Why does the amount of energy transferred depend on distance rather than time? [duplicate]

Question

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?

Answer 1

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).

Answer 2

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.

Answer 3

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?