Talk:Work (physics)

From Wikipedia, the free encyclopedia
Jump to navigation Jump to search

Wikipedia Version 1.0 Editorial Team  
WikiProject iconThis article has been reviewed by the Version 1.0 Editorial Team.
 ???  This article has not yet received a rating on the quality scale.
 ???  This article has not yet received a rating on the importance scale.
Note icon
This article is Uncategorized.
Note icon
This article was included in the 2006 Wikipedia CD Selection, or is a candidate for inclusion in the next version (see Work (physics) at Wikipedia for Schools). Please maintain high quality standards and, if possible, stick to GFDL-compatible images.

Conflict? Work/energy equivalence vs work against of done by gravity?[edit]

The article explains that if you apply a force over a distance, that's work. Apply the same force over twice the distance, and you've done twice the work; that seems pretty clear. So if you lift a weight a certain height above the ground, you've done a certain amount of work; if you lift it twice as high, you've done twice the work.
And the article says that if you drop an object a certain distance, gravity does a certain amount of work on the object; if you drop it twice the distance, gravity does twice as much work.
But then the article talks about the equivalence of work and energy.
If you drop the object twice a certain distance, it will be going twice as fast by the time it hits the ground (disregarding air resistance and the drop-off of gravity when the distance is large). That means it'll have four times the kinetic energy, because kinetic energy is 1/2mv2.
Can someone have the article sort this out? Or is it already done, and I'm just missing it? Uporządnicki (talk) 02:08, 23 January 2015 (UTC)

I think I've begun to realize the answer to my question, and I'm actually feeling a bit stupid now. I didn't want just to delete it because that would leave an odd trail of disembodied signatures. So I decided to be honest and just strike it out. In short, what I'm realizing is, dropping the body from twice the height won't double the speed. To double the speed (and thus, quadruple the kinetic energy), you have to drop it from such a height that it falls for twice the length of time. That's considerably more than twice the height. In fact, it turns out to be four times the height. Uporządnicki (talk) 02:57, 23 January 2015 (UTC)
I agree that dropping the body from twice the height won't double the speed. But it will double the kinetic energy which is related to the work done; and work is force times distance (or force times height).
You say that to quadruple the kinetic energy you have to drop the body from such a height that it falls for twice the time. I disagree because time plays no part in determining the work done or the change in kinetic energy. To quadruple the kinetic energy requires four times as much work done by the weight of the body, and work is force times distance, so it requires the body to be dropped from four times the height.
One part of the work-energy theorem states that the change in kinetic energy of a body is equal to the work done by the resultant force acting on that body. When a body is falling under the influence of no force other than its own weight, the resultant force on the body is its weight. Dolphin (t) 07:27, 23 January 2015 (UTC)
Four times the height IS twice the time. Yes, it's Force times the Distance. But you can express the distance as a function of the time: Distance fallen = acceleration x time2. To quadruple the kinetic energy (of ANY particular moving body), you double the speed. To double the speed of a free-falling (or ANY constantly accelerating) body, you double the time falling. When you double the time falling, you quadruple the distance.
In short, when you drop an object and it falls freely: Double TIME falling = double DISTANCE fallen FINAL SPEED = quadruple (i.e. double2) FINAL SPEED DISTANCE fallen = quadruple (i.e. double2) KINETIC ENERGY.
Notice, by the way: TRIPLE time falling = TRIPLE distance fallen final speed = NINE TIMES (i.e. triple2) final speed distance fallen = NINE TIMES (i.e. triple2) kinetic energy.Uporządnicki (talk) 15:28, 23 January 2015 (UTC)
The concept of work is very useful because of its wide applicability in all general situations, including applicability to bodies that are subject to varying forces and varying accelerations. Our article on work is written with the assumption that a body’s acceleration won’t be constant. Your statements are true but only in the special situation where a body is experiencing a constant acceleration.
If you find it easier to think about the time for which a force is acting you might find it more useful to think about the impulse given by the resultant force, causing a body to change its momentum. In this way your thoughts won’t be constrained to a constant acceleration.
In the situation where the resultant force is constant you could make the following statement: To double the speed (and thus double the momentum), you have to drop it so that it falls for twice the length of time. Dolphin (t) 05:22, 24 January 2015 (UTC)
Actually, I see I still had it garbled (although I was getting close; I think my error came in the translation of thought to keyboard). I've made the corrections in my last statement--presented in such a way as to show where I was wrong and what is corrected.
As for your comment, well, yes! I was particularly addressing the case of lifting/dropping a given body to/from a particular height or twice that height. So, yes, constant acceleration. It wasn't that I particularly preferred thinking in terms of the time; it's that I'd confused it with the point of how far the body is falling. I was thinking: 1) Drop the body twice as far. 2) It falls twice as long (DING, DING, DING, DING! ERROR! ERROR!). 3) It finishes twice as fast (ERROR IN CONSEQUENCE OF STEP 2), 4) thus, finishing with four times the kinetic energy (DITTO). And how could that be, if you've done twice the work lifting it that high in the first place? In fact, 2) it will fall √2 times as long, 3) finish √2 times as fast, 4) thus finishing with 2 times the kinetic energy. Problem solved. Uporządnicki (talk) 05:55, 24 January 2015 (UTC)

