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Your friendly physics retard is at it again

  1. Nov 30, 2007 #1
    Alright, I'm doing a homework problem and for the love of holy things, I can't seem to find the solution to this problem, though I feel I have a pretty 'solid' physical feel for it (which, I admit, is better then a complete lack of understanding).

    Anyways, the 'question' is:
    My understanding is that the constant friction force makes it unlikely that the ball is going to be going its fastest at the end of the barrel, and that it is more likely to hit its "top speed" somewhere near the end, but not quite.

    My reasoning is this, as the spring approaches its state of equilibrium (not stretched, nor compressed) the force it inflicts on the ball lessens as it moves down the tube. At some point the force of friction overcomes the strength of the spring, and begins slowing it down.

    My problem is in "finding" this position.

    I reason that the maximum potential energy, at some point, becomes equal to the force of friction at some point, this happens while the spring loses its energy and the friction remains constant.

    According to my math (and an initial guess):

    Potential Energy of a Spring at position x = Force of Friction acting over distance X
    1/2kx^2 =F_f*x

    This produced 3/100 or 0.03 when I solve for it. Which is a nice understandable, and seemingly correct answer. Not so, I guess. Which then had me thinking - okay, if not at half, then when? Is it still accelerating at some point? Do I have the wrong idea entirely?!

    Thanks for the attention,
  2. jcsd
  3. Nov 30, 2007 #2
    the spring is being compressed 6cm which is the distance of the barrel so it doesn't meet equilibrium until its at the end of the barrel

    it's been awhile since i took physics so i might be wrong but that's how i see it
  4. Nov 30, 2007 #3
    Very cool problem, haven't seen one asked like this. Anyway there are some basic facts:

    Net Work = Change in Kinetic Energy
    Net work done by the friction = Force*Distance
    Net work done by the spring = .5*k*(x_i^2-x_f^2)

    see if you can solve it using this info
  5. Dec 1, 2007 #4
    Think carefully about 'balancing' forces against energies. If I were you I'd think about the connection between force and energy. Then talk about only forces or about only work done (energies). You can work out the speed of the ball (as a function of x) using either (although I'd guess that work done might be neater, faster and easier). Then working out the place where this occurs is just an easy bit of calculus.
    Hope that helps!
  6. Dec 2, 2007 #5
    I've reworked this a million times - is anyone getting an answer that is different from 0.03? I'd also like to note that the speed here using 0.03 isn't as great as the final velocity when it leaves the barrel (4.90 m/s, which by the way was the correct answer to a question prior to this regarding the final velocity)

    Am I really messing up here?

    I went so far as to graph out my velocity function as well:

    As you can see the velocity just increases as it moves down the barrel...

    Oddly - it begins at 0.3 though...
    Last edited: Dec 2, 2007
  7. Dec 2, 2007 #6
    Firstly, I think from looking at the equation you have on your chart that you need to re-read Bishopuser's post. The work done on the ball by the spring is not 1/2kx^2!
    Secondly, do you have an answer? (so I can check I've done this right! I get x=1.5cm)
  8. Dec 2, 2007 #7
    muppet: Your result is correct.

    I would say that the 'safest' approach to solve this problem would be to consider the equation of motion for the ball.

    1. Set up your coordinate system. The easiest thing is to set the origin to coincide with the equilibrium position of the spring, ie at the end point of the gun.
    2. Find out what forces are acting on the ball.
    3. Use Newtons law to write out the equations of motion.

    You can manipulate the equation a little, so it will be easy to solve, once you do that, you should see that the solution is only 'shifted'. You know where max velocity occurs wthout force, so where would it appear when the 'shift' is applied?
  9. Dec 2, 2007 #8
    your equation is almost correct, but you have to take into account the springs initial compression into your spring work terms.
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