Q. A block is kept on an inclined plane of inclination $\theta$ of length $l$. The velocity of particle at the bottom of inclined plane is (the coefficient of friction is $\mu$)
A) $\sqrt{2gl(\mu\cos\theta - \sin\theta)}$
B) $\sqrt{2gl(\sin\theta - {\mu\cos}\theta)}$
C) $\sqrt{2gl(\sin\theta + {\mu\cos}\theta)}$
D) $\sqrt{2gl(\cos\theta + \mu\sin\theta)}$
Explanation
(b) Acceleration $(a) = g(\sin\theta - \mu\cos\theta)$ and $s = l$
$v = \sqrt{2as} = \sqrt{2gl(\sin\theta - \mu\cos\theta)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.