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

a = `mu g = 0.5 xx 10 = 5 m//s^(2)`
u = 2 m/s
Stopping distance (S) = `(u^(2))/(2a)`
`(S) = (4)/(5 xx 2) = 0.4` m