Question

A conveyor belt is moving at a constant speed of $2\, ms^{-1}.$ 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. 0.4 m
• B. 1.2 m
• C. 0.6 m
• D. Zero

Answer 1

Answer: (A)

0.4 m

Answer 2

retardation of the block on the belt
$a =\frac{ F }{ m }=\mu g$
From $v^{2}=u^{2}+2 a s$
$0=2^{2}-2(\mu g ) s$
$s =\frac{4}{2 \times 0.5 \times 10}$
$=0.4\, m$