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A bullet is fired in a rifle

  1. Nov 7, 2012 #1
    1. The problem statement, all variables and given/known data

    smMBN.png
    Need help on question f.

    2. Relevant equations

    1)KE = 1/2 mv2
    2)PE = mgy
    3)Wnet = KE + PE
    4)W = Fdcosθ
    5)Vf2= Vo2 + 2a(X-Xo)
    6) F= ma

    3. The attempt at a solution

    A) Equation 1 = 4563J
    B) Equation 3 = 4563.10584J
    C) Equation 4 = 0
    E) Equation 5 for accerleration = 422500 m/22. Then used Equation 6 = 6337.5N

    F) Not sure if they should be the same and my answers are wrong, or they are just different.
     
  2. jcsd
  3. Nov 7, 2012 #2
    The answers for average force should be the same which is 6337.5 J for the work-energy or Newton formulations of the problem. For part C, use the fact that work done on the bullet equals the change in kinetic energy. From that you compute the force from work-energy theory.
     
  4. Nov 7, 2012 #3
    That's what I did.

    W = Fdcosθ
    W/dcosθ = F
    4563.10584J/(.72m)cos90 = 0

    Cos90 = 0, and you cant divide by 0, I am stuck on this
     
  5. Nov 7, 2012 #4
    The angle is 0, not 90 degrees. The force due to pressure is coincident with the direction of the bullet. So the formula reduces to Fd=W.
     
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