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

Presenting concepts as simply as possible
while maintaining full correctness, has been my major concern in more than 30 years of teaching physics. Now, we may have differing views on what "simple" is, but there are common sense limits. For example, any reader who came to work article to learn something about work will very likely be scared away after merely scrolling down the page; or, if he could digest so many different symbols and notations, such reader would already know everything that this elementary article might contain.

So, as I have repeatedly argued, the most reasonable approach to defining and explaining general expressions for mechanical work would be to address very early the "tangential component of force" and its effect on the motion of an otherwise free particle: it changes its speed - and thereby its kinetic energy - and the work done is equal to that change in kinetic energy. A "physical" argument, not too difficult to follow, which should be rather helpful for understanding of any subsequent formal math. And then it would be easy to chip in another "physical" argument (here, or later in the work-energy theorem): tangential force gives tangential acceleration, which is the rate of change (i.e. time derivative) of speed. And none of these "physical" arguments need to substract from formal math and its correctness - they are just introductory explanations of "what is really going on" under cover of dot products and similar.

And as for mathematical correctness/completness of the curvilinear work-energy derivation in terms of tangential components, Prof McCarthy this time went too far from impartial justification of preferences. Quote: "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." First, he overlooks that the symbol for speed is introduced in the article subheading "Work-energy theorem" before any derivations begin:
 * $$W=\Delta E_k=\tfrac12mv_2^2-\tfrac12mv_1^2$$,
 * where $$v_1$$ and $$v_2$$ are the speeds of the particle before and after the change and m is its mass

Second, even if that were not the case, in the above mathematical description I presented power dot product as
 * $$\mathbf{F}\cdot \mathbf{v} = F\cos\theta\,v = F_t \,v  $$

which leaves no doubts about the speed symbol.

I could now proceed by discussing unnecessary or inadequately justified lines in the above derivation of "Work-energy using tangential and normal components", but I do not see the purpose (could we at least strike a bet for moivation?). In my wiev, Prof McCarthy prefers lengthy math formalism to simpler physically and intuitively oriented explanations (without loss of mathematical correctness). And he believes that readers will not mind different notations in the same article.

I irrevocably disagree. But will not force my view withou a broader consensus.--Ilevanat (talk) 02:09, 27 July 2012 (UTC)


 * I am sorry, but as I have said, I do not see the dispute that our colleague finds irrevocably disagreeable.  It is my opinion that appeals to intuition can be in error when considering more than the simplest cases, and I feel Work-Energy for a curvilinear path is not a simple case.  I have presented two different derivations that at a fundamental level are identical but which have differing advantages in the eye of the beholder.  One is already in the article.  The second is simply my attempt to provide a mathematically correct formulation of what I believe he is talking about.  He understands well what he means and feels it is easily understood by others, but, as perhaps I have made clear, it is not obvious to me.   Prof McCarthy (talk) 05:27, 27 July 2012 (UTC)


 * Ilevanat, please do the changes you propose. As I have said before I agree with you. My understanding of Prof McCarthy is that he does not disagree with you but do not exactly understand what you want to write in the article. I believe that will clear once you have written it. Ulflund (talk) 07:26, 27 July 2012 (UTC)

OK Ulflund, let me test your hypothesis in the following way: I shall add derivation of Work-energy (under the curvilinear case) using tangential acceleration (it will take me a few days); so we shall have two alternative derivations, which is not uncommon in wiki articles. And then I can reconsider other edits.

