# What actually is 'Work Done'?

1. Jul 16, 2014

### Jo Bloggs

Does "work done" actually have any real meaningful sense or value?

I've got no problem with "work done" when moving an object in a conservative field. This makes sense - you've got to put in energy when, for example, lifting an object in a gravitational field, and this energy is the work done, and you get this energy back if the object then falls back to its original position. In this case, the work done is simply another name for the energy required to move 'up' in the field, and is equal to the (change of) the potential energy of the object.

But what if you're not talking about moving in a conservative field? If you just push against a wall, for example, no work is done because the wall doesn't move. But there are all sorts of energy processes going on and energy is expended. If you push a car, there is some "work done" if the car moves. But, again, you have to expend energy to overcome friction, and so on. In both cases, the "work done" is a small fraction of the energy input, and it doesn't immediately seem to have any value.

So, what actually is the value of "work done"? What's the point of it? Wouldn't it make more sense to dispense with it entirely and just say that the kinetic energy of the wall or the car that you're pushing is one part of the energy of the system? Why does "work done" have any value over and above kinetic energy?

Wikipedia doesn't seem to help here, because its derivation of kinetic energy is in terms of work done when lifting an object in a gravitational field, which seems to me to be rather circular.

2. Jul 16, 2014

### Mr-R

Welcome to PF

Well, in the case of the wall. I think that the work done ON the wall is zero because it did not move right? It is true that you used you energy (lets say kinetic and chemical) to try moving the wall But the wall did not move and the energy you spent turn into sound and thermal energy we shall say. If you calculate the work done on the wall, then basically you will have the amount of energy you have put in the WALL, hence you can calculate the WALL's speed or kinetic energy, anything related to the wall using the energy that was put into it (work done).

Lets wait for more members to reply that are more knowledgeable than me :tongue:

Sorry for using caps and I will kindly take any criticisms from my fellow members.

3. Jul 16, 2014

### Jo Bloggs

Well, you've transferred some energy to the wall, and that energy was primarily expended in moving molecules closer to each other, so it's essentially currently stored as electrostatic potential. You haven't moved the wall (not in any 'macro' sense, anyway), so the "Work Done" is zero, but the energy expended is non-zero.

So my point is that, in this context, "Work Done" is meaningless. In the more general case, where something (the car, for example) does actually move, "Work Done" is still fairly meaningless, because the Work Done is still only some part of the energy input (the rest going in overcoming friction, and so on), and the Work Done part of the overall result can still be measured as kinetic energy.

So, to rephrase my question, what value does "Work done" actually have (apart, possibly, from the case of moving objects in conservative fields?) Is it more or less fundamental than kinetic energy? Is it actually fundamental in *any* sense? Surely the fundamental concept is momentum and/or kinetic energy, and "Work Done" is just a red herring?

4. Jul 16, 2014

### Mr-R

But the molecules would revert back to their original positions right? Anyway, then can't we say that the work done in moving the wall is zero and the work done in moving the molecules is not zero (if the molecules did not revert back)?

I think you can view "Work" as energy needed or expended in doing a particular thing. By Newton's laws, you can view work as a change in kinetic energy $W=ΔKE$. Therefore, you have related work to the movement of an object (The whole wall). If you want to talk about it's molecules. Then you can relate work to the internal energy and added heat (from wiki) by $dE=\delta Q-\delta W$ (internal energy equals the added heat minus the work performed by the system)

i know that you have read Wikipedia, try reading it again as it have some words on the various situation where you can use the concept of work. Might clear up some stuff. Note again that I am learning just like you so don't take my explanations very seriously

5. Jul 16, 2014

### CWatters

You (Jo) appear to be suggesting that "work done" isn't a useful term because energy is always conserved?

Work done is normally used to describe the transfer of energy between parts of a system not the total energy of a system. It matters a lot of you are interested in the efficiency of a process.

6. Jul 17, 2014

### atyy

The relation between work and kinetic energy holds in general, not just in a gravitational field.
http://ocw.mit.edu/courses/physics/...echanics-fall-1999/video-lectures/lecture-11/

The work done is a line integral, and in general depends on the path taken. A conservative force is an important special case in which the work done is not dependent on the path taken.

Last edited: Jul 17, 2014