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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.

2007-08-23 21:29:25 · 2 answers · asked by balaji.k 2 in Science & Mathematics Physics

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..??

balaji.k, tirupur,
t.n., india....

2007-08-23 21:35:12 · update #1

2 answers

Answer 1 is correct. 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.

2007-08-26 01:03:50 · answer #1 · answered by kirchwey 7 · 1 0

How did you calculate that the terminal velocity of the links of the chain is 14m/s?

By having the chain in a heap, I presume you are saying that the falling links are not moving or accelerating the links that are yet to fall. But if that is the case, then wouldn't they just fall at 'g'? And so the final velocity will be 20m/s rather than 14m/s, and so solve your problem?

If the chain is laying straight, the first links will accelerate at slower than 'g' because they are accelerating the whole length of the chain. However, as the chain falls the acceleration will increase. When the last link leaves the table, it will already be traveling at a considerable velocity, and will accelerate at 'g' from then on, and so will reach a velocity higher than 20m/s.

I suspect that we will find that energy is conserved. Surprise!

2007-08-24 04:57:47 · answer #2 · answered by Anonymous · 1 0

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