Why is kinetic energy proportional to v^2?

I answered this question here, channeling this. That seems like the best answer – kinetic energy is proportional to v^2 because of Galilean relativity.

The more common answer, though, is to say that work is force times distance. Then because f=ma, we can show that the work done is proportional to the change in v^2. Ron, at least, is not satisfied with this approach, saying

The previous answers all restate the problem as “Work is force dot/times distance”. But this is not really satisfying, because you could then ask “Why is work force dot distance?” and the mystery is the same.

I’m not sure if this is the case. Suppose we assume that acceleration is some function of position, so

\mathbf{\ddot{x}} = a(\mathbf{x})

Then we have

\mathbf{\dot{x}}\mathbf{\ddot{x}} = \mathbf{\dot{x}} a(\mathbf{x})

\frac{1}{2}\frac{d}{dt}(\mathbf{\dot{x}}^2) = \frac{d}{dt} \int_{x_0}^{\mathbf{x}} a(\mathbf{x'}) ds

This shows that we can create a conserved quantity, namely \frac{1}{2}\dot{x}^2 - \int_{x_0}^{\mathbf{x}} a(\mathbf{x'}) ds, using just the knowledge f=ma, which doesn’t seem to be begging the question at all. It does, however, assume that the line integral is independent of path. So if we’re told f=ma and told that f is a gradient, we can find that kinetic energy is proportional to v^2. I suppose we ought to be able to, since from f=ma it is manifest that physics has Galilean invariance, and the thought experiment linked up top shows that energy depends on v^2.

So both explanations seem correct, but the one based on invariance certainly feels better.


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One Response to “Why is kinetic energy proportional to v^2?”

  1. Paul Murray Says:

    One tack would be using conservation of energy (which is the main thing about energy). Demonstrate that it takes the same amount of energy to accelerate something at rest to 3 m/s as it does to accelerate something at 4 m/s to 5 m/s.
    It seems counter-intuitive. After all, from the POV of someone moving at 4 m/s, in the second situation the mass is at rest. Why does an increase of 1 m/s take only 1/2 a joule from the POV of someone moving along with the mass, but 4.5 joules from the POV of someone moving at -3 m/s with respect to the mass?
    Well – in order to accelerate the mass, you have to eject reaction mass in the opposite direction. Conservation of momentum. From the POV of a person travelling along with the mass, both the mass and the reaction mass acellerate. From the POV of the person travelling at -3m/s with respect to the mass, the reaction mass slows down (because it is ejected backwards).
    No matter what the mass of the reaction mass used to accelerate the original unit mass, it all works out that only 1/2 mv^2 works.
    But frankly I’m too drunk at the moment to take the idea further than that.
    Oh – except that we can all agree that one joule of energy will lift 1kg by one meter (in a gravity of 1m/s/s). Whatever “energy” might be, we can all agree that lifting a mass by one unit of height takes one unit of energy (to put it another way, we can be comfortable with that as a definition of energy – lifting things and then those things falling down). Perhaps that, conservation of momentum, and coordinate transformations will suffice to create a thought experiment demonstrating the proportionality of KE to the square of the velocity.

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