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Math Help - Hooke's Law

  1. #1
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    Hooke's Law

    A light elastic string, of natural length 0.8m and modulus of elasticity 35.4N, has one end A attached to a fixed point and the other end B attached to particle P of mass 3kg. Initially P is held at rest at A. It is then released and allowed to fall. Calculate the speed of P when the length of the string is 1.2m

    OK this is how i've been taught to answer the question.

    energy at start = potential energy + kinetic energy + energy stored

    = 0 + 0 + 0

    energy at greatest extension = potential energy + kinetic energy + energy stored

    P.E. = mgh

    [I dont know what h is]

    K.E. = 1/2 mv^2

    i believe energy stored is 'Hooke's Law' integrated

    \int \frac{\lambda x}{l}

    but not sure about all of this.


    if this is all correct would i do (0+0+0) = mgh + 1/2 mv^2 + \int\frac{\lambda x}{l}

    then solve for v?
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  2. #2
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    That approach looks correct. h should be 1.2 metres, because the particle P has dropped that distance. The potential energy in the string should be \tfrac12\lambda x^2, where \lambda is the modulus and x is the distance stretched (0.4m in this case).
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  3. #3
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    Thanks for helping but i think i sorted it now


    The potential energy in the string should be
    do you mean the energy stored?

    potential energy = mgh

    energy stored = \frac{1}{2} \frac{\lambda x^2}{l}

    looks more like what your getting at?
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