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Problem involving work.

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data
    A 100-g bullet is fired from a rifle having a barrel 0.600 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 15 000 + 10 000x -25000x^2, where x is in meters. (a.)Determine the work done by the gas on the bullet as the bullet travels down the length of the barrel. (b.) What if? If the barrel is 1.00m long, how much work is done, and how does this value compare with the work calculated in part (a.)?

    m=100g
    x=0.600m
    F=15 000N +10 000x-25 000x^2, x is meters


    2. Relevant equations
    F=kx
    w=.5k[itex]\Delta[/itex]x


    3. The attempt at a solution
    I did tried to find k, using F=kx.
    k=F/x=15 000N+10 000x-(25 000x^2/x)
    =(15 000N/x)+10 000-25 000x

    I also tried to find work, using the w=.5k[itex]\Delta[/itex]x equation.
    w=.5(15 000N/x)+10 000-25 000x*.600

    I am having trouble working this out problem.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 28, 2011 #2
    Hi,
    Make sure you take the integral of the force for the work, in that:
    [itex]
    \displaystyle W = \int_0^x{Fdx}
    [/itex]
     
  4. Sep 28, 2011 #3
    You are given the force as a function of x.
    F(x) = ( 15000 + 10000x -25000x^2 )N

    Work = Force * Distance --> W = Fd
    Distance is simple, d = 0.6m

    Force depends on the position of the bullet in the rifle. Thus you integrate the force function over distance 0m -> 0.6m.

    That should give you about 9kN.

    Then you just calculate W = 9000N * 0.6m which is aroundish 5400Nm.
    The b part is simple after that. Hopefully i did it right :).
     
  5. Sep 28, 2011 #4
    Thank you so much. One of my problem was that I didn't do was the integral of work.
     
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