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Energy work calculations (please check over)

  1. Nov 5, 2006 #1
    A 2-kg block is attached to a horizontal ideal spring with a spring constant of 200N/m. When the spring has its equilibrium length the block is given a speed of 5 m/s. What is the maximum elongation of the spring?

    1/2mv= 1/2KX^2
    max PE equals starting KE.
    then follows
    x= v/sqrt k
    x=.25 , that is the answer I am getting. But how is the book getting .05m

    my second question:
    A 6.0kg block is released from rest 80m above the ground. When it has fallen 60m its kinetic energy is:
    I got 1200J as my answer
    mgh=(6.0*9.81* 60m)=3531.6
    PE + KE= (ME + KE)
    3531.6 + KE= (6.0*9.81*80m)
    KE= 1177.2
    is that right
  2. jcsd
  3. Nov 5, 2006 #2


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    You posted #1 earlier. your method is correct, but your equation and/or math is incorrect.
    1/2mv^2 = 1/2kx^2
    mv^2 = kx^2
    mv^2/k = x^2
    2(5)(5)/200 = x^2 = 0.25
    x = 0.5 m (I don't know why the book says .05m).

    for #2 its
    PE_initial + KE_initial = PE_final + KE_final
    For simplicity, choose the 60m point as the reference height for PE (PE =0 at this point). There will then be a couple of 0 terms in that equation, and you can easily solve for v, without worrying about the 80 m height at all..
    Last edited: Nov 5, 2006
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