1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    d/v=t
    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes (though you forgot g in the second line).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Power required to operate a ski tow
  1. Ski tow (Replies: 36)

Loading...