Talk:Work (physics)/Archives/2012/July

Move
I object to your moving Work (physics) to Mechanical work. I was actually just about to perform the reverse move when I discovered your previous action. There are other types of work, such as electrical work (energy lost by a system due to pushing electrons through a circuit). -Smack 00:16, 10 Jan 2004 (UTC)


 * Electrical work can be reduced to mechanical work, since it involves moving bodies (electrons or ions) against a force (electrical field). Additionally, the term "work" is not usually used outside of Newtonian mechanics - in other fields of physics, one generally speaks of energy rather than work.
 * That said, I don't feel strongly enough on the matter to make a fuss if you and others prefer "Work (physics)". Mkweise 23:10, 17 Feb 2004 (UTC)

In modern physics, the term energy is generally used to refer to all forms of energy, including mechanical work. -- I don't think this statement is correct. Work is not energy, work is the transfer of energy. Bensaccount 04:06, 23 May 2004 (UTC)


 * I think "Mechanical work" is the proper name for this article. That is what the lede starts out with. I would expect an article named "Work (physics)" to talk also about electrical work and work in thermodynamics since electricity and thermodynamics are subtopics of physics. Ulflund (talk) 17:02, 6 July 2012 (UTC)

The article is becoming an even greater mess
I revisited it after a while (about half a year), and now I am beginning to regret I ever made any contributions and proposals. Before, I was considering the possibility to make further improvements, such as conceptual clarification, consistent use of same symbols for the same physical quanitities, removal of trivia or matters better addressed elsewhere (e.g. in work-thermodynamics article) etc., but now it is hopeless.

Why? There is too much mess in there. And too little conceptual understanding. The second sentence in the lead (one of the good ones) states: "Only the component of a force in the direction of the movement of its point of application does work." How so? Because only this component (called tangential component Ft) changes the amount of velocity (i.e. speed) of a particle on which the force acts, thereby changing its kinetic energy, thereby doing work. And this is the underlying basis for the very definition of the "work of force".

Whereas another contributing author (Prof McCarthy) says: "I understand why the symbol Ft is proposed, but in my opinion this simply hides well-known and important calculations behind unnecessary notation." What calculations, they are mere trivia! And then he continues: "More importantly the integration of velocity is limited to one dimension. The only way to make this calculation in two and three dimensions is using the vector notation provided." Which is simply not true. Just as it is not true that the first law of thermodynamics can be interpreted to "equate work and energy" (if anything, it explains the difference between the two concepts).

Faced with such differences in opinion, I definitely give up on this article.--Ilevanat (talk) 02:52, 1 July 2012 (UTC)
 * And why are you telling us all this? Dolphin  ( t ) 03:01, 1 July 2012 (UTC)
 * Colleagues, as I am sure that you know a Wikipedia article, even on a subject as precise as Work(physics), is always a compromise. It is the nature of the activity.  Each of us chooses to try to achieve some level of quality, recognizing that one person's contribution is another person's mess.  In my opinion, there is no best way to summarize work generated by a force along a trajectory in the lead, though it seems that different editors would like to achieve this.  While I am willing to give my opinion, I have not reverted edits that are correct in principle.  If we want to move through a phase were Ft is preferred then so be it.  As for work equal change in energy, I cannot understand how equality defines how things are different, when it clearly identifies how things are the same.Prof McCarthy (talk) 17:38, 3 July 2012 (UTC)

First, thank you for the prompt responses, which came sooner than I expected. And next, to clarify my general motivation for contribution/discussion in some wiki articles, I was hoping (and still mildly am) that differing opinions at the talk page would lead to involvement of a greater number of people with reasonable understanding of the subject, and result in some kind of a "qualified consensus" - after all, this is one of the most basic concepts in physics and many of its applications. However, that did not happen.

One thing I never do (I am much more engaged in another wiki), I do not change contributions of other authors without some consensus at the talk page, preferably with the previous authors (unless the article was more or less a stub).

As for the question raised by Dolphin (t): When I look up a wiki article (looking for information on something I do not know), I always check its talk page, to see if there might be any information on its reliability. Most people probably do not, but still I feel I should leave a comment when appropriate, just in case (for those who understand what wiki is)... And, by the way, "On Wikipedia, the User who does the work gets to choose the notation" sounds reasonable in principle - unless there already was and still remains different notation for same quantities in the same article.

