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)  1.2 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,    \[{{v}^{2}}={{u}^{2}}+2as\]                 \[0={{(2)}^{2}}-2(\mu g)s\]                 \[s=\frac{4}{2\times 0.5\times 10}=0.4\,m\]