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Rifle Force question

  1. Feb 17, 2009 #1
    1. The problem statement, all variables and given/known data

    A rifle shoots a 4.20 g bullet out of its barrel. The bullet has a muzzle velocity of 955 m/s just as it leaves the barrel.

    Assuming a constant horizontal acceleration over a distance of 43.0 cm starting from rest, with no friction between the bullet and the barrel, what force does the rifle exert on the bullet while it is in the barrel?
    2. Relevant equations

    f=ma Vx^2=Vox+2ax(x-xo)

    3. The attempt at a solution
    I did (955)^2/ 2*.43m to get the acceleration of 1029.8m/s^2
    when i do F=ma I get 4.3218 which the online homework is saying is incorrect.
     
  2. jcsd
  3. Feb 17, 2009 #2

    LowlyPion

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    You might want to recalculate your acceleration.
     
  4. Feb 17, 2009 #3
    Hi, knight4life.

    The second relevant equation looks misquoted. Nevertheless, the substituted values in your computation of the acceleration are correct, so you did something wrong when entering them into the calculator.

    I noticed how you solved the problem: First, you solved for the acceleration, and then you substituted this value, along with the given mass, into the equation which expresses Newton's Second Law.

    A better approach is to substitute the equation, a = F/m, into the other equation and solve for the unknown, which is the force F, and then substitute the values. This will allow you to see how changing one variable, such as the mass, affects the unknown (a generalization of the problem--does a bullet with more mass require more force to obtain the given velocity? this is also a means to check your work), and among other things, it will reduce numerous calculator computations, which could lead to large round-off errors if your not careful. Try it and compare with your approach.

    Good luck with your studies. =)
     
  5. Feb 17, 2009 #4
    I did (955)^2/ 2*.43m then too square root to get 1029.8 m/s2 should i not have taken the square root of 1060494.186?
     
  6. Feb 17, 2009 #5

    LowlyPion

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    What do you think if

    a = V2/2x
     
  7. Feb 17, 2009 #6
    No, I redid my calculations, without taking the square root and I got the correct answer.
     
  8. Feb 17, 2009 #7

    LowlyPion

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    All's well that ends well then.
     
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