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Energy in Simple Harmonic Motion

  1. Aug 23, 2009 #1
    Problem:

    A 2.4-kg object on a frictionless horizontal surface is attached to one end of a horizontal spring of force constant k=4.5 kN/m. The other end of the spring is held stationary. The spring is strecthed 10 cm from equilibrium and released. find the system's total mechanical energy.

    Attempt at solution:

    It seems as if I should be able to just substitute values into the equation [tex]E_{sys}=\frac{1}{2}kA_o^2[/tex] where [tex]A_o[/tex] is just the amplitude of the motion. The given units seem inconsistent, I need to convert k into centimeters by mulitplying by the factor [tex]\frac{1m}{100cm}[/tex] which gives [tex]k=0.045 kN/cm[/tex]. I get 2.25 joules. My book gives an answer of 23 joules. Where am I going wrong?
     
  2. jcsd
  3. Aug 23, 2009 #2

    Tom Mattson

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    If you convert the spring constant to [itex]kN/cm[/itex] then your answer will be in [itex]kN\cdot cm[/itex], which is not the same as Joules. If you convert your answer to Joules you'll have it right.

    But it would be a lot easier to convert [itex]A_0[/itex] to meters instead...
     
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