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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

1.2 m

B

0.6 m

C

zero

D

0.4 m

Text Solution

Verified by Experts

The correct Answer is:
D
Frictional force on the box `f=mu mg`
`:.` Acceleration in the box
`a= mug=5 ms^(-2)`
`v^(2)=u^(2)+2as`
`rArr 0=2^(2)+2xx(5) s rArr s=-(2)/(5)` w.r.t. belt
`rArr` distance =0.4 m
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