# The physical definition of work seems paradoxical [duplicate]

So this is possibly a misunderstanding of the meaning of work, but all the Physics texts, sites, and wiki that I've read don't clear this up for me:

In the simplest case with the simplest statement, work is force times distance. If you push with a force $F_{1}$ on an object that doesn't budge because of friction, you do no work. If your friend helps push and you still apply the same force $F_{1}$ and the thing moves, all the sudden you're doing work and it's not really because of what you're doing. Moreover, if you continue applying the same force, and your friend increases her force so that the thing moves faster, and covers a greater distance, again you're doing more work and by no fault of your own.

This just seems paradoxical, and maybe the only sensible answer to this paradox is "Well, the physical notion of work is not the same as the everyday notion of work," but I'm wondering if anyone can say anything about this to make it feel more sensible than just accepting a technical definition for a word that doesn't seem like the right word to use.

• In point of fact your muscle tissue is doing work when you strain against a immobile object (because individual fibers are contracting and relaxing), but all that energy goes into heat instead of into a larger scale mechanical change. Jul 20 '14 at 5:13
• Muscles are inefficient. Instead of thinking about pushing the truck, think about leaning sideways on the truck. Jul 20 '14 at 9:10
• Actually, the everyday notion of work is quite similar to the physics notion. You are just confusing the everyday notion of effort with that of work. In your first case you expended effort, but accomplished no work. Because even in our everyday understanding, work is an accomplishment, and not merely an attempt nor an expenditure. Jul 20 '14 at 13:32
• @RBarryYoung Thank you for clearing that up. It was helpful to remember that there is a distinction between effort and work. Jul 21 '14 at 13:24
• Possible duplicate of Why does holding something up cost energy while no work is being done? Jun 21 '17 at 8:19

If you're pushing a 10-ton truck and it's not moving, you are not doing any work on the truck because the distance $ds=0$ and the nonzero force $F$ isn't enough for the product $F\cdot ds$ to be nonzero.

Your muscles may get tired so you feel that you're "doing something" and "spending energy" but it's not the work done on truck. You're just burning the energy from your breakfast by hopelessly stretching your muscles. The energy gets converted to heat and your body is really losing it, but when we talk about "work", we usually mean "mechanical work" done on an external object, and it is zero.

If someone loosens the brakes and you suddenly manage to move the truck, your perception how "hard" it is may be the same as before. You may be spending the same amount of energy obtained from the breakfast. But there is a difference. A part of this energy is converted not to useless heat of your muscles but to the kinetic energy of the truck.

Your impression that the work changes "not because of what you're doing" is an artifact of the fact that a big part of the energy is spent on heat in the muscles in one way or another. But it's really the usefully spent part of the energy, however small, that does the mechanical work. It may be a small part so it may be hard to notice it.

Physical terms often deviate – and they are more accurate than – their counterparts in everyday English (or another language). But I would argue that the physics definition of (mechanical) work does agree with the everyday life usage. If you're hired to do some work with the truck and move it and the truck doesn't move an inch, your boss will conclude that you haven't done your work and you won't be paid a penny, just like what physics seems to calculate. You may have spent your energy by stretching and heating muscles but that's not called (mechanical) work. Work is actually supposed to be something useful – both in everyday life and in physics. In both cases, the conversion of energy into useless heat isn't included to "work".

Just to re-emphasize this insight. There are many forms of energy and work and many "quantities with the units of one joule". But the words denoting them are not synonymous. So energy isn't quite the same thing as work and it isn't the same thing as heat or mechanical work or something else (also, debt and profit aren't the same despite the same unit of one U.S. dollar). The energy conservation law says that the sum of several quantities of this kind are zero or equal etc. but the different terms have to be distinguished and in these contexts, "work" really means "mechanical work".

