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Error in equation for kinetic energy

  1. Jan 7, 2009 #1
    Okey, lets say we got an object with no start velocity (V_i = 0) which is being pushed by a constant force F, under a certain distance s (during a time intervall dt).
    We get through F = ma that
    F*s = ma*s = ma*v*dt
    We also know that when a is constant (which it is due to the constant force) we got:
    a*dt = 2v
    which gives us that the work, F*s, equals
    2mv^2 instead of (1/2)mv^2
    What is wrong?
  2. jcsd
  3. Jan 7, 2009 #2
    The object does not travel a finite distance s during the infinitesimal time dt. You are mixing your concepts here.

    F*ds = m*a*v*dt

    Try taking it from here.
  4. Jan 7, 2009 #3
    Ah, dt says that t_2 - t_1 approaches zero. But lets just use (t_2 - t_1), I still get it wrong :(
    F*s = m*a*v*(t_2 - t_1) = m*v*2v = 2mv^2

    I assume that over our given time and with our constant force we have a * (t_2 - t_1) = 2*v.
  5. Jan 7, 2009 #4
    Did you know that

  6. Jan 7, 2009 #5


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    You're using "v" to mean the average velocity (= s/dt)

    When we say that work = (1/2) m v^2, "v" is the final velocity.

    The final velocity is twice the average velocity, they are not equal.
  7. Jan 7, 2009 #6

    Doc Al

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    You here assume that s = v*Δt, but that should be s = vave*Δt, where vave is the average speed. For constant acceleration starting from rest, vave = v/2, where v is the final speed. So really:
    F*Δs = ma*Δs = ma*(v/2)*Δt

    No, for constant acceleration, starting from rest: a*Δt = v.
    Nope, you get:
    F*Δs = ma*Δs = ma*(v/2)*Δt = m*(v/2)*a*Δt = 1/2mv^2.

    As expected. :wink:
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