Q. A conveyor belt is moving at a constant speed of $2\ m/s$. A box is gently dropped on it. The coefficient of friction between them is $\mu = 0.5$. The distance that the box will move relative to belt before coming to rest on it, taking $g = 10\ ms^{- 2},$ is
A) Zero
B) $0.4\ m$
C) $1.2\ m$
D) $0.6\ m$
Explanation
(b) $a = \mu g = 5$
$v^{2} = u^{2} + 2as$
$0 = 2^{2} + 2 \times (5)s$
$s = - \frac{2}{5}$ w.r.t. belt
Or distance $= 0.4\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.