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Energy Proof

  1. Oct 16, 2004 #1
    I am in grade 12 physics, and i have to practice equation proofs. I am currently studying work, kinetic energy, springs, and potential energy (gravity and elastic).

    Does anyone have a good proof?
     
  2. jcsd
  3. Oct 16, 2004 #2

    arildno

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    Proof of what?
    Your question is much too vague..
     
  4. Oct 16, 2004 #3
    Sorry, i mean proving an energy-related equation.
     
  5. Oct 16, 2004 #4

    arildno

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    Which of them?
    Specifics, please.
     
  6. Oct 16, 2004 #5
    i know its a simple equation, but i suck at proofs.

    how about the general equation for gravitational potential energy?
     
  7. Oct 16, 2004 #6

    arildno

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    Ok, I'll derive the conservation of mechanical energy for you, in the case of a point particle under the influence of the force of gravity.
    I write newton's 2.law in vector form:
    [tex]-mg\vec{j}=m\vec{a}[/tex]
    where [tex]\vec{a}=\frac{d^{2}\vec{x}}{dt^{2}}[/tex] is the acceleration, and
    [tex]\vec{x}(t)=x(t)\vec{i}+y(t)\vec{j}[/tex]
    and the velocity is given by:
    [tex]\vec{v}=\frac{d\vec{x}}{dt}[/tex]
    We also have:
    [tex]\vec{a}=\frac{d\vec{v}}{dt}[/tex]
    1. Form the dot product
    between [tex]\vec{v}[/tex] and Newton's 2.law:
    [tex]-mg\frac{dy}{dt}=m\frac{d\vec{v}}{dt}\cdot\vec{v}[/tex]
    2. Integrate this equation between 2 arbtriray points of time:
    [tex]\int_{t_{0}}^{t_{1}}-mg\frac{dy}{dt}dt=\int_{t_{0}}^{t_{1}}m\frac{d\vec{v}}{dt}\cdot\vec{v}dt[/tex]
    3. The left-hand side is easy to compute:
    [tex]\int_{t_{0}}^{t_{1}}-mg\frac{dy}{dt}dt=-mgy(t_{1})+mgy(t_{0})[/tex]
    4. We note the identity:
    [tex]\frac{d\vec{v}}{dt}\cdot\vec{v}=\frac{d}{dt}(\frac{\vec{v}^{2}}{2})[/tex]
    where [tex]\vec{v}^{2}\equiv\vec{v}\cdot\vec{v}[/tex]
    5. Hence, the right-hand side in 2.) may be computed:
    [tex]\int_{t_{0}}^{t_{1}}m\frac{d\vec{v}}{dt}\cdot\vec{v}dt=\frac{m}{2}\vec{v}^{2}(t_{1})-\frac{m}{2}\vec{v}^{2}(t_{0})[/tex]
    6. Collecting insights from 3. and 5., 2. may be rewritten as:
    [tex]-mgy(t_{1})+mgy(t_{0})=\frac{m}{2}\vec{v}^{2}(t_{1})-\frac{m}{2}\vec{v}^{2}(t_{0})[/tex]
    7. Or, rearranging 6., we gain:
    [tex]mgy(t_{1})+\frac{m}{2}\vec{v}^{2}(t_{1})=mgy(t_{0})+\frac{m}{2}\vec{v}^{2}(t_{0})[/tex]
    8. Or, noting that [tex]t_{1},t_{0}[/tex] were ARBITRARY, every mechanical energy amount must remain the same at all times, so we get, by eliminating the specific time parameter:
    [tex]mgy+\frac{m}{2}\vec{v}^{2}=K[/tex]
    where K is some constant for the whole motion.
    That is, the mechanical energy is conserved for the particle
     
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