And I agree with general considerations by Prof McCarthy; and it was in this respect that I hoped some more people would join in, and express their preferences. As it is, we are in a stalemate, and I am certainly not going to re-write his contributions, which are generally correct at the formal level. Though, he does avoid attention to detail, e.g. in this talk he says "As for work equal change in energy...", wheras in the article he wrote "This equates work and energy.", and appears to have forgotten a discussion we had some number of paragraphs above. And has no comment on my explicite claim "Which is simply not true." regarding some his statements quoted above.--Ilevanat (talk) 00:25, 5 July 2012 (UTC)


 * Ok how about this, work has the same units as energy. It can transform into potential and kinetic energy and it can transform into heat energy, and potential and kinetic energy and heat energy can transform into work. Work is simply a different form of energy. Prof McCarthy (talk) 05:21, 5 July 2012 (UTC)

No, I do not think it would be a most appropriate improvement. Though I do understand your apparent intention to communicate to the novice in the field that work is almost the same thing as energy. This similarity, including the same units, is didactically a crucial notion to impress on a young student when he first encounters these physical quantities. When he does work on a system, it does not get lost, it becomes energy contained in the system. However, soon the student will learn that, besides this huge similarity, there are important differences:

Energy is a state of a system, whereas work is a process. Work is one of the processes of energy transfer between two systems, in addition to heat flow or matter exchange. Most energy in the universe will not (cannot) ever be exchanged in the form of work.

Therefore, I do not think that an ecyclopedia article should contain "loose" (not fully correct) and unnecessary statements such as "This equates work and energy." or "...and heat energy can transform into work. Work is simply a different form of energy."

After all, what would be missing if the last paragraph of the lead contained only the following:

Work is a process of energy exchange. When work is done to a system, the system's energy state changes by the same amount of the work input (if there are no other energy exchange processes). That is why work and energy are measured in the same units. The first law of thermodynamics describes how work and heat exchange affect energy of a system. Work-energy theorem describes how work done on a rigid body changes its kinetic energy.--Ilevanat (talk) 17:31, 5 July 2012 (UTC)


 * I have no trouble with you writing any of this if you wish. I understand what you are saying and why you express it in this way.  The problem is, however, that Work as it is defined in Physics is not a process, it is the result of a process.  The correct word for what you consider to be a process is Power.  The integration of Power from beginning to end results in a quantity of Work, which is exactly equal to the change in the Energy state of the system to which the Power is applied.  Now this is a distinction that I consider to be important because the consideration of power flow through a machine is how you calculate mechanical advantage, though most elementary analyses consider Work.  Work is incorrect for calculating mechanical advantage for exactly the reason we are discussing here.  Mechanical advantage is defined at instants throughout the process of movement, not as the integration over time like work is defined.  None-the-less, edit this page however you want. Prof McCarthy (talk) 21:36, 5 July 2012 (UTC)

The good thing is that we both try to understand the point of view of the other; and that we do not insist on imposing own point of view into the article. The bad thing is that we cannot quite reach an agreement ...

Looking back at the paragraph I proposed, I see now (after your remarks) that the first sentence should be corrected into "Doing work is a process of energy exchange." (or something like that). The result of the process is the work done, and the change in energy, and that is described in the next sentence (which is basically yours).

And you are right that definition of work refers to "the work done", i.e. to the result of the process. Still, that work is the change in energy - and not the energy itself, in the sense in which the two terms could be used interchangeably. But doing work takes time, and that is a process. Whereas power gives instantanous values expressing the rate at which the process proceeds, and of course any mechanical advantage that may change in time (during the process) must be defined based on power. So I do not see why: "None-the-less, edit this page however you want."--Ilevanat (talk) 23:38, 5 July 2012 (UTC)
 * You understand my concerns, please edit the page the way you want. Prof McCarthy (talk) 04:19, 6 July 2012 (UTC)

lede paragraph about thermodynamic work
The last paragraph of the lede currently states: "The first law of thermodynamics states that when work is done to a system (and no other energy is subtracted in other ways), the system's energy state changes by the same amount of the work input. This equates work and energy. In the case of rigid bodies, Newton's laws can be used to derive a similar relationship called the work-energy theorem." This is feels very missplaced since thermodynamic work is a more general concept than mechanical work and is not mentioned elsewhere in the article. The sentence equating work and energy is also wrong. If noone presents a good reason for keeping this paragraph I will remove it. Ulflund (talk) 17:30, 6 July 2012 (UTC)

Repetition in lede
Am I the only one who thinks the lede is a bit repetitative. The definition is repeated four times with slightly different wordings.
 * "...can be described as the product of..."
 * "...is defined as the product of..."
 * $$\text{Work} = \text{Force} \times \text{Distance} $$
 * $$W = Fd$$

I would prefere if the second (or second and third) repetition was removed and the more general definition for a variable force acting along a general curve was included in the end. Ulflund (talk) 18:10, 6 July 2012 (UTC)


 * Welcome, Ulflund! The article may need a fresh editor! I generally agree with your arguments, in particular about removing the "second and third" repetition. However, on other proposals, you may choose to consider recent history of the article and some talk sections above.--Ilevanat (talk) 22:58, 6 July 2012 (UTC)


 * Thanks Ilevanat. From the discussion above I see that you have put quite some time into finding a lede that everyone can agree on. Instead of changing it I have therefore written a proposal for a new lede section below. I hope we can discuss which of the changes you consider to be improvements so that those can be inserted. The main changes are the following. I would appreciate a thumbs up or down on each from anyone who has an opinion.


