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How to derive the kinetic formula

  1. Jan 13, 2012 #1
    1. The problem statement, all variables and given/known data
    There are other ways to derive the kinetic motion equation. show an alternative way using the equations of motion


    2. Relevant equations

    Ek= 1/2mv^2


    3. The attempt at a solution

    W= FD
    W= MAD
    = m( v2-v1/t)(V1+V2/2) T
    = M/2 ( V2^2-V1^2)
    = EK= MV^2/2

    they need another one with the formulas of motion but i can't solve it. Thanks for the help
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 14, 2012 #2

    Curious3141

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    What formulae of motion have you been taught? List them out (there are three (sometimes four) "standard" ones), and you can see which is the relevant one to use.

    Note that those all assume uniform acceleration (as did your solution). The derivation of the equation for kinetic energy can be made without this assumption, but it involves basic calculus. I don't think your question is asking you for this method, so we'll stick with the more basic one.
     
  4. Jan 14, 2012 #3
    yes it's grade 11 physics. These are the ones they are talking about
    v2= v1+ at
    d= (v1+v2/2)t
    d= v1t+ 1/2a(t)^2
    d= v2t+1/2a(t)^2
    v2^2=v1^2+2ad
     
  5. Jan 14, 2012 #4
    Maybe something along this, since it is a no-calc approach;

    W=F*s
    W=m*a*s with angle of attack =0, as with kinetic energy
    we have from equation of motion that s=1/2*a*t^2
    and also v=a*t
    t=v/a
    inserting in motion equation;
    s=1/2*a*(v/a)^2
    inserting this in work formula;
    W=m*a*1/2*a*v^2/a^2
    W=m*1/2*a^2*v^2/a^2
    W=1/2*m*v^2

    Just my initial guess, never seen it done :)
     
  6. Jan 14, 2012 #5
    but isn't w= fd? where did fs come from?
     
  7. Jan 14, 2012 #6
    I'm sorry, here in denmark we use F for force, and s for distance. I take it you use f for force and d for distance of course .. in my previous calculations, s=d :)
     
  8. Jan 14, 2012 #7

    Curious3141

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    The last equation is what you want. Set v1 (initial velocity) = 0 (since the particle starts from rest, when KE is zero). Set v2 to be equal to V, the final velocity. Manipulate a little till you have a*d on the right hand side in terms of V^2 on the left hand side.

    From W (work which is the same as energy) = Fd and F = ma, you can immediately see W = m*a*d.

    Now put that expression for a*d from the previous equation into this one and you'll see that W equals the expression you want. Simply conclude by stating that the work done on the object to bring it from rest to speed V is equal to the KE gained, so KE = (1/2)mV^2.
     
  9. Jan 14, 2012 #8
    Thanks guys it worked out!
     
  10. Jan 14, 2012 #9

    Curious3141

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    Of course, if you can use calculus, this is the best way to do it:

    [itex]dW = Fdx = ma.dx = m\frac{dv}{dt}.dx = m\frac{dx}{dt}.dv = mv.dv[/itex]

    The second last step involved a change of variable of integration.

    Now integrate:

    [itex]\int_0^{E_k} dW = \int_0^V mvdv \Rightarrow E_k = \frac{1}{2}mV^2[/itex].

    And it's actually a shorter proof despite it being a more general result (acceleration does not have to be constant here).
     
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