MatinSAR said:
Thanks for your time @kuruman . So it's not pulling down when the lower parts at rest. That makes sense.
But does it push up the upper parts? In post #2 you've mentioned that it cannot. Can I say that because this rope doesn't act like a spring it cannot push up the upper parts?
No it cannot push on the upper parts. Mass element ##dm## which is in free fall collides with the table and stops. The information that it stopped cannot be communicated to the part of the string that is above and is still falling. The energy is lost to heat and vibrations of the table as would be the case in a perfectly inelastic collision.
If you want to put some math in this, the relevant equation to use (derived
here) is $$m~\frac{dv}{dt}=(u-v)\frac{dm}{dt}+F_{\text{ext}}.$$In this equation
##m=~##the mass of the rope still falling
##v=~##the velocity of the mass still falling relative to the lab frame
##u=~##the velocity of mass ##dm## that is stopped in time ##dt## relative to the lab frame
##F_{\text{ext}}=~##the external force in this case ##-mg##.
Because all parts of the rope that are still in motion move as one at all times, ##u=v## and the equation becomes $$m~\frac{dv}{dt}=-mg.$$ If the rope is released from rest,the solution is $$v(t)=-gt.$$ This is what one would get if one modeled the continuous rope as consisting of
separate mass elements ##dm## arrayed in a continuous vertical line and released from rest all at the same time.
You can clearly see now why a mass element starting at height ##x## above the table has speed given by ##v^2=2gx## when it hits the table as claimed by the textbook.
. It is a rope that's in free fall. It doesn't act like a spring. So