 * 1) I removed the sentance "Only the component of a force in the direction of the movement of its point of application does work.". I do not feel strongly about this, but argues that the start of the lede is not directly discussing vectors.
 * 2) I insterted the sentence "The SI unit of work is the joule (J)." since I think units are important, and a corresponding sentence appears in the first paragraph of many articles about physical quantities.
 * 3) I removed the sentance "The work done on an object by an applied force is defined as the product of the force and the distance through which the object is moved (when the force is constant and the motion is in a straight line in the direction of the force)." since it says almost the same thing as the first sentence and exactly the same thing as the sentence following it.
 * 4) I removed $$\text{Work} = \text{Force} \times \text{Distance} $$. I do not feel strongly about this but don't see the point of it when it has just been said in words and a notation has just been defined to write it properly.
 * 5) I have shortened the paragraph starting with "For example, if a force of 10 newtons ..." slightly. I am ready to shorten it further if others agree. Since this paragraph seems to be devoted to the reader who is not very familiar with multiplication I have linked the word "product" in the first sentence.
 * 6) I have rewritten the paragraph about thermodynamic work so that it focuses on mechanical work and its relation to energy. I think this is important to mention since this distinction often is not very clear. At the end of this paragraph I have added one sentence about electrical work and one about thermodynamic work. If you do not think these are appropriate they can be removed.
 * 7) I have inserted a paragraph giving the generalization with vector notation and ending with the general definition. In the discussion above I see that this have been a part of the lede before but removed because it was too technical. Too mee it doesn't seem like a consensus was reached so I raise the question again. My argument for including the general definition is that it in my opinion is the single most important part of the article and that since it is short enough to be expressed in one sentence it deserves to be in the lede.
 * 8) I would like to move the article to "Mechanical work". I do not understand Smacks argument for "work (physics)" in Talk:Work_(physics). Ulflund (talk) 14:27, 7 July 2012 (UTC)

Mechanical work
In physics, mechanical work is a scalar quantity that can be described as the product of a force and the distance through which it acts in the direction of the force. The term work was first coined in 1826 by the French mathematician Gaspard-Gustave Coriolis. The SI unit of work is the joule (J).

If a constant force of magnitude F acts on a point that moves a distance d in the direction of the force, then the work W done by this force is calculated as
 * $$W = Fd.$$

For example, if a force of 10 newtons (F = 10 N) acts along a path of 2 metres (d = 2 m), it will do work W = (10 N)(2 m) = 20 N m = 20 J. If two loads are lifted one story up, twice as much work is done as in lifting one load the same distance, as the force required to lift twice the weight is twice as much. Similarly, if a load is lifted two stories instead of one story, twice as much work is done because the distance is twice as large.

The concepts of work and energy are closely related. According to the work-energy theorem the work done by all forces acting on a rigid body is equal to that body's change in kinetic energy. Work is a transfer of energy between systems mediated by a force. Work thus has the same physical dimension and units as energy. Electrical work is a special case of mechanical work where an electric field is excerting a force on a charge. In thermodynamics work is a generalization of mechanical work encompassing all energy transfer except heat.

In the more general case where the constant force vector F is not parallel with the displacement vector d the definition above is generalized with the dot product W = F&middot;d = Fdcosθ, where θ is the angle between the vectors. For a variable force the work must instead be calculated with an integral over the curve of motion
 * $$W = \int \mathbf{F} \cdot \text{d}\mathbf{x}.$$



Ulflund (talk) 14:27, 7 July 2012 (UTC)


 * You certainly have a number of interesting proposals. And your cautious approach, which takes into account the history of the article, exemplifies what should be a true wiki procedure (or at least I believe so). But let me proceed to specific issues:


 * I feel strongly against incorrect definitions, even in the lead. Admittedly, the second sentence, which you would remove, may be a bit too technical. But it may be replaced by something "milder", such as ...through which it acts, "provided that the point on which it acts moves in the direction of force". Or the previous formulation should be kept as it was. (New formulation, however, would require general definition at the end of the lead.)


 * I certainly feel no need for the expression $$\text{Work} = \text{Force} \times \text{Distance} $$. Perhaps the load-lifting example may be kept, but it is a bit clumsy.


 * The paragraph on work-energy relationship definitely deserves a special accent. In the view of other elementary considerations, it may deserve an introductory sentence such as "The concepts of work and energy are closely related" (or something like that). And please be careful about formulation of the work-energy theorem: it should say "the work done by all forces acting on a rigid body"... (you may take a look at my article on "Net force" that was vandalized by an "ivy league somebody" - where I might have overreacted...).


 * In principle, I have no serious objections to the last paragraph, though details could be better tuned to the rest of the article (notation etc) - or vice versa. Here we come to the issue of the article as a whole.


