So the formula for work is$$ \left[\text{work}\right] ~=~ \left[\text{force}\right] \, \times \, \left[\text{distance}\right] \,. $$

I'm trying to get an understanding of how this represents energy.

If I'm in a vacuum, and I push a block with a force of $1 \, \mathrm{N},$ it will move forwards infinitely. So as long as I wait long enough, the distance will keep increasing. This seems to imply that the longer I wait, the more work (energy) has been applied to the block.

I must be missing something, but I can't really pinpoint what it is.

It only really seems to make sense when I think of the opposite scenario: when slowing down a block that is (initially) going at a constant speed.

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    $\begingroup$ The key is that distance in that formula refers not to the distance moved by the object, but to the distance over which the force is applied. You can push an object and have it coast due to inertia, but the formula only cares about how far you actively pushed it. $\endgroup$ Sep 13, 2018 at 17:10
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    $\begingroup$ ... in a vacuum, -> ... in a frictionless environment, ... $\endgroup$
    – CramerTV
    Sep 13, 2018 at 19:29
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    $\begingroup$ In your example, the block's energy only changed when you were pushing it. After that it has a constant unchanging speed, so a constant unchanging kinetic energy. The change in energy from before the push to after is the work done on the block. $\endgroup$
    – usul
    Sep 13, 2018 at 19:39
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    $\begingroup$ Work is not force times distance. That's just a simplified version used to solve textbook problems with constant force. See Dale answer for the real definition of work. $\endgroup$
    – Bakuriu
    Sep 13, 2018 at 21:21
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    $\begingroup$ Maybe the counterintuitive part is why it scales with distance, not time. $\endgroup$
    – user126527
    Sep 14, 2018 at 6:50

11 Answers 11


You have to put in the distance on which the force acts. If you release the force, there will be no work done since there is no force acting on the body.

  • $\begingroup$ Your deserved vote. $\endgroup$ Feb 5, 2023 at 15:34

Often it is important to know if a given formula is a simplification of a more general equation and, when you encounter a conceptual problem, check the general formula. In this case it is a simplification of this formula: $$W=\int_S F\cdot ds $$ Where $S$ is the path over which we are interested in the work and $ds$ is an infinitesimally small segment of $S$.

So back to your question, wherever $F=0$ the integrand is $0$ regardless of how long that segment of the path is. So it is only that first segment where you are applying the 1N that work is done. Once you stop pushing the distance increases but the work does not.

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    $\begingroup$ Now what I'm curious about a followup question. If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives? $\endgroup$
    – orlp
    Sep 15, 2018 at 10:07
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    $\begingroup$ @orlp Why not ask that as a separate question? $\endgroup$
    – Tommi
    Sep 15, 2018 at 13:11
  • $\begingroup$ I agree. I have a good answer to that, but it won’t fit in the constraints of a comment $\endgroup$
    – Dale
    Sep 15, 2018 at 13:32
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    $\begingroup$ @orlp Asked over here: physics.stackexchange.com/questions/428952/… $\endgroup$ Sep 15, 2018 at 19:50

If I'm in a vacuum, and I push a block with a force of 1N, it will move forwards infinitely

and accelerate the block ie change the block's velocity and hence change the kinetic energy of the block.

The longer you apply the force, the n more work the force does resulting in a greater change in the kinetic energy of the block.

  • $\begingroup$ The phrase "the longer you apply the force" could imply a longer time, rather than a longer distance. (At least, that's my intuitive understanding.) The original question could easily arise from a confusion of time and distance, so I think it would be worth clarifying your answer. $\endgroup$
    – craq
    Sep 15, 2018 at 20:19

Work is a definition, so the reason is "because it is defined that way".

However, we can ask why it makes sense to define it this way. Intuitively, you'd want to think of "work" as being a measure of what you do when you push, say, a box up a ramp, which causes you to get tired. In doing this, you apply a force to the box, and you also move a distance, and if the box is heavier (i.e. you need to use more force) or the distance you have to push it (the length of the ramp) is longer, then you would like to say the work is larger. If I have to push twice as hard for the same distance or I have to push twice as long, "intuitively" I should expect to do twice the work, and thus we get

$$\mathrm{Work} = \mathrm{Force} \cdot \mathrm{Distance}$$

And this simple, intuitive idea, it turns out makes a lot of physical sense when we actually use it, well beyond whatever limitations the original intuition might have (e.g. the biological inefficiency of our own bodies in doing "work", for example) thus we keep it. In particular it leads us to the concept of kinetic and potential energy, and their total ends up being conserved, thus showing we hit upon a core physical concept in the Universe. There isn't really any more of a "why" than this - it's science. Science is about the application of intuition or imagination, evidence, and reasoning, together, to understand how the world works. Intuition and imagination generate ideas for what is going on from which we can reason out consequences, and then we use evidence to see if those consequences are borne out and so whether or not that our ideas connect with reality.


This is a conceptual answer for learners, not a rigorous answer.

How do you know that you moved something?

  • You (not something else) must have pushed it. That's force.

  • It must have actually gone somewhere. That's distance.

So we define "work" as the product of these two things.

