Question

# An object is gently placed on a long conveyer belt moving with a uniform speed of 5 ms−1. 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 ms−2.

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=2as⇒s=v22a=v22μg=(5)22×0.5×10=2.5 m Hence, the correct choice is (b).

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