# Law of conservation of energy wrong?

1. Aug 25, 2007

### Lakshya

Law of conservation of energy wrong???????

a problem.....pls check..
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...

now what will the terminal velocity be when the planeA is above 40 metres...?
i guess its around 19.6m/s instead of 28m/s......

ur thoughts pls correct me if im wrong...

2. Aug 25, 2007

### nealh149

One of your first premises is that there is not gravitational field, but then you use gravitational potential energy to calculate energy.

3. Aug 25, 2007

### cyruskoras

Here when you consider plane A you said it has no gravity , but you have used g = 9.8 I dont know how is it possible?

4. Aug 26, 2007

### Lakshya

I said that on the plane no effect of gravity can be felt. This can be achieved by accelerating it upward with an acceleration of 9.8 m/s. Then it will have no effect of earth's gravity.

5. Aug 26, 2007

### Staff: Mentor

The stuff about "gravity has no influence" sounds mysterious, but I assume you just mean that the chain is resting on a frictionless horizontal plane.

Please show your calculation. What do you mean by terminal velocity? The speed of the chain when the first link hits the floor? When the last link hits the floor?

Realize that when the chain runs down like a thread, it loses energy as it hits the floor inelastically--kinetic energy is transformed into thermal energy. Could that be what you are concerned with?

6. Aug 26, 2007

### AlephZero

No, that's logically impossible.

Part of the chain is moving at a finite speed and the other part is at rest. However you try to arrange it, when the next link starts moving it has to accelerate "instantaneously" from rest, to match the speed of the moving part.

It is impossible for moving part to accelerate at 9.8m/s, because part of its KE will used to accelerate the stationary part to the same speed. It will only accelerate at 9.8m/s when all the chain has left the table.

This has nothing to do with imagining a "real" chain with finite sized links - the same applies to an idealized flexible string.

Last edited: Aug 26, 2007
7. Aug 26, 2007

### Idjot

Accelerating upward at 9.8 m/s would not cancel out the g's. It would double them. Free fall is the way to cancel g's.

This hypothetical experiment would be better conducted in a vacuum with the chain hanging from its top link. Then the chain would fall without bunching up and you wouldn't need the confusing anti-gravity plane to find the suspected difference from the bunched up chain. Of course there will be no difference other than that which might be measured if the bunched up chain is released at the wrong altitude required to compensate for the incremental altitudes of the links on the hanging chain. To make it easier - how about a twenty pound ball compared to twenty, one pound balls released one after the other? Same thing, right?

8. Aug 27, 2007

### Lakshya

Please clear yourself. If you will perform the experiment mentioned by me you will get the law of conservation of energy wrong.

9. Aug 27, 2007

### Staff: Mentor

10. Aug 27, 2007

### txshender

PE is not constant

Supposing that the distance to fall is 20 meters (it matters not that this chain is 100 meters long or 200 meters long the moment it passes through the "opening" from the plane) and that all of the chain drops together...

x=.5*a*t^2
-20=.5*9.8*t^2
t~= 2.02031 s

v(final)=a*t + v(initial)
v=9.8*2.02031 + 0
v~=19.8 m/s

KE=1/2 * m * v^2
KE=.5*100*19.8^2
KE=19602 ~= 19600 J which is the same as PE=mgh=100*9.8*20

However, this is done ideally and is only a free fall problem...

If I am not mistaken... F=ma... When just one link of that chain passes through the "hole," there will only be the force being exerted on that one chain by gravity, and that force will be transfered to the entire mass... Assuming one link of the chain weighed 1 kg, and the whole chain weighed 100 kg... 1*9.8= 100 * a... That means that the whole chain will come out of the plane eventually, and it will do so in a jerking accelerating (I don't know how else to describe non-uniform acceleration) manner until all chains are out of the plane... Another point is that energy will be used accelerating the other chains out of the other plane, so I do not believe it should be expected that total KE should equal PE (which is what you used in your first calculation, and the concept of using it here is wrong as I will explain later)...

Besides, what is your definition of KE in this case? Since each part of the chain will hit the ground with a different terminal velocity, is it correct to calculate KE as if it were one mass like you did? And what is the terminal velocity that you used? Did you use the last ring? Because if so, it would be expected to be somewhere near (not exactly, but close to) 19.8 m/s at the moment before it touches the ground and experiences an impulse.

I am sure somebody could model this with an equation... But it would be... very messy because you have to deal with the non-uniform acceleration (1/6 * j * t^3?), the mass outside of the plane is the only one experiencing force, and that force is applied to the entire mass of the object, which causes the acceleration at any moment to be completely governed by how much chain has already been let out... Run-ons for the lose...

But my final words are that potential energy is CONSTANTLY CHANGING and you will not be able to calculate it using m*g*h. Potential energy is merely defined as E=F*d with F being the weight force... The weight of that entire chain is constantly changing until the last chain leaves the plane...

Edit: Apologies for the lack of focus throughout this post

Last edited: Aug 27, 2007
11. Aug 27, 2007

### Idjot

To the OP:

I'm wondering if there is any possible scenario where you might prove this theory. Otherwise the calculations you're doing are entirely dependent upon the links of the chain magically entering Earth's gravity as if through a doorway one by one. Don't get me wrong. I have nothing against hypothetical props, even impossible props, if the theory itself can at least be proven somehow in a possible way.

But I reiterate:
The "working" theory here seems to be wholly dependent on an impossible prop. Please correct me if I'm wrong - anyone.

Last edited: Aug 27, 2007
12. Aug 27, 2007

### DaveC426913

"...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..."

The rest of the entry is merely gilding the lily.

Let me see if I can summarize:

"Imagine we can nullify the effects of gravity at will. Under these circumstances, we can violate the law of conservation of energy."

I would say YEP! With your imaginative premise, there's no reason why CoE couldn't be violated. But why bother with a gravity nullifier? Why not ghosts or tooth fairies?

13. Aug 27, 2007

### Gokul43201

Staff Emeritus
Actually, Dave, there's no reason that the OP's construction should result in energy violation. There's nothing wrong with requiring potentials that vary differently than 1/r. Even a discontinuous potential is fair game, and will not lead to non-conservation.

If you wish, replace the gravitational field with an electrostatic field created by a giant, parallel plate capacitor (field is uniform everywhere inside and zero everywhere outside) and make the chain be a charged insulator.

What is missing in this thread, however, is the calculation performed by the OP in determining the "terminal velocity". Unless we see this calculation, we can't say where it's wrong.

Last edited: Aug 27, 2007
14. Aug 27, 2007

### jhat21

I don't know how you got 14m/s for the velocity instead of 20m/s, but here is the calculation. it looks like a straightforward high school physics problem.

h = 1/2 g t^2 + v0 t
20m = 1/2 10m/s^2 t^2 + 0

t = sqrt( 2h/g )
t = sqrt( 2*20/10) = sqrt(4) = 2

vf= g t
vf= 10m/s^2 * 2 = 20m/s

1/2 m v^2 = mgh
1/2 v^2 = gh
v^2 = 2gh
so
20^2 = 2 * 10 * 20 = 400

mass is irrelevant and

i don't see any problem with energy conservation here, not today

15. Aug 27, 2007

### jhat21

If you calculate for each link on the chain, you will get the same equality.

Now let's assume that the link is horizontally placed, so that mgh is the same for each link.
It appears that the potential energy is the same for all links, and it also appears that the later links will hit the ground with more velocity than the first as it will accelerate over a longer time. This seems like kinetic energy does not equal potential energy.
But this is wrong.
Because we cannot express the potential energy as mgh anymore! The potential energy is due to the force of gravity acting upon the link, F, over a distance, d, it is equal to work the integral of F*d. Therefore we do need to consider the length of chain that the link must traverse horizontally also! since it is being acted upon with the same force of gravity over this distance as the link that is falling.
You would then write mg(h+L) as the potential energy, where L is the horizontal distance of the link from the drop point.

16. Aug 30, 2007

### Lakshya

And why is it necessary to take the horizontal distance.

17. Aug 30, 2007

### Staff: Mentor

The explanation was clear about that: w=fd

18. Aug 30, 2007

### Staff: Mentor

Until Lakshya clarifies the scenario he has in mind--by explicitly showing his calculations--this thread is rather pointless.

As long as the stretched out chain doesn't hit the ground--thus introducing dissipative forces--its speed as it slides off the (frictionless) surface can be found at any point using conservation of energy. The chain loses PE and gains KE; it's as simple as that.

Having the chain stretched out, as opposed to being bunched up, simplifies things greatly. The chain moves as a unit with a single speed.

19. Aug 30, 2007

### Gokul43201

Staff Emeritus
There are many things incorrect here and in the previous paragraph.

$$U = \vec{F} \cdot \vec{d} = Fd cos \theta$$

The horizontal distance, L, is traversed perpendicular to the gravitational force F. So the change in gravitational PE is 0 over that section.

Lakshya has had ample opportunity to show his/her calculations, and hasn't done so. I recommend the thread be locked.

Last edited: Aug 30, 2007
20. Aug 30, 2007