- #1

- 33

- 0

## Homework Statement

A particle moves on a curved path from (x

_{1}, y

_{1},z

_{1}) to (x

_{2}, y

_{2},z

_{2}). At the start, the particle has a velocity of v = v

_{1x}i+v

_{1y}j+v

_{1z}k. This curved path can be divided into segments infinitesimally, which are, dl = dxi +dyj +dzk. It is acted on by a net force F = Fxj + Fyi + Fzk. The force components Fx, Fy, and Fz, are in general functions of position. Prove the work energy theorem for this general case. That is prove that

W

_{tot}=K

_{2}-K

_{1}

where

W

_{tot}=[itex]\int[/itex]F dl = [itex]\int[/itex] F

_{x}dx + F

_{y}dy + F

_{z}dz

where the limits of integration are from (x

_{1}, y

_{1},z

_{1}) to (x

_{2}, y

_{2},z

_{2}) for each.

## Homework Equations

W = Fd

∫F=W

a

_{x}= [itex]\stackrel{dvx}{dt}[/itex] = [itex]\stackrel{dvx}{dx}[/itex] *[itex]\stackrel{dx}{dt}[/itex] = v

_{x}[itex]\stackrel{dvx}{dx}[/itex]

## The Attempt at a Solution

I don't really have any idea where to start but what I did was I took the velocities and made them into accelerations then changed the Fx, Fy, and Fz, into max, may, and maz values, then I'm confused now because I have an integral which looks like this..

[itex]\int[/itex] m*v

_{x}*dv

_{x}+ m*v

_{y}*dv

_{y}/dx dy + m*v

_{z}*dv

_{z}/dx dz

which makes no sense because the dx only canceled out for the X direction, and I need to prove that W

_{tot}= 1/2 mvf^2 - 1/2mvi^2

If anyone could help it'd be appreciated because I've got really no idea what I am doing