I'm confused! I wonder if there is a provable answer?

BR-549

Joined Sep 22, 2013
4,928
I have not done it with a super magnet. But I have dropped magnets thru wire coils.

I didn't realized that I hijacked the thread. Pardon me.
 

Thread Starter

recklessrog

Joined May 23, 2013
985
You got it.
Hi Tony, that part was not intended as a question as it is well understood what is happening between the magnet and copper tube. The part in question being the use of gravity as the motive force acting on the magnet to cause it to fall. :)
 
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Thread Starter

recklessrog

Joined May 23, 2013
985
Just to be clear, and I appreciate all the comments, Let us for a moment consider this........ Lets do the experiment in a vacuum, or have good clearance between the magnet and tube walls so any air compressive effect can be disregarded. The only thing of consequence causing the motion of the magnet to travel through the tube being gravity.
So, if we reverse the experiment, and apply a force to move the magnet from the bottom of the tube through to the top at the same rate of acceleration that gravity would in the downward direction, (32 ft per sec per sec) the amount of "work" required is greater than if the tube were not there. This because of the braking effect caused by the magnetic interaction. That "work" has been provided buy imparting energy by some means to lift it. If for instance, it where lifted by a cable to a petrol powered winch, the the fuel used by the winch is consumed to provide the work.
This brings me back to the question posed by the young chap. Gravity has acted on the magnet causing it to fall, magnetic interaction has "opposed" the rate of acceleration and slowed the fall, so "gravity has done "work" is that work energy "consumed"?
That was the part that I could not answer :)
 

WBahn

Joined Mar 31, 2012
33,072
Have you guys ever even done the experiment (magnet in a copper pipe)? If you have, you'll realize how stupid the conversation about air resistance is. It takes about 5 to 10 seconds for a good magnet to fall through a 1-foot pipe.

Your concern and prolonged hijacking of the thread to discuss air resistance on the magnet's fall through the pipe is like discussing how insects hitting your windshield lower fuel economy. If you guys don't have anything to add for the OP, sit quietly instead of hijacking.
Why do you feel the need to hijack the thread and appoint yourself as the TS's spokesman?

Yes, I have done the experiment. I have done it with a thick OFHC copper pipe cooled to liquid nitrogen temperature using a rare-earth magnet in which it took nearly two minutes to fall through the three foot pipe.

The TS is asking about how to account for the energy aspects of his observations and relay that to a bright, inquisitive ten year old. The discussion comes down to the conservation of energy and thus there is nothing wrong with discussing other places where the energy might end up, including what happens to it when the magnet is simply dropped in a vacuum or how the energy can get spread out among a variety of other places.

Those discussions with that bright young lad might well extend into either thought experiments or actual experiments to explore this. In which case, being able to at least discuss and then reasonably dismiss various factors could well enrich the experience.

If the TS feels that this amounts to hijacking his thread, then he can certainly say so.
 

WBahn

Joined Mar 31, 2012
33,072
Just to be clear, and I appreciate all the comments, Let us for a moment consider this........ Lets do the experiment in a vacuum, or have good clearance between the magnet and tube walls so any air compressive effect can be disregarded. The only thing of consequence causing the motion of the magnet to travel through the tube being gravity.
So, if we reverse the experiment, and apply a force to move the magnet from the bottom of the tube through to the top at the same rate of acceleration that gravity would in the downward direction, (32 ft per sec per sec) the amount of "work" required is greater than if the tube were not there. This because of the braking effect caused by the magnetic interaction. That "work" has been provided buy imparting energy by some means to lift it. If for instance, it where lifted by a cable to a petrol powered winch, the the fuel used by the winch is consumed to provide the work.
This brings me back to the question posed by the young chap. Gravity has acted on the magnet causing it to fall, magnetic interaction has "opposed" the rate of acceleration and slowed the fall, so "gravity has done "work" is that work energy "consumed"?
That was the part that I could not answer :)
The "work" done by gravity in either case is identical. Work is a force acting through a distance. In either case, the force of gravity on the magnet is the same and the distance through which it fell is the same. The energy associated with that work has to appear in some other form. Previously it appeared not as "gravity", but as gravitational potential energy as a consequence of the magnet being at a certain location within a gravitational field. When it is at the lower level, it has less gravitational potential energy. That different MUST appear as some other form. It might appear as kinetic energy in the falling magnet, it might appear as electrical energy that gets stored in a battery or that gets converted to heat in a resistance, it might get converted to sound and heat when the magnet hits the floor, it might get converted to some combination of these. But the total energy associated with all of those other forms (almost certainly including some minor ones that got dismissed or overlooked completely) exactly adds up to the work done by gravity as the magnet fell through that same distance, with or without the metal tube.

