Talk:Work (physics)/Archives/2011/March

Work in a gravitational field
Suppose I lift a weight of mass m off the ground at a constant velocity. The total force on the weight is zero, my upward lifting force matches the downward gravitational force. So the work done on the weight is zero? Or is it mgh where h is the distance I have lifted it? PAR (talk) 18:41, 12 March 2011 (UTC)
 * The work-energy theorem tells us that the work done on a body equals its change in kinetic energy. The body is moving upwards at constant velocity, and therefore constant kinetic energy.  This tells us the change in kinetic energy has been zero, so the work done has also been zero.  Work done is the net force on the body multiplied by the distance travelled in the direction of the force.  The net force on the body is zero because weight acting downwards is exactly matched by the force applied upwards by your hands.
 * mgh is the increase in potential energy. The significance of this is that kinetic energy plus potential energy are called mechanical energy.  When the only forces acting on a body are conservative forces (gravity and springs) it is true to say that mechanical energy is conserved.  However, the force applied by your hands is not a conservative force so mechanical energy is not constant, and mgh is not relevant.  Dolphin  ( t ) 12:36, 24 March 2011 (UTC)


 * Thanks, I came to that same conclusion, but was unsure. Would you say the first sentence of the Gravitational potential article needs to be changed? PAR (talk) 16:13, 24 March 2011 (UTC)
 * I agree that something appears to be missing from the first sentence. The sentence would make more sense if it was stated that the only forces acting on the body must be conservative forces.  If a non-conservative force acts, such as friction or air resistance or a person's hand or thrust from a rocket, then the first sentence is not applicable.  Dolphin  ( t ) 21:35, 24 March 2011 (UTC)


 * I modified it. Could you check it? PAR (talk) 22:51, 24 March 2011 (UTC)


 * I don't agree that kinetic energy won't change. If the only forces acting on the body are conservative forces the kinetic energy will change as potential energy changes in order that mechanical energy is conserved.  Dolphin  ( t ) 06:10, 25 March 2011 (UTC)
 * I am having second thoughts about my post immediately above. Perhaps kinetic energy is not particularly relevant, and it can remain constant during the displacement, so long as the body doesn't arrive (at the point whose gravitational potential is being determined) with any significant kinetic energy.  Dolphin  ( t ) 11:52, 25 March 2011 (UTC)


 * LOL - I'm having second thoughts about mine. I don't think it matters what the kinetic energy is unless you are worrying about total work. The potential is the work done by the gravitational force - integral of F.dx, which is independent of the initial and final kinetic energies.

Frame of reference
Does anyone has a reference to support the section "Frame of reference". In my opinion, the displacement is between the body that applies a force and the one on which the force is applied. There is no consideration about a frame of reference. I suspect that this statement may come from AIEEE (Indian) books like http://books.google.com/books?id=N_vXN1IJRPQC&lpg=SA1-PA41&dq=physics%20work%20of%20force%20frame%20of%20reference&hl=fr&pg=SA1-PA41#v=onepage&q=physics%20work%20of%20force%20frame%20of%20reference&f=false This leads to other nonsenses like saying that friction can do positive work, http://en.wikipedia.org/wiki/Friction#Work_of_friction GuillaumeMillet (talk) 03:21, 24 March 2011 (UTC)


 * The current text is pretty horrible, isn't it? My College level introductory physics textbook taught me that the significance of mechanical work is its relationship with kinetic energy, as expressed in the work-energy theorem.  The important thing is that providing we choose an inertial reference frame, the choice of reference frame does not diminish the validity of the work-energy theorem.  In different reference frames the work done by the force will be different, and the change in kinetic energy will be different, but the work done will always equal the change of kinetic energy.
 * Intuitively, it would appear very odd to choose a reference frame in which friction does positive work, but such reference frames are just as legitimate as any other. The reason we can say such reference frames are legitimate is that, despite oddities like friction doing positive work, the work done still equals the change in kinetic energy so we can say the work-energy theorem remains valid.
 * If no-one gets to it before me I will look at refining the wording of the section on reference frames. I can always cite my physics textbook, Physics by Resnick and Halliday (1966), round about Section 8-3.  Dolphin  ( t ) 12:21, 24 March 2011 (UTC)
 * Done. Dolphin  ( t ) 05:05, 28 March 2011 (UTC)

Simplification of lede (discussion moved from Ashershow1's talk page)
On 22 March 2011, User:Ashershow1 added the following to the lede of this article: "In other words, the net work done on an object is equal to its change in kinetic energy. Note that the work in the work-energy theorem is the work done on an object by a net force. It is the algebraic sum of work done by all forces."

