Talk:Work (physics)/Archives/2014/February

Why rigid body?
The article says "Also, from Newton's second law for rigid bodies it can be shown that work on an object is equal to the change in kinetic energy of that object." This is either too strong or too weak depending on your pedagogical viewpoint. For simplicity I would simply delete "for rigid bodies" as being an unnecessary distraction. But when I did so DavRosen reverted the edit on the ground (presumably, based on reading between the lines of his edit summary) that the object might vibrate mechanically (so some of the energy would go back and forth between kinetic and potential energy, though their sum would always remain equal to the original work done) or undergo plastic deformation (so some of the energy might dissipate as thermal energy).

While I don't mind such pedantry when it's accurate and complete, in this case the requirement of a rigid body opens a whole can of worms. First of all Newton's laws of motion don't hold for rigid bodies in general but only for point masses. Hence for the pedantry to be justified one must appeal instead to Euler's laws of motion, but do we really want to do this here? Second, the same concerns one can raise about a non-rigid body can equally be raised for a rigid body, which might be a bullet in a viscous medium such as air for example, requiring the statement to be hedged around with further restrictions aimed at making "rigid body" a sufficient condition for the truth of the statement.

Rather than have the requirement "rigid body" with its attendant can of worms, I propose that for simplicity "for rigid bodies" be deleted as being unnecessarily distracting. Strengthening the requirement to "for a point mass" avoids getting involved with Euler's laws, but is still a distraction. Vaughan Pratt (talk) 18:03, 31 October 2013 (UTC)
 * Some equations on here are wrong for the basic understanding of how work works. I would if I had the time go through and correct them all. Go to any good physics book and see what they say "Physics for scientists and engineers with modern physics eighth edition by serway jewett" is a good book.. Cky2250 (talk) 18:43, 31 October 2013 (UTC)

Vaughan Pratt, again you removed the end of the sentence: the equation just after the comma and before the period:
 * Also, from Newton's second law for rigid bodies it can be shown that work on an object is equal to the change in kinetic energy of that object,
 * $$W = \Delta KE.$$

So kinetic energy is specified twice in this sentence. If it's not rigid, then it could be, for example, a gas having work performed on it by a piston. In this case we can't gloss over the fact that it has internal energy in the form of potential energy of its compression as well as thermal energy, both which may increase due to the work performed, rather than only the macroscopic kinetic energy, and which will not necessarily "go back and forth" with macroscopic kinetic energy, particularly in the case of thermal energy since the process isn't necessarily reversible, but even in the case of the compression since there may be mechanical constraints inside the system that keep it compressed. You could remove the "rigid body" qualification, but then you'd have to broaden "kinetic energy" in both places as well. Mechanics of a rigid body is a standard way of limiting the discussion to the movement of a system having three translational and three rotational degrees of freedom (all macroscopic). A "body" without any qualification could be any system, even perhaps an engine (since I mentioned pistons), which can have lots of other internal degrees of freedom that we want to declare are not in the scope of this discussion. DavRosen (talk) 21:46, 31 October 2013 (UTC)
 * Fair enough, though I still claim the equation doesn't follow from Newton's laws, which say nothing about angular momentum or rotational energy. Vaughan Pratt (talk) 00:54, 1 November 2013 (UTC)
 * I agree with the viewpoint expressed by Vaughan Pratt. In particular, I disagree with using the expression Newton's second law for rigid bodies because such a sentence construction implies that there is also a Newton's second law for non-rigid bodies. I have been working with Newton's laws of motion for over forty years and I don't recall ever seeing them referred to as anything other than Newton's first, second, third laws, until now. Elsewhere on Wikipedia, we talk about Newton's laws of motion without needing to qualify them as referring to any particular type of body. Also, Wikipedia talks about the work-energy theorem, saying the change in kinetic energy of an object is equal to the work done on that object by the resultant force acting on the object. There is no good reason why Work (physics) should depart from that terminology by being the only Wikipedia article to talk about Newton's second law for rigid bodies. Dolphin  ( t ) 23:57, 4 November 2013 (UTC)


 * The word "rigid" was not intended create a new variant of Newton's second law, but rather to exclude changes in internal energy (for example for a gas or an inelastically-compressible object) as a precondition to the assertion being made by this particular sentence, which is about a change solely in the kinetic energy of the motion of the body in space, specifically the sign of this change in relation to that of the work performed. I've taken a shot at moving and clarifying "rigid" so that it doesn't inadvertently create the subphrase "Newton's second law for rigid bodies". DavRosen (talk) 14:15, 5 November 2013 (UTC)