So, yes, the "work energy" is "consumed". But it is consumed regardless of whether the tube is used or not. The work energy is just consumed in a different way.
 

MrChips

Joined Oct 2, 2009
35,017
This brings me back to the question posed by the young chap. Gravity has acted on the magnet causing it to fall, magnetic interaction has "opposed" the rate of acceleration and slowed the fall, so "gravity has done "work" is that work energy "consumed"?
That was the part that I could not answer :)
You are not paying attention and you're back to square one.
Gravity does not do any work. You have done the work already.

If you push a cannon ball up a hill and allow it to roll down the hill, gravity is not doing the work.
The work was already done by pushing the ball up the hill against gravity.

When you lift the magnet to a height of h, the energy E you have given the magnet is
E = mgh

where,
m = mass of the magnet
g = acceleration due to gravity
h = height

If you allow the magnet to free fall, the potential energy would be converted into kinetic energy. After falling the height h, all of the potential energy is lost. The kinetic energy gained would be:

E = ½ m x v x v

where, v = velocity of the magnet after falling height h.

From kinematics, we can calculate the velocity v using the formula:
v^2 = 2gh

Substituting into the kinetic energy equation:

E = ½ m x v x v = ½ m x 2gh = mgh

which equals the original potential energy that was imparted to the magnet in the first place.

The energy E = mgh was not from gravity but from the human who had to lift the magnet to a height h.
 

WBahn

Joined Mar 31, 2012
33,072
You are not paying attention and you're back to square one.
Gravity does not do any work. You have done the work already.

If you push a cannon ball up a hill and allow it to roll down the hill, gravity is not doing the work.
The work was already done by pushing the ball up the hill against gravity.
Work was done in both directions. Work was done AGAINST gravity as you moved the cannonball up the hill. Work was done BY gravity as the ball came rolling back down.

If the ball was at rest at the top of the hill and was rolling along at the bottom of the hill, then there was a force acting on that ball in order to effect the linear and rotational acceleration of the ball. Whatever provided that force did the work on the ball. That force was provided by gravity. Hence gravity did the work.

In moving the ball up the hill, the net work done on the ball was zero (assuming it was at rest at the bottom and at rest at the top). For discussion sake, consider that the ball is initially at rest. It is then placed in motion and accelerated to a constant speed, perhaps by a winch or someone picking it up and walking up the hill. It then moves up the hill at constant speed along a straight line. Finally, it is brought back to rest at the top. The net force acting on the ball was zero at all times except for a brief period initially when it was accelerated to its travelling speed and at the end when it was accelerated back to rest from its traveling speed. The net force on the ball in the initial phase is in the direction of motion, and so positive work was done by the provider of the force on the ball. The net force on the ball in the final phase is opposite the direction of motion so work was done on the provider of the force by the ball. These two cancel out. During the bulk of the motion, no net force was acting on the ball since it is moving in a straight line at the same speed. One way to account for it is that the person did work on the ball but, at the same time and in the exact same rate, the ball did work on the gravitational field (which indirectly means that it did work on the Earth).

Unlike a compressed spring that stores mechanical energy, the ball is not storing gravitational energy. It is not a property of state. The energy is stored in the gravitational field, just like the energy in a capacitor is stored in the electric field and the energy in an inductor is stored in the magnetic field.
 

MrChips

Joined Oct 2, 2009
35,017
If we put two masses in space and pull them apart, isn't the gravitational energy thus created analogous to the energy created by the stretching of a tension spring?

Just asking.
 

WBahn

Joined Mar 31, 2012
33,072
If we put two masses in space and pull them apart, isn't the gravitational energy thus created analogous to the energy created by the stretching of a tension spring?

Just asking.
Quite analogous, which emphasizes what I just said (and I'm glad you mentioned it because it might clear things up a bit). Consider the ball on a spring. You pull on the ball and it extends the spring. You did work on the ball and the ball did an exactly equal amount of work on the spring (again, assuming the ball starts and ends at rest). The BALL doesn't contain ANY additional energy since the net work done it is was zero. But YOU did work, so where did that energy end up? The energy is stored in the spring. When the ball is released, work is now done ON the ball BY the spring. The spring transfers energy that was stored in it to the ball.

The same with gravity. When you pull the ball up the energy is stored IN the gravitational field, which is the spring in the analogy. The BALL doesn't contain ANY additional energy since the net work done on it was zero. But YOU did work, so where did that energy end up? The energy is stored in the gravitational field. When the ball is released, work is done ON it BY the gravitational field. The gravitational field transfers energy that was stored in it to the ball.
 

MrChips

Joined Oct 2, 2009
35,017
Unlike a compressed spring that stores mechanical energy, the ball is not storing gravitational energy. It is not a property of state. The energy is stored in the gravitational field, just like the energy in a capacitor is stored in the electric field and the energy in an inductor is stored in the magnetic field.
Then why did you suggest that the gravitational field is unlike a compressed spring?
 

