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Homework Help: Power required to operate a ski tow

  1. Mar 6, 2007 #1
    "A ski tow operates on a slope of angle 14.3 of length 350m. The rope moves at a speed of 12.4km/hr and provides power for 49 riders at one time, with an average mass per ride of 73.0kg. Estimate the power required to operate the tow."

    The eqn I used to sum the forces in the x-direction is:

    EFx= Fcos(theta) - mgsin(theta) = 0
    F = (mgsin(theta))/cos(theta) = 8935.3N

    12.4km/h = 3.44m/s

    350m/3.44m/s = t = 101.74s

    W = F*s = (8935.3N)(350m)
    W = 3127356.9 J

    Power = W/dt = 3127356.9J/101.74s
    P=3.07*10^4 W

    However, this is not right. I feel like I need to include the force in the y-direction in my work equation, but then again perpendicular forces do now work. Does anyone have any idea what I may be overlooking here?
  2. jcsd
  3. Mar 6, 2007 #2


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    As I think you suspect, the component of the force perpendicular to the slope does no work. Get rid of it.
  4. Mar 6, 2007 #3
    Leaving, then,

    F=(49)(73.0kg)sin(14.3) = 8658.5N
  5. Mar 6, 2007 #4


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    Yes (though you forgot g in the second line).
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