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 $I=3.0 \mathrm{m}$. 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 released at point A which is at height $h=1.5 \mathrm{m}$ above the flat part of the track. Where does the block finally comes to rest?

As initial mechanical energy of the block is (m g h) and final is zero, so lose in mechanical energy = (m g h). This mechanical energy is lose in doing work against friction in the flat part.

So, loss in mechanical energy = work done against friction or

$$\begin{gathered} \text{ mgh }=\mu mgd \\ d=\frac{h}{\mu }=\frac{1.5}{0.2}=7.5 \mathrm{m} \end{gathered}$$

After starting from B, the block will reach C and then will rise up till the remaining K E 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 block will come to rest at E such that

BC + CB + BE = 7.5

3 + 3 + BE = 7.5 

BE = 1.5

So, the block comes to rest at the center of the flat part.