In the meantime, let me invite Prof McCarthy to analyze the text appearing immediately below his derivation in the article: "It is helpful to consider the special case when the trajectory of a body is a straight line. In this case, let the component of the force in the direction of the movement of the body be written as F and we have..." There is a flaw that can be removed, and the subsequent brief derivation reinterpreted. Or one can reconsider whether it should be there at all.--Ilevanat (talk) 01:15, 28 July 2012 (UTC)


 * I have examined that text as you ask, and I do not see a need to revise what is there. Prof McCarthy (talk) 05:00, 28 July 2012 (UTC)

The text obviously refers to net force (because it deals with the work-energy theorem), and addresses straight line motion. By saying "let the component of the force in the direction of the movement of the body be written as F" the author declares (implies, if you prefer) that the entire net force is not in the direction of movement. In that case, however, this would not be rectilinear motion because the normal component of the net force would change direction of velocity. So much about the flaw.--Ilevanat (talk) 01:06, 29 July 2012 (UTC)


 * If this is meant to explain that there is an error in what is written, then I am sorry I do not see any need for revisions. The movement of the point is a straight line which means the velocity is along that line and the component of the force in the direction of this line is specified to be constant.  These are the conditions that underlie the computation that is presented.  Prof McCarthy (talk) 06:10, 29 July 2012 (UTC)

Oh, my! This is getting worse and worse. I clearly specified (quoted) the disputed text appearing immediately below your "curvilinear" work-energy derivation in the article, and let me quote it again: "It is helpful to consider the special case when the trajectory of a body is a straight line. In this case, let the component of the force in the direction of the movement of the body be written as F and we have..." You now introduce an irrelevant invention that component of the force "is specified to be constant", which is not contained in the text, and the subsequent derivation uses integration clearly because this component is not assumed to be constant... I fail to see what might be the purpose of this irrelevant diversion you came up with.

But let us forget this diversion. The original issue regarding the disputed text was that it specifies the component of the force in the direction of the movement i.e. the tangential component. Thereby it implies that the net force may also have a normal component. Or do you think such wording does not imply that the net force may also have a normal component? Before we can proceed, would you care to give a simple and clear answer to this question?--Ilevanat (talk) 01:34, 30 July 2012 (UTC)


 * The case of linear movement under a constant force that follows the curvilinear discussion is what I assumed you were referring to. The wording is fine and there is no error.  As you state there may well be a component to the applied force that is normal to the linear movement, but this does not change the analysis.  I assume you are aware that this normal force would be supported by the constraint that ensures the linear movement.  Prof McCarthy (talk) 01:51, 30 July 2012 (UTC)

The force involved in derivation of the work-energy theorem for a particle is the net force acting on the particle. Any forces provided by any constraints are included in the net force. The symbol F refers to the net force; otherwise that would not be derivation of the work-energy theorem, nor would F=ma be true. There can be no "exempted" constraints that are not already included in F, and which would "support" or "ensure" anything acting somehow "from the outside" (as some kind of "addition" to F). F contains all forces acting on the particle, constraints or no constraints. And this is very elementary physics.

In order that the particle moves along a straight line, both the initial velocity and the net force F must be parallel to that line. Any statement about the component of F along that line reveals lack of elementary physics understanding and misleads the reader, because it implies that F may have a normal component. And that is an indisputable flaw.

When the flaw is corrected, the case is seen to be a minor generalization of the section above your curvilinear derivation. Therefore, in order to avoid further unproductive discussion, I have merged the rectilinear sections under new title. Please feel free to add whatever you think is missing, or improve it otherwise.--Ilevanat (talk) 01:19, 31 July 2012 (UTC)


 * I am sorry but there is no indisputable flaw. The fact that constraint forces do not generate work is a fundamental tool in the analysis of mechanical systems.  The normal force is eliminated in the calculation of power not because it does not exist, rather because the constraint that supports it does not generate work.  Consider the forces pushing a vehicle on a horizontal and straight track.  The forces on the vehicle include gravity and others that hold the vehicle on the track, but the work done on the vehicle does not include gravity or any other normal force because these forces are not in the direction of movement.  Only those components of the forces parallel to the direction of the track generate work as the vehicle moves.  Despite what you may think, there is no flaw in the discussion that is, or at least was, presented in this article.. Prof McCarthy (talk) 04:42, 31 July 2012 (UTC)


