The Work-Energy Theorem

Here’s something not everybody learns about in their all-too-short career in school physics. Did you know about it? About the Work-Energy Theorem?

Something that kinda bothers me about it though, is that it’s called a Theorem, when it’s not quite up to snuff with other theorems I’m all about, like the Binomial Theorem (beginners), Fermat’s Little Theorem (modular!), Euclid’s Theorem (so many Primes!), and Pythagoras’ Theorem (ofc). The feature that ties those together is that they require a Proof, and in my mind, a theorem requires a proof when the statement is not immediately obvious. I think it was Timothy Gowers from the University of Cambridge that said it best. He’s actually quoting an undisclosed colleague here, but oh well:

An aspect of advanced mathematics that many find puzzling is that some of its theorems seem too obvious to need proving… [a good response to this] is that a statement is obvious if a proof instantly springs to mind.

Now I’m not saying this is “obvious” by any means, but it’s more obvious than the others – just my two cents.

Moving on, it’s really not a theorem – ahem – but a rearranging of some mathematics describing the work and kinetic energy of a dynamical system. That’s all. This’ll be a short post, in terms of derivations and such, but this so-called “theorem” has some pretty nice applications in all kinds of places that Newton probably never even dreamed of! (*cough cough* Special Relativity *cough*)

Here’s a simple 1D derivation of the “theorem” – using the SUVAT Equation: v_{final}^2 = v_{initial}^2+2as where s is the displacement and a is the uniform acceleration.

By definition, in 1D, Work can be defined as W=Fs in the direction of the Force.

But force can be defined (in 1D) as F=ma so W=mas

Rearrange the SUVAT equation to get the acceleration a=\frac{v_{final}^2-v-{initial}^2}{2s}

So what happens when we substitute? Hm, I wonder, so let’s find out.

W=Fs=mas=m(\frac{{v_{final}^2}-{v_{initial}^2}}{2s})s=m(\frac{{v_{final}^2}-{v_{initial}^2}}{2})=\frac{1}{2}mv_{final}^2-\frac{1}{2}mv_{initial}^2

Hey, isn’t Kinetic Energy K=\frac{1}{2}mv^2 ? Hey, yeah it is!

W = \Delta{KE}

The ultimate result we get from shifting the variables around, is that the Work done on an object is equal to the change in the object’s Kinetic Energy. Work can therefore be thought of as energy transference, and as we all know… Energy is the ability to do work.

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3 thoughts on “The Work-Energy Theorem

  1. Pingback: Work – Delblg

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