Hello Ashershow1. Thanks for your simpler explanation added to Work (physics). I have reverted your edit for a couple of reasons.
 * The introductory paragraphs (and perhaps the whole article) is written in terms of the work done by the net force. No mention is made of the net work or the algebraic sum of the work done by all the external forces.  I think your explanation was potentially confusing because it added new concepts - namely the idea of the work done individually by all external forces, and the algebraic sum of the work done by all these forces.  In my experience, introductory physics text books including Tipler, the one cited in our article, define work in terms of the work done by the resultant force (Wikipedia calls it net force), not the algebraic sum of the work done by all external forces acting on the body.
 * Secondly, your edit began In other words ... I think this expression is colloquial and not appropriate in an encyclopedia.

Regards, Dolphin  ( t ) 01:24, 23 March 2011 (UTC)


 * Thanks for the explanation of your revert. The reason I added a simpler explanation to the article was because I felt the lead was confusing, especially to someone who has little or no prior knowledge of physics.  The equation stated in the article; $$W = \Delta E_k = E_{k_2} - E_{k_1} = \tfrac12 m (v_2^2 - v_1^2) \,\!$$, can be simply stated: The net work done on an object is equal to its change in kinetic energy.  That is effectively what the equation represents.  I don't see any reason not to call a spade, a spade.  Perhaps "in other words" can be replaced with "in essence" or "essentially," but, frankly, many readers of this article will be high school students and will appreciate the use of a minor colloquialism to demonstrate a point.  Let me know what you think. --Ashershow1talk • contribs 01:42, 23 March 2011 (UTC)


 * The concept of mechanical work and its application to the work-energy theorem can be explained in either of two ways:
 * the work done by the net force acting on a body is equal to the change in kinetic energy of the body, or
 * the algebraic sum of the work done by all external forces acting on a body is the net work and is equal to the change in kinetic energy.


 * These are two significantly different ways of explaining the work-energy theorem. Any explanation of the subject should take one of these ways and stick to it, not mix the two together and assume they are the same.  After your recent edit to the article I felt the lead was presenting readers with a mixture of the two.
 * Wikipedia already has an article on net force but no article on net work so readers can't refer to an existing article to find an explanation of net work. (There is a link called net work but it is a redirect to Network and has nothing to do with physics.)
 * I agree with calling a spade a spade, but I don't think that is what your recent edit was doing. It was confusing two different ways of explaining the work-energy theorem.  I have no problem with a separate section of the article talking about the work done by individual forces (such as weight whose work is directly related to gravitational potential energy), or introducing the concept of net work, but it should be in a separate section, not in the lead.
 * I also agree that the lead to an article should be relatively simple so that it is accessible to newcomers to the subject, such as high school students, but remember that WP:NOTTEXTBOOK explains that Wikipedia is not a textbook. Cheers.  Dolphin  ( t ) 02:23, 23 March 2011 (UTC)


 * How about:
 * --Ashershow1talk • contribs 02:39, 23 March 2011 (UTC)


 * The cited source is Paul Tipler's Physics for Scientists and Engineers, page 138, 3rd edition. I don't have a copy but I understand it (and certainly my own College physics textbook) says:
 * the work done by the net force is equal to the change in kinetic energy
 * You are proposing another sentence following immediately afterwards that says:
 * In essence, the work done by all the forces acting on an object is equal to the change in kinetic energy
 * These two will appear contradictory until there is a separate explanation that the work done by the net force is in fact equal to the algebraic sum of the work done by all the forces.
 * The lead to the article should present an unambiguous explanation of the work-energy theorem, not a mixture of two. I don't particularly care which of the two explanations is presented, so long as it is uniform through the lead and is supported by a cited source.  Tipler is a suitable source for the net force explanation.  Do you have a source for the proposed algebraic sum of all forces explanation?  Dolphin  ( t ) 03:24, 23 March 2011 (UTC)


 * Yes, Paul Zitzewitz's Physics: Principles and Problems, page 220. --Ashershow1talk • contribs 18:20, 24 March 2011 (UTC)
 * I don't have access to a copy of Zitzewitz's book. Are you able to quote the relevant sentence or paragraph here on the Talk page?  Thanks. The reason I ask is because I can see some problems, albeit minor problems, with this approach so I am keen to see how Zitzewitz avoids these problems when using the above explanation.   Dolphin  ( t ) 12:00, 25 March 2011 (UTC)
 * The second paragraph on page 220 in Physics Principles and Problems:
 * --Ashershow1talk • contribs 20:27, 25 March 2011 (UTC)
 * --Ashershow1talk • contribs 20:27, 25 March 2011 (UTC)

