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Calculating Work Done by Variable Force

  1. Jan 23, 2007 #1
    1. The problem statement, all variables and given/known data
    I need To find the work done by a force that varies with position like this:
    F(x)=2x^3 -5x^2 +10

    with x at values between 0 to 10 meters.

    2. Relevant equations

    see above, and maybe w=fd

    3. The attempt at a solution
    I found the area underneath the curve by putting the values from 0.001 to 10 into Excel, and putting in the equation for all values. it looks like this:

    x (in meters) f(x) area of each increment
    0.1 9.952 0.009952
    0.2 9.816 0.009816
    0.3 9.604 0.009604
    0.4 9.328 0.009328
    0.5 9 0.009
    0.6 8.632 0.008632
    0.7 8.236 0.008236
    0.8 7.824 0.007824
    0.9 7.408 0.007408
    1 7 0.007
    1.1 6.612 0.006612
    1.2 6.256 0.006256


    area of all increments(using 0.001, not 0.1)
    3434.083375

    ^thats just a sample and is only using .1 increments.

    would 3434.083375 actually be my final answer for the work performed, or would that be the force, and if so, what distance would I multiply it by?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 23, 2007 #2

    berkeman

    User Avatar

    Staff: Mentor

    If you're going to solve it (approximately) nunerically instead with integration, then for each delta-X, you multiply the force in that delta-X by the delta-X, and add up all those pieces of the work to get the total.
     
  4. Jan 23, 2007 #3
    I beleive thats what I did, because I took my force, of say 6.612, and * it by .001 and did that for all of them, and added it up. Im wanting to know if that answer is the work done or not.
     
  5. Jan 23, 2007 #4

    berkeman

    User Avatar

    Staff: Mentor

    In your relevant equations, you listed W=fd. So it would seem you have calculated the correct thing.

    There is a subtlety, though in doing numerical calculations like this. You will get small errors for each delta-X section, because the delta-Work that you are calculating is not exact. That's why you will later learn to use integration to calculate the exact total work -- integration basically uses infinitesimally small delta-X sections (called "dx"). One improvement you could make to your numerical calculation is to use the y value for the middle of each delta-X section, rather than the value at the end of the delta-X section as you are doing now.
     
  6. Jan 23, 2007 #5
    Yeah I'm not in Calculus yet, so Integrals wouldn't be an option even IF he hadn't specifically said we couldn't use it.
     
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