A conveyor belt is moving at a constant speed of `2m//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=10ms^(-2)` is:

Options

A. zero
B. `0.4 m`
C. `1.2 m`
D. `0.6 m`

Answer

B

Explanation

`v^(2) = u^(2) + 2 as = 0 + 2 mu gs`
`(2)^(2) = 2 xx 5 xx 10 xx s rArr s = 0.4 m`