A conveyer belt is moving at a constant speed of 2 `ms^(-1)` 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 ms^(-2)`, is

Options

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

Answer

A

Explanation

Force of Friction ` f^(-) mu mg `
` therefore a = (f)/( m ) = (mu mg)/( m) = mu g = 0.5 xx 10= 5 ms^(-2)`
using `v^2 - u^2 = 2aS `
` 0^2 -2^2 =2 (-5) xx S implies S= 0.4 m`