# Why is energy calculated with respect to distance and not time? [duplicate]

Okay, so looking at the basic definition of energy the force is summed over the distance it is applied. Why exactly is it taken over the distance applied and not the time applied? I understand that the impulse and hence the change in momentum is what we call this summation over time, but it's not exactly clear to me why we chose to do it the way we did.

Side note - I've seen the examples where there's an object at rest and if you took the force over time you would get infinite energy, but if you took the sum of the force (both holding it up and pushing it down (gravity) you would get 0, and thus an integral over the net force would be 0, right? Thanks

-edit

Okay, so really the crux of what I'm getting at that I can't seem to find an answer for is: Why did we choose to do it this way? What experiment or thought experiment led us to believe that momentum isn't in fact energy, but a separate quantity? The problem I keep having is that when you want to sum up this quantity we know as force, you have two options, sum it over the time its applied or the distance. I just don't understand the idea behind choosing one over the other.

• Because it works. Simple as that. – garyp Oct 16 '16 at 21:12
• Because force time the time over which it is applied is impulse (the transfer of momentum just as work is the transfer of energy) which is also justified because it works. Both concepts are useful and used, but your text has to introduce one at a time. – dmckee --- ex-moderator kitten Oct 16 '16 at 21:18
• To the addition: read the link that Sean gives you below about the vis viva debate. But the short answer is that you can set of experiments that are explained by one and not the other as well as experiment explained by the other and not the one. This is science after all, and no amount of thinking and calculating overrules simple truth on the ground. – dmckee --- ex-moderator kitten Oct 16 '16 at 22:52
• I've rolled back the addition of yet another version of the question. The model we use here supposes that there is a well defined and fixed question to which people can prepare good answers. If you keep mutating the question you break that model. Please take some time to digest what is written here and then ask another question. (And please take a little more time to craft it, the version you posted here was not entirely clear to me.) – dmckee --- ex-moderator kitten Oct 17 '16 at 0:37
• Okay, I do feel that every edit was built upon the same question. If there was an answer to any of the edits, I feel like it would have answered the question at hand. So far, none of these answers have done more than say "it is what it is". – J. LeMoine Oct 17 '16 at 0:50

Because the integral of force over time is impulse, the change in momentum. Momtentum is also conserved, but it is a very different concept from energy, inasmuch as momentum in one direction can cancel momentum in the opposite direction, but energy never cancels out, to my knowledge, it just gets transferred around/changes form.

The recognition of energy as a conserved quantity that is conserved separately from momentum is an interesting chapter in the history of physics that involved experiments where the penetration depth of a ball falling into clay more closely corresponds with energy than momentum (See the vis viva debate).

Your body is acted on by the force of gravity all the time, but unless you jump out of the window your body does not gain any kinetic energy. Summing force over time doesn't give a meaningful quantity, whereas summing force over distance gives us the increase in kinetic energy.

• Force summed over time is called impulse and net impulse is most definitely a meaningful quantity. Nor is the necessity of computing the net special to impulse/momentum. You must use the net work to learn about changes in speed in the work-energy theorem as well. – dmckee --- ex-moderator kitten Oct 16 '16 at 22:50
• @dmckee - that is the integral of a resultant force or net force, a force which is doing work. The integral of a force over time is a meaningless quantity. Refer to the first sentence in my answer for a good example. – Suzu Hirose Oct 16 '16 at 22:54
• I understand fully where you are going with this: we don't factor the impulse-momentum theorem and define a bunch of "potential momenta" as we factor the work-energy theorem to define potential energies. Agreed. But that doesn't give the answer to this question, because the answer is that we do use $\int F \,\mathrm{d}t$. – dmckee --- ex-moderator kitten Oct 16 '16 at 23:03

When you lift a book to a high shelf, the distance makes the difference. The bigger the distance to the floor, the more energy is stored.

No matter how much time that passes, this stored energy is the same. It doesn't depend on time, it depends on distance.

• Say I have a variable acceleration, a=5t (as an example). If the function was allowed to stay steady before and after the acceleration (force) was applied (so constant velocity before t=0 and after time = t passed, but exponential in between) then how would we describe the kinetic energy? The distances would be different but the final velocities would be the same. Furthermore, if you only looked at the object in a window of time past when all of this occured, how can you say the kinetic energy is 1/2 mv^2? – J. LeMoine Oct 16 '16 at 23:27
• @JLeMoine Here we have to look closely at the way you are using words (something that I tried to intimate above). Force times distance is not "energy"; it is "work". What's the difference? Work is a transfer or change of energy, and it is very important to keep the two ideas separate. When no work is being done the energy of the system remains the same, so when your particle passes beyond the region where it is subject to net force of course it's kinetic energy remains the same. – dmckee --- ex-moderator kitten Oct 17 '16 at 0:31
• Oh, I feel like my understanding of the concept of "work" vs energy is lacking after reading that. I guess I'll have to really think deep on your words. Outside of that, with your reference to KE, I agree that the KE outside of the bounds would be constant - that I never disagreed with. – J. LeMoine Oct 17 '16 at 1:16
• What I was trying to say was that the work done between the two scenarios SHOULD be considered different (summing force over the distance, since the distance achieved between the two scenarios was different for each). This led me to ask about the KE. If we say the KE is given by the equation 1/2mv^2 and look at the velocity of the particle before and after the experiment, they both would also have the same change in KE. – J. LeMoine Oct 17 '16 at 1:17
• If they both have the same change in KE, but two different distances achieved when the forces were applied, wouldnt their resulting KE be different? Or is this situation somehow not possible that I've contrived. – J. LeMoine Oct 17 '16 at 1:18