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Work along a line integral

  1. May 19, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle is attracted towards the origin by a force proportional to the cube of its distance from the origin. How much work is done in moving the particle from the origin to the point (2,4) along the path y = x^2 assuming a coefficient of friction [tex]\mu[/tex] between the particle and the path?

    3. The attempt at a solution

    For the path, set x = t. Thus, y = t^2

    [tex]\vec{r}(t) = t\hat{i} + t^2\hat{j}[/tex]
    [tex]\vec{r}'(t) = \hat{i} + 2t\hat{j}[/tex]

    Because we're going from the origin to (2,4), t goes from 0 to 2.

    For the force, we know that:

    [tex]\left\|F\right\| = {k(x^2 + y^2)}^{3/2}[/tex]

    I want this ideally in vector form, but the best I could come up with:
    [tex]{\left\|F\right\|}^{1/3} = kx\hat{i} + ky\hat{j}[/tex]

    My main problem is I can't really use [tex]{\left\|F\right\|}^{1/3}[/tex] to calculate [tex]\int_{C} {F\cdot dr}[/tex]. If I could just find a way to express F in vector form, I'm pretty much off to the races. (Because I can then calculate the work done by friction by finding the normal to [tex]\vec{r}(t)[/tex], multiplying it by [tex]\mu[/tex], and dotting it with [tex]\vec{r}'(t)[/tex] from [tex]0\leq t \leq 2[/tex], etc. etc.)
     
    Last edited: May 19, 2009
  2. jcsd
  3. May 19, 2009 #2

    Dick

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    The vector (x*i+y*j)/sqrt(x^2+y^2) points away from the origin in the direction of (x,y) and has length 1. Just take that vector and adjust it's length and direction.
     
  4. May 19, 2009 #3
    How would I adjust for the "cube of the distance" though ? Looking at the numerator of that unit vector, there isn't much, I think, that one can do with [tex](x\hat{i} + y\hat{j})^3[/tex]
     
  5. May 19, 2009 #4

    Dick

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    The cube of the distance is r^3=(x^2+y^2)^(3/2), like you said. If u=(x*i+y*j)/sqrt(x^2+y^2) has length 1, shouldn't r^3*u have length proportional to r^3? What does that vector simplify to? (x*i+y*j)^3 doesn't even mean anything.
     
    Last edited: May 19, 2009
  6. May 19, 2009 #5
    (facepalm of realization)

    I'd probably multiply r^3 and u together, wouldn't I.

    Then slap on the k at the front of the product, and equate it to F.
     
  7. May 19, 2009 #6

    Dick

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    Sure, where k is going to be negative if you want it to point towards the origin.
     
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