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Can one derive F=ma from kinetic energy ?

  1. Jan 9, 2012 #1
    I start with the expression for kinetic energy

    [itex]\Large E = \frac{1}{2}mv^2[/itex]

    Differentiate both sides by [itex]x[/itex]

    [itex]\Large \frac{dE}{dx} = m v \frac{dv}{dx}[/itex]

    Substitute the following expression for the velocity [itex]v[/itex]

    [itex]\Large v = \frac{dx}{dt}[/itex]

    To get

    [itex]\Large \frac{dE}{dx} = m \frac{dv}{dx} \frac{dx}{dt}[/itex]

    Using the differentiation chain rule:

    [itex]\Large \frac{dE}{dx} = m \frac{dv}{dt}[/itex]

    If I apply a force F to the mass m then the differential work done by the force is

    [itex]\Large dE = F dx[/itex]

    Substituting this expression we find:

    [itex]\Large F = m a[/itex]
     
    Last edited: Jan 9, 2012
  2. jcsd
  3. Jan 9, 2012 #2
    I have never seen this before but it does come up with the correct relationship.
    From a dimensional point of view it is correct dE/dx has units of joule/metre.which is 'force'
    You are quite right to note that field (force) is the negative gradient of potential (energy)
     
  4. Jan 9, 2012 #3
    Sorry to have re-edited under you!
     
  5. Jan 9, 2012 #4

    D H

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    You have found a simplified version of Hamilton dynamics,
    [tex]\begin{aligned}
    \dot p = - \frac{\partial H}{\partial q} \\
    \dot q = \phantom{-} \frac{\partial H}{\partial p}
    \end{aligned}[/tex]
    where H is the Hamiltonian, p is generalized momentum, and q is generalized position. In the simplest form, H is the total mechanical energy, p is the canonical momentum vector, and q is the canonical position vector.
     
  6. Jan 9, 2012 #5
    Here you are shifting the meaning of your symbols.
    "E" was kinetic energy up to this point.
    You can say that Fdx is the work done by the force F (by definition) but the fact that the work is equal to the variation of kinetic energy is not so by definition. You may prove it by using Newton's second law, for example (see work-energy theorem).
    On the other hand, if you assume the work-energy theorem, you can prove Newton's 2nd starting with it. Everything before this point becomes quite un-necessary.
     
  7. Jan 9, 2012 #6
    Here's another go to derive F=ma from the total energy of a 1-d particle:

    Let the total energy E be given by

    [itex] \Large E = \frac{1}{2}m v^2 + V(x)[/itex]

    Now assume that the total energy is zero (*crucial assumption*):

    [itex] \Large E = 0[/itex]

    So that we have:

    [itex] \Large -V = \frac{1}{2}m v^2 [/itex]

    Differentiate both sides by x

    [itex] \Large -\frac{dV}{dx} = m v \frac{dv}{dx} [/itex]

    Substituting the following expression for the velocity [itex]v[/itex]

    [itex]\Large v = \frac{dx}{dt}[/itex]

    To get

    [itex] \Large -\frac{dV}{dx} = m \frac{dv}{dx} \frac{dx}{dt} [/itex]

    Using the differentiation chain rule

    [itex] \Large -\frac{dV}{dx} = m \frac{dv}{dt} [/itex]

    Now I use a definition for force in terms of the gradient of the potential:

    [itex] \Large F = - \frac{dV}{dx} [/itex]

    to finally obtain

    [itex] \Large F = m a [/itex]
     
    Last edited: Jan 9, 2012
  8. Jan 9, 2012 #7
    I know advanced techniques are way more fun but for those of us who like to keep things as simple as possible, does this work?

    E=1/2*m*v2;
    Wk=F*s (F*distance) delta work equals final energy, so we can say that E=F*s
    a=v2/2*s
    So,
    F=m*v2/2*s
    and,
    F=m*a
     
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