Talk:Work (physics)

First sentence needs work?

The first sentence states: "In physics, a force is said to do work when acting on a body there is a displacement of the point of application in the direction of the force."

Consider gravity and a ball going upwards. The ball will slow down. Is gravity doing work there? Yes, of course. But the displacement is not in the direction of the force. Possible re-write:

"In physics, a force is said to do work when acting on a body there is a displacement of the point of application in the direction of, or opposite the direction of, the force."

But I still don't like it. Consider the gravity between the earth and moon. Where is the "point of application"? Gravity acts on every atom in the other body; there's no single point of application.

Matt DrMattB (talk) 18:28, 22 August 2014 (UTC)

Consider that negative work is done when a force is acting in a direction opposite to the displacement. Similarly, stopping a baseball in a glove does negative work on the ball. One could also say the ball did work on the glove; if you don't like the concept of negative work. But either way work is done.

Gravity doesn't do work if we consider that when stopping a ball's upward motion the ball's potential energy increases in the same amount as the kinetic energy decreases. In other words, gravity is considered a conservative force, i.e., no mechanical work is done. Dger (talk) 19:36, 22 August 2014 (UTC)

Gravity is certainly doing work on the ball moving upward. It is a force acting through some distance. The integral of F dot ds is not zero. Also, for there to be work, the force doesn't have to be in the direction of the motion; only a component of the force has to be along the line of motion. DrMattB (talk) 12:16, 23 August 2014 (UTC)

I agree with DrMattB's initial proposal. There is an absence of scientific rigor in saying "gravity does work". The definition of work only acknowledges work being done by a force, not by any other entity such as gravity. The work-energy theorem states that, when the weight of an object (ie a force) does positive work, the object's potential energy decreases; and when the weight of an object does negative work the object's potential energy increases. In both cases the change in potential energy changes by an amount equal to the work but opposite in sign.

I think the words "of the point of application" are redundant and should be removed. The essential criterion is that the center of mass of the object suffers a displacement. Whether the point of application suffers the same displacement, and even whether there is an identifiable point of application, are not relevant. Dolphin (t) 13:26, 23 August 2014 (UTC)

The first sentence is an introduction to the concept of work and therefore does not need to consider every variation. The focus is on work done by a force, which does have a point of application and a line of action, which allows consideration of the displacement of that point of application for the purposes of this definition. A person who lifts an object against gravity is fully aware of the point of application of the resultant force of their efforts, and its displacement in the vertical direction. This is why it is a good introduction to the concept. If the subtleties of various situations are to be explored, then provide a section that allows examination of these details. I recommend this introductory sentence remain unchanged. Prof McCarthy (talk) 00:07, 24 August 2014 (UTC)

The sentence is correct as it stands since if the displacement in the direction of the force is negative the work will be negative. In general I prefer the first sentence to be as simple as possible, but I see how the new version could be clearer to someone new to the concept. Either way is ok for me. I think "point of application" needs to be kept, though, to keep the definition accurate. In the example with the moon, there is not a single force and therefor not a single point of application. Assuming the moon to be a rigid body we can model the system with a single force with the point of application in the center of mass of the moon. Every force has a point of interaction. Ulflund (talk) 00:36, 24 August 2014 (UTC)

Thank you for leaving the first sentence alone. I find the change in the example to be a problem, but maybe that is just my own issue. To me the work of a person lifting a weight against gravity is obviously the work of the force applied by the person on the weight which therefore achieves positive work. It did not occur to me that this would be confused with the work of a gravity field. I understand the appeal of computing work of gravity on a falling ball, but I think this introduces the opportunity for confusion. Gravity is a conservative force so in principle work results from the difference in values in a scalar field rather than the result of a force acting on a point of application through a distance. This may be too fine a point however. Prof McCarthy (talk) 23:09, 24 August 2014 (UTC)

It is unusual for an example to be given in the lead paragraph of an article in an encyclopedia so I think the example given here may be short-lived. (It was added on 6 Oct 2013, but then mangled on 11 Feb 2014.) The example of the suitcase seemed naïve to me, for a couple of reasons:

• According to the Work-Energy Theorem, there are only two kinds of force whose work has any significance - the resultant force acting on a body, and conservative forces such as weight; work done by a resultant force is equal to the change in kinetic energy of the body, and work done by a conservative force is equal to the change in potential energy. The force exerted by a person on a suitcase is neither of these.
• If the work done on the suitcase was equal to its weight multiplied by the height through which it was lifted, its kinetic energy would increase by this amount, and clearly the kinetic energy of the suitcase starts at zero and ends at zero.

If you explain why you think the example of a falling ball introduces the opportunity for confusion, I might be able to refine the example to eliminate that confusion. Dolphin (t) 04:56, 25 August 2014 (UTC)

The only force that does not generate work is a constraint force because in this case its point of application either does not move or moves perpendicular to the direction of the force. The work done by lifting weights is correctly calculated and easy to grasp, but if you do not like it that is fine. To me describing work in terms of gravity and a falling body is awkward because the body is actually on a constant energy orbit and work is not generated until the impact with the ground, but go ahead. Prof McCarthy (talk) 14:03, 25 August 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)

Wrong symbol for Work????

Hello everyone,

Maybe I am mistaken but a symbol for work is J, not W.

See https://en.wikipedia.org/wiki/SI_unit#Derived_units — Preceding unsigned comment added by Jiri.podsednik (talk • contribs) 08:25, 8 August 2014 (UTC)

J is the unit, but W is the general symbol for the quantity itself. We say W = 10 J for example. It's just like we say F = 100 N.--Jasper Deng (talk) 10:58, 8 August 2014 (UTC)

Since work is not a state function, the appropriate symbol is w, not W.