 * As for work (physics) or mechanical work, I am not sure about your proposal. One needs to consider what the readers will look for, and the dissambiguation page on work, even perhaps the other languages (Swedish and Russian are "mechanical", German and many others are "work (physics)", French has a nice way out by "work of force" etc.)--Ilevanat (talk) 00:35, 8 July 2012 (UTC)


 * Thanks for the comments. I have made a few changes to my proposed lead above to reflect your comments.
 * I added "in the direction of the force" to the end of the first sentence to make it more correct. I'm not sure which description is easiest to understand, taking the component of the displacement vector in the direction of the force or vice versa. They are equivalent and this I found easier to write consicely. What do you think?
 * I added your sentence "The concepts of work and energy are closely related.". Of cource the work-energy theorem is only valid for a rigid body and should include torque. I have corrected this mistake above. In my defence I see that the erronous formulation with the net force is also used later in the article.
 * I changed the integration variable to x to be more coherent with the rest of the article. Do you think notation for the curve should be introduced here? You are right that the rest of the article needs a lot of work as well. I started doing some improvements but then realized the whole article should be restructured. I will try to find time for this in the future but don't want to promise anything. Ulflund (talk) 07:57, 8 July 2012 (UTC)


 * I agree that the edit you propose would be an improvement to the present article: "light" introductory sections of the lead (where "in the direction of the force" may not be fully uderstood by a novice reader, but it will not perplex him/her) - and a full general definition at the end of the lead, to which I would add the subscript "C" to the integral symbol, and an explanation below the integral, something like this: "where C denotes the curve along which the line integral is taken, and x denotes the position of the force application point along that curve" (I do not see any other reference to the application point in the lead, and this one would suffice).


 * But the overall outcome will very much depend on editting the rest of the article, though there is no hurry there. However, the change to "Mechanical work" does require simultaneous additional interventions in the section of "Work" page reproduced below:

Physics
 * Work (physics), the amount of work done by, or energy transferred by, a force acting through a distance
 * Work (electrical), the work done by, or energy transferred by, a force from an electric field acting on a charge though a distance
 * Work (thermodynamics), the quantity of total energy transferred from one system to another, not counting energy transferred by heat
 * Mechanical work, a subset of work (physics) with the connotation that the force doing work is a "mechanical," pressure, or contact-type force

Clearly, new first line should be some combination of the present first and last lines. And any search for "work physics" should redirect to this section. Good luck!--Ilevanat (talk) 14:31, 8 July 2012 (UTC)


 * I have now inserted the new version of the lead into the article including a version of your last suggestion. I will now start working on the rest of the article and think about the name of it later. Ulflund (talk) 22:32, 8 July 2012 (UTC)


 * Good job, and a reasonable decission. By the way, looking back at the lead, it does mention all aspects of work in physics that are mentioned at the "Work" page. So, there may be no need for any changes there, or in this article title. Except, perhaps (in that case), for a minor additional specification in the sentence preceeding the integral, something like this: "Most generally, for a variable force the mechanical work must instead be calculated with the integral"--Ilevanat (talk) 23:39, 8 July 2012 (UTC)

New structure
I have now corrected most of the false statements in the article, and would like to improve on the structure. This is not what I am best at, so I appreciate any comments. My sugestion for new sections would be I am also thinking about moving the sentence about Coriolis in the lead to its own section History just after the lead. — Preceding unsigned comment added by Ulflund (talk • contribs) 10:21, 14 July 2012 (UTC)
 * Lead: As it is now.
 * Definition: Containing most of what is now in Work (physics). I'm not sure about the title for the section, but I don't like Mathematical calculation.
 * Zero work: A combination of the paragraph in Work (physics) on zero work and the section Work (physics).
 * Torque and rotation: With the same content as Work (physics), but also including the case where torque and rotation are not parallell.
 * Units: Same as current section Work (physics), but with the paragraph about heat removed.
 * Work-energy theorem: Same content as Work (physics) but with a correct proof also for the simple case (not requiring $$v_1=0$$).
 * Power: Same content as in Work (physics) but with the integrals and derivatives written out. Maybe this should instead be merged into the Definition section.
 * Frame of reference: Keep as is. Maybe write that the work-energy theorem only works in an inertial frame of reference unless fictitious forces are included.


 * I fully agree that the lead should be followed by a title containing the word "definition" rather than "calculations". I would prefer something like "Definition and explanations" (since the general definition itself is now already mentioned in the lead). And that "Units" should be moved downwards; and, for same reasons, a section on "History" (if it is written) may be better placed downwards.


 * However, our views on the contents of this "Definition..." section may be quite different. For more than a year now (with some breaks) I have been trying to promote a concept which would provide motivation for the work of force definition, and received no support (though no counter arguments were given). Why only tangential component of force does work, why only the movement of its application point (rather than of center of mass)... etc. Clearly, in order that we may have work-energy theorem (still incorectly formulated in some instances in the article, it should be "the work of all forces acting on a rigid body..." or "the summ of works of all forces...", and "net force" is incorrect for extended body, and "net work" would need to be defined, and why would anybody reasonable need to use any of these terms?).


