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Math question related to the work-energy theorem

  1. Dec 1, 2014 #1
    There’s a mathematical physics question I have that’s been bugging me lately. I’m not a mathematician so I don’t know if my logic is mathematically “legal” or sound.

    Part 1

    1. Say we restrict ourselves to one dimension and define a spatial coordinate, x. Then we square it, so now we have x^2.

    2. Now we make an equation of it, x^2 = x^2.

    3. Now I want to divide each side by 1/t^2, yielding (x^2)/(t^2) = (x/t)(x/t) = v^2, where v=velocity.

    Does that look ok so far, can I do that operation?

    4. Now I want to redo step 3 in a different manner. Instead of what I wrote above, I want to do the following: divide each side by 1/t^2, yielding (x^2)/(t^2) = (x/t^2) (x) = (a)(x), where a=acceleration and x=spatial displacement along the x coordinate.

    My question is, can I take my choice as to how I can split up the time variables on the RHS of the equation in both 3 and 4 given what’s on the left?

    5. What about the differential form of 3 and 4? Can I write the RHS version of 3 as (dx/dt)(dx/dt), and version 4 as (d^2x/dt^2)(dx)?

    Part 2

    I’m interested in the math aspect of the equation but it was the work energy theorem that got me thinking about it. Through a few mathematical tricks I can comfortably follow, the work energy theorem gives you a kinetic energy term, ½ mv^2 from the work term, or, Force times distance (m)(a)(x) = ½(m)(v^2), where m is mass, v is velocity, a is acceleration, and x is spatial displacement.

    Now, if we add a mass term to my equation 3 above in part 1, you get mv^2, and if you add that same mass term to equation 4 you get (m)(a)(x), or the work term.

    So, if it isn’t obvious already, my conundrum here is that, from my mathematical tinkering in numbers 1-4 above in part 1, I came up with an equivalence (m)(a)(x) = mv^2, but the work-energy theorem tells us that (m)(a)(x) = 1/2mv^2. Where did that extra ½ term come from?
  2. jcsd
  3. Dec 1, 2014 #2
    You've assumed constant speed and a non zero acceleration. In general ##\frac{x}{t} \neq v ## and similarly ##\frac{x}{t^2} \neq a ##
  4. Dec 1, 2014 #3


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    Staff: Mentor

    Loosely speaking (very loosely!) the factor of 1/2 appears because the speed of an object undergoing constant acceleration from rest for a time ##t## is ##v(t)=at## but its average speed over that time is ##at/2##.

    The problem with your calculation is that it's not correct to say that ##v=x/t## - if it were, the speed of an object would change according to which point you choose to be ##x=0##. You can say that the average velocity is ##\Delta{x} / \Delta{t}## but that doesn't tell you anything about the speed at any given moment.
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