any physicsts here...? pls check .thanx..?

Question

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}

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.

i guess u can use a long chain and try this experiment in space ....

Answer 1

An interesting problem. But why do you think the velocity = sqrt(Xg) for the link-by-link case and not for the whole-chain case? Do these materializing links have zero initial velocity? If so, how is this consistent with their being attached to the moving chain they are part of? Do they instantly accelerate to the chain speed? Does this involve a tensile "collision"? Is the collision elastic? You can see where the "materializing" scenario leads to unnecessary complexity and difficult questions. I suggest that you model your system as follows:
The chain lies along a horizontal straight line on a 20 m high table and slides along the line frictionlessly to the edge from which it falls. The weight of the hanging 20 m of chain accelerates that 20 m plus all the chain still on the table. Thus acceleration increases as the chain amount of chain remaining decreases. At the beginning the speed is less than the free-fall speed, but toward the end it exceeds the free-fall speed. Every descending link contributes to the kinetic energy of the remaining chain. The total impact energy will equal the initial potential energy of the elevated chain mass. An analysis via calculus will show this.