1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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


    User Avatar
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: How would you calculate the work done