Thanks for that. Textbooks on this subject devote a fair amount of space to explaining resultant force or net force. Presumably Zitzewitz does too. Does Zitzewitz explain net work, and assuming he does, what does he say about it? Dolphin ( t ) 23:37, 25 March 2011 (UTC)
 * Yes, on page 89 Zitzewitz explains (or attempts to explain) net force. He begins by stating Newton's first law:
 * Thanks for that. What does Zitzewitz say about net work?  Dolphin  ( t ) 11:15, 26 March 2011 (UTC)
 * Zitzewitz does not have a specific section about net work-- it appears to be an impromptu term used to describe "all work being done on an object." He does, however, explain how net work operates:
 * --Ashershow1talk • contribs 00:39, 27 March 2011 (UTC)
 * --Ashershow1talk • contribs 00:39, 27 March 2011 (UTC)
 * --Ashershow1talk • contribs 00:39, 27 March 2011 (UTC)


 * It looks to me that Zitzewitz gives two meanings to net work:
 * It is the algebraic sum of work done by all forces
 * It is the work done by the net force
 * Through the eyes of a student or a newcomer to mechanics these will appear to be two different meanings until there is some proof or explanation that the two are, in fact, the same. For this reason, I don't believe that introducing the notion of the net work into the lede is simplifying or clarifying.  It actually complicates the narrative by introducing a new term that hasn't previously been defined or explained, and for which Wikipedia has no article or section within an article.  In my view it would neither clarify nor simplify the lede, so it should not be added to the lede because it would not be appropriate to use it without the necessary definition, and it wouldn't be appropriate to insert the necessary definition into the lede.  I have no objection to a new section being added to Work (physics) to explain the concept of net work and show how it is relevant to the work-energy theorem.   Dolphin  ( t ) 08:03, 27 March 2011 (UTC)

A force can change both translational and rotational kinetic energy of a rigid body
and this total change will be equal to its work. Imagine striking a football ball, in the air, somewhere off its center, by a baseball bat. Therefore


 * $$W = \Delta E_k = E_{k_2} - E_{k_1} = \tfrac12 m (v_2^2 - v_1^2) \,\!$$

can be true only for particles and translational motion. So, if you talk about rigid bodies in general, better stay with a descriptive statement, such as "change of kinetic energy".

Many textbooks in English disregard the role of the point where the force acts on the body. For a more detailed approach you may want to consult e.g. Eshbach's Handbook of Engineering Fundamentals.--Ilevanat (talk) 20:07, 28 March 2011 (UTC)
 * I agree with Ilevanat that the explanation of work and work-energy theorem presented in our article is correct when applied to a particle, or to a rigid body where the force acts at the center of mass of the body.
 * The rotational analogue of force is torque $$\tau$$ (or moment of force.) The rotational analogue of displacement is angular displacement $$\theta$$, and velocity is angular velocity, $$\omega$$.  The rotational analogue of mass is polar moment of inertia, I.  Where the angular velocity and polar moment of inertia are significant, the work-energy theorem is:


 * $$\tau . \theta = \tfrac{1}{2} I (\omega_2^2 - \omega_1^2)$$
 * The concepts of work, kinetic energy and the work-energy theorem are introduced to physics courses during coverage of rectilinear motion. Angular motion is a more advanced concept and is invariably introduced after work, kinetic energy and the work-energy theorem.  I suggest we don't attempt to introduce the rotational analogues of work and the work-energy theorem in the lede of Work (physics).  Instead, it should be made clear that the lede applies only to particles, and rigid bodies in rectilinear motion.  It should also be made clear that the net force is applied at the center of mass of the rigid body.  Dolphin  ( t ) 22:04, 28 March 2011 (UTC)

It is easy to present the work-energy theorem in the lede simply and correctly without reference to rotation (I took the liberty). However, the issue of the rotational energy change is not as simple as suggested above (there is a thorough account in the rigid rotor article). And the question remains whether this article should be made more complete by considering the role of the point where the force acts on the body.--Ilevanat (talk) 16:34, 29 March 2011 (UTC)
 * Thanks for refining the lede to show that it only applies to translational motion. Feel free to add an extra section or two to describe the role of the point of application of the net force.  Dolphin  ( t ) 22:01, 29 March 2011 (UTC)

I made changes I considered most necessary, and without affecting the scope of the article. Feel free to improve my English (I am not a native speaker). If you generally approve, you may wish to check the end of the lede (to avoid repetitions etc). I tend to be more thorough than this, but I am using most of my spare time for the Croatian Wiki (Hrvatski), trying to prevent my students from citing incorrect definitions.--Ilevanat (talk) 23:28, 30 March 2011 (UTC)