Calculating Power Requirements for a Ski Lift

[elsternj]

Homework Statement

A ski tow operates on a slope of angle 14.3 of length 350m . The rope moves at a speed of 12.1km/h and provides power for 49 riders at one time, with an average mass per rider of 75.0kg Estimate the power required. to operate the tow.

Homework Equations

W=mgh
P=W/T
x-x₀=(V_0x+V_x/2)t

The Attempt at a Solution

First I converted my speed to m/s
12.1km/h = 3.36m/s
Then I had to find the height of this ski lift. I figured the height would be the y component of the ski lift. 350Sin(14.3)=86.45
m = 75*49 = 3675
W=mgh
W=(3675)(9.8)(86.45)=3113496.75

Then I had to find the time.
x-x₀=(V_0x+V_x/2)t
350=(3.36/2)t
t = 208.3
P = 3113496.75 / 208.3 = 14947.18 (not the right answer)

Where did I go wrong?

 

[PhanthomJay]

You assumed that the the rope and riders were accelerated from rest to 3.36 km/s. Instead, the problem is (somewhat unclearly) stating that the speed is constant at 3.36 km/s.

 

[elsternj]

okay so if the speed is constant then acceleration is 0.

3.36=86.45/t
t=25.73
P=w/t
P=3113496.75/25.73=121006.48 not the right answer

i feel that I am particularly unsure about my W=mgh equation in this problem.

 

[elsternj]

oops

 

[PhanthomJay]

okay so if the speed is constant then acceleration is 0.

3.36=86.45/t
t=25.73
P=w/t
P=3113496.75/25.73=121006.48 not the right answer

i feel that I am particularly unsure about my W=mgh equation in this problem.

Your Work done is OK, but your time is incorrect. The rope moves 350 m , not 86.45 meters.
Incidentally, if you can calculate the force in the rope, F, you can use P =Fv as a check on your answer.

 

[elsternj]

ah and there's the right answer! of course it should have been 350! I don't know what made me use the height. Thank you so much for the help! I love this site. I see tutors at my college but it is currently our spring break and we have a test the week we get back! This has been just as helpful as my tutors! Much appreciation to you and everyone else.