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Non conservative force of a wave on a surfer

  1. Sep 24, 2007 #1
    1. The problem statement, all variables and given/known data

    A surfer is catching a wave. Suppose she starts at the top of the wave with a speed of 1.93 m/s and moves down the wave until her speed reaches 12.3 m/s. The drop in her vertical height is 2.95 m. If her mass is 72.3 kg, how much work is done by the (non-conservative) force of the wave?

    2. Relevant equations

    Wnc = Ef-E0
    Wnc = .5mvf^2 + mghf - (.5mv0^2 + mgh0)

    3. The attempt at a solution

    Wnc = .5*72.3*12.3^2 + 72.3*-9.8*0 - (.5*72.3*1.93^2 + 72.3*-9.8*2.95)
    .5*72.3*12.3^2 - (.5*72.3*1.93^2 + 72.3*-9.8*2.95)
    5469.1335 - (134.66 + (-2090.19))
    = 7424.66 J

    This should become negative because the non conservative force is opposing displacement (since the surfer is moving down the wave). However, this answer is not correct. What should be changed?
  2. jcsd
  3. Sep 24, 2007 #2


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    Staff: Mentor

    Increasing velocity from 1.93 m/s to 12.3 m/s represents a change in kinetic energy of 5334.5 J

    And the change in height of 2.95 m represents a change in grav. potential energy of only 2092 J.

    So from where does the extra energy/work come?
  4. Sep 24, 2007 #3
    The wave, I suppose... so do I take the total of those two and subtract the number I found?
  5. Sep 24, 2007 #4


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    Staff: Mentor

    Well the difference in change in kinetic energy and change in potential energy is 3242.5J and presumably that is work done by the wave.
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