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?

Explanation

The particle will finally come to rest on the flat part. Hence, displacement of the particle along vertical is h. If `W_8` be the work done on the particle by the gravity, then
`W_g=mgh`(i)
where m is the mass of the particle.
If distance travelled by the particle on the flat part is `x`, the work done on the particle by the friction is
`W_f=-mumgx` (ii)
Since, initially, particle was at rest and finally it comes to rest again. Hence, change in its `KE` is zero.
From work-energy theorem, `W_g+W_f=DeltaKE`
`implies mgh-mumgx=0`
`impliesx=h/mu=1.5/0.2mimpliesx=7.5m`
Since `xgtl`, 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 descent to C and will have the same kinetic energy as it had when ascending but now will move from C to B. At B, same thing will be repeated (because `7.5lt2l`), and finally, the particle will stop at E such that
`BC+CB+BE=7.5`
`BE=7.5-6=1.5m`