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Modern Physics - Work Energy Theorem

  1. Sep 23, 2014 #1
    1. The problem statement, all variables and given/known data
    The work-energy theorem relates the change in kinetic energy of a particle to the work done on it by an external force: [itex] \triangle K = W = \int F\, dx [/itex]. Writing Newton's second law as [itex] F = \frac{dp}{dt} [/itex], show that [itex] W = \int v\, dp [/itex] and integrate by parts using the relativistic momentum to obtan equation 2.34.

    (this is a 2 part problem. one part is showing that [itex] W = \int v\, dp [/itex] and the second is integrating that equation. i am using modern physics 3rd edition kenneth kramer. i have no idea what equation 2.34 as i can not find it in the chapter.)


    2. Relevant equations

    the equations listed in the problem

    relativistic momentum (in [itex] \frac {\text{kg} \cdot \text{m}}{\text{s}}[/itex] ) [itex]\vec{p} = \frac{m\vec{v}}{\sqrt{1-\frac{v^2}{c^2}}} [/itex]

    relativistic momentum (in MeV) [itex] pc = \frac{mvc}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{mc^2(\frac{v}{c})}{\sqrt{1-\frac{v^2}{c^2}}} [/itex]
    3. The attempt at a solution

    first part, not quite sure. i know i have to make the substitution for F so [itex] \int{\frac{dpdx}{dt}}\ [/itex] after that i dont know.

    second part probably gonna need help with that integral once i figure this first part out. i need to use the second equation to isolate v and than integrate whats left somehow.

    thanks for the guidance.
     
  2. jcsd
  3. Sep 23, 2014 #2

    TSny

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    Integration by parts for two functions ##u## and ##v## can be expressed as $$\int{udv} = uv - \int{vdu}$$

    I suspect that equation 2.34 in your text is the expression for relativistic kinetic energy.
     
  4. Sep 23, 2014 #3

    vela

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    Use the chain rule to express the force as
    $$ F = \frac{dp}{dt} = \frac{dp}{dx} \frac{dx}{dt} $$ and plug that into the integral.
     
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