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Math Help - Kinetic, Gravitational Potential and Elastic Potential Energy

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    Question Kinetic, Gravitational Potential and Elastic Potential Energy

    I have a question involving the extension of an elastic string that has me a little stumped.

    The set up of the question is as follows:

    Particle P of mass 0.5kg is attached to one end of a light, elastic string that has a natural length of 2m, and a modulus of elasticity of 20N.

    The other end of the string is attached to a ceiling, and P is held 1.5 metres directly below the point of attachment, before being released.

    What is the length of the string when P reaches its lowest point?

    To me, the question relies on calculating the Kinetic Energy of P as the string becomes taught (i.e. how much Kinetic Energy is gained in the 0.5m before the tension in the string begins to play a role), and then determining the amount of potential energy stored in the string when the Kinetic Energy becomes zero (at which point I expect the energy in the string to be equal to the loss in gravitational potential energy since the release of P). Hence:

    M&=0.5kg, l&=2m, \lambda&=20N

    K.E.&=Mgh&=0.5kg * 9.8ms^{-2} * 0.5m

    P.E.&=Mg\left(y-2\right)&=0.5kg * 9.8ms^{-2} * \left(y - 2m\right)

    E.P.E.&=\frac{\lambda\left(y - 2\right)^{2}}{2l}&=\frac{20N * \left(y - 2m\right)^{2}}{4m}

    E.P.E.&=K.E. + P.E.

    From here I just need to solve for y (i.e. the extension of the string) and add on the requisite 2m, but I'm not coming up with the right answer (which should be 3.34m). I seem to be missing something in the set up of the problem, can anyone see what it is? Thanks very much.
    Last edited by britmath; June 11th 2012 at 12:20 PM.
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  2. #2
    mfb
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    Re: Kinetic, Gravitational Potential and Elastic Potential Energy

    In your equations, y is the length of the string, not the extension (otherwise you should not subtract 2m).

    Using your ansatz: 5N/m * (y-2m)^2 = 1/2kg*9.81m/s^2 * (y-1.5m)
    (y/m-2)^2 = 0.981*(y/m-1.5)
    This has the solutions y=1.64 (unphysical) and y=3.35, which is the right answer.
    Thanks from britmath
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    Re: Kinetic, Gravitational Potential and Elastic Potential Energy

    Quote Originally Posted by mfb View Post
    In your equations, y is the length of the string, not the extension (otherwise you should not subtract 2m).
    Not quite sure how I managed to make that error and not notice it! Thanks a lot.
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