Why is this Experiment violating the conservation of Energy?

[Maximillian]

No it is not.

Stop ignoring reality.

This

PE = mgh

is wrong.

So what gravitational potential energy does a body of 1kg and 10m from the Earth's surface have?

[John Cuthber]

So what gravitational potential energy does a body of 1kg and 10m from the Earth's surface have?

Well, if it is 1 Kg of polystyrene foam (let's say it's density is a tenth that of water) and it's 10M from the surface, but in a water tank 20 metres deep then the conventional gravitational potential energy is negative- because it you let go of it, it will "fall" upwards.

 

If we say that local gravitational acceleration is 10 m/s/s then the answer to your question can be calculated thus

The volume displaced by the polystyrene is 10 litres.

And, from Archimedes' principle, we know that it experiences an upthrust equal to the weight of the water it displaces so that's 100 newtons.

Set against that is the weight of the block- that's 10 Newtons. So the net force acting on it is 90 Newtons (upwards).

If you let go of it the block will rise by 10 metres to the surface of the tank.

So the potential energy of the block is 900 Joules

 

Now do you see why mgh has sod all to do with the calculation?

[imatfaal]

The problem is I had to use 100J of energy to carry the ball and drop it into the tank of water.

The ball then sinks to the bottom of the water.

Then this time I only use 20J to achieve the 100J I used previously.

 

Listing inputs and PE

 

A. Start / finish at the floor

1. So you lift it to the top and if you dropped it off the side it would fall ten metres

2. it drops through the water

3. You expend energy to make it expand

4. It floats to the surface and if you dropped it off the side it has ten metres to fall

 

 

1. -100 joules

2. 0 joules (ball loses/water gains)

3. -20 joules

4. +100 joules

 

seems to me you have expended 20 joules to end up where you were. If you are thinking of inserting ball through some gate at the base of the tank think how much energy that would take - wanna guess?

[Maximillian]

 

Listing inputs and PE

 

A. Start / finish at the floor

1. So you lift it to the top and if you dropped it off the side it would fall ten metres

2. it drops through the water

3. You expend energy to make it expand

4. It floats to the surface and if you dropped it off the side it has ten metres to fall

 

 

1. -100 joules

2. 0 joules (ball loses/water gains)

3. -20 joules

4. +100 joules

 

seems to me you have expended 20 joules to end up where you were. If you are thinking of inserting ball through some gate at the base of the tank think how much energy that would take - wanna guess?

I don't have to always carry it up, lets assume its a river or lake at sea level we shall not need the -100J.

0-20+100 = 80J which is created energy.

For the tank of water at 10m up,

Lets say the densities I have stated are composite.

Combining a pocket of air and some dense material.

In the first instance when the ball sinks the air pocket is at atmospheric pressure. I then reduce that density to let the ball float. At the top I unlock the structure holding the air pocket so that the atmospheric pressure pushes it back to its original volume without me using any energy which makes it more dense hence it sinks.

You now see that I can perform this repeatedly.

So you will get -100+0-20+100+0-20+100 = 60J of energy have been created in just two sinks.

The more you repeat the process you will create more and more energy.

Well, if it is 1 Kg of polystyrene foam (let's say it's density is a tenth that of water) and it's 10M from the surface, but in a water tank 20 metres deep then the conventional gravitational potential energy is negative- because it you let go of it, it will "fall" upwards.

 

If we say that local gravitational acceleration is 10 m/s/s then the answer to your question can be calculated thus

The volume displaced by the polystyrene is 10 litres.

And, from Archimedes' principle, we know that it experiences an upthrust equal to the weight of the water it displaces so that's 100 newtons.

Set against that is the weight of the block- that's 10 Newtons. So the net force acting on it is 90 Newtons (upwards).

If you let go of it the block will rise by 10 metres to the surface of the tank.

So the potential energy of the block is 900 Joules

 

Now do you see why mgh has sod all to do with the calculation?

So are you saying bodies never posses gravitational potential energy?

Edited September 12, 2016 by Maximillian

[imatfaal]

I don't have to always carry it up, lets assume its a river or lake at sea level we shall not need the -100J.

0-20+100 = 80J which is created energy.

For the tank of water at 10m up,

Lets say the densities I have stated are composite.

Combining a pocket of air and some dense material.

In the first instance when the ball sinks the air pocket is at atmospheric pressure. I then reduce that density to let the ball float. At the top I unlock the structure holding the air pocket so that the atmospheric pressure pushes it back to its original volume without me using any energy which makes it more dense hence it sinks.

You now see that I can perform this repeatedly.

So you will get -100+0-20+100+0-20+100 = 60J of energy have been created in just two sinks.

The more you repeat the process you will create more and more energy.

....

 

If it is at sea level you don't have 100J of PE.

 

Do your calcs bringing the ball back to the same point. It is pretty simple if you are making a claim of breaking the conservation of energy. You have a 10 m tank - start the ball either 10m up or on the floor outside the tank (if you start the ball inside the tank then you need to audit the water too); show a simple energy audit that you have returned to the same point and been able to extract energy

[swansont]

 

So you will get -100+0-20+100+0-20+100 = 60J of energy have been created in just two sinks.

The more you repeat the process you will create more and more energy.

 

 

So build an engine from this concept. (Spoiler alert: it won't work)

[Strange]

 

For more information please follow this link to the filed patent https://www.researchgate.net/publication/307865505_CREATION_OF_ENERGY.

 

 

1. Where was this "patent" filed?

 

2. When was this filed?

 

And for bonus points:

3. How quickly was it rejected?

[swansont]

 

Listing inputs and PE

 

A. Start / finish at the floor

1. So you lift it to the top and if you dropped it off the side it would fall ten metres

2. it drops through the water

3. You expend energy to make it expand

4. It floats to the surface and if you dropped it off the side it has ten metres to fall

 

 

1. -100 joules

2. 0 joules (ball loses/water gains)

3. -20 joules

4. +100 joules

 

seems to me you have expended 20 joules to end up where you were. If you are thinking of inserting ball through some gate at the base of the tank think how much energy that would take - wanna guess?

 

To make step #2 easy to mentally visualize, consider a tube the same width as the object, and make the object a cylinder that neatly fits inside the tube. Just a tiny amount of a gap to let water pass by. When your drop it in, it displaces the water, as we have known since Archimedes. The object ends up displacing the water at the bottom, so you have effectively moved that volume of water up to the top.

 

It's not actually even, though, since the volume displaces less than 1 kg of water. The volume is 0.9 liters, so that's ~90 J (remember we're approximating g as 10 here, from the OP). The object will have some kinetic energy as it falls, and the extra 10J is lost when it hits the bottom or to friction. When the object expands, the only direction the water can go is up — the volume is now 1.1L. So the PE of the water increases even more, by 20J. Now the object rises. to the top, recovering its initial PE of 100J, but again losing any KE it has gained.

[Maximillian]

 

1. Where was this "patent" filed?

 

2. When was this filed?

 

And for bonus points:

3. How quickly was it rejected?

What makes you think it was rejected when you can't disapprove it?

 

If it is at sea level you don't have 100J of PE.

 

Do your calcs bringing the ball back to the same point. It is pretty simple if you are making a claim of breaking the conservation of energy. You have a 10 m tank - start the ball either 10m up or on the floor outside the tank (if you start the ball inside the tank then you need to audit the water too); show a simple energy audit that you have returned to the same point and been able to extract energy

Lets start at the top of the tank.

In this case we don't need to carry it up since the ball is at the same level with the water surface so no -100J needed.

So we started from the water surface, sunk the ball and it rose back to the water surface.

0-20+100 = 80J which is created energy.

 

 

So build an engine from this concept. (Spoiler alert: it won't work)

Even if it didn't work the question would be are our equations sufficient?

Since theoretically it seems to work but practically it doesn't.

Anyway try it out practically for yourself you will be amazed.

Edited September 13, 2016 by Maximillian

[swansont]

 

Lets start at the top of the tank.

In this case we don't need to carry it up since the ball is at the same level with the water surface so no -100J needed.

So we started from the water surface, sunk the ball and it rose back to the water surface.

0-20+100 = 80J which is created energy.

Even if it didn't work the question would be are our equations sufficient?

Since theoretically it seems to work but practically it doesn't.

Anyway try it out practically for yourself you will be amazed.

 

 

No your equations are not sufficient. You have ignored the energy of the water. It only "seems" to work theoretically because you did it wrong.

[imatfaal]

Lets start at the top of the tank.

In this case we don't need to carry it up since the ball is at the same level with the water surface so no -100J needed.

