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.
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).