Work done by chain multiplied by number of links[edit]

If I do work by lifting an object with a chain, and I pull with one Newton of force on the chain over one meter, I've done one Joule of work on the first link of the chain. But the first link does the same amount of work on the second link, and so on. If there are 100 links then 100 Joules of work is done, even though I only did one Joule pulling on the chain. How does one explain what is lacking in this idea? What search terms could one use to find discussion of this on the web? Should we add an explanation of this to the article? Mindbuilder (talk) 20:22, 13 April 2017 (UTC)

The first and second links achieve an identical displacement and share forces of identical magnitude. However, the two forces are opposite in direction so the first link does positive work on the second; and the second does negative work on the first. Net work is zero. Dolphin (t) 22:41, 13 April 2017 (UTC)
I had thought of the negative work, however it left me unsatisfied. But as I think of it more, maybe that is as good an explanation as is needed. I had thought that we should probably put something about this in the article. At least one other person I know of had a problem with the idea. But now I'm leaning towards leaving it out. It seems like it may not cause much practical confusion to many people and may just be complication of little value. Mindbuilder (talk) 04:42, 14 April 2017 (UTC)
There are some oblique references to negative work in the article but I think there should be a more direct comment. I will find a citable source and add a suitable comment. Dolphin (t) 11:44, 14 April 2017 (UTC)
Some further thoughts: Forces between links of the chain should be regarded as internal forces. With a rigid body it is only external forces that are capable of altering the energy of the body. In your example it is only the weight of the object and the chain, and the force you apply, that are external forces. All the forces between links are internal. (A chain is not a rigid body but the principle remains valid.) WP doesn't have articles on internal and external forces but there is good information at Rigid body and Rigid body dynamics. Dolphin (t) 22:01, 14 April 2017 (UTC)
Internal forces. That's an interesting way of looking at the forces between links. I hadn't thought of that since the links are separate objects. But that still may be a valid way of looking at it. Especially since I was going to bring up the point that every layer of atoms in the chain does another joule of work on the next layer. And so with billions of layers of atoms in the chain between my pull and the load, that would be billions of joules of work done (and billions of joules of negative work). But my chain example was just the simplest I could easily describe. We've also got situations like: the water from behind a dam does a joule of work on a turbine, the turbine does a joule of work on an electric generator, the electric generator does a joule of work on the electrons in its wires, the electrons do a joule of work in a motor, etc. etc. Thus the original work of the water coming from behind the dam can be multiplied many times. It seems a bit of a stretch to say the turbine, generator, wires, motor, etc. are all internal components with only internal forces between them. I still haven't found a link to anyone discussing this issue of work being multiplied and/or canceled in this way. Mindbuilder (talk) 09:04, 15 April 2017 (UTC)
The concepts of internal and external forces are particularly important in any discussion of the momentum of a system of bodies - where a system is simply two or more bodies (that may or may not be in contact.) If two bodies share a pair of forces (as guaranteed by Newton's third law) then the impulse (force x time) given by those two forces must be equal in magnitude but opposite in direction, so they cancel. Consequently we have the Law of Conservation of Momentum saying if the resultant of all the external forces acting on a system is zero, the momentum of the system will be conserved; internal forces can be ignored because, as they always occur in pairs, they do not contribute to the resultant force on the system. If you look at Momentum and search for external force you will find four uses of this expression. Unfortunately, the expression internal force is not used in the article. Any suitable text book on the subject should introduce the concept of internal forces in its coverage of conservation of momentum.
Internal and external forces are less relevant to the study of work and energy but they are the key to my answer to your original question. A popular illustration of the concept involves a yacht bobbing in the water alongside a pier. A man with a barge pole is standing on the pier and he uses the barge pole to push on the mast of the yacht - he is able to move the yacht in whichever direction he chooses. The force the barge pole is applying to the mast is external to the yacht system. But when the man jumps on board the yacht and pushes on the mast he is unable to move the yacht one millimetre because now he is part of the yacht system and any force he applies is only an internal force and therefore incapable of influencing the all-important resultant force acting on the system.
You have written "It seems a bit of a stretch to say the turbine, etc. are all internal components ..." No, that isn't a stretch. If we define a system consisting of all these components the forces between them are all internal forces and so do not influence the resultant force on the system. We can see positive work being done on the system by the water from the dam; and positive (electrical) work being done on all the resistances that constitute the electrical load. The load does negative work on the system so the net work done on the system is zero. Zero work on the system must be the result because the energy of the system of turbine, alternator etc. doesn't change: the amount of water in the dam is reduced and the accumulated work/heat generated in all those electric resistances increases and matches the reduction of energy of the water in the dam; but the energy of the system of turbine, alternator etc. does not change until it is shut down and the kinetic energy of the spinning components falls to zero. All the forces within the system of turbine, alternator etc. are internal forces and so can be ignored; and all the work they do simply cancels because every joule of positive work is matched by a joule of negative work. If this were not true, the turbine and alternator would spin faster and faster as time passed, never reaching an equilibrium speed!
The system also does negative work on the water coming from the dam. This negative work causes the water to have much smaller kinetic energy than it would have in the absence of a turbine to slow it down. Dolphin (t) 13:24, 15 April 2017 (UTC)
It's a good point that we can consider the whole electrical system from turbine to motor a single system. So then of course the forces between the devices would be considered internal forces. Indeed, sometimes the whole earth is considered a single system. But I don't think anyone would claim that even when the whole earth is considered a single system, that all those motors and turbines and generators and such do no work, due to their forces being internal forces. They might do no work into or out of space, but they still do work on earth. So to clarify, do you think that at least 100 Joules of work is done in the hundred link chain (along with a corresponding amount of negative work)? Not 100 Joules into or out of the chain system, but internally within the chain system? Are we just saying that that 100 Joules of work should be ignored because it is canceled internally or are we actually saying that that 100 Joules of work does not happen, or what? I should be clear that I don't think for a second that the 100 Joules work implies any violation of conservation of energy or work coming from or being multiplied from nowhere. I understand it is canceled by the negative work. Mindbuilder (talk) 19:00, 15 April 2017 (UTC)
I wrote that the concepts of internal and external forces are always relevant in the study of momentum of a system, but less so in the study of work and energy. This is because internal forces can do work in a system which can draw on an energy source other than mechanical energy. A skyrocket uses combustion of its propellant to do work on the body of the rocket and on the escaping gas; the kinetic energy of the body and the gas ends up being substantial whereas it was zero before the propellant was lit. But the momentum of the body and the gas remains zero because momentum is a vector and velocity of the body is upwards and of the gas is downwards.
The Earth is very like a skyrocket. Internal forces can do positive work when reserves of chemical, nuclear, potential energy are depleted. There is nothing that can be done to give a non-zero impulse to the Earth and alter its momentum.. Dolphin (t) 21:51, 15 April 2017 (UTC)
So what about the chain? Does each link do a Joule of work on the next link for more than 100 Joules done in the chain (plus a corresponding amount of negative work)? I think I know your answer, but I'm not sure. Consider the whole chain a single system so that the forces between links are internal. Or draw system boundaries around each link for the work done between 100 separate systems, if you prefer. Mindbuilder (talk) 04:58, 16 April 2017 (UTC)