So we started from the water surface, sunk the ball and it rose back to the water surface.

0-20+100 = 80J which is created energy.

 

No. The ball is where it started from and no useful energy has been extracted - however you have spent 20 joules

[swansont]

 

No. The ball is where it started from and no useful energy has been extracted - however you have spent 20 joules

 

 

 

Indeed. The only place the alleged addd energy could show up is in KE (which will be lost to friction of the water), but the net force on the object has the same magnitude in this problem for both paths. You are either displacing an extra 0.1L of water or you are short by that amount, so the force on the object is 1 N. Over 10m, that's 10 J of work. The same up as down. IOW, that's the 20 J you've lost. Everything else is symmetric, and cancels.

 

So you have an energy balance analysis and a work-energy analysis that tell you the same thing. You lose 20 J of energy each cycle.

[imatfaal]

 

 

 

Indeed. The only place the alleged addd energy could show up is in KE (which will be lost to friction of the water), but the net force on the object has the same magnitude in this problem for both paths. You are either displacing an extra 0.1L of water or you are short by that amount, so the force on the object is 1 N. Over 10m, that's 10 J of work. The same up as down. IOW, that's the 20 J you've lost. Everything else is symmetric, and cancels.

 

So you have an energy balance analysis and a work-energy analysis that tell you the same thing. You lose 20 J of energy each cycle.

 

I do like it when two different analyses match up - it gives you that nice feeling that you have arrived in the right place; lots of different ways to get lots of different wrong answers but all correct routes lead to just one answer. And in this case it keeps Homer Simpson happy

[Maximillian]

 

To make step #2 easy to mentally visualize, consider a tube the same width as the object, and make the object a cylinder that neatly fits inside the tube. Just a tiny amount of a gap to let water pass by. When your drop it in, it displaces the water, as we have known since Archimedes. The object ends up displacing the water at the bottom, so you have effectively moved that volume of water up to the top.

 

It's not actually even, though, since the volume displaces less than 1 kg of water. The volume is 0.9 liters, so that's ~90 J (remember we're approximating g as 10 here, from the OP). The object will have some kinetic energy as it falls, and the extra 10J is lost when it hits the bottom or to friction. When the object expands, the only direction the water can go is up — the volume is now 1.1L. So the PE of the water increases even more, by 20J. Now the object rises. to the top, recovering its initial PE of 100J, but again losing any KE it has gained.

Correct, I added 20J of gravitational potential energy to the water by expanding the ball.

From your example this gives the water 110J which is enough energy to help the ball attain 100J.

I can see you are also coming to the conclusion that we input only 20J but have got out 100J.

Edited September 13, 2016 by Maximillian

[Strange]

I can see you are also coming to the conclusion that we input only 20J but have got out 100J.

 

No one is coming to that conclusion, because it is wrong.

 

Surely, you must know that you have the wrong answer (because it is physically impossible) therefore the only problem is for you to understand why you have got it wrong.

[swansont]

Correct, I added 20J of gravitational potential energy to the water by expanding the ball.

From your example this gives the water 110J which is enough energy to help the ball attain 100J.

I can see you are also coming to the conclusion that we input only 20J but have got out 100J.

 

 

No, I am not. You lost that 100 J when you dropped the ball. You get that 100 J back when it rises. There is no energy to be gained when the ball stops and starts in the same position.

[Maximillian]

 

 

 

Indeed. The only place the alleged addd energy could show up is in KE (which will be lost to friction of the water), but the net force on the object has the same magnitude in this problem for both paths. You are either displacing an extra 0.1L of water or you are short by that amount, so the force on the object is 1 N. Over 10m, that's 10 J of work. The same up as down. IOW, that's the 20 J you've lost. Everything else is symmetric, and cancels.

 

So you have an energy balance analysis and a work-energy analysis that tell you the same thing. You lose 20 J of energy each cycle.

Then when it gets to the top what is its gravitational potential energy?

 

 

No, I am not. You lost that 100 J when you dropped the ball. You get that 100 J back when it rises. There is no energy to be gained when the ball stops and starts in the same position.

Yes you get the 100J back but how much energy did you input to acquire the 100J at the rise?

 

No one is coming to that conclusion, because it is wrong.

 

Surely, you must know that you have the wrong answer (because it is physically impossible) therefore the only problem is for you to understand why you have got it wrong.

