Work done is defined as the dot product of force and displacement.

However, intuitively, should it not be the product of force and the time for which the body is acted upon by the force (force * time) because while time is independent of force applied, displacement is not.

Were these formulae (for work and energy) actually derived based on some physical understanding or are they just constructs to understand forces better?

  • $\begingroup$ I don't think it should. What changes in a body that is compressed by a pair of forces for 10 seconds in comparison to, say, an year? Nothing I guess. Also, it's nice to accept an answer. $\endgroup$ Nov 10, 2014 at 15:47

3 Answers 3


Work done is defined as the dot product of force and displacement. ... should it not be the product of force and the time

Were these formulae (for work and energy) actually derived based on some physical understanding or are they just constructs to understand forces better?

Neither of the two. Most formulas and definitions have an historical motivation. The issue is too complicate to fully discuss it here. There are lots of definitions which are not 'rational' in science, as the first explanations of phenomena where based on misconceptions and some are the cornerstones on which a skyscraper has gradually been erected.

You surely know the names of the orbitals of an atom: s, p, d, f. They are derived from the description by early spectroscopists of certain series of alkali metal spectroscopic lines as s harp, p rincipal, d iffuse, and f undamental. It would be more simple and rational to call them in any other way: a, b, c, d, or 1, 2, 3, 4 or , even simpler, to identify them with $l$, the angular momentum quantum number: 0,1,2,3. It would be rather easy and unpainful to change these definitions (but also physicists are subject to the 'force of inertia': see the comment by John Rennie here), whereas it's extremely difficult to alter the fundamental definitions/ derivations, just like changing the left-hand traffic in England: you ought to change all road signs overnight and, what is worst, scrap all LHD vehicles.

You may find a full and detailed description of how energy was first discovered and neglected in my post here and feel proud that you have had the same ideas of Leibniz.

Leibniz had suggested the more logical, rational and natural integration on time. Now, as you surely know, that is not possible anymore since $F * t$ is defined as momentum.

Another historical reason/ justification related to this is that the concept of energy was understood very late, suffice it to say that, to date, it is not yet considered a fundamental concept and has no own unit: you probably know that the SI has seven base units and energy is not among them. It is really amazing and disconcerting: *the most important concept in the whole universe is a [derived unit]*(http://en.wikipedia.org/wiki/SI_derived_unit) derived in various ways from derived units.

The term work was introduced in 1826 by the French mathematician Gaspard-Gustave Coriolis as "weight lifted through a height", which is based on the use of early steam engines to lift buckets of water out of flooded ore mines. The SI unit of work is the newton-metre or joule (J). (wiki)

Another reason is that the definition was modelled upon the force of gravity, which was for a long, long time the only force that was understood and math- described, that definition, $F*s$, of course, is appropriate to describe it: if you want lift 1 kg by 2m you spend energy that's double the energy to lift 1Kg by 1m. This peculiar definition produced also the 'paradoxical' [see here] (The physical definition of work seems paradoxical) consequence that, in some circumstances, you have spent a lot of 'energy calories' but you have really spent no 'energy work'

But, coming back to the derivation of the formula, after all 'integration on time' is not even necessary as there is not really anything much to be integrated, it is the simple geometric formula you need to find the side of a square: if you consider the unitary mass the real, 'true' relation between entities reveals itself in all its stunning semplicity: $$ v = p = \sqrt{E}$$ We might even drop the 'k' since not only $E_k = W$, but it is also same energy as thermal, and EM etc. energy.

Physicist are probably accustomed to it and do not bother, but it is really puzzling to see the energy of light being related in some way 'equivalent' to the dot product of $F *s$. It would be sensible to replace/integrate this definition distinguishing between work done (energy spent) and work done on something (mechanichal work), and call difference wasted energy/work: $W_d = W_{mech} + W_w$

You'll find more details about wasted energy and wasted energy in a torque at the question linked

I hope this is enough to appease your disconcert.