BR-549

Joined Sep 22, 2013
4,928
Two 6 ft high ramps. One at 30 degrees, one at 15 degrees. Identical masses...or use the same mass.

What's the difference in work done?
 

WBahn

Joined Mar 31, 2012
33,072
Then why did you suggest that the gravitational field is unlike a compressed spring?
Different situation. The post you are quoting here didn't have a ball -- it had just a spring (instead of a ball). The proper analogy needs both. We don't have JUST gravitational fields (in this context, since they need to interact with masses or other things). We have an object (the ball) interacting with a gravitational field (the spring).
 

WBahn

Joined Mar 31, 2012
33,072
Two 6 ft high ramps. One at 30 degrees, one at 15 degrees. Identical masses...or use the same mass.

What's the difference in work done?
Work done by what on what? Frictionless ramps? Are the masses at rest both before and after the experiment? If so, how are they brought to rest?

If you are asking about the work done by gravity on the mass as they slide down the ramps through a vertical change of six feet, the answer is that there is no difference.

Now, power is a different thing.
 

GopherT

Joined Nov 23, 2012
8,009
The "work" done by gravity in either case is identical. Work is a force acting through a distance
I could forgive thus phrasing because the quote marks are implying something - not clear but I won't push it. But this next sentence is the dumbest thing I have ever heard... The capitalization takes away all question of an implied meaning of work origin - you flat-out claim that gravity does work!?!?

Work was done in both directions. Work was done AGAINST gravity as you moved the cannonball up the hill. Work was done BY gravity as the ball came rolling back down.


Work = force x distance
Force = mass x acceleration
Therefore,
Work = mass x acceleration x distance.

Since gravity is an acceleration, claiming gravity is doing work ("work was done BY gravity") is comperavle to saying something like Distance is done BY Velocity. Ignorant.
 
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MrAl

Joined Jun 17, 2014
13,761
Hi,

I had a conversation with someone about this a little over two years ago but i presented the argument in a slightly different form where the actions were not all localized and thus using gravity as a means to perform some action might be thought of as using something up.

The discussion centers around trying to figure out if gravity can be "used up" somehow, and what i think is a better discussion for this is to think of systems that are relatively independent of each other. One way to do this is to think of a space ship traveling in a straight line through space until it comes close to a sizable planet.
When the space ship gets close, the gravitational field pulls on the ship and thus the speed increases. It gets a "free ride" for a while. Once the ship gets just past the planets center, it turns toward the planet slightly and continues on a curve in such a way that it does not get pulled in but flies off into empty space again, but now going faster.

The question now is that the ship started out at a certain speed and with no extra energy of it's own, but is now going faster, even though in a different direction, so where did that extra energy come from. It must have come from the "gravitational field" of the planet. So the obvious question now is, what was taken from that planet that supplied the energy to speed up the ship.

The solution he gave was unacceptable to me. His view was that the amount of whatever it was would be so small that it would not be practical. My view was that in this kind of discussion we want to know every last bit of what changed, no matter how small, because if we integrate that 'error' we will eventually find something has changed. For example, if we fly one such ship like that, no big deal, but what if we fly 10 such ships, or 100, or 1000, or 100 trillion ships, or even more.

From what we know about gravity, we probably cant figure this out. We can think about it a little more though.
If the ship gets pulled on by the planet, then the planet gets pulled on by the ship. The planet changes orbit slightly, which pulls on it's star. The star changes it's path in the universe, which pulls on other planets and changes their orbit slightly, and all these pull on other star systems. This all means the universe expands in a very slightly different way.

So now the question is, did anything get used up? If so, then the planet would have had to have gotten smaller. I dont think that's the case because of the view of space time and that i dont think there is any theory that talks about mass converting into gravity. There are actions and reactions, but some of them are lossless. It's only when an energy dissipator appears somewhere that we see a true loss where it gets converted to some other form.
In both systems (pipe/magnet and planet/ship) we can imagine no friction, but in the pipe/magnet we also have to imagine a perfect conductor (pipe) or else we have to consider losses due to current flow in the metal. If we consider a pipe with resistance, then there was heat dissipation, and so we have to ask where that energy came from. The only mover in this case was the gravity, so some of the movement caused heat and the only force present was gravity. The thing is, the view on this is that it took work to move the ball up to the top of the pipe and so the energy it looses on the way down is just the energy that it gained from the initial movement to the top. If part of that turns into heat, then it probably falls at a different rate than if there was no heat (perfect conductor).

The spacetime view suggests that gravity is not a force anyway it just acts like one to us. To understand this better though i think we'd have to understand nature down to very small scales like around 1e-35 meters (or something like that) which i dont think anyone understands exactly yet.

