0.36/2A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart, and each chair can seat three people. