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Question

An object is gently placed on a long conveyer belt moving with a uniform speed of 5 ms1. If the coefficient of friction between the block and the belt is 0.5, how far will the box move on the belt (relative to belt) before coming to rest? Take g=10 ms2.


A

2.0 m

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B

2.5 m

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C

3.0 m

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D

3.5 m

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Solution

The correct option is B

2.5 m


The force of friction between the block and the belt is f=μmg, where m is the mass of the object. This force produces an acceleration of the block which is given by
a=fm=μmgm=μg
The block will move relative to the belt until its speed (v) becomes equal to the speed of the belt. Since u = 0, we have
v2=u2+2as=2ass=v22a=v22μg=(5)22×0.5×10=2.5 m
Hence, the correct choice is (b).


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