# Skiing away, away the day

1. Aug 17, 2009

### harrinj4

1. The problem statement, all variables and given/known data

Physics and Skiers and Rope and Hills?
A 80 kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 24 degrees hill. The skier is pulled a distance x= 200m along the incline and it takes 2.2 min to reach the top of the hill.

If the coefficient of kinetic friction between the snow and skis is Uk = 0.12, what horsepower engine is required if 30 such skiers (max) are on the rope at one time?

2. Relevant equations

F=umgcos(theta)
P=F(change in x)/t
HP=746?

3. The attempt at a solution

Since the tow rope moves at constant speed, the tension in the rope must equal the force of kinetic friction, which is:

f = μmgcosθ
= 0.12(80kg)(9.8m/s²)cos24°
= 86N (rounded)

Power is work over time and work is force times distance, so:

P = FΔx / t
= 86N(200m) / 132s
= 130 W

In horsepower:

130W = 130W(1.00HP / 746W)
= 0.175HP

For 30 skiers then:

30 x 0.175HP = 5.3HP (rounded)

I got the wrong answer when I put this in. Everything seems in order. Did I miss something? I've double checked my work so its not a calculator error. Any help=thanks!

2. Aug 17, 2009

### kuruman

This would be the case if the skiers were on a horizontal surface. You did not include the power required to raise up to a certain height that you can calculate from the slope and the distance along the incline, i.e. Δ(mgh)/Δt.

3. Aug 17, 2009

### harrinj4

So insert that when trying to figure out power?

P = FΔx / t + Δ(mgh)/Δt??
or P=Δ(mgh)/Δt??

4. Aug 17, 2009

### kuruman

Insert. The engine works against both friction and gravity.

5. Aug 17, 2009

### harrinj4

So...=( 86N(200m) / 132s) + (80*9.8*200cos(24)/132)??

6. Aug 17, 2009

### kuruman

Draw yourself a right triangle of hypotenuse 200 m making an angle of 28o with respect to the horizontal. What is the vertical side?

7. Aug 17, 2009

### harrinj4

It's sin24*200=h

Boom got it

Last edited: Aug 17, 2009