Power required to operate a ski tow

  • Thread starter ph123
  • Start date
  • #1
41
0
"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?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619
As I think you suspect, the component of the force perpendicular to the slope does no work. Get rid of it.
 
  • #3
41
0
Leaving, then,

F=mgsin(theta)
F=(49)(73.0kg)sin(14.3) = 8658.5N
 
  • #4
Dick
Science Advisor
Homework Helper
26,260
619
Yes (though you forgot g in the second line).
 

Related Threads on Power required to operate a ski tow

Replies
3
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
9K
Replies
2
Views
1K
Replies
3
Views
3K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
7
Views
7K
  • Last Post
Replies
1
Views
414
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
2K
Top