27
• D. 0.33

Answer 1

Answer: (C)

Frictional force acts in a direction opposite to direction of motion.
The various forces acting on the block are as shown.
Let a be the acceleration with which the block is moving upwards.
Then from Newtons
law, we have
$mg\sin 60^{\circ }+f=ma$ ... (i)
$R-mg\cos 60^{\circ }=0$ ... (ii)
$f_k=\mu R$ ... (iii)
From Eqs. (i), (ii) and (iii), we get
$ma=g\left(\sin 60^{\circ }+\mu \cos 60^{\circ }\right)m$
$ma=10\left(\frac{\sqrt{3}}{2}+\frac{\mu }{2}\right)mma$
$=5(\mu +\sqrt{3})m$
$\Rightarrow a=5(\mu +\sqrt{3})$
Since, block stops, hence $v=0$.
From equation of motion $v=u-at$
$\therefore 0=u-at$
$\Rightarrow 20=5(\sqrt{3}+\mu )\times 2$
$\Rightarrow \mu =2-\sqrt{3}$
$\Rightarrow \mu =2-1.732$
$\mu =0.27$