 * The derivation as it stood was not explicitly wrong (but a bit confused), but I think it is considerably better now. Prof McCarthy, if the net force had a component not parallel with the direction of motion the particle would not follow a straight path and it would not be rectilinear motion. In your example gravity and other forces that hold the vehicle on the track do not contribute to the net force since they cancel each other out.
 * The reference is to Hewitt (2006) which is not very easy to find. Could anyone having access to that book add the isbn using e.g. template:cite book. I don't think it is up to wikipedia to say what is easily derived (see WP:EDITORIAL). Maybe the first sentence can be changed from "The second version of the theorem can easily be derived for rectilinear motion." to "The work-energy theorem is most easily derived in the special case of rectilinear motion and for the second formulation stated above."? Ulflund (talk) 11:52, 31 July 2012 (UTC)


 * Thank you for this comment. I guess I have not appreciated the regular use of the term net force.  Clearly from your discussion above net force is determined by isolating the body from its environment and summing the various active and constraint forces to obtain the resultant force which is then used to formulate Newton's second law.  This is well-known to be a tedious, and often unnecessary, procedure in the study of the movement of mechanical systems, because it requires the determination of all the constraint forces in order to calculate the movement of the system.  The importance of the work-energy principle is that it provides an analytical formulation that specifically avoids this resolution of forces and eliminates constraint forces because they do not do work..  The work-energy principle specifically eliminates the need to evaluate the net force as it is called in this discussion. Requiring evaluation of this net force is not an elementary approach to describing the work-energy principle, and it is misleading because it assumes the ability to resolve and cancel opposing forces that is difficult in most mechanical systems.  Prof McCarthy (talk) 15:36, 31 July 2012 (UTC)


 * Prof McCarthy, I think I see your point but do not agree fully. The derivation as it stands is quite abstract and it might not be obvious that it applies to any rectilinear motion. The first sentence could be rewritten to make this clear. It now says "In the case the net force F is constant in both magnitude and direction, and parallel to the velocity of the particle, the particle is moving with constant acceleration a along a straight line:" and could be rewritten into "For a particle moving with constant acceleration a along a straight line the net force F will be constant and in the direction of motion:". This does not show that you can disregard constraint forces and and the normal component of the force doing work, but I think it clarifies that the special case used in the derivation is not purely theoretical.
 * The general form of the work-energy theorem is only valid for the net force so I can't see how that concept could be avoided. The work-energy principle can also be applied in the case of a single force doing work and other (constraint) forces not doing work, but I don't think that is where to start. That is theoretically more complicated and less general although it is often more practically usefull. A discussion about this and about constraint forces could improve the article but that would fit better in the Concept subsection in my opinion. Ulflund (talk) 16:42, 31 July 2012 (UTC)


 * The problem is the definition of net force. The work-energy principle is not formulated in terms of net force, but in terms of active forces. Prof McCarthy (talk) 16:58, 31 July 2012 (UTC)


 * In this article it is formulated in two ways, either with the work of all forces or the work of the net force. None of the definitions I found and presented in the section mention active forces although I see how this could sometimes be practical. If you want to introduce an alternative definition please provide reliable sources. Ulflund (talk) 17:35, 31 July 2012 (UTC)


 * I am not introducing a new definition, I am trying to provide the correct definition. The concepts of all forces and net force are both awkward and misleading in the formulation of the work-energy principle.  Maybe I can explain it as follows.  Consider the traditional free body diagram of a body, and for convenience lets not consider rotation, though the formulation is basically the same.  We can separate the forces acting on the body into applied forces F, which in the case of my vehicle would be the driving force and gravity, and constraint forces R, which would be the reaction at the road.  The resultant force on the body is clearly F+R, however if we compute the power associated with these forces we obtain (F+R).v = F.v, because constraint forces by definition are orthogonal to the movement.  Integration of this power, yields the work on the body.  Thus, the analysis of the work-energy principle allows us to focus on the forces applied to the system, often called active forces, and ignore the constraint forces.  This is of profound importance in the use of the work-energy principle in the analysis of mechanical systems and is the foundation of Lagrangian mechanics. I hope this makes sense.  Prof McCarthy (talk) 21:48, 31 July 2012 (UTC)

Of course the orthogonal forces do not contribute to work and kinetic energy change (let me restrict this discussion to particle motion in order to avoid questionably defined terms)! And it is for this reason that I insisted on discussion of normal and tangential components early in the definition of mechanical work...

