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

A)  1.2m            

B)  0.6m

C)  Zero              

D)  0.4 m

Correct Answer: D

Solution :

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