The change in energy of an object can be determined by the work equation, where work is the change in energy:

$$ W = F \cdot d $$

I conceptualize the transfer of energy as simply a series of small "packets" of energy being transferred at every Planck length. These small "packets" of energy add up to the total energy transferred (i.e. work). I'm not sure if this conceptualization is correct, so correct me if I am wrong.

However, it makes me wonder why the amount of energy transferred is dependent on distance and not time.

$$ m_1 = 10~kg \\ m_2 = 20~kg \\ W_1 = (10~N)\cdot(5~m) = 50~J \\ W_2 = (10~N)\cdot(5~m) = 50~J \\ W_1 = W_2 \\ t_1 \neq t_2 $$

If I apply a constant force on an object, why isn't the energy transferred at a constant rate with respect to time? The energy transfer rate varies dependent on how long it takes to cover the set distance.

In other words: why is the energy transferred consistent per unit of distance, and not per unit of time?

  • $\begingroup$ Energy is a conserved quantity, and no matter the rate at which it is transferred, you would only be interested in the difference between the final state and the initial state. Power is energy transfer rate. $\endgroup$ – mcodesmart Oct 3 '13 at 18:34
  • $\begingroup$ I'm probably using the wring terminology to describe my question. I can see that I can derive the rate at which the energy is transferred by dividing the work by the time it takes to move the object that set distance -- but I'm having trouble understanding why energy arrives in portioned packets determined by distance. That is, what is it about the nature of the universe which causes this phenomenon? $\endgroup$ – Vilhelm Gray Oct 3 '13 at 18:43
  • $\begingroup$ are you referring to the quantaization of energy?? $\endgroup$ – mcodesmart Oct 3 '13 at 18:45
  • $\begingroup$ I can see that the portions vary depending on the time it takes to travel the distant, but I'm confused why they vary if there is constant force. What is quantization of energy? $\endgroup$ – Vilhelm Gray Oct 3 '13 at 18:45
  • $\begingroup$ @ProgrammingEnthusiast After doing a quick search, I do believe I am referring to the concept of quantums. I've only had a Newtonian background in physics, so perhaps this might be the source of my confusion. $\endgroup$ – Vilhelm Gray Oct 3 '13 at 18:48

There is a name for the quantity $F\cdot t$, it's called the impulse. The impulse tells you how much momentum is transferred to the system in a given time interval if you apply a constant force, much as how the work tells you how much energy is transferred over a given distance interval if you apply a constant force.

So, what's up?

Newton's 2nd law can be expressed as

\begin{equation} F = \frac{dp}{dt} \end{equation} where $p=mv$ as usual.

A constant force thus means that momentum is introduced into the system at a constant rate. That is what the force is measuring, the rate at which you are introducing momentum into the system.

Now if we only have kinetic energy, then $p = \sqrt{2 m E} = mv$. So... \begin{equation} 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} \end{equation}

As you can see, the extra factor of velocity between energy and momentum is crucial. That factor of $dx/dt$ converts the time derivative $dp/dt$ to a space derivative $dE/dx$.

Off the top of my head I can't think of a simple physical reason for the difference. It's just a matter of which variables are more convenient, it turns out to be more meaningful in many problems to study how the momentum changes with time instead of the energy, and so physics is set up to talk about forces measuring the rate of change of momentum, instead of the rate of change of energy. One reason is that momentum is a vector whereas energy is a scalar, so the momentum has more information and so is more useful to track in general.

Incidentally, rate at which you pump energy into the system is called the power. It is related to the force by $P=F\cdot v$ (at least over time intervals short enough that the velocity doesn't change by much).

  • 1
    $\begingroup$ I think the fault in my logic was conceptualizing constant force as introducing energy at a consistent rate, when it was in fact introducing momentum at a consistent rate! $\endgroup$ – Vilhelm Gray Oct 3 '13 at 19:03
  • 2
    $\begingroup$ Exactly. It's one of those subtle things, that doesn't really have an intuitive 'layman' explanation, you really have to look at the formalism to see which one of energy and momentum you are talking about. As an aside, I'm intrigued that the force measures the spatial derivative of the energy, which is conserved because of a time translation symmetry, and conversely measures the time derivative of the momentum, conserved because of space translations. I don't know if it's significant or not but it's sort of striking, as I look over this again. $\endgroup$ – Andrew Oct 3 '13 at 19:07
  • $\begingroup$ Is your math right? there's not a lot of explanation of what you did, I assume you multiplied by 1 on $\frac{\sqrt{2m}}{2\sqrt{E}} \frac{dE}{dt}$, but that would have been $\sqrt{2m}$ which would be $\frac{\sqrt{2m}*\sqrt{2m}}{2\sqrt{E}*\sqrt{2m}}$, which then should be $\frac{2m}{2\sqrt{2mE}}$ ie $\frac{2m}{2p}$ $\endgroup$ – Krupip May 13 at 18:13
  • $\begingroup$ @hythis Wow it has been a long time since I wrote this answer! I think you are right -- does it look better now? $\endgroup$ – Andrew May 13 at 20:14
  • 1
    $\begingroup$ @Andrew Yes I think that's right $\endgroup$ – Krupip May 13 at 23:15

It might be delivered at a constant rate of time or it might not, but the formula has to be set up with distance, because time has nothing to do with force. It takes a certain amount of force to move a box from one side of the room to the other in a straight line. It will always take that amount of force to move it that distance by pushing. On the side of time, you could apply the force in one long push across the room, or you could push it halfway, take a break and push it the other half. The fact of the matter is, unless you are a robot equipped with a force sensor on your hands and the ability to adjust your force and movement speed if there is a bump or something, you will never push at a perfectly consistent time rate. On the other hand, given the same floor and same box it will always take the same amount of force to move it. Time is by its very nature inconsistent, and we are forever unable to perfectly execute perfect packets of energy over a perfect time rate. If humans could move perfectly like that, there would be no reason for the Olympics or any sports for that matter.


Suppose energy is transferred through a distance of 'd' m in t sec, then energy transferred through a distance of '1' m will be in t/d sec. This is the energy transferred per unit length.

Therefore, the rate of energy transferred per unit length will be E*d/t.

Is that you are looking for? Am I correct?


Not the answer you're looking for? Browse other questions tagged or ask your own question.