• # question_answer A conveyor belt is moving at a constant speed of 2 m/s. A box is gently dropped on it. The co- efficient 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 mC)  zero          D)  0.4 m

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