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Finding Work Done

  1. Nov 21, 2009 #1
    1. The problem statement, all variables and given/known data
    If c represents a constant force, find the work done if the point of application of c moves along the line segment from P to Q.
    (Hint: Find a vector b = (b1, b2) such that b is a vector of PQ)

    c = -i+7j P(-2,5) Q(6,1)

    2. Relevant equations
    Components...but I am not sure. Dot Product.


    3. The attempt at a solution
    We went over components and dot product in class, but not how it relates to work. I'm not even sure to begin with this one.
     
  2. jcsd
  3. Nov 21, 2009 #2

    tiny-tim

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    Hi TrueStar! :smile:

    work done = force "dot" displacement (of the point of application of the force). :wink:
     
  4. Nov 21, 2009 #3
    Thanks. :) I've researched it myself..I tried the problem and the answer I got was 32. I found many examples where one of the vectors was at the origin, but not two vectors with different components.

    This is what's throwing me off. Then again, maybe I got the right answer. I don't know yet.
     
  5. Nov 21, 2009 #4

    tiny-tim

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    Maybe you've used the wrong vector for PQ.

    Show us how you got 32. :smile:
     
  6. Nov 21, 2009 #5
    Thanks Tim:

    I made a mistake. I think it's 36. I kind of did it my own way. I knew how to get the answer if one of the vectors was at (0,0). So, I picked -2,5 and decided that I needed to add two to x and subtract 5 from y. I did this to the other vector which changed its components from 6,1 to (8,-4).

    I then used dot product with (8,-4) and -i+7j and got -36. I noticed similar problems to this had positive answered, so I put down 36.

    I know..I'm horrible. I'd rather put something down instead of nothing at all.
     
  7. Nov 22, 2009 #6

    tiny-tim

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    D'oh! :rolleyes:

    You got it right, and then you changed it so as to "conform".

    It is -36 … this particular one comes out negative because the force is in (roughly) the opposite direction to the displacement …

    eg what work does gravity do when a roller-coaster goes up? Obviously, negative! :smile:

    This is the way to do it …

    (btw, don't talk about "changing the components" of a vector … you can't do that to vectors (well, unless you're changing the frame of reference, or the basis)! … talk about subtracting them … that is the official word … they form a "vector space" under addition subtraction and scalar multiplication :wink:)

    P = (-2,5) Q = (6,1), so PQ = Q - P = (8,-4) or 8i - 4j

    c = (-1,7) or -i + 7j

    Then you can find c.PQ either by doing (8,-4).(-1,7) or (8i - 4j).(-i + 7j),

    (both are perfectly valid)

    and either way it's -8 -28 = -36 :wink:
     
  8. Nov 22, 2009 #7
    Oh wow..I just can't believe I got as much correct as I did! The instructor never talked about the concept of work in relation the the topic (components). Our textbook has examples of this where one was at the origin, but not like this one.

    I apologize for my incorrect terminology, this is all new to me and I have yet to take a physics class.

    Thank you so much for helping me out and explaining everything so clearly. I really appreciate it!
     
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