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Determine the speed of the rock as it leaves the spring.

  1. Dec 27, 2014 #1
    Here is the question:
    A spring is compressed 20cm. When it is released, it exerts an average force of 57.2 N on a 7.3 kg rock, shooting it across a frictionless floor. Determine the speed of the rock as it leaves the spring.


    2. Relevant equations
    W=change in energy
    Work energy theorem
    W=Fd

    3. The attempt at a solution
    W=Fd
    =(0.20cm)(57.2N)
    =11.44J

    W= change in EP elastic
    =1/2kx^2
    Rearrange formula to solve for k (spring constant)
    =10.69579357 N/m
     
  2. jcsd
  3. Dec 27, 2014 #2
    I think you're going the wrong direction with this. You have the force applied and the mass affected, so you can solve for the average acceleration of the rock (F=ma, a=7.85616438 m/s2). With the acceleration and Δx (20 cm or .2 m), you can apply the formula vf2 = vo2 + 2(a)(Δx). Since vo is 0, 2 times a times Δx = your final velocity squared.
     
  4. Dec 28, 2014 #3

    haruspex

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    This is awkward.
    Unless stated otherwise (e.g. as "average force with respect to distance") average force means the average over time. ##F_{avg} = \frac{\int F.dt}{\int dt} = \frac {\Delta p}{\Delta t}##, where p is momentum. On that basis, you need to use the fact that the spring will expand according to simple harmonic motion.
    However, it's 10 to 1 that the problem setter has screwed up and expects you to take the average force as being an average over distance: ##F_{distance avg} = \frac{\int F.ds}{\int ds} = \frac {\Delta E}{\Delta s}##, so I'll answer below on that assumption.
    So far so good, but you no longer care about the spring. You know how much work has been done on the rock. Deduce its speed from that.
     
  5. Dec 28, 2014 #4
    These give you the same answer... Work in = Kinetic Energy Final, right?
     
  6. Dec 28, 2014 #5

    NascentOxygen

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    This formula is not useful to you here, because you are told the spring equation is not that of a normal spring. In a normal spring, F=kx. But in this exercise, you are told to use F=constant, independent of x.
     
  7. Dec 28, 2014 #6

    NascentOxygen

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    That should work.
     
  8. Dec 28, 2014 #7

    haruspex

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    What give the same answer? Are you saying that both definitions of average force give the same answer?
    This is annoying.. I posted a long response to this last night and it has disappeared. Maybe some site monitor mistakenly thought I was providing you with a complete answer to the OP, so I'm not going to try typing it all again.
    The upshot was that I demonstrated that using the correct definition of average force gives a different answer The ratio between the answers is sqrt(pi/4).
     
  9. Dec 28, 2014 #8
    I solved it the way I showed in my original post and with the equation (Work = kinetic energy final) and both most certainly gave me the exact same answer, accurate to like twelve or fifteen sig-figs, however many my calculator allowed.
     
  10. Dec 28, 2014 #9
    Okay, I think I see what you're saying. But this is a high school physics question, I'm sure the question is implying a force over a distance, so it would be an Energy problem, as indicated by the formula for potential energy in a spring. It does say to find the speed as it leaves the spring, and the spring was compressed .2m, so I believe it would be indicating that the average force over the distance from the initial position to the end of the spring, .2m away.
     
  11. Dec 28, 2014 #10

    haruspex

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    I agree that this is almost certainly what the problem setter intends here, but I hate the thought of students being taught that this is a valid interpretation of average force. It leads to the bizarre situation that mass * average acceleration does not always equal average force.
     
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