And its for that reason that I asked the question in the first place.

 

No. The ball is where it started from and no useful energy has been extracted - however you have spent 20 joules

How much energy did I input to the ball to get 100J gravitational potential energy?

Edited September 13, 2016 by Maximillian

[Strange]

Then when it gets to the top what is its gravitational potential energy?

 

The same as it was when you started. In other words, you have expended 20 J to get no change in energy.

 

 

Yes you get the 100J back but how much energy did you input to acquire the 100J at the rise?

 

-100J + 20J = a net loss of 20J

 

 

How much energy did I input to the ball to get 100J gravitational potential energy?

 

-100J + 20J = a net loss of 20J

[swansont]

Then when it gets to the top what is its gravitational potential energy?

 

Yes you get the 100J back but how much energy did you input to acquire the 100J at the rise?

 

 

How much energy did I input to the ball to get 100J gravitational potential energy?

That's not the issue. That 100J is not extra energy. The PE went down by 100 J, and then went back up. You haven't extracted any energy from the system. The ball dropped, and then it came back — it hasn't changed potential energy at the end of the cycle. You can't just look at half the cycle.

 

Where was the 100 J when the ball dropped? 90J was in the water column's increased height. 10J was lost.

 

To extract energy only from the rise of the ball requires that it magically appear at the bottom of the column.

[Maximillian]

 

The same as it was when you started. In other words, you have expended 20 J to get no change in energy.

 

 

-100J + 20J = a net loss of 20J

 

 

-100J + 20J = a net loss of 20J

I can see we are coming to a conclusion.

Its obvious that when the ball returns to the surface it has covered a zero displacement, and from energy is equal to force times displacement you get zero energy with zero displacement. That's elementary physics.

What your not understanding is giving a body 20J and it rises to a height that has an energy more than I input.

To put it in simple words is imagine I kick a ball with 20J I will expect it to rise to a maximum height equal to 20J but weirdly it reaches the mountain top equal to 100J.

[Strange]

But you are not gaining any energy.

 

It is like you have taken a ball to the top of a hill and then you say to people: "look if I give it a small push it rolls down the hill and we can use that to generate more energy than the push I gave it." While ignoring the energy taken to carry the ball up the hill in the first place.

 

 

I can see we are coming to a conclusion.

 

Does that mean you now see where your error is?

[John Cuthber]

 

So are you saying bodies never posses gravitational potential energy?

Nonsense.

How can you come up with that as a reply to a post that calculates the gravitational potential energy of something as 900 Joules?

 

Did you actually read what I posted?

 

i showed you how to calculate the potential emery of a body that is immersed in water. Why don't you do that calculation with the ball you are "experimenting" on?

Then you will realise the mgh isn't the right answer.

[Maximillian]

But you are not gaining any energy.

 

It is like you have taken a ball to the top of a hill and then you say to people: "look if I give it a small push it rolls down the hill and we can use that to generate more energy than the push I gave it." While ignoring the energy taken to carry the ball up the hill in the first place.

 

 

Does that mean you now see where your error is?

This is getting harder to explain than I thought but I will try to be clear as possible.

If this can work in water it means it works in air.

Now imagine the ball is lying on the Earth's surface I give it a pressure potential energy of 20J and it rises to a gravitational potential energy of 100J.

With the energy I input I expected it to rise to a height not more than 2m but it reaches 10m above the Earth.

Isn't that violating energy conservation?

[Strange]

This is getting harder to explain than I thought but I will try to be clear as possible.

 

 

Because you are unable to see where your error is. (And it is you that must be wrong, not anyone else. For obvious reasons.)

 

 

 

Now imagine the ball is lying on the Earth's surface I give it a pressure potential energy of 20J and it rises to a gravitational potential energy of 100J.

With the energy I input I expected it to rise to a height not more than 2m but it reaches 10m above the Earth.

Isn't that violating energy conservation?

 

No.

 

For one thing, this is a made up thought experiment that does not appear to be physically realistic. What makes the ball rise to 10m? Magic?

[Maximillian]

 

 

Because you are unable to see where your error is. (And it is you that must be wrong, not anyone else. For obvious reasons.)

 

 

No.

 

For one thing, this is a made up thought experiment that does not appear to be physically realistic. What makes the ball rise to 10m? Magic?

That's what exactly is beating my understanding.