A ski lift carries 900 skiers per hour up a slope with a constant incline of 15

Question

A ski lift carries 900 skiers per hour up a slope with a constant incline of 15 degrees and a vertical drop (difference in altitude between top and bottom) of 240 meters. It takes 3.5 minutes for the lift to bring each skier from the bottom to the top of the slope.

a) If there are no losses due to friction, what is the power required to run the lift? Make reasonable estimates for any quantities you need that are not given in the problem, and explain how you arrived at those estimates. (I used 70 kg for the weight of each skier.)

b) Assume the lift works by dragging the skiers along the snow and that the coefficient of kinetic friction between the skis and the snow is 0.04. Also, let

Explanation / Answer

a)

Now, the weight of the skier be 70 kg. (say)

He was lifted to an elevation of 240m. Work required of that is

Work done per skier = m x g x H = 70 x 9.81 x 240 J = 164808 J = 164.8 KJ

900 skiers climb the slope per hour. So, power required = 900 x 164.8 / 3600 = 41.202 KW

b)

Now, including friction, additional work has to be done.

Distance of slope = 240/ sin(15 degree) = 927.289 m

Kinetic friction co-efficient = 0.04

Normal force on each skier = 70 x 9.81 x cos(15 degree) = 663.301 N

Frictional force = 0.04 x 663.301 = 26.532 N

Work done against friction = Frictional force x Distance of slope

                                                = 26.532 N x 927.289 m = 24.603 KJ

Total power against friction required for 900 skiers = 24.603 x 900/3600 = 6.150 KW

Power required till now = 41.202 + 6.150 = 47.352 KW

10% of total power is lost in heat which is assumed as 0.1*TP.

TP = 0.1*TP + 47.352 KW

TP = 47.352/0.9 = 52.613 KW is the total power required to run the lift with frictional and heat losses.

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