question_answer

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 \[g=10\,\,m{{s}^{-2}},\] is [AIPMT (M) 2011]

A) 1.2 m

B) 0.6 m

C) zero      

D) 0.4 m

Correct Answer: D

Solution :

Force, \[F=\mu \,mg\] Retardation of the block on the belt

\[a=\frac{F}{m}=\frac{\mu \,mg}{m}=\mu g\]

Form,    \[{{v}^{2}}={{u}^{2}}+2as\]

\[0={{(2)}^{2}}-2(\mu g)s\]

\[s=\frac{4}{2\times 0.5\times 10}=0.4\,m\]