A block of mass $ m=0.1 \mathrm{kg} $ is released from a height of $ 4 \mathrm{m} $ on a curved smooth surface. On the horizontal surface, path AB is smooth and path BC offers coefficient of friction $ \mu=0.1 . $ If the impact of block with the vertical wall at C be perfectly elasti

Question

A block of mass $ m=0.1 \mathrm{kg} $ is released from a height of $ 4 \mathrm{m} $ on a curved smooth surface. On the horizontal surface, path AB is smooth and path BC offers coefficient of friction $ \mu=0.1 . $ If the impact of block with the vertical wall at C be perfectly elastic, the total distance covered by the block on the horizontal surface before coming to rest will be : $ \left(\operatorname{take} g=10 m / s^{2}\right) $

Accepted Answer

49 $ \mathrm{m} $. The block starts with potential energy mgh = 0.1×10×4 = 4 J, so its speed at B gives 4 J of kinetic energy. On the rough segment BC, friction force is μmg = 0.1×0.1×10 = 0.1 N, so the block loses 0.1 J per meter; accounting for the elastic bounce at C and repeated passes, the total horizontal distance adds up to 49 m before stopping.

Solution

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