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Kinetic Energy Theorem

  1. Jan 24, 2005 #1
    Bonny Blair skated 5.00 x 10^2 m with an average speed of 12.92 m/s. Suppose she corssed the finish line at this speed and then skated freely until her speed was 8.00 m/s. If her was mwas 55.0 kg, how much work was done by friction?

    Can anybody explain the kinetic energy theorem or help me work through this problem?
     
  2. jcsd
  3. Jan 24, 2005 #2

    quasar987

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    The kinetic energy theorem says her variation in kinetic energy is equal to the work done on her.

    If you can find by how many Joules her kinetic energy varied, then this same number is equal to the work done on her:

    [tex]\Delta K = W[/tex]
     
  4. Jan 24, 2005 #3
    [tex]W_{friction}=F_{friction}d\cos{\theta}[/tex]

    [tex]F_{net}=ma[/tex]

    Can you figure it out using those (careful what you use for [itex]d[/itex])?
     
  5. Jan 24, 2005 #4
    Oh, yes, quasar's method may be easier.
     
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