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Work-Energy Theorem with Line Integrals

  1. Dec 6, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem is to prove the work-energy theorem: Work is change in kinetic energy.


    2. Relevant equations

    Line integral stuff, basic physics stuff.

    3. The attempt at a solution

    I'm given the normal definitions for acceleration, velocity and I'm given Newton's second law. I'm asked to show that ##\int_c{F \cdot dr}## leads to ##\frac{1}{2}mv(b)^2-\frac{1}{2}mv(a)^2## along an arbitrary path from a to b defined by ##\vec{r}##(t).

    I'm stuck. I got to ##Work = m \int_c{ \frac{d \vec{v}}{d \vec{r}} \, \vec{v} \cdot \vec{v} \, dt}##, but I don't know how to proceed with that pesky dot product in the integrand, and without explicit functions to help me simplify, can anyone help?

    Thanks a billion.


    I'll post my work if it seems like I'm going the wrong way, but I don't know where I would be.
     
  2. jcsd
  3. Dec 6, 2012 #2

    micromass

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  4. Dec 6, 2012 #3
    But without knowing an explicit formula for Force, how am I going to find a gradient field?


    Also, I don't know that my Force field will be conservative, so I can't assume there will be a gradient curve.
     
  5. Dec 6, 2012 #4

    micromass

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    The statement of the result that you must prove makes me think that if a potential function [itex]\phi[/itex] exists, then [itex]\phi(c(t))=\frac{mv(t)^2}{2}[/itex]. Maybe you can use this as definition of your potential function??
     
  6. Dec 6, 2012 #5

    micromass

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    It has to be conservative since the line integral will be the same regardless of the path you take.
     
  7. Dec 6, 2012 #6

    Dick

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    The force doesn't have to be conservative. It's still true. F=ma. a=dv/dt. v=dr/dt. So dr=v*dt. F*dr=m*(dv/dt)*(v*dt). Integrate it. You are making a great deal out of nothing.
     
  8. Dec 6, 2012 #7

    micromass

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    Listen to Dick and ignore my posts. I should have known better than to post in a thread that involves physics. :redface:
     
  9. Dec 6, 2012 #8
    This is more or less what I wrote down, but can I get away with acting like those dot products are multiplications?


    To Micro: Don't worry :)
     
  10. Dec 7, 2012 #9

    Dick

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    Not exactly. But if v is a vector then d(v.v)/dt=2*(dv/dt).v. You'll have to put in the dot products where I omitted them.
     
  11. Dec 7, 2012 #10
    Ah, interesting, I can see where that would lead me. Thank you to the both of you :)
     
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