# Find work on a straight line path

1. Oct 29, 2008

### steph3824

1. The problem statement, all variables and given/known data

In an experiment, one of the forces exerted on a proton is F=<-ax2,0> where a =12N/m2. Calculate the work done by this force as the proton moves along a straight-line path from the point ri=<0.10m,0> to the point rf=<0.30m, 0.40 m>

2. Relevant equations
I'm not sure if I would use W=F*change in x (don't know how to type in the delta symbol) since it says it's a straight line path, or if I would use W=F*change in r since the problem contains r's in it

3. The attempt at a solution
I'm not too sure what to do here. I see that I'm given a Fx and Fybut I don't know exactly what to do with the multiple r's. Thanks for any help!

2. Oct 29, 2008

### Hootenanny

Staff Emeritus
Where the force is not constant as is the case here, you must use the general defintion of the work done be a force. That is

$$W = \int_\gamma \bold{F}\left(\bold{r}\right)\bold{\cdot}d\bold{\gamma}$$

Where $\bold{\gamma}$ is the path.

3. Oct 29, 2008

### steph3824

Oh ok I didn't realize that the force wasn't constant so I understand why that equation would be used. I still need help though..I'm not sure where to go with that equation

4. Oct 29, 2008

### Hootenanny

Staff Emeritus
Okay. Generally with these types of questions, it is best to determine the path first. You have already figure out that the path is a straight line, so the next thing to do is to parametrise it. Can you do that?

5. Oct 29, 2008

### steph3824

No, I'm not sure what parametrise means.

6. Oct 29, 2008

### Hootenanny

Staff Emeritus
Okay, it simply means write the equation of the straight line in terms of a parameter. You know that the general form of a straight line is y=mx+c, but writing the path in this form doesn't really make it easy to evaluate the path integral. Instead we want to write the x and y coordinates separately in terms of a parameter, say t.

In this case we know that the path is a straight line and can be written in the form

$$y=mx+c = \frac{0.4}{0.2}x - 0.2 = 2x - \frac{1}{5}$$

Now, we need to parametrise this line. Lets start by letting

$$x = t \hspace{1cm}\text{where }\frac{1}{10}\leq t \leq \frac{3}{10}$$

This will give us the change in x as we follow the path. Now we need to obtain an equation for y. All we need to do now is substitute x=t into the equation for our straight line to obtain

$$y = 2t - \frac{1}{5}$$

Hence we have parameterise our path

$$\gamma = \left\{\begin{array}{l} x = t \\ y = 2t - \frac{1}{5}\end{array}\right.\hspace{1cm}\frac{1}{10}\leq t \leq \frac{3}{10}$$

Do you follow?