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The concept of work and types of energy

  1. Sep 28, 2012 #1
    The concept of "work" and types of energy

    I have read the other posts regarding the concept of work and still don't understand why distance has anything to do with it. Anyway, I thought that maybe an explanation of the following might help:

    Suppose you have a 1kg space-ship in an empty universe, with zero velocity in some frame of reference. Then it consumes its rocket, which is able to push with 1 Newton force for 1 second**. The ship is now going 1meter / second and has a total energy of 1/2*mv^2 = 0.5 Joules kinetic energy.

    Now suppose the same scenario, but the ship initially has a 1 meter / second velocity. The final velocity is 2 meters / second = 1/2*mv^2 = 2 Joules.

    So in the first case, the transition of chemical energy to kinetic energy results in a gain of 0.5 Joules for the ship, whereas in the second case, a gain of 1.5 Joules occurs.

    **Is it safe to say that a rocket can provide the same impulse, regardless of starting velocity?
     
  2. jcsd
  3. Sep 28, 2012 #2
    Re: The concept of "work" and types of energy

    You are correct. The same rocket will provide the same impulse.

    Here's a simple derivation from basic kinematics that should help convince you that W = Fd. It's not a perfect derivation, because I wont factor in any possible potential energy, but here it goes:

    [itex]d = \frac{v_2^2-v_1^2}{2a}[/itex]

    therefore,

    [itex]\Delta v^2 = 2ad[/itex]
    [itex]\frac{1}{2}\Delta v^2 = ad[/itex]

    multiplying both sides by mass:
    [itex]\frac{1}{2}m\Delta v^2 = mad[/itex]

    F = ma, and (1/2)mΔv2 = ΔE, so:

    [itex]\Delta E = Fd[/itex]

    and ΔE is defined as the Work.

    If you look at equations for, say, graviational potential, you'll very easily see that the change in potential energy is the work:

    ΔU = mgh is the same as saying ΔE = W = mad = Fd where g→a, and h→x. This being the case where only the potential changes, and not the kinetic. This should convince you as to why distance is necessary in work calculations.
     
    Last edited: Sep 28, 2012
  4. Sep 28, 2012 #3

    DrDu

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    Re: The concept of "work" and types of energy

    Your model is too simplistic. The rocket moves by expelling hot gas at a certain velocity w so that the momentum of the gas expelled equals the change of momentum of the spaceship and the kinetic energy of the gas plus the kinetic energy of the spaceship equals the energy stored in the fuel burned.
     
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