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Speed Squared

  1. Sep 29, 2004 #1
    I'm trying to tackle the work-energy theorum.

    It has occured to me that the distance in work: Force x distance

    is determined by the acceleration in the force variable, and time of force as well.

    The distance, creates a halving of the final product of the Joule unit, because the distance is half the acceleration. I also see the distance is also a result of the time of force. A longer distance implies more time the force occured. So, acceleration and time result in the distance, but time is sort of embedded in the distance.

    I've read, Work = KE: Force x distance = 1/2 mass x velocity ^2.

    I see the 1/2ing property in Work, because the distance traveled at an acceleration rate is half, but how is the speed squared explained for the other part of KE representation?

    I hope that made sense.
    Last edited: Sep 29, 2004
  2. jcsd
  3. Sep 29, 2004 #2


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    Energy does not make sense unless its units are the same as mass times speed squared. If you are not happy with the "squared", then you need to develop more intuition about energy. I'm not kidding; energy is an abstract quantity, not like volume (you can see), or force (you can feel).

    A simple way to derive the result you want is using the formula from elementary physics:[tex]\Delta v^2=2ad[/tex]. This works for a constant acceleration.

    A better way that does not require a to be constant is the following (but you need to know calculus): [tex]W=\int Fdx=m\int adx=m\int {dv\over dt}dx=m\int {dv\over dx}{dx\over dt}dx=m\int vdv=m\Delta v^2/2[/tex]
  4. Oct 2, 2004 #3
    [tex]\Delta v^2=2ad[/tex], seems a bit cumbersome. My textbook explains something similar by comparing momentum to energy, but saying that kinetic energy is equal to momentum squared divided by twice the mass. I just don't get the round about way that shows the corelation.

    Since I've asked the question, something occurred to me in looking at the equation, that seems to be a bit more directly representative.

    If I think in terms of KE by isolating the the magnitude of acceleration and the magnitude of distance traveled, I see the work acceleration and distance magnitudes corelate exactly.

    In terms of there respective magnitudes:

    In Work = mass(acceleration)(distance).

    In KE = mass (velocity) and either (distance) or (average velocity)

    The acceleration magnitude is multiplied by the distance or average velocity magnitude the acceleration produces. Why they are multiplied is my next question. Do you know why?
  5. Oct 2, 2004 #4


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    You could use lineal momentum. krab used the chain rule.

    [tex] W = \int F dx [/tex]

    Newton's 2nd Law

    [tex] F = \frac{dP}{dt} [/tex]

    Constant Mass

    [tex] F = m \frac{dv}{dt} [/tex]


    [tex] W = m \int \frac{dv}{dt} dx [/tex]

    You know

    [tex] v = \frac{dx}{dt} [/tex]


    [tex] vdt = dx [/tex]

    Substituing again:

    [tex] W = m \int \frac{dv}{dt} vdt [/tex]

    [tex] W = m \int vdv [/tex]

    [tex] W =\frac{m \Delta v^2}{2} [/tex]

    Work done is equal to the change in the kinetic energy

    [tex] W = \Delta K [/tex]

    For your books way:

    [tex] \frac{P^2}{2m} = \frac{mv^2}{2} \frac{m}{m} [/tex]

    Same thing.
    Last edited: Oct 2, 2004
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