 * As for "the equation doesn't follow from Newton's laws, which say nothing about angular momentum or rotational energy", I disagree. Angular motion and rotational energy merely serve to account for all the linear motion (at a given point in time) of all the infinitesimal masses making up the rigid body, each of which observes the 2nd law with respect to its acceleration due to the forces on it, and each of which contributes to the kinetic energy by its mass traveling at some (linear) velocity.  On the other hand I suppose we *could* talk about a single point mass rather than a rigid body, but I'm not sure what we'd gain. DavRosen (talk) 14:48, 5 November 2013 (UTC)
 * Thanks DavRosen. I think your changes make an improvement. I have some difficulty with the expression kinetic energy of the velocity and rotation of that body. The kinetic energy associated with linear velocity and that associated with angular velocity can be added to yield the total kinetic energy of the body so I see your extended expression to be unnecessarily cumbersome. We can simply talk about the kinetic energy of the body. The same isn't true of linear momentum and angular momentum which can't be added - they don't even have the same units.
 * I would prefer to see a simple statement of the work-energy theorem: From Newton's second law, it can be shown that work on a body is equal to the change in kinetic energy of that body, perhaps then followed by qualifications about no fields, no internal degrees of freedom etc., rather than trying to insert the whole truth about work & energy into one sentence. Dolphin  ( t ) 23:20, 5 November 2013 (UTC)
 * Agreed. The current wording is an improvement but cumbersome as one sentence.  Moving the caveats to a separate sentence seems cleaner.
 * Incidentally at what scale does the concept of internal energy kick in?  I'm fine with calling vibrations in molecular bonds internal energy, since their statistical mechanics is the basis for thermodynamics, but what if you had a nanomachine with a million linkages, springs, etc?  Would its oscillations be classified as internal energy or work?  Where do you draw the boundary?  Vaughan Pratt (talk) 00:44, 7 November 2013 (UTC)

How much Work is done lifting 2kg. of book into a shelf of 1.5m? — Preceding unsigned comment added by 112.198.82.1 (talk) 21:09, 10 February 2014 (UTC)
 * This looks like a homework question. Wikipedia doesn't answer homework questions because, if we did, the full benefit would not be available to the student. If you read the first few paragraphs of our article Work (physics) the answer will be clear. Dolphin  ( t ) 00:53, 11 February 2014 (UTC)

Work by gravity error?
In the Work by gravity section, the equation yields "W = .. = WΔz"

Something seems fishy! 206.47.231.195 (talk) 15:42, 17 September 2013 (UTC)

It is correct-ish. they defined z as height and W as weight or also known as mg. I will reword this section to be W = Fh = mgh. Cky2250 (talk) 18:40, 17 September 2013 (UTC)


 * it's just plan wrong, and the idea that proceeds it is wrong . Gravity is Not doing work in that example because the motion is not on the direction of the gravitational force.  If it were otherwise the energy equation wouldn't balance.  The force up times the distance up equals the potential energy.  That's balanced energy equation.  But, if you add to it, what , the gravitation force times the same distance?  Then get work = 2 x weight x distance ,,,but the potential engery is 1x weight x distance...........they don't equal,!!!! Big hint that your doing something wrong.     Ok so maybe you think (despite the definition and common sense) that the gravity force is doing NEGative work,   Then you have work= weight x distance (the force up) - weight x distance (gravity)=zero!!!! Yet the potential energy gained is still weight x distance!!!!!! More nonsense !!!!!!!!! No my friends,,,, work is force x distance IN THE DIRECTION OF the force!!! TheSAME direction! not opposite it.  Physics 101!!!!  — Preceding unsigned comment added by 108.69.52.251 (talk) 04:14, 11 February 2014 (UTC)

As to' it doesn't need to cause the displacement'. Where to begin,,,, if Force and no displacement, then, no work. If there is force and the object moves in the direction of the force, i cannot imagine how you could describe that force as not in some way causing that movement! If a bat hits a ball do you think it's just a coincidence that the ball moves? — Preceding unsigned comment added by 108.69.52.251 (talk) 04:38, 11 February 2014 (UTC)


 * I have refined the section on work done by gravity. See my diff.
 * I agree with 108.69.52.251 that if the vector representing the weight of an object, and the vector representing the displacement of the object point in opposite directions, the work done on the object is negative work. Work is formally defined as the dot product of force and displacement. (According to the Work-Energy Theorem, when the weight of an object does negative work, the gravitational potential energy of the object increases.) Dolphin  ( t ) 02:23, 12 February 2014 (UTC)