Work-Energy Theorem with Line Integrals

[Vorde]

Homework Statement

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

Homework Equations

Line integral stuff, basic physics stuff.

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.

 

[micromass]

What about the fundamental theorem of line integrals? http://en.wikipedia.org/wiki/Gradient_theorem

 

[Vorde]

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.

 

[micromass]

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.

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

 

[micromass]

Also, I don't know that my Force field will be conservative, so I can't assume there will be a gradient curve.

It has to be conservative since the line integral will be the same regardless of the path you take.

 

[Dick]

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.

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.

 

[micromass]

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.

Listen to Dick and ignore my posts. I should have known better than to post in a thread that involves physics.

 

[Vorde]

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.

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 :)

 

[Dick]

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 :)

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.

 

[Vorde]

Ah, interesting, I can see where that would lead me. Thank you to the both of you :)