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.

SIMPLY PUT THERES DIFFERANCE IN ENERGY RECOVERED WHEN U TAKE 100 METRES OF CHAIN TO HEIGHT OF 10 METRES AND DROP IT AS WHOLE TOGETHER THAN WHEN YOU ARRANGE IT SO THAT IT RUNS DOWN LIKE A THREAD , EACH LINK PULLING DOWN ON THE OTHER LINK IT IS CONNECTED TO...

am a bit curious abt this...i guess i can explain this thing better through phone.....anyone frm india..?

some ppl say that the velocity of links when they hit ground would be 20 m/s. i dont think so..
then goin by that when the planeA is about 40 metres from surface does that mean the links would have a velocity of 40m/s ..? acc to me it would be only 19.6 m/s instead of 27.6m/s...

HERE EVERY LINK STARTS WITH A VEL OF XM/S AND CONTINOUS WITH IT UNTIL HITTING THE GROUND.....

Answer 1

Right. And your question is...?

Answer 2

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 3

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.

Answer 4

Great question. Must've taken years of insight to formulate.

Answer 5

Stop posting nonsense