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Acceleration of a brick, involving integrals

  1. Oct 1, 2012 #1
    1. The problem statement, all variables and given/known data
    A 15 kg brick moves along an x axis. Its acceleration as a function of its position is shown in Fig. 7-32. What is the net work performed on the brick by the force causing the acceleration as the brick moves from x = 0 to x = 8.0 m?

    2. Relevant equations
    W = FΔx = max

    3. The attempt at a solution
    I haven't taken calculus prior to the class(am taking it now), and I was just introduced to integrals today by a friend. He showed me how to integrate another Work problem, where F = -6x. I understood that it would integrate into -3x2 + C, and I found C. My question is how would I go about this one?

    Would I start as ∫F dx = ∫ (15 kg)a dx
    W = ∫ 15a dx
    W = 15ax
    W = 15ΔaΔx ... I'm guessing this is what I should do...
    W = 15(24)(80) = 2400 J

    But apparently that's wrong.
    I would be soooooo thankful for anyone to help me with the integration process from here.

    ***************************************************************
    And just in case anybody tells me to ask my teacher for help, I asked both my Physics C teacher AND my calc teacher, but both refused.
     
  2. jcsd
  3. Oct 1, 2012 #2
    There is no picture in your post.
     
  4. Oct 2, 2012 #3
  5. Oct 2, 2012 #4
    You don't need to know integration techniques. It is enough to understand that the value of the (definite) integral of a function represents the area under the graph of the function.
    Here is a lot more straightforward to calculate the area (as all the segments are straight lines).
     
  6. Oct 2, 2012 #5
    oh, wow. I'm an idiot.
    I made a typo in typing the url of the picture
    here's the correct one:
    http://www.webassign.net/hrw/07_34.gif
    perhaps I should start using Ctrl C more often...
     
  7. Oct 2, 2012 #6
    I mean, it seems rudimentary that the net work is the area under the curve times mass, since
    the area under the curve is aΔx, and multiplying by m gives you maΔx, which equals work.
    And yet, that was wrong.
     
  8. Oct 2, 2012 #7
    So did you get the right answer with the new figure or not? I don't understand.
     
  9. Oct 2, 2012 #8
    ahh.... I just viewed the answer key. It has 1200 J, while I kept getting 2400 J.
    I think I just forgot to throw the 1/2 in there. Damn. A point lost.
    But thanks for helping anyways!
     
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