Q. The upper half of an inclined plane with inclination $\phi$ is perfectly smooth, while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if coefficient of friction for the lower half is given by
A) $2\sin\phi$
B) $2\cos\phi$
C) $2\tan\phi$
D) $\tan\phi$
Explanation
(c) For the smooth portion $BC$,
$u = 0,s = l,a = g\sin\phi$
$u = ?$
From $v^{2} - u^{2} = 2as$
$v^{2} - 0 = 2g\sin{\phi \times l}$
For the rough portion $CO$
$u = v = \sqrt{2g\sin{\phi \bullet l}}$
$v = 0,a = g(\sin{\phi - \mu\cos\phi})$
$s = l$
From $v^{2} - u^{2} = 2as$
$0 - 2gl\sin{\phi = 2g}(\sin{\phi - \mu\cos\phi})l$
$- \sin{\phi\mathbf{=}\sin\phi - \mu\cos\phi}$
$\mu\cos\phi = 2\sin\phi$
$\mu{= 2\ tan}\phi$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.