  • $\begingroup$ The "paradox" you describe is not related to $F\cdot \Delta x$ at all, nor is it a paradox. Or could you please how it is so? $\endgroup$ Nov 10, 2014 at 15:58
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    $\begingroup$ @AndréNeves, the paradox was presented in the link: you can spend a lot of energy and yet do no work. Read all related links. $\endgroup$
    – bobie
    Nov 10, 2014 at 16:15
  • $\begingroup$ I surely did read them, and the paradox is wrong. The energy spent is due to the biological mechanism of producing force. There are many simpler ways to exert force on the wall, none of them requiring an alive being, ergo not spending any calories or joules unless work is produced. Would you like examples? $\endgroup$ Nov 10, 2014 at 18:15

If you spend ages pushing against a brick wall, it won't budge, and you haven't actually done any mechanical work; the wall has the same energy at the end. But do the same to a car (take the handbrake off first!), and you'll get it moving: the force you are applying is giving the car motion, and hence kinetic energy. In other words, you are doing work on the car.

Work is defined to be the change in energy, which is defined as it is because it is a conserved quantity. If your intuition of work or energy don't conform to these ideas, you should try to develop it by thinking about problems like these. (Take care since the precise physics definition of terms might not quite match with the way the same words appear in common usage!)

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    $\begingroup$ If you push against the wall, the wall has not the same energy, has same kinetic energy but has increased its thermal energy. $\endgroup$
    – bobie
    Sep 14, 2014 at 15:32
  • $\begingroup$ @bobie, only if the wall underwent deformation, in which case you did produce some work. If it is absolutely rigid, I believe nothing at all will change. $\endgroup$ Nov 10, 2014 at 15:33

Yeah I've asked the same question. I've a feeling that in reference to the Holographer's comment, I do know that when a clay ball smacks into a wall, kinetic energy is made but in the heat sense. I'm not entirely sure that no energy is added to the wall. One asks oneself, what is the difference between a clay ball hitting a wall and a clay ball pressing against a wall for a long time?

One can see where the idea is coming from in that energy, in terms of energy found from motion (kinetic energy), is not macroscopically apparent in the motion of the wall. If you're not causing an object to have energy, then you aren't giving energy, and so F*t should not be the appropriate energy definition.

If we are trying to define energy, it must be transferable. F*d is transferable in that if I apply force over a distance I can know that resultant energy has been produced. Examine a ball having force times a distance, the ball gains kinetic energy that can be used to destroy a clay figure. Examine a crate pushed along the floor. That energy is transferable to the floor in heat and sanding effects upon the floor. Both of these situations break down if energy is defined as impulse and here's why:

If we press the ball on both sides equally, the impulse is greater than the work done, even more so for greater times. But, if we let the ball go, it has gained no abililty to do anything at all, unless it is elastic! The same goes for the crate. If a crate is pushed on both sides with an equal force for a long time, the crate goes nowhere and yet has had much impulse. Try and transfer impulse. The more thought experiments you make the more convinced you'll be. Try and do the same for work, and it 'works' better.

  • $\begingroup$ I think you are making a mistake when referring to lots of work being done and impulse being had when a force is applied for a long time. Because in the case of work done and impulse the force being referred to should be the resultant force acting on a body or on a system $\endgroup$
    – jerry
    Aug 19, 2014 at 15:24
  • $\begingroup$ Difference is that when a clay ball huts a wall, it stops. That means that the wall, or the wall and the building it is in, or the wall, the building, and the earth, have absorbed the momentum that the clay ball lost when it stopped. So something has moved. $\endgroup$ Apr 5, 2018 at 13:54
  • $\begingroup$ @CharlesBretana Good point, but consider a case in which identical clay balls hit a plate on opposite sides at the same time. Where does the energy go then? $\endgroup$ Jul 23, 2018 at 18:35
  • $\begingroup$ What energy? Energy is not a vector. It is not directional. The two balls rebounding have exactly the same energy as they did before they hit the wall. Unless some of the energy is converted to heat. Then they will rebound to a slightly slower velocity and everything becomes a bit warmer. $\endgroup$ Jul 23, 2018 at 22:41
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    $\begingroup$ Yes, that is of course correct. I missed the implications of your use of a "clay" ball. You meant, of course, soft clay that deforms on impact and does not rebound. Even in the case where only one ball hits a wall, and the wall and whatever it is attached to absorbs some momentum, some energy will be transformed into heat. $\endgroup$ Jul 25, 2018 at 2:07

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