The curved portion is smooth and the horizontal portion are rough.The block is released from P.At what distance from A the block will stop. ( μ=0.2) ?

During the fall from `1 m` height,the block will gain kinetic energy and the total energy will be purely kinetic on coming to point `A`,which was purely potential energy while sitting on top of the curve.

Applying law of conservation of energy,

we can say, `mgh = 1/2mv^2` (where, `v` is the velocity at `A`)

So, `v^2=19.62`

Now,the frictional force that will be acting on the block is `mu N=mu mg` (`mu` = coefficient of friction=`0.2`)

So,deceleration while going along the horizontal part will be `(mu mg)/m = mug=1.962 ms^-2`

So,if it stops by going distance `x` then applying, `v^2=u^2 - 2ax`

we get, `0^2 = 19.62 - 2*1.962 x`

So, `x=5 m`

But,givn length of the horizontal part is `2 m`, so let us see what will be the velocity,after travelling through `2m`,

So, `v'^2 = 19.62 - 2*1.962 *2=11.76`

So,at point `B` total energy = kinetic energy = `1/2m v'^2`

Now,if it goes up by height `s` then, again applying law of conservation of energy,

`1/2 mv'^2 = mgs`

So, `s=0.6 m`

Now,after reaching that point it will again come down and gain the same velocity at point `B` while it was going up,so now if it moves distance `r` towards `A`, we can check like previously,

`0^2 = 11.76 - 2*1.962 *r`

So, `r=3`

So,it will still not come to rest still now, so lets find velocity after moving `2m` to reach at `A`

so, `u^2 = 11.76 - 2*1.962*2=4`

So,after rising up when coming down,it will have same velocity at `A`,

So,this time if it stops after going `ym`,we can say,

`0^2 =4 - 2*1.962*y`

or, `y=1.01`

Almost about `1 m` from `A` after these above set of rides the block will stop.

The initial total energy is just the potential energy : `mgh` - where `h` is given as 1 m. For the body to come to a complete halt, this energy has to be dissipated in the form of work done by friction against the motion of the block. Since the force of friction is given by `mu mg` in magnitude, the distance that the block has to slide over the rough ground is `{mgh}/{mu mg} = h/mu = 5` m

Thus the block will cover the 2m stretch of rough ground twice and then travel an extra distance of 1 m before coming to a halt (in between it will climb up and down the two curved surfaces, but since these are smooth, this will not dissipate any energy ).