Modern Physics - Work Energy Theorem

[giraffe]

Homework Statement

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.)

Homework Equations

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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]

The Attempt at a Solution

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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 don't know.

second part probably going to 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 what's left somehow.

thanks for the guidance.

 

[TSny]

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.

 

[vela]

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.