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Intergal? work done by a non constant force

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A force F = (4.38 x i + 2.84 y j) N acts on an object as it moves in the x direction from the origin to x = 5.52 m. Calculate the work done on the object by the force.


    2. Relevant equations
    intergal of F from 0 to 5.52m

    3. The attempt at a solution

    not sure if im suppose to intergrate, or what im supposed to do to start the process
     
  2. jcsd
  3. Oct 8, 2009 #2
    [tex]W=\int\limits_{x=0}^{x=5.52}\vec{F}.{\rm d}\vec{x}[/tex]

    where [tex]\vec{x}=x\vec{i}[/tex] is the position vector.
     
  4. Oct 8, 2009 #3
    ok so u intergrate the sum of X and Y right?
     
  5. Oct 8, 2009 #4
    Not exactly. By the looks of the problem, the object only moved in the x direction. So there is no net work in the y-direction. You only integrate the x-component.
     
  6. Oct 8, 2009 #5
    So then it would be the intergal of 4x?
     
  7. Oct 8, 2009 #6
    Yes.

    [tex]{\rm d}\vec{x}={\rm d}x \vec{i}[/tex]

    so

    [tex]\vec{F}.{\rm d}\vec{x}=\left(4.38x \vec{i}+2.84y \vec{j}\right). \vec{i} {\rm d}x=4.38x{\rm d}x[/tex]

    [tex]W=\int\limits_0^{5.52} 4.38x{\rm d}x[/tex]
     
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