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 ....


CAPS THERES ANOTHER WAY OF PUTTIN THIS QUESTION...


TAKE A LONG CHAIN MADE OF STEEL RINGS(SAY 200 METRES) AND TAKE IT TO A PLANE 5 METRES ABOVE AND DROP ONE END OVER THE EDGE SO THAT IT HITS THE GROUND BELOW AND KEEPS PULLIN THE REST OF THE CHAIN LIKE A RUNNING THREAD FROM A THREAD CONE....

NOW THE DYNAMMICS IS DIFFERENT FROM WHAT IT WILL BE WHEN U DROP THE WHOLE CHAIN AT ONCE FROM 5 METRES...(IT WILL HIT THE GROUND IN 1 SEC )
BUT IN THE OTHER CASE IT WILL TAKE MANY SECONDS TO RUN DOWN LIKE A THREAD

IN THIS CASE WHEN THERES NO AIR RESISTANCE AND NO FRICTION THE ENERGY RECOVERED WILL BE ONLY HALF OF P.E..

BCOZ THE LINKS HAVE A TERMINAL VELOCITY THAT IS LESS THAN THE ONE DURING FREE FALL...

THATS AGAIN BECOZ GRAVITY FORCE ACTS ON THE INDIVIDUAL LINKS FOR A LESSER TIME...

IN THE FIRST PROBLEM THE LINKS CANNOT HIT THE GROUND AT 20M/S..IF IT DOES SO IT COULD VIOLATE LAW OF CONS OF MOMENTUM...

Answer 1

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.

Answer 2

i have no i idea waht u asked

Answer 3

yes.. that is right i think... but u confuse me i am not sure


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