Quote:

the mass is no longer being forcefully accelerated , it is only passing through the turnaround.

You really really need to understand the terminology. _All_ accelerations are forceful, and anything changing its direction of motion is being accelerated - with force. The speed can remain a constant 40m/s, but it is still accelerating. I can't believe you still haven't looked this up. Again I have a sudden realization that I'm talking in a foreign language to you.


Quote:

if the turnaround is 40 meters in lenght the mass will pass
through the turnaround in 1 second because the mass has a velocity of 40 m/s.


speed=length/time
40m/s = 40m / 1s
Indeed. So a 40m, 1s turnaround is possible. And it can only have one constant force being applied to the mass.

Quote:

if the turnaround is 20 meters in lenght the mass will pass
through the turnaround in 0.5 seconds because the mass has a velocity of 40 m/s.

40m/s = 20m / 0.5s
Yep, so a 20m, 0.5s turnaround is also possible.

Quote:

I think we should use a turnaround lenght of 40 meters.

OK. Let's stick to the 40m, 1s turnaround.

The mass spends 1s turning around.
It enters with a _velocity_ of -40m/s
It leaves with a velocity of +40m/s
It's velocity changes by (+40m/s) - (-40m/s) = 80m/s
It does that change in 1s.
Acceleration = (change in velocity) / (time taken)
Acceleration = 80m/s / 1s = 80m/s^2
The _mass_ must experience a +ve force to turn around. That force is:
F=ma
F=100kg * 80m/s^2
F=8000N
Furthermore, the pipe must experience the same force that the mass does, but in the opposite direction. That force is
F_pipe = -8000N

Again, please either agree or point out the exact part(s) of my calculation that you don't agree with. You didn't last time I asked, but we're not going to get anywhere unless we can isolate our disagreement to its exact root.