Question

A block of mass $10kg$ is moving up an inclined plane of inclination $30{}^{\circ }$ with an , initial speed of $5m/s$. It stops after $0.5s$, what is the value of coefficient of kinetic friction?

• A. 0.5
• B. 0.6
• C. $\sqrt{3}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

Answer: (D)

For equilibrium, normal to plane
$N=mg\cos \theta$ ... (1)
Net force along the plane downward
$F=mg\sin \dot{\theta }+f_k$ ...(2)
where $f_k$ is kinetic friction
but $f_k=\mu N=\mu mg\cos \theta$ ...(3)
from eq. (1), (2), and (3) we get
$\therefore F=mg\sin \theta +\mu mg\cos \theta$

According to Newton's II ${}^{\text{nd }}$ law
$F=ma$
$\therefore ma=mg\sin \theta +\mu mg\cos \theta$
$\therefore$ Retardation, $a=g\sin \theta +\mu g\cos \theta$
From equation $v=u+at,$ we have
$0=u-(g\sin \theta +\mu g\cos \theta )t$
$\Rightarrow g\sin \theta +\mu g\cos \theta =\frac{u}{t}$
$\Rightarrow 10\times \sin 30^{\circ }+\mu \times 10\cos 30^{\circ }=\frac{5}{0.5}$
$\Rightarrow 10\times \frac{1}{2}+10\mu \times \frac{\sqrt{3}}{2}=10$
$\Rightarrow 5\sqrt{3}\mu =5$
or $\mu =\frac{1}{\sqrt{3}}$