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Finding Work Done by Compressing A Spring

  1. Apr 16, 2012 #1
    1. The problem statement, all variables and given/known data
    I did a lab today in Physics in which we launched ball from a spring loaded cannon directly into a pendulum that captured the ball, held it, and swung upwards with it (representing a totally inelastic collision). One question that confuses me:

    > How much work did you do in joules in compressing the spring of the spring gun for the long-range case? Which law of conservation is your answer based upon?

    Some additional information: the long-range case is where we fired the ball at it's maximum speed (which I've calculated to be 4.79 m/s^2.

    2. Relevant equations
    E = K + U = 0
    initial momentum = final momentum
    m(v_initial) = (m + M)(v_final)
    K = (1/2)(m + M)v^2
    U = (1/2)kx^2 [for the spring]

    3. The attempt at a solution

    I'm guessing the law of conservation to use here is for that of energy, in which K + U = 0, where K is kinetic and U is final energy. The system has all potential energy (no kinetic) when the spring is compressed, which is known as (1/2)kx^2, but I don't know k...

    If any other information is needed, let me know. Thanks for the help!
     
  2. jcsd
  3. Apr 16, 2012 #2
    The potential energy stored in the spring is there because you put it into it. That's the work you did.
     
  4. Apr 16, 2012 #3
    Yes, well, this means 2 things to me:

    That the answer is (1/2)kx^2, because thats the energy I put into it (which I can't find - I don't know k, and I also don't know how far the spring got compressed [aka x]).

    Or, we use the law of conservation of energy so that:
    K + U = 0
    K_f - K_i + U_f - U_i = 0 [there is no initial kinetic energy or final potential, so...]
    U_i = K_f
    So, the work I did is equal to K_f, which is (1/2)mv^2.

    I did out this calculation and got some ridiculous number .0246 J... I definitely did more work than that, so I'm still so confused.
     
  5. Apr 16, 2012 #4
    The method is correct. I don't know how heavy that ball is, but when I run the calculation for a 10g marble with a final speed of 5 m/s I get 0.125J. So unless you are using something lighter and/or moving slower, you'll probably have made a mistake in the calculation.
     
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