Feel free to add to this. I felt that the planet/ship discussion better isolates the parts of the experiment so i added that to the discussion.
 
The work is done by the potential energy which is stored in the magnet when you lift it up. Or in other words, the energy comes from your hand. And it has no relation with air resistance. Here is a good video which explains the effect at the end:

 

Thread Starter

recklessrog

Joined May 23, 2013
985
The work is done by the potential energy which is stored in the magnet when you lift it up. Or in other words, the energy comes from your hand. And it has no relation with air resistance. Here is a good video which explains the effect at the end:


That is exactly how I do the demonstration, thanks for posting the video :)
 

WBahn

Joined Mar 31, 2012
33,072
I could forgive thus phrasing because the quote marks are implying something - not clear but I won't push it. But this next sentence is the dumbest thing I have ever heard... The capitalization takes away all question of an implied meaning of work origin - you flat-out claim that gravity does work!?!?





Work = force x distance
Force = mass x acceleration
Therefore,
Work = mass x acceleration x distance.

Since gravity is an acceleration, claiming gravity is doing work ("work was done BY gravity") is comperavle to saying something like Distance is done BY Velocity. Ignorant.
Perhaps you should consider taking a high school physics course (although this is usually covered in middle school, so perhaps you need to go back further).

Gravity is NOT an acceleration, it is one of the four fundamental FORCES of physics (the so-called "forces-at-a-distance"), the others being the electromagnetic force, and the strong and weak nuclear forces). What you appear to be referring to is 'g' (often referred to as "little-g"), which is the strength of the gravitational field and has units of force per unit mass (the value is approximately 9.81 N/kg). An object located at a point where this is the force of the gravitational field experiences a FORCE due to gravity in direct proportion to its mass, independent of the magnitude and direction of that body's acceleration. IF this is the ONLY force acting on it, the acceleration of the object will then be 9.81 m/s², which is dimensionally equivalent with N/kg.

By Newton's Universal Law of Gravity, the FORCE of one body acting on another is

\(
F_{12} \; = \; \frac{Gm_1m_2}{r^2}
\)

The force acting on m1, per unit mass of m1, is independent of m1 and hence a property of the point is space where m1 is located relative to m2.

\(
g \; = \; \frac{F_{12}}{m_1} \; = \; \frac{Gm_2}{r^2}
\)

This force has units of force per unit mass.

An object sitting on a table is not accelerating (in the reference frame of the Earth, which is m2 in this case), but it is experiencing a force due to gravity. That force is the product of little-g and the mass of the object and has units of force (such as newtons or lbf).

This is directly analogous to the strength of an electric field, which has units of force per unit charge, and which interacts with charge to produce a force on that charge.
 

WBahn

Joined Mar 31, 2012
33,072
The work is done by the potential energy which is stored in the magnet when you lift it up. Or in other words, the energy comes from your hand. And it has no relation with air resistance. Here is a good video which explains the effect at the end:
While the energy comes from your hand, it is not stored in the magnet. It is stored in the gravitational field. Now, you might be asking how this is possible when the gravitational field is the same both before and after you lift the magnet up. The answer is that it ISN'T the same. The gravitational field is a function of ALL the mass in the universe and, in particular, its distribution relative to the point in question. Move any of it, and you change the gravitational field at all points. But moving objects within the field requires force and that force either does work ON the field, thus increasing the energy stored in it, or resists work done BY the field, thus extracting energy from it. We don't perceive a change in the gravitational field when we move the magnet only because the change is so small in comparison to the strength of the field.

Thinking of the energy as being stored in the magnet is generally a useful view, but it isn't what is actually happening. It's merely a useful shortcut.
 
While the energy comes from your hand, it is not stored in the magnet. It is stored in the gravitational field. Now, you might be asking how this is possible when the gravitational field is the same both before and after you lift the magnet up. The answer is that it ISN'T the same. The gravitational field is a function of ALL the mass in the universe and, in particular, its distribution relative to the point in question. Move any of it, and you change the gravitational field at all points. But moving objects within the field requires force and that force either does work ON the field, thus increasing the energy stored in it, or resists work done BY the field, thus extracting energy from it. We don't perceive a change in the gravitational field when we move the magnet only because the change is so small in comparison to the strength of the field.

Thinking of the energy as being stored in the magnet is generally a useful view, but it isn't what is actually happening. It's merely a useful shortcut.
Well, that is a more detailed description of what happens actually. But while describing simple phenomena with Newtonian physics that "storing of gravitational potential energy" is more commonly used. For example, we use Newtonian gravity instead of Einstein's curved space-time when we describe the dropping down of an apple, although curved space-time is more true and rightly explained.:)
 
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