However, the work-energy theorem is standardly formulated for the net force (or for "all forces" if it is not for a particle). Not for non-othogonal forces, active forces, tangential components or anything similar... The reason for this is that definition of net force is conceptually simple, regardless of practical difficulties one may have in determining net force in specific examples (classical mechanics is not concerned only with operating various mechanisms, although that is one of its most important applications).

Any reader of the work article, who understands the role of normal-tangential force components in mechanical work, will immediately grasp that the orthogonal constraints do not need to be included in the work-energy calculations. The net force can be reduced to the sum of active forces for that purpose. And, Prof McCarthy, you are welcome to elaborate on this if you wish. But you did not choose to do so in your contribution. Instead, you wrote

"consider a particle P that follows the trajectory 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."

That can be correct only if "a force F acting on it" is the net force (with all constraints included).

In the wiev of all above, Prof McCarthy, I am now getting really sick and tired of your "sophisticated" arguments. There is no way the two of us can work on the sane article.

As for the edits Ulflund proposed, I have no objections (at least until--Ilevanat (talk) 02:42, 1 August 2012 (UTC) i see the result).


 * I am not sure how to respond to these comments. It seems to me that the fact that constraint forces are orthogonal to the movement of a system, which allows the work-energy principle to be formulated in terms of the applied forces rather than net force or all forces, is a rather fundamental concept.  I am sorry to be slow to realize that this is the source of some misunderstanding.  Furthermore, it is not a matter that is addressed by resolving the components of forces in tangential and normal components along a trajectory, but perhaps this is getting tiresome.  Prof McCarthy (talk) 06:45, 1 August 2012 (UTC)


 * I think I can say even more firmly that the use of net force or all forces, which are identified by separating a body from its environment so that constraint forces are exposed and summed with the applied forces, is contrary to the very purpose of the work-energy formulation. The goal of this formulation is to analyze a mechanical system in its entirety without separation into free bodies.  This is done by using the fact that constraint forces do not contribute to work so their calculation can be avoided, and that the applied forces can be resolved relative to the movement of the system by computing power.  I have to say that I am grateful for the patience of our colleagues, whose persistence has made it possible to expose this potential misunderstanding at the core of the derivation of the work-energy principle. Prof McCarthy (talk) 07:14, 1 August 2012 (UTC)

The general purpose of the work-energy theorem is to reveal (and prove) whatever is contained in its statement (which is the purpose of any theorem). The general statement of the work-energy theorem is: "The change in kinetic energy of a rigid body is equal to the work done by all forces acting upon it." Not necessarily in that order. The theorem explains what this total work goes into: it goes into the change of kinetic energy. And, knowing that, one can contemplate that the very definition of mechanical work is devised so as to make this theorem true. (Not in the actual historical chronology; many physics concepts were initially limited to contemporary specifics; but in the end result.) So much for elementary physics.

As for various practical implementations, the theorem (and not only this one) can be used in such specific ways that may lead to different practical perception of its purpose. In mechanical systems with complicated constraints, any calculations based on work and power (and not necessarily on this theorem) will obviously eliminate these othogonal forces. From the physics point of view, this is because such forces do not contribute to work - and there is no need to invoke properties of dot product or invent some new fundamental "purpose of the work-energy formulation".

