A block of mass m slides along a frictionless surface as shown in the figure. If it released from rest at A, what is its speed at B?

v = gh

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v = 2gh

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Solution

The correct option is C

Take the block + the earth as the system. Only the block moves, so only the work done on the block will contribute to the gravitational potential energy. As it descends through a height h between A and B, the potential energy decreases by mgh. The normal contact force N on the block by the surface does no work as it is perpendicular to its velocity. No external force does any work on the system. Hence,

Increase in kinetic energy = decrease in potential energy or, $\frac{1}{2} m v^{2} = m g h$ or , $v = \sqrt{2 g h}$

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