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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

0.4 m

B

1.2 m

C

0.6 m

D

Zero

Text Solution

Verified by Experts

The correct Answer is:
A
`v^(2)=(2mug)xxS`
`2^(2)=2xx((1)/(2)xx10)xxS`
`4=10S`
`rArr S=0.4m`
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