Ulflund, I am writing these lines for you, should you choose to remain with this article. As for Prof McCarthy, I can admire his politeness, and proficiency in producing various formulas, but that is all. I am to tired of his selective approach to topics discussed and pointless diversions, which all lead to supposed "misunderstandings". Perhaps he only reads last few lines of a discussion and neglects the rest. So I ask you in these last few lines: Can you reconcile his recent crucial statement (in this talk)
 * "The concepts of all forces and net force are both awkward and misleading in the formulation of the work-energy principle."

with his earlier use of net force in the theorem derivation (in the article):
 * "$$ \mathbf{F}=m\ddot{\mathbf{X}}$$"?

I can not. Nor did he care to explain.--Ilevanat (talk) 01:41, 2 August 2012 (UTC)


 * If this is your only concern, then let me put you at ease. I would be pleased to rewrite this equation as:
 * $$ \mathbf{F}+\mathbf{R} =m\ddot{\mathbf{X}},$$
 * where F represents the applied forces and R the constraint forces. The remainder of the derivation remains the same.  Prof McCarthy (talk) 05:01, 2 August 2012 (UTC)


 * Ilevanat, I agree with you and dislike Prof McCarthy's diversions, but I might be a bit more patient. It seems like Prof McCarthy also understands the contradiction between those statements, but I had really hoped that he would retract the former one.
 * Prof McCarthy, I do not think that this was Ilevanats only concern, but lets take this one first. Above I have stated the 6 different formulations of the work-energy theorem I first found. After reading these, can you honestly say that the work-energy theorem should not include either all forces or the net force? If you say yes, please provide a reference.
 * I am going to China in an hour and don't think wikipedia is availible there. Therefore I will not be able to respond within the until august 12. Prof McCarthy, I have not had time to read your latest contributions to the article, but will do that soon. Ulflund (talk) 08:42, 2 August 2012 (UTC)
 * Most books on Analytical Dynamics will explain that the work-energy principle involves the applied forces because the constraint forces do no work. Here are some that I have found that are accessible on-line:
 * 1. The Variational Principles of Mechanics.  On page 76 is the statement "The virtual work of reaction forces is always zero for any virtual displacement which is in harmony with the given kinematic constraints."
 * 2. Engineering Mechanics: Dynamics, which states on page 123 that one of the advantages of the work-energy principle is "only forces that do work need be considered, nonworking forces do not appear in the analysis."
 * 3. Here is a 1904 book by Whittaker, A treatise on the analytical dynamics of particles and rigid bodies. It states on page 37, "These are known as Lagrange's equations of motion.  It will be observed that the unknown reactions (e.g. constraints) do not enter into these equations.  The determination of these reactions is a separate branch of mechanics known as Kineto-statics: so we say that in Lagrange's equations the kineto-statical relations are altogether eliminated."
 * 4. Here is a 1998 book Classical Dynamics: A Contemporary Approach, which on page 65 has the quote: "Lagrange's equations have been derived from Newton's laws. They are in fact a restatement of Newton's laws written out in terms of appropriate variables that allow constraint forces to be eliminated from consideration." Prof McCarthy (talk) 06:25, 3 August 2012 (UTC)


 * None of the statements you have quotet give any definition of the work-energy theorem. References 1, 3, and 4 are about analytical mechanics which is not the subject of this article. Reference 2 defines the work-energy theorem as "The principle of work and kinetic energy (also known as the work-energy principle) states that the work done by all forces acting on a particle (the work of the resultant force) equals the change in the kinetic energy of the particle.", which agrees perfectly with the definition used. Can we agree on the definitions as stated in the article? Please answer with yes or no so that there is no confusion. I do not think you will be able to find a good reference giving any contradicting definition. Ulflund (talk) 10:47, 4 August 2012 (UTC)


 * Yes. Prof McCarthy (talk) 13:39, 4 August 2012 (UTC)

