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Greatest Speed

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data
    The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with mass 0.0300 kg is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 cm long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal.

    1) Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 N acts on the ball as it moves along the barrel.

    2) For the situation in part (1), at what position along the barrel does the ball have the greatest speed? (In this case, the maximum speed does not occur at the end of the barrel.)

    2. Relevant equations
    KE = 1/2mv^2


    3. The attempt at a solution
    I solved the first part already and I have 4.90m/s but I'm stuck on the second one. When does it have the greatest speed anyway? I'm thinking that's when the KE is the greatest and positive? Is this correct?

    ...That's probably wrong because I tried it and had the wrong answer. :redface:
     
  2. jcsd
  3. Feb 12, 2007 #2

    berkeman

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    Staff: Mentor

    The speed (and KE) will be greatest at the instant when the positive acceleration goes down to zero. Can you tell us why?
     
  4. Feb 12, 2007 #3
    Erm, I honestly have no idea. :( I'm going to blindly take a guess. When a = 0, doesn't that mean there's no force? And doesn't that mean there's no work? So when there's no work, that's when the speed and KE is greatest...because it doesn't have to do anything? :x
     
    Last edited: Feb 12, 2007
  5. Feb 12, 2007 #4

    berkeman

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    No, you should be thinking of what's going on in this question in a different way, not in terms of "no work".

    Remember that the sum of F=ma, so you add up the positive force from the spring and the negative (retarding) constant force, and you will get a net force and a corresponding net acceleration. The ball will be accelerating (gaining speed, right?) until the two forces are equal, and after that it will be decelerating (losing speed). So to figure out the place where the speed is the greatest, figure out where the two forces become equal. Does that make sense? The spring force decreases as it extends, but the retarding force is constant....
     
  6. Feb 12, 2007 #5
    So that means the spring force has to be 6.00N right? The same as the constant force? This is what I come up with:

    (6.00N * .06m) = 1/2kx^2 ... where I plug in k and then solve for x? Is this correct?
     
  7. Feb 12, 2007 #6

    Doc Al

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    Right!

    No. Why did you bring spring PE into it? As berkeman suggests, forget energy and think in terms of forces.

    Just do what you said you'd do: Find the point where the spring force (given by Hooke's law) equals the 6 N resisting force.
     
  8. Feb 12, 2007 #7
    I got this part now. It should be resisting force/force constant right? I have .015 for the answer. I'm trying to figure out the next part but I'm stuck again. It's asking for the greatest speed and the eqn should be:

    (resisting force)(-.015) + (1/2)(force constant)(.015^2) = 1/2mvf^2 right? But I'm not getting the greatest speed. Instead I'm getting a velocity much lower and negative also...is there something wrong with my equation?
     
  9. Feb 13, 2007 #8

    Doc Al

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    That's correct.
    That equation is incorrect. Think in terms of accounting for the initial total energy (which was initially all spring PE). When the ball is at x = 0.015 m, the total energy must equal: Work lost to resistive force + spring PE + KE. Solve for the KE and then the velocity.
     
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