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How would you calculate the work done

  1. Jan 11, 2006 #1
    if the force depended on the DIRECTION of the VELOCITY
    that is [tex] \vec{F} = - F \hat{v} [/tex].
    so suppose it was dragged along a path - say from (0,0) to (1,1) i na straight line what is the work done?
    It going in a striaght line so we could parametrize the curve into r(t) = (t,t)
    and r'(t) = (1,1) right
    and thus v hat = (1,1) as well then?
    now the problem is the work depends on the direction of the velocity

    so then the work done is [tex]W = -F \int_{0}^{1} (1,1) (1,1) dt [/tex] is that right?

    ok so lets say it was dragged along the curve y = x^2
    then r(t) = (t,t^2)
    and r'(t) = (1,2t)
    and [tex] \hat{v} = \frac{(1,2t)}{\sqrt{1+4t^2}} [/tex]
    ok but what the dirrefetial quantity fo the function here? Is it (t,t^2) dt?
    so then is
    [tex] W = -F \int_{0}^{1} \frac{(1,2t))}{\sqrt{1+4t^2}} (t,t^2) dt [/tex]
    is that right?

    Please help!
  2. jcsd
  3. Jan 11, 2006 #2


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    Homework Helper

    You should use:

    [tex]W=\int \vec F \cdot d\vec x = \int \vec F \cdot \vec v \mbox{ } dt[/tex]

    If [itex]\vec F = -|F| \hat v [/itex], then:

    [tex]W= \int -|F|\hat v \cdot \vec v \mbox{ } dt = -|F|\int |v|\hat v \cdot \hat v \mbox{ } dt [/tex]

    [tex]= -|F| \int |v|dt = -|F| \int ds [/tex]

    So you don't have to do most of that work you did.
    Last edited: Jan 11, 2006
  4. Jan 11, 2006 #3
    oh ok that seems much simpler
    but was i did correct by teh way?

    doing it the simpler way though...
    for the straight line from (0,0) to (1,1)
    the integrand would be
    [tex] \int_{0}^{1} \int_{0}^{1} (1,1) \bullet (1,1) dx dy [/tex] ?
    and for hte line from (0,0) to (1,1) along y = x^2
    [tex] \int_{0}^{1} \ int_{y=0}^{y=x^2} (1,1) \bullet (1,1) dy dx [/tex]

    are thos correct?
  5. Jan 11, 2006 #4


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    Homework Helper

    No, there aren't any double integrals in this. 'ds' means the differential arclength. Note that I substituted it in for |v|dt. The integral of ds over a segment of a curve is the length of that segment of the curve. For a curve y(x) in 2D, ds2=dx2+dy2=dx2(1+(dy/dx)2). For example, for the straight line from (0,0) to (1,1), [itex]\int ds =\sqrt{2}[/itex].
    Last edited: Jan 11, 2006
  6. Jan 11, 2006 #5
    ooooo i see i understnad now
    there was another part of this problem that involved proving what you proved in your first post. I totally disregaraded the meaning of dS. SO the integral of dS represents the arc length.

    so for the y=x^2 function the arc legnth would be for r(t) = (t,t^2) and r'(t) = (1,2t)
    [tex] \int_{0}^{1} \sqrt{1 + 4t^2} dt [/tex] yes?
  7. Jan 12, 2006 #6
    can i get a confirmation on my answer? Is it right? I m not sure on how to integrate it though. I tried using mathematica and got an answer that involved sinh which i haven ot come across as of yet.
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