1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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:
    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook