Answer
The particle stops at $20$ $cm$ from the left end of the flat part.
Work Step by Step
Its Potential energy at point $A$ is $m\times g\times h=mgh$
Because the surface is frictionless, the energy is conserved and is converted to kinetic energy at bottom, just before the flat part.
So,
$Potential$ $Energy =Kinetic$ $Energy$
$mgh=\frac{1}{2}mv^{2}$
$ v^{2}=2gh$
Substituting the known values in this expression gives:
$v^{2} =2\times g\times 20$
$v^{2}= 40g $
Now, acceleration due to friction is $\frac{Frictional force} {mass}$
$\frac {0.2 mg}{m}=-0.2g$ $m/s^{2}$
Negative sign denotes opposite direction.
We know,
$v^{2}=u^{2}+2ad$,
Plugging the known values and solving gives:
$0^{2}=40g+2(-0.2g)d$
$d=100$
This means it will travel a total distance of $100cm$ throughout its motion along the flat part.
Thus, it will go to the other end and comeback covering $40cm\times
2=80cm$
and then cover the remaining $20cm$ from the the left edge.
Therefore it stops at $20$ $cm$ from the left edge.
