# Modern Physics - Work Energy Theorem

1. Sep 23, 2014

### giraffe

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: $\triangle K = W = \int F\, dx$. Writing Newton's second law as $F = \frac{dp}{dt}$, show that $W = \int v\, dp$ and integrate by parts using the relativistic momentum to obtan equation 2.34.

(this is a 2 part problem. one part is showing that $W = \int v\, dp$ 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 $\frac {\text{kg} \cdot \text{m}}{\text{s}}$ ) $\vec{p} = \frac{m\vec{v}}{\sqrt{1-\frac{v^2}{c^2}}}$

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

first part, not quite sure. i know i have to make the substitution for F so $\int{\frac{dpdx}{dt}}\$ 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. Sep 23, 2014

### 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.

3. Sep 23, 2014

### vela

Staff Emeritus
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.