Power required to operate a ski tow

1. Mar 6, 2007

ph123

"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. Mar 6, 2007

Dick

As I think you suspect, the component of the force perpendicular to the slope does no work. Get rid of it.

3. Mar 6, 2007

ph123

Leaving, then,

F=mgsin(theta)
F=(49)(73.0kg)sin(14.3) = 8658.5N

4. Mar 6, 2007

Dick

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