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Somewhat basic work problem

  1. Mar 6, 2006 #1
    A single conservative force acts on a 5.00 kg particle. The equation Fx = (2x + 4) N describes this force, where x is in meters. As the particle moves along the x axis from x = 2.60 m to x = 6.20 m, calculate the following.
    (a) the work done by this force
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    Alright this seems pretty straight forward....W=FD, distance equals the change in X which is (6.2-2.6) which equals 3.6....The force acting on the particle is where I get a bit confused. F=2x+4...do you do it with x=6.2 or 2.6? Or do you find the value of 6.2 and subtract is from 2.6? Thanks for any help
     
  2. jcsd
  3. Mar 6, 2006 #2

    Fermat

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    In the general case, you would do an integration.

    WD = int, from x=2.6 to x=6.2, of Fx dx

    However, in this particular case, since Fx is linear, then you can just take the average Force.

    F_av = (F_6.2 + F_2.6)/2.
     
    Last edited: Mar 6, 2006
  4. Mar 6, 2006 #3

    Hootenanny

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    The force is not constant. You need to integrate to find the work done.
     
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