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A conveyor belt is moving at a constant ...

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:

A

0.4m

B

1.2m

C

0.6m

D

0

Text Solution

Verified by Experts

The correct Answer is:
A

retardation of the block on the belt
`a=(F)/(m)=mug`
From `v^(2)=u^(2)+2as`
`0=2^(2)-2(mug)s`
`s=(4)/(2xx0.5xx10)=0.4m`
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