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Help with Work (very easy problem)

  1. Oct 23, 2006 #1
    I honestly feel like a complete idiot asking this, since I know it's really easy, but I just can't get the answer. The question is...

    A 3.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown in Figure 7-38 (attached). How much work is done by the force as the block moves from the origin to x = 8.0 m?

    I thought this problem was as simple as finding the area under the curve, because it's force versus distance. Doing this, I got -45 because the bottom of the graph goes into the negatives (or 105 if negatives don't mean anything). Neither of these answers are correct (using WebAssign for homework), and I can't figure out why. Am I just being really stupid with the math? Does the mass even matter, I thought it didn't because you're just integrating? Sorry for the stupid question, I just don't have anyone else to ask... :shy:
     

    Attached Files:

  2. jcsd
  3. Oct 23, 2006 #2

    radou

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    I can't see the attachment, but basically, the work should be [tex]W = \int_{0}^8 F(x)dx[/tex], where F(x) is your graph. The integral actually equals the change in potential, U(8) - U(0), where the potential U(x) is the antiderivative of F(x).
     
  4. Oct 23, 2006 #3

    rsk

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    Can you give us an expression for F - given that we can't see the attachment yet?

    You must have one, to be able to integrate? Or is your graph just straight lines?
     
  5. Oct 23, 2006 #4
    It is simply a graph with straight lines. I thought that you simply break them up into rectangles and triangles and just add the areas..I really hope my simple math is just wrong, because otherwise I have no idea..Heres the direct link to the graph.

    http://www.webassign.net/hrw/07_33.gif
     
  6. Oct 23, 2006 #5

    rsk

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    It must be your maths. I can't get 105 or 45 no matter what.

    Remember you´re taking the areas around the F=0 line.
     
  7. Oct 23, 2006 #6
    Oh, around the 0? Okay, 25, I got it. I thought you just took it to the bottom of the graph itself. Sorry for the stupid question, Im a freshman in high school in physics, so I don't pick up on this stuff easily by myself. Thank you very much. :biggrin:
     
  8. Oct 23, 2006 #7

    rsk

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    Easy mistake from the way that graph is drawn!
     
  9. Oct 23, 2006 #8

    radou

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    There are no stupid questions. :smile:
     
  10. Oct 23, 2006 #9

    rsk

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    But there are, at times, stupidly presented graphs.... ;)
     
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