• The emphasis on mechanical work does a lot to sooth my troubled mind, thank you! Jul 20 '14 at 5:45
• @Addem, good to hear. Just to re-emphasize this insight. There are many forms of energy and work and many "quantities with the units of one joule". But the words denoting them are not synonyma. So energy isn't quite the same thing as work and it isn't the same thing as heat or mechanical work or something else (also, debt and profit aren't the same despite the same unit of a dollar). The energy conservation law says that the sum of several quantities of this kind are zero or equal etc. but the different terms have to be distinguished and in these contexts, "work" really means "mechanical work" Jul 22 '14 at 6:46
• " ... are not ' synonyma ' . to se řekne česky, :) Nov 12 '14 at 9:57
• You're right, @Bobie! I thought it was international because -a is the right declension in plural in Latin. But the 1st case in Latin is synonymūs, I guess, not synonyma! ;-) So it's indeed just some Czech idiosyncrasy based on declension of "město" (town). You can still find "are not synonyma" via Google on 92 pages, including Wikipedia main pages. ;-) Nov 13 '14 at 7:35
• "..based on declension of "město" ..." I do not think so, I think you were influenced by the English curriculum . Down to MiddleEnglish the word conserved the Latin form synonymum and had therefore the Latin plural, like curricula. Then the word was aglicized and shortened to synonym and takes 's'. The Latin form (and plural) survives in specialized texts and in old books (which are retrieved by google-books). It was a joke, anyway. :) Nov 13 '14 at 7:57

I'm sure everyone has had that concern when we encountered the definition for the first time, in school.

There is a valid reason why this definition is still persisted with, despite the deficiency that you hit on. The most popular (and simple) forces in physics (also the ones with which we begin learning physics) are conservative forces, implying that the (mechanical) 'work' done depends only on the end state, and not on the path followed in getting there. (Imagine a magic force where you spend the same energy in going from your table to the kitchen via the shortest path, or going to Mars first and then going to the kitchen!) In this sort of a situation, it makes sense to be concerned with the displacement (and not the distance covered) under the influence of the force. That's encapsulated in the defining relation $W = \int \vec{F}\cdot\vec{ds}$.

If you account for the fact that both (Newtonian) Gravitational force relation, as well as electrostatic forces both fall into this category of conservative forces, you can imagine that these definitions are sufficient to provide a description of a huge range of known phenomena. However, most forces that you encounter in everyday life aren't eligible for such a simplified description, since they are vastly more complicated. Especially when you interface with biological systems. So, while the physical definition of work looks paradoxical here, it actually isn't, if you adopt this perspective:

(Let me construct a new term to keep things distinct from the physical work.) The ''un-physical'' work $W_{\rm unp}$ would still be the negative of the energy you spent biologically, minus the heat energy you are contributing to the universe. i.e. we have $\Delta E = H + W_{\rm unp}$. This work could refer to, e.g. the gravitational potential energy gained by lifting something up, in which case, it really is the convenient 'work' of ours. However, even when it isn't, you can easily see that this is a ''one-way definition'', since you always keep spending energy to do work, unlike conservative systems (e.g. when you throw a ball up, it gains energy on the way up and loses that much on the way down. That doesn't happen here). If you walk around your building $n$ times, you are accumulating $\Delta E$, because both $H$ and $W_{\rm unp}$ increase. (For imagining the second one, suppose I fold your circular path into a straight path of length = total distance covered. Then you are doing work even as per the above definition. Notice that the sign of this $W_{\rm unp}$ won't reverse like the gravitational analog. So, it upstream the physical force, is as good as downstream this force). Clearly, apart from verifying your intuitions, there is nothing else physically useful that can be drawn from this definition.

If your friend's energy+ yours= F1, then you would see your own energy expenditure halved, which we know cannot be the case. If your friend helps you push the object, then you are no longer applying the same force, or (lazy answer) the force is no longer localised and motivates the part of the object most subject to friction.

So, once the object is first motivated, it requires less energy to keep it moving than it does from rest. You are right to ask the question, because from your example the total sum of work is itself composed of 'a number' of different calculations. Good question friend!

It appears that you are confusing the force you contribute (F1), with the force necessary to overcome friction (F2). The force necessary to overcome friction, is fixed by the mass of the object and the surface friction. As an example let F2 be 100N, lets assume you can only provide 80N, then you will not be able to move the object. If you get a friend and she too can exert a force of 80N, together you can apply a total (F) of 160N. Now, the first 100N will be spent in overcoming friction and the remaining 60N will go towards making the object move.

In the equation for calculating work (W = force x distance), it is the NET force (F3 = F - F2, = 160 - 100 = 60) that needs to be used, because this is the part that does "useful" work.