Find work on a straight line path

[steph3824]

Homework Statement

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 r_i=<0.10m,0> to the point r_f=<0.30m, 0.40 m>

Homework 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

The Attempt at a Solution

I'm not too sure what to do here. I see that I'm given a F_x and F_ybut I don't know exactly what to do with the multiple r's. Thanks for any help!

 

[Hootenanny]

Homework Statement

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 r_i=<0.10m,0> to the point r_f=<0.30m, 0.40 m>

Homework 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

The Attempt at a Solution

I'm not too sure what to do here. I see that I'm given a F_x and F_ybut I don't know exactly what to do with the multiple r's. Thanks for any help!

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.

 

[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

 

[Hootenanny]

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

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?

 

[steph3824]

No, I'm not sure what parametrise means.

 

[Hootenanny]

No, I'm not sure what parametrise means.

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. Let's 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?