Any good physicsts here.?a problem in conservation of energy....?

Question

imagine theres a horizontal planeA bove which the effects of gravity are not felt. any object above the plane just floats and gravity has no influence on it.
now take a chain(for e.g:made of small steel ring) of length hundred metres above the planeA and arrange it so that its in a heap just above the planeA. but take care the chain is not entangled and should be easy to stretch along its length.

now pull the lower end of the thread below the planeA and the chain runs down like a single thread and hits the ground and as links pull down on the ones they are linked to , the whole length of the chain will run like a thread and hit the ground in some time. now take planeA is 20metres above the ground surface. the terminal velocity of links when they hit ground will be around 14 metres .
now consider hundred metres of the chain weighs 100 units. the potential energy of the chain just above the planeA will be m*g*h =100 *9.8*20
=19600 units.

but kinetic energy recovered would be 1/2 * m * v*v = 0.5 * 100* 14 *14 = 9800 units. so in this method only half the energy is recovered hence violating the law of conservation of energy.

Now Imagine theres a point 'A' vertically above the surface of earth. the distance of point A from the horizontal plane is given by X metres .also, imagine theres no atmosphere.

Now Imagine from within that point A , a chain {made of identical oval shaped rings linked sequentially} keeps on appearing {materializing}

9 hours ago
and it starts to fall down due to gravity and hits the ground.the liks which appear from within point A are static and are pulled down by the links below them i.e., the faster you pull the faster the new links materialize. now what will be the velocity of the chain { particularly the rings} when they hit the ground if X = 20 metres .(remember the chain keeps on materializing from the point A}.

i think the terminalvelocity of the chain can be given by Vt = squarerootof (Xg)

where g = acceleration due to gravity(take g= 10 m/s^2).
now sqrroot of(20*10) = 14.14.so , here Vt = 14.14.

but when you take an object to a height of 20 metres above the ground and
drop it , the velocity when it hits the ground would be 20 m/sec .

so the kinetic energy when ,

v = 14.14 = 1/2(M* 14 .14 *14.14) = 100M (M = mass)
v = 20 = 1/2(M * 20 * 20) = 200M (M = mass).

SOTHIS WAY K.E RECOVERED IS ONLY HALF THAT OF P.E.

Answer 1

Certainly in the real world there would be horizontal considerations. But bg is trying to pose the question in a way that eliminates horizontal considerations so we can understand what's happening in the chain. So lets go along with that.
But Prof. Zikzak said it all. "There's a tension in the chain." At any given point in the chain (as it falls) there's a tension slowing down the links below that point. But there's also an equal and opposite tension speeding up the links above that point. For example, think about the last link in the chain. There's a non-zero tension, in addition to the acceleration of gravity, tending to speed up that link, and nothing to slow it down. That tells me that that link will hit the floor faster that it would if it were falling all by itself. So some links are accelerating slower than g and some links are accelerating faster than g. And, no doubt, they are all cooperating to preserve the law of conservation of energy.

Answer 2

The links do not all hit the ground at 14 m/s or at 20 m/s. Each hits the ground at a different speed because each has been accelerating by different amounts for different lengths of time. You must consider each link individually, find its speed when it hits, and calculate the resulting total KE, which will equal the initial PE.

The speed at the ground will NOT be sqrt(hg) because gravity is not the only force acting on the link. there will be a tension force acting upwards as well. The amount of tension varies as each successive link falls off the table. Most links will be in motion long before they fall off the table, and thus will have horizontal velocity as well, carrying more kinetic energy that way too.

Answer 3

I believe that all the variations on this problem (9 so far and counting) that you have posted fall into either of two basic scenarios:
1. The descending part of the chain accelerates the (finite) not-yet-falling part (e.g., the stretched-out chain that firctionlessly slides over the edge). In this case I claim that the not-falling part gets accelerated by the falling part so velocities become greater than free-fall impact velocity as the process continues, and that energy is conserved.
2. The descending part is essentially unconnected to the (potentially infinite) not-falling part, and a demon is sitting up there adding links to the top of the falling part as needed. In this situation I claim that the falling part is in a steady-state situation, falling at a speed greater than 0 and less than free-fall impact speed, so kinetic energy is indeed less than potential energy. The missing energy goes into the inelastic collision involved in attaching a link at 0 velocity to the moving chain. This link is jerked up to speed with no rebound. Momentum, but not energy, is conserved, and all the energy is accounted for.