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Work done by a varying force

  1. Oct 31, 2005 #1
    A force F = (4xi + 3yj) N acts on an object as the object moves in the x direction from the origin to x = 5.00m. Find the work W = (Integral)(F)(DOT)dr done on the object by the force.

    i figure that to solve this i would integrate 4xi +3yj from 0 to 5.00

    my main question here is how do i integrate 4xi + 3yj? if im not doing this right...could some1 direct me in the right path. thx
     
  2. jcsd
  3. Oct 31, 2005 #2
    Take the dot product first. If it moves along the x-axis there is no displacement in the y direction.
     
  4. Oct 31, 2005 #3

    cepheid

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    Normally, [itex] d\mathbf{r} [/itex] is expressed as [itex] dx \mathbf{\hat{i}} + dy \mathbf{\hat{j}} [/itex]. However, in this case, the displacement is entirely in the x direction, dy = 0 along the entire path, and so an infinitesimal displacement vector along that path becomes simply [itex] dx \mathbf{\hat{i}} [/itex]. Dotting this with the force vector of course eliminates the y component, giving you a scalar that is the magnitude of the force times the magnitude of the infinitesimal displacement in the x direction:

    [tex] \mathbf{F} \cdot d\mathbf{r} = (4x \mathbf{\hat{i}} + 3y \mathbf{\hat{j}}) \cdot (dx \mathbf{\hat{i}} + 0 \mathbf{\hat{j}}) [/tex]

    [tex] = 4xdx (\mathbf{\hat{i}} \cdot \mathbf{\hat{i}}) = 4xdx [/tex]

    [tex] \therefore \ W = \int{\mathbf{F} \cdot d\mathbf{r}} = \int{4xdx} [/tex]

    Of course, we could have arrived directly at this result without doing any dot product explicity if we had just thought to ourselves, "well, all the displacement is in one direction, and we know that only the component of the force that acts in that direction will do any work." So we can arrive directly at the result that applies to the special case of work done by a varying force with a displacement in a straight line:

    [tex] W = \int{F_x dx} [/tex]
     
    Last edited: Oct 31, 2005
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