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Find work on a straight line path

  1. Oct 29, 2008 #1
    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. jcsd
  3. Oct 29, 2008 #2

    Hootenanny

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    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

    [tex]W = \int_\gamma \bold{F}\left(\bold{r}\right)\bold{\cdot}d\bold{\gamma}[/tex]

    Where [itex]\bold{\gamma}[/itex] is the path.
     
  4. Oct 29, 2008 #3
    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
     
  5. Oct 29, 2008 #4

    Hootenanny

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    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?
     
  6. Oct 29, 2008 #5
    No, I'm not sure what parametrise means.
     
  7. Oct 29, 2008 #6

    Hootenanny

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    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

    [tex]y=mx+c = \frac{0.4}{0.2}x - 0.2 = 2x - \frac{1}{5}[/tex]

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

    [tex]x = t \hspace{1cm}\text{where }\frac{1}{10}\leq t \leq \frac{3}{10}[/tex]

    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

    [tex]y = 2t - \frac{1}{5}[/tex]

    Hence we have parameterise our path

    [tex]\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}[/tex]

    Do you follow?
     
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