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Work done on a force using dot product

  1. Oct 6, 2006 #1
    The question with all the info is:
    A force F = (6x i + 5y j) N acts on an object as it moves in the x direction from the origin to x = 5.04 m. Find the work (work = integral of F (dot product) dx done on the object by the force.

    I am confused as to what dx would be in this equation, and how to find it.


    The work that I've done so far is (using an equation from my book) W= F(dot product) change in x. First off, will this equation even work? Its a little different than the first one I wrote. But second, since the dot product is used for two vectors i wrote the change in x part of that equation as a vector <5.04, 0>. So taking <5.04, 0> (dot product) <6x, 5y> = 30.42x. I don't think there should be a variable in my answer so I tried to answer with 30.42 but that isn't correct.

    I have a feeling that I am far off base with what I've done so far. Please help, thank you.
     
  2. jcsd
  3. Oct 6, 2006 #2

    Doc Al

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    Staff: Mentor

    The displacement element is just "dx i", since it only moves in the x direction. Now integrate F*(dx i) over the range of x.
     
  4. Oct 6, 2006 #3
    I don't know what "dx i" stands for, or how to calculate it. Is it the displacement in the x direction?
     
  5. Oct 6, 2006 #4

    Doc Al

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    Yes, that's all it is. Displacement is a vector; the element magnitude is "dx" and the direction is "i". Now take the dot product and integrate.

    In general the displacement would be: dx i + dy j + dz k

    But in this case, it only moves in the x direction.
     
  6. Oct 6, 2006 #5
    Thanks for the quick replies DocAl!
    Here are my calculations:

    F=<6x, 5y> delta x = <5.04, 0>
    F*delta x = 30.42x
    Integral from 0 to 5.04 of 30.42xdx = (30.42x^2)/2 from 0 to 5.04 = 386.23

    Webassign says that this is incorrect, and that my answer differs from the correct one by 10% to 100%

    Ive triple checked my calculation, I'm not sure where my error is..
     
  7. Oct 6, 2006 #6

    Doc Al

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    Your error is putting dx = 5.04. 5.04 is the range of x; dx is an infinitessimal element along the x axis. You must integrate over the range of x. Do this:
    F = 6x i + 5y j
    ds = dx i

    What's F*ds? First figure that out, then integrate from x = 0 to x = 5.04.
     
  8. Oct 6, 2006 #7
    F * ds = 6xdx
    So the integral over the range of x is 3x^2 from x=0 to x=5.04 so the answer is 76.2048 right?
     
  9. Oct 6, 2006 #8

    Doc Al

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    Looks good to me.
     
  10. Oct 6, 2006 #9
    Thanks very much! I appreciate it
     
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