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Integral problem.

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data

    A 80 g bullet is fired from a rifle having a barrel 0.550 m long. Assuming the origin is placed where the bullet begins to move, the force (in newtons) exerted by the expanding gas on the bullet is 13000 + 5000x - 26000x2, where x is in meters.

    (a) Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

    (b) If the barrel is 1.15 m long, how much work is done, and how does this value compare to the work calculated in part (a)? (work done with the 1.15 m barrel / work done with the 0.550 m barrel)

    2. Relevant equations

    13000 + 5000x - 26000x2

    13000x + (5000/2)x2 - (26000/3)x3

    3. The attempt at a solution

    I know I have to get the integral. So that would be 13000x + (5000/2)x2 - (26000/3)x3. So when I substitute .550 m in I get 3580.5 J which is then 3.5805 Kj. Then for 1.5 m i get -62625 J, which is -6.2625 Kj. Did I do this right? I have a feeling I didn't, since I know almost nothing about integrals.
  2. jcsd
  3. Nov 18, 2008 #2
    Complete Solution Removed
    Last edited by a moderator: Nov 18, 2008
  4. Nov 18, 2008 #3


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    unscientific, please check your PM's
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