Q.
A body of mass $M$ was slowly hauled up a rough hill by a force $F$ which at each point was directed along a tangent to the hill. Work done by the force
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Solution:
If the body slips over a rough surface such that normal reaction of the surface has to balance only the normal component of weight of the body, then the energy lost against friction depends only upon the horizontal component of displacement and is equal to $\mu m g x$. It does not depend upon the shape of the surface.
If the vertical component of displacement of the body is equal to $y$, increase in its gravitational potential energy will be equal to $m g y$.
Hence, total work done by the force will be equal to $\mu m g x+m g y$. Hence, options (1) and (3) are correct.
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