 * I still do not see why the role of the tangential component and the application point should not be described in the context of kinetic energy change. And why the work-energy theorem should have so many repetitious and unpalatable lines. And why dot product properties should be so fascinating to people who try to explain physics of work and energy: math is just a tool... and I am not going to take part in any different approach.--Ilevanat (talk) 00:24, 15 July 2012 (UTC)


 * I do not think that our views on the contents of the "Definition and explanations" section are different (at least not from what I hear so far). I have not pushed for introducing the "tangential component of the force" because to me the scalar product says just that. Now that you mention it I do agree that using this concept will probably make the article easier to understand for those not too familiar with the dot product. Reading the discussion above I would say that you have gotten support. Prof McCarthy has ended several replys with "Please make the changes you want", which I would take as encurraging. Now you have also my support, so please make the article easier to understand using the tangential component of the force (as long as you also keep the dot product definition). Ulflund (talk) 07:58, 16 July 2012 (UTC)

Work and kinetic energy
The derivation of work equal change in kinetic energy provided in the Work and kinetic energy section does not make clear that this applies to a system modeled as a particle moving under the action of a constant force. This is an important constraint on the force, because Work is a sum of power over a duration of time. If the force changes during this time, then the velocity changes, and the derivation of both work and change in kinetic energy are incorrect. Prof McCarthy (talk) 16:14, 15 July 2012 (UTC)
 * The situation with a variable force appears to me to be adequately covered at Work (physics). Do you see any problem with this coverage? Dolphin  ( t ) 22:39, 15 July 2012 (UTC)


 * It is simply a matter of clearly stating the assumption of linear motion under a constant force in the introduction to the derivation, and perhaps mentioning the a more general derivation is provided later in the article. Prof McCarthy (talk) 15:55, 16 July 2012 (UTC)
 * I agree with you.
 * I saw what I believe was a significant error in the derivation, so I have changed it. Previously, the derivation relied on the assumption that displacement d was: $$d = \tfrac{1}{2} at^2$$. This is only true when the initial velocity is zero. In deriving the work-energy theorem we don't want to be burdened by the constraint that initial velocity (and therefore initial kinetic energy) is zero. I have changed the derivation - see my diff. It now uses the following equations:
 * $$(v_2)^2 = (v_1)^2 + 2ad$$
 * and therefore
 * $$d = \frac{(v_2)^2 - (v_1)^2}{2a}$$
 * Dolphin ( t ) 01:35, 17 July 2012 (UTC)

I have now amended the section in an attempt to clarify that the net force must be constant in both magnitude and direction; and it must also act parallel to the velocity. If this is the case the resulting acceleration will also be constant in both magnitude and direction; and will act parallel to the velocity. See my diff. Dolphin ( t ) 06:31, 17 July 2012 (UTC)


 * I had noticed that error as well and was planning on fixing it. Good that you did this though, Dolphin. I didn't know the equation $$v_2^2 = v_1^2 + 2ad$$, so a link to e.g. Equations_of_motion would probably be good. As Ilevanat has mentioned it should also be stated that this is only valid when there is no net torque, or the derivation can be done for a particle instead of a rigid body. I don't see the reason for all the parantheses around $$v_1$$ and $$v_2$$. I will fix these things eventually if noone else does. Ulflund (talk) 15:14, 17 July 2012 (UTC)
 * Thanks. I only used the parentheses because it worked that way. Seeing it can be made to work without them I have no objection to them being removed. Dolphin  ( t ) 02:27, 18 July 2012 (UTC)
 * I have removed the redundant parentheses - see the diff. Dolphin  ( t ) 03:56, 18 July 2012 (UTC)

One dimensional case
These derivations are interesting, and I offer the following only because it makes clear what assumptions are necessary. Starting with Newton's Law for movement in one-dimension, F=ma, we can compute the power at any instant to be
 * $$Fv = mav.$$

Integrating this power from time t1 to time t2 we obtain the work
 * $$W=\int_{t_1}^{t_2}Fvdt = \int_{t_1}^{t_2}mavdt.$$

The right side of this equation simplifies to be the change in kinetic energy
 * $$\int_{t_1}^{t_2}mavdt = \int_{t_1}^{t_2}m\frac{d}{dt}(\frac{v^2}{2}) dt = \frac{mv^2(t_2)}{2} - \frac{mv^2(t_1)}{2} .$$

If the force F is constant, then the work is computed to be
 * $$W=\int_{t_1}^{t_2}Fvdt =\int_{t_1}^{t_2}F\frac{dr}{dt}dt =F(r(t_2)-r(t_1)).$$

or
 * $$W = F\Delta r = \Delta KE.$$

I realize that the use of integration is unfortunate, and that the current derivations finesse this for elementary understanding, so I am not recommending changes. I just regret that the fundamental relationship between power and work seems to be lost in the process. Prof McCarthy (talk) 15:43, 17 July 2012 (UTC)
 * Thanks for restoring the parenthesis that I carelessly removed!
 * Your derivation above leaves me with a couple of questions. You say F=ma is applicable to movement in one direction; is that simply because it isn't in vector notation? (I am happy to apply F=ma (where F and a are vectors) to all movements, including curved trajectories.) Is r simply displacement; or a special class of displacement? Dolphin  ( t ) 01:36, 19 July 2012 (UTC)
 * You are correct a scalar version of Newtons second law is restricted to straight-line movement. I believe this is preferred for this case.  A general vector formulation for curved paths is in a following section of the article.  No, there is nothing special about r, I used it to avoid defining velocity as dd/dt. Prof McCarthy (talk) 15:36, 19 July 2012 (UTC)

