A small block slides along a track with elevated ends and a flat central part as shown in figure. The flat portion BC has a length `l=3.0 m`, The curved portions of the track are frictionless. For the flat part, the coefficient of kinetic friction is `mu_(k) = 0.20`, the particle is releast at point A which is at height `h = 1.5 m` above the flat part of the track. Where does the block finally comes to rest?

Explanation

As initial mechanical energy of the particle is mgh and final is zero, so loss in mechanical energy = mgh. This mechanical energy is lost in doing work against friction in the flat part,
So, loss in mechanical energy = work done against friction
or mgh = `mu`mgs i.e., `" " s = (h)/(mu) = (1.5)/(0.2) = 7.5` m
After starting from B the particle will reach C and then will rise up till the remaining KE at C is converted into potential energy. It will then again descend and at C will have the same value as it had when ascending, but now it will move from C to B. The same will be repeated and finally the particle will come to rest at E such that
BC + CB + BE = 7.5
or `" " 3 + 3 + BE = 7.5`
i.e., `" " BE = 1.5`
so, the particle comes to rest at the centre of the flat part.