A particle slides along a track with elevated ends and a flat central part as shown in figure. The flat part has a length `l=3m`. The curved portions of the track are frictionless. For the flat part, the coefficient of kinetic friction is `mu_k=0.2`. The particle is released at point A which is at height `h=1.5m` above the flat part of the track. Where does the particle finally come to rest?

Answer

A

Explanation

As initial mechanical energh of the block 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 `mght=mu mgd` i.e. `d=h/mu = (1.5)/(0.2)=7.5m`
After starting from, B, the block will reach C and then will rise up till the remaining (KE) at (C) is converted into potential inergy. It will then again descend and at C will have the some value as it had when ascending, but now it will move from C to B. The same will be repeated and finally the block 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 block comes to rest at the centre of the flat part.