(i) There is a vertical loop of radius R. A small block of mass m is slowly pushed along the loop from bottom to a point at height h. Find the work done by the external agent if the coefficient of friction is `mu`. Assume that the external agent pushes tangentially along the path. (ii) A block of mass m slides down a smooth slope of height h, starting from rest. The lower part of the track is horizontal. In the beginning the block has potential energy U = mgh which gets converted into kinetic energy at the bottom. The velocity at bottom is `v=sqrt(2gh)`. Now assume that an observer moving horizontally with velocity `v=sqrt(2gh)` towards right observes the sliding block. She finds that initial energy of the block is `E=mgh+1/2 mv^(2)` and the final energy of the block when it reaches the bottom of the track is zero. Where did the energy disappear?

Answer

`(i) mgh+mumg sqrt(2Rh-h^(2))`