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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

Zero

B

0.4 m

C

1.2 m

D

0.6 m

Text Solution

Verified by Experts

The correct Answer is:
B
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
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