Physicists soon discovered that this definition is actually useful for calculating the behavior of systems. When you do work on an object, the same amount of work is taken away from you. Thus, the total amount of work that has been done (or could be done) inside a system remains a constant amount.

From this simple concept, you can elaborate it into the more rigorous definition:

  • The distance is only meaningful during the interval to which your force is applied. Any continued motion by inertia doesn't count as your work.
  • Positive work means "adding" to an object's movement. Negative work means "taking away" an object's movement.
  • The force can be at an angle to the distance (introduce dot product).
  • The force can vary (introduce integral over distance).
  • We can talk about work that has been done in the past, and the ability to do work in the future (energy).
  • We can calculate the work done in various scenarios, leading to some generally-useful formulas ($mgh$, $\frac{1}{2}mv^2$, etc.).
  • We can more rigorously examine conservation of energy.

When a weight is sitting on the floor, the floor is applying a force to the weight (and vice versa), but no distance. And it should make intuitive sense that there's no work being done.

For your example of a weight in a vacuum: if you push it with force 1N for distance 1m, and then stop pushing, it will move forever at constant speed. If you push another block with force 1N for distance 2m, it will move forever at a higher constant speed. You did more work to it, so it has more kinetic energy.


Think of a unit of work as what you do when you lift a kilogram to a height of a metre. How do you do 2 units of work? You lift that kilogram 2 metres. 3 units? 3 metres. Etc. If you lift something the work you do is the force you exert times the distance traveled (height raised).

For your example of pushing a body in a vacuum you are only doing work while you are actually pushing, and so making it go faster. If you let it coast you are not doing work because the force during this time is zero.

The work you put in is the change of the energy of the object. When you lift the weight the work you have done becomes potential energy (stored energy). When you push the object in space the work you do becomes kinetic energy (energy of movement). You can change the potential energy to kinetic energy by letting the thing fall, and obviously it will be falling faster after it has fallen for a greater height.

  • $\begingroup$ Peter, can you please remove whichever parts of this post are not actually part of your answer? A lot of this seems to consist of contributions from other people, and it's hard to follow exactly which bits are yours. $\endgroup$ Sep 15, 2018 at 2:20
  • $\begingroup$ Rahul, thank you for removing the additional material $\endgroup$
    – Peter
    Sep 15, 2018 at 11:49

I see several answers that all seem to explain it, but for someone trying to understand the why, perhaps it's best answered simply.

It is going to be a lot more "work" for me to push a heavy trashcan out the door and down the driveway than the amount of "work" for me to just push the trashcan out of the house.

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    $\begingroup$ The OP is thinking of an object pushed in outer space. If you apply a force to the object for 1 second it would, potentially, move forever. He's wondering how you calculate the 'forever' movement. The answer is, you only pushed for 1 second - the movement due to inertia in a frictionless environment has nothing to do with the calculation of applied work. $\endgroup$
    – CramerTV
    Sep 13, 2018 at 19:33

Other answers have covered the misunderstandings around the $W=F \cdot d$ equation. I think it's also worth noting that "distance travelled" is not a form of energy. In physics, the phrase "doing work" means converting energy from one form to another, so you don't need to do any work to travel infinite distance. You only need to do work at the beginning to get some kinetic energy.


As a simple and naive argument for a constant force, I would first convince myself that the kinetic energy equals $mv^2/2$ with Ron's excellent symmetry arguments:

So once you believe in that, let's show that this definition of work matches the change in energy of a particle.

$$ \Delta E = \frac{mv_{f}^2}{2} - \frac{mv_{0}^2}{2} $$

and we also have:

$$ L = v_0t + \frac{F}{m}\frac{t^2}{2} \\ v_f L = v_0 + \frac{F}{m}t $$

By playing with those two last expressions, we can conclude that:

$$ LF = \Delta E $$

F: force, L: distance, m: mass, $v_f$: final speed, $v_0$: initial speed, t: final time.

Another good more differential explanation of this is: https://physics.stackexchange.com/a/79529/31891

$$ F = \frac{dp}{dt} \\ F = \frac{d}{dt}(\sqrt{2 m E}) = \frac{\sqrt{2m}}{2\sqrt{E}} \frac{dE}{dt} = \frac{m}{p} \frac{dE}{dt}=\frac{1}{v}\frac{dE}{dt}=\frac{dt}{dx}\frac{dE}{dt}=\frac{dE}{dx} $$


Force, according to Aristotle, is that which applied to something that can change, actually does change.

Since change is primarily motion, this means that motion goes from natural motion (ie inertial motion) to violent motion, the motion that occurs when force is applied (non-inertial or accelerated motion).

More, we see that force, when applied to something, and which doesn't cause change is not a force. In our world, there are no such forces, forces always cause change even if the change is imperceptible or in equilibrium.

Now, when we apply a force to a block of wood, it moves. If we want a measure of this change then there are only two variables we can look at, the distance moved and the velocity. From the first we can build work and from the second we can build kinetic energy. It turns out that in our world the change in kinetic energy of the block equals the work done on the block. Now, in any situation that changes, it is always worth looking for that which does not change. This equality does not change, it is an invariant. It is also the precursor to the principle of the conservation of energy - one the main underlying principles of physics. And this is one good reason to look at these concepts as they are traditionally defined.


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