─────────────────────────One element of the work-energy theorem tells us that the change in kinetic energy of a rigid body is equal to the work done on that body by the resultant force acting on it. In your example of using a chain to raise an object, the kinetic energy of the object and chain remains constant; therefore we must conclude that zero work is being done by the resultant force. We can conclude that the resultant force must be zero; or alternatively that all the positive work done by all the internal forces within the chain and the object is exactly matched by negative work.

When learning about work we contemplate the work done by all manner of forces but the work-energy theorem tells us there are only two kinds of work that are mentioned in the laws of mechanics:

  • the work done by the resultant force acting on a body because it is equal to the change in kinetic energy of that body (positive work on a body is accompanied by an increase in kinetic energy.)
  • the work done by the weight of a body because it is equal to the change in potential energy of that body, multiplied by negative one (if the weight does positive work on the body it is accompanied by a decrease in potential energy; and if the weight does negative work it is accompanied by an increase in potential energy.)

The internal forces within the chain and the object being lifted can be considered to do work, positive and negative, but these elements of work cancel to zero and the laws of mechanics say nothing about them. Dolphin (t) 12:34, 16 April 2017 (UTC)

That section on the work energy theorem that you linked to, held the key idea that clarified this all for me. The key is the last sentence. It reads "Work transfers energy from one place to another or one form to another." That is one of the key ideas of what work is and does. I'm going to put it in the lead. It makes sense. By the law of conservation of energy, the energy I put in with my pull is not destroyed, it is just repeatedly transferred link to link down the chain. So of course each link does that one joule of work on the next, because that's what work does, transfer the energy from one link to the next. I'll bet this was in my physics textbook. I must have just forgotten it. Mindbuilder (talk) 20:53, 16 April 2017 (UTC) I've added it to the lead. So that answers my original question and settles this issue as far as I'm concerned. Mindbuilder (talk) 20:57, 16 April 2017 (UTC)
I'm glad you found it useful. I like your addition to the lead! Dolphin (t) 21:03, 16 April 2017 (UTC)

This article is badly written.[edit]

There are far more categories than are strictly necessary. All categories depicting formulae should be merged into a single formulae category, and then be subcategorized by type. C. J. T. T. Wilson (talk) 11:56, 24 January 2019 (UTC)