Recent changes in Integration for curvilinear path
This is a partly continuation of the discussion in the previous section. Prof McCarthy, I'm glad we can agree on the definition. I like some of your recent changes and dislike some. The new section about the moving car gives a good example but could be made simpler by directly using the work-energy theorem and the formula for work by a constant force in the direction of motion. The discussion about constraint forces in the begining could be longer and the second paragraph of the next section could be moved here. I will do this changes when I get the time (not the coming week) if you do not do them before me. I do not like the changes to the derivation. Spliting the forces into constraint forces and "applied forces" only complicates the derivation. How this discinction is made would have to be discussed and that would take us away from the topic. Do you agree? The discussion about constraint forces doing no work can just as well be after the derivation (making the derivation simpler). Ulflund (talk) 15:36, 4 August 2012 (UTC)
 * No. Prof McCarthy (talk) 16:02, 4 August 2012 (UTC)
 * At the risk of a distasteful diversion let me add that the distinction between applied forces and constraint forces is readily apparent to the point of being obvious. Applied forces are externally applied to the system and constraint forces are internal to the system.  Applied forces are usually explicitly defined such as gravity or a push or pull by a person, while constraint forces must be exposed by separating a component of the system to create a free-body.  It is the step of decomposing the system into free-bodies and identifying the constraint forces that is complicated and worth avoiding when possible. Prof McCarthy (talk) 17:46, 4 August 2012 (UTC)
 * A note in passing, to Ulflund. It is beyond my understanding that anybody would want to waste space (lines, formulas etc.) to demonstrate (by dot product, ha ha...) that normal forces do no work. This should have been clarified somewhere within the very definition of work (a normal force cannot affect the speed, because it is neither forward nor backward etc...). Then, of course, within the work-energy theorem context one could at will discuss its usefullness in applications regarding elimination of constraints from actual calculations.
 * Advantages of the work-energy theorem application extend a bit beyond elimination of constraints (though they are particularly important). Not only to other normal forces, but in the context of rigid bodies (invoked in the article) also to forces acting at some instantaneous pole of rotation (such as is static friction acting on the rolling ball),etc.
 * As for my "by dot product, ha ha" remark: dot product is convenient, but let us not overdo it. It is just a mathematical tool, and certainly not the reason why normal forces do no work. After all, dot product might not have been invented, and normal forces would nevertheless do no work! Therefore, this discussion may not really be about "diversions", but about real understanding of physical concepts. --Ilevanat (talk) 02:35, 6 August 2012 (UTC)

Revisions to article
I have made a large number of revisions to this article. Thanks to all of you for your patience. I hope this is found to be useful. Prof McCarthy (talk) 02:16, 17 August 2012 (UTC)

0 Parallel
The recent revisions to this article introduce the term 0 parallel, which seems to be an odd phrase. Could it be the editor means perpendicular? And I do not mean to be picky but is tauter a word? I would have thought that a string is either taut or not taut? Last but not least, is stuck to a slope correct? I think stuck implies some adhesive. Prof McCarthy (talk) 20:32, 21 August 2012 (UTC)

Work and energy
This section has been completely rewritten in a way that has introduced a number of errors. The relationship of work to energy is easily misunderstood and these revisions show why. While it may be that all energy can be represented as resulting from kinetic and potential energy in some way, this is far from being well-known and it is not the best way to present elementary material. It confuses the internal energy, work and heat transfer in general systems with the kinetic energy and potential energy of rigid body systems. This also confuses the notion of the work of forces that are derivable from a potential and the work of forces that are not derivable from a potential. I look forward to someone revising this to correct its errors, I am biased toward the original version. Prof McCarthy (talk) 15:52, 22 August 2012 (UTC)

Persistence of errors
The statement that work equals the negative change in potential energy is correct only for the work of forces that are not path dependent. Because the work of a force is generally path dependent this statement is wrong and misleading. Even worse is the statement that the sum of potential and kinetic energy is always zero, because this depends on how the system is defined and is not true in general. Even after several days, over 5000 views and 80 plus watchers these errors persist. Prof McCarthy (talk) 19:05, 26 August 2012 (UTC)