Work-energy theorem
A few moments ago the definition of the work-energy theorem was the following "According to the work-energy theorem, if one or more external forces act upon a rigid object, causing its kinetic energy to change from Ek1 to Ek2, then the work (W) done by the net force is equal to the change in kinetic energy. In equation form, this is:
 * $$\text{work} = \Delta E_k$$"

As Ilevanat has pointed out previously this statement is not correct since rotational kinetic energy added to the rigid object is not taken into account by the net force. The solution to this is to not use the net force but instead talk about the work done by all forces. I therefore rewrote the definition to

"The work-energy theorem, states that if one or more external forces act upon a rigid object, causing its kinetic energy to change from Ek1 to Ek2, then the work W done by these forces is equal to the change in kinetic energy. In equation form, this is:
 * $$W = \Delta E_k$$"

I also wanted to add a reference, but didn't find this general statement anywhere. My two textbooks on the subject both restict the theorem to particles, thereby removing the problem with rotational energy. One says "The work done by the net force on a particle equals the change in the particle's kinetic energy: The other one says essentially the same thing but in swedish. A google search gives some statements about particles, some about the net force and some more general. Here are my first 5: I like the last definition the best although I don't think it can pass as a reliable source. I will now include a compromise containing both definitions into the article like this
 * $$W_\text{tot}=K_2-K_1=\Delta K$$".
 * "The energy associated with the work done by the net force does not disappear after the net force is removed (or becomes zero), it is transformed into the Kinetic Energy of the body. We call this the Work-Energy Theorem."
 * "net work done by forces on a particle causes a change in the kinetic energy of the particle"
 * "The work done on an object by a net force equals the change in kinetic energy of the object"
 * "the net change in Kinetic Energy of a particle (system) is equivalent to the sum of work done by all types of external forces"
 * "The change of kinetic energy of a body is equal to the work of all the forces acting upon it."

The work-energy theorem states that
 * The change in kinetic energy of a body is equal to the work done by all forces acting upon it.

A less general formulation not taking rotational energy into acount is often used. This can be written as
 * The work W done by the net force on a particle equals the change in the particle's kinetic energy $$E_k$$,
 * $$W=\Delta E_k=\tfrac12mv_2^2-\tfrac12mv_1^2$$,
 * where $$v_1$$ and $$v_2$$ are the speeds of the particle before and after the change and m is its mass. Ulflund (talk) 13:46, 20 July 2012 (UTC)


 * I appreciate the detailed analysis, but there may be confusion regarding the assumptions underlying the derivation in this section of the article. This section focuses on the work-energy theorem for one dimensional movement of a body under a constant force.  This means that there is no rotational movement and no force other than a constant net force along the line of movement.  Prof McCarthy (talk) 14:29, 20 July 2012 (UTC)


 * Yes, the derivation is for a very simple case, and its first sentence is now clear on that. I still think we should present the general theorem. Ulflund (talk) 22:19, 20 July 2012 (UTC)


 * I have inserted "rigid" in front of "body" in the introductory definition; otherwise some work will be spent on deformation. As for this "straight line" thing: derivation given under the subtitle "Derivation" is fully applicable to particle moving along any curvilinear path, provided that force of constant magnitude is parallel to velocity (and "displacement" is changed to "path length"). A common example such force is sliding friction (though antiparallel to velocity).


 * Which reminds me of my unresolved dispute with Prof McCarthy, regarding a derivation I (repeatedly) reproduce below. He claimed: "More importantly the integration of velocity is limited to one dimension. The only way to make this calculation in two and three dimensions is using the vector notation provided."


 * $$W = \int_{t_1}^{t_2} \mathbf{F}\cdot \mathbf{v}dt =  \int_{t_1}^{t_2} F_t \,v dt = m \int_{t_1}^{t_2} v \,{dv \over dt}\,dt = m \int_{v_1}^{v_2} v\,dv = \tfrac12 m (v_2^2 - v_1^2) $$


 * I do not insist on using this derivation. But how can we seriously proceed without resolving that dispute?--Ilevanat (talk) 23:38, 20 July 2012 (UTC)
 * I am not aware of any dispute. But I can say that the derivation provided in this section relies fundamentally on one dimensional motion and a constant force.  Otherwise the power cannot be integrated to obtain the formula W= Fd that appears in this equation. Prof McCarthy (talk) 06:28, 21 July 2012 (UTC)


 * What I have been disputing is the meaning of your sentence quoted above (refering to the above derivation). It implies that the derivation is not fully general, but somehow "one dimensional". And that some other vector notation is needed to cover "other dimensions". Which is just not true for the line integral of work. Once you transform it into time integral of power, you have introduced parametrization of the curve (line), which covers all three dimensions.--Ilevanat (talk) 00:34, 22 July 2012 (UTC)


 * It is my understanding that this discussion is about the derivation presented in the article under the heading "Derivation" which only makes sense for one-dimentional motion under a constant force. The derivation of the Work-Energy relationship for a general three-dimensional trajectory is provided in the section under the heading "Integration for a curvilinear path." Prof McCarthy (talk) 16:59, 22 July 2012 (UTC)


 * In general, you and I have had a fairly good communication regarding the work article, and I would say a tolerant one. After all, the choice of approach and methodology in any presentation is largely a matter of personal preference... (about which, however, we did not explicitly agree).


 * But there remain missunderstandings about hard facts, which have nothing to do with preferences. For some reason, you seem to tend to effectiviely confuse the topics (from my point of view), or perhaps I was not clear enough, so let me spell them in more detail:


 * My account of "our dispute" immediately above was not about the section under the heading "Derivation" - which I thought was rather obvious. But let me now first address this section, as it may be helpful for a more general argument. Here, you seem to use the phrase "one-dimensional motion" in the meaning of "rectilinear motion". If there is any difference, please, tell me what it might be! (A curve, such as the path of the work line integral, is by definition a "one dimensional" geometrical object, though it may "curve" through all three spatial dimensions.) Regarding this section, I only noted (further above) that the derivation presented holds for any shape of the path (curve), provided that the force is tangential and of constnant magnitude, and that displacement is substituted by path length. If, for that purpose, we need a change of notation, and if the path length is denoted by "s", then the power is integrated to obtain the formula W= Fs, whatever you may think about that. And this is a very common case in circular motion (not to mention friction again), which has been used in the derivation of work of a constant torque.


 * But to return to the more general issue, which I believe is crucial to the presentation of the entire work article, my account of "our dispute" immediately above was related to your refutal of the use of the tangential component of force in the general derivation of the work-energy theorem. This is the last time I will mention it here (because at my age I am getting tired of this kind of evasive discussion), and I am quoting your statement in full: "I understand why the symbol Ft is proposed, but in my opinion this simply hides well-known and important calculations behind unnecessary notation. More importantly the integration of velocity is limited to one dimension. The only way to make this calculation in two and three dimensions is using the vector notation provided." And, unless you explicitly and specifically agree that these last statements (in bold type) are simply not true, I do not see how we can co-operate in editting the same article. (Please, do reconsider my contribution to this talk immediately above; and, perhaps, consult the wiki article on line integral, it is reasonably well written.)--Ilevanat (talk) 00:49, 23 July 2012 (UTC)


 * The derivation under the heading "Derivation" simplifies the calculation using the formula:
 * $$d = \frac{v_2^2 - v_1^2}{2a}.$$
 * This formula is only true for straight-line movement under constant acceleration. This is the reason that the derivation is only true for one-dimensional movement under constant force. As for the rest, if you provide the mathematical definition of the terms in your formulas, then I have no concerns.  Prof McCarthy (talk) 14:29, 23 July 2012 (UTC)


 * Thank you for a polite answer: my style above might have been a bit irritable. Though I would have prefered a more specific answer to politeness. So, let me consider your suggestion about "providing the mathematical definition of the terms in my formulas". If we assume that the reader should know or seek elsewhere in wiki the basics about vectors and derivatives, including definitions of velocity, speed and acceleration (e.g. enough to understand the present text of the article), the rest is easy to explain/define.


 * From F=ma it should be clear that a similar equation holds for any Cartesian component (no matter what the subscritp t might be), i.e.:
 * $$F_t=ma_t$$
 * Then $$\scriptstyle F_t $$ is defined as "the scalar tangential component", i.e. the component along the direction of velocity, which is equal to $$\scriptstyle F\cos\theta $$ and which appears in the power dot product
 * $$\mathbf{F}\cdot \mathbf{v} = F\cos\theta\,v = F_t \,v  $$
 * And this should be clear to anybody assumed to know what dot product is. There only remains to understand that
 * $$a_t=\frac{dv}{dt}$$
 * which is shown in the acceleration article. I do not see that any other mathematical definitions are needed in order to demonstrate full generality ("curvilinear") of the work-energy derivation based on $$\scriptstyle F_t $$.


 * However, in my opinion the roles of tangential components of force and acceleration (regarding change in speed) should be addressed early in explanation of the mechanical work formulas, and starting with "physical" arguments which would lead to formal mathematical expressions.


 * As for what I consider to be less relevant dispute about the derivation under the heading "Derivation", let me appologize for some hasty statements and correct myself: you are right that the derivation is applicable only to rectilinear motion. My statements would be correct only if F was not the net force (which it must be in this theorem, and there was my oversight), i.e. if there were an additional normal force to curve the path. And my motivation was to point out that same formulas for uniformly accelerated motion can also be used for curvilinear case (provided that acceleration actually meant the tangential acceleration).--Ilevanat (talk) 23:34, 24 July 2012 (UTC)

More on Work-Energy

 * The notation provided by our colleague above identifies velocity as a vector, but uses the symbol v that seems to denote velocity as a scalar. It would seem that v refers to the magnitude of the velocity vector, but I do not see this definition.  Then the time derivative of this magnitude of velocity yields the tangential component of acceleration, but this could use some elaboration.  I believe one would normally expect velocity to be a vector and its time derivative to be the acceleration vector.  Is it too much to provide these details? Prof McCarthy (talk) 05:35, 25 July 2012 (UTC)

Work-Energy reproduced from the article

 * Perhaps I can help. Denote the trajectory of a particle as X(t) with a force F acting on it.  Newton's second law provides a relationship between the force and the acceleration of the particle as
 * $$ \mathbf{F}=m\ddot{\mathbf{X}}, $$
 * where m is the mass of the particle.


 * The scalar product of each side of Newton's law with the velocity vector yields
 * $$ \mathbf{F}\cdot\dot{\mathbf{X}} = m\ddot{\mathbf{X}}\cdot\dot{\mathbf{X}},$$
 * which is integrated from the point X(t1) to the point X(t2) to obtain
 * $$ \int_{t_1}^{t_2} \mathbf{F}\cdot\dot{\mathbf{X}} dt = m\int_{t_1}^{t_2}\ddot{\mathbf{X}}\cdot\dot{\mathbf{X}}dt. $$


 * The right side of the first integral of Newton's equations can be simplified using the following identity
 * $$ \frac{1}{2}\frac{d}{dt}(\dot{\mathbf{X}}\cdot \dot{\mathbf{X}}) = \ddot{\mathbf{X}}\cdot\dot{\mathbf{X}}, $$
 * which is obtained using the product rule and which can be integrated explicitly to obtain the change in kinetic energy,
 * $$\Delta K = m\int_{t_1}^{t_2}\ddot{\mathbf{X}}\cdot\dot{\mathbf{X}}dt = \frac{m}{2}\int_{t_1}^{t_2}\frac{d}{dt}(\dot{\mathbf{X}}\cdot \dot{\mathbf{X}}) dt = \frac{m}{2}\dot{\mathbf{X}}\cdot \dot{\mathbf{X}}(t_2) - \frac{m}{2}\dot{\mathbf{X}}\cdot \dot{\mathbf{X}} (t_1), $$
 * where the kinetic energy of the particle is defined by the scalar quantity,
 * $$ K = \frac{m}{2}\dot{\mathbf{X}}\cdot \dot{\mathbf{X}}. $$ Prof McCarthy (talk) 06:00, 25 July 2012 (UTC)

Work-energy using tangential and normal components

 * Suppose one does not like this notation and prefers to resolve the velocity and acceleration vectors into tangential and normal components, such that
 * $$ \dot{\mathbf{X}}=v \mathbf{T},$$
 * and
 * $$\ddot{\mathbf{X}}=\dot{v}\mathbf{T} + v^2\kappa \mathbf{N}. $$
 * where
 * $$ v=|\dot{\mathbf{X}}|=\sqrt{\dot{\mathbf{X}}\cdot\dot{\mathbf{X}}}.$$
 * Then, the dot product of velocity with Newton's second law takes the form
 * $$ F_t v = m\dot{v}v.$$
 * Integrating both sides yields
 * $$ \int_{t_1}^{t_2} F_t v dt = m\int_{t_1}^{t_2}\dot{v}vdt. $$
 * The right side of this first integral of Newton's equations can be simplified using the following identity
 * $$ \frac{1}{2}\frac{d}{dt}v^2 = \dot{v}v, $$
 * which allows integration to define the change in kinetic energy,
 * $$\Delta K = m\int_{t_1}^{t_2}\dot{v}vdt = \frac{m}{2}\int_{t_1}^{t_2}\frac{d}{dt}v^2 dt = \frac{m}{2}v^2(t_2) - \frac{m}{2}v^2(t_1), $$
 * where the kinetic energy of the particle is defined by the scalar quantity,
 * $$ K = \frac{m}{2} v^2 = \frac{m}{2} \dot{\mathbf{X}}\cdot\dot{\mathbf{X}}. $$Prof McCarthy (talk) 06:17, 25 July 2012 (UTC)


 * The two derivations presented above are identical. The first derivation uses velocity and acceleration vectors explicitly while the second packages these vectors into the magnitude v of the velocity vector and its derivative.   The use of either formulation is a personal preference.  I prefer the explicit use of the velocity and acceleration vectors because it clearly distinguishes the straight line and the curvilinear cases. Prof McCarthy (talk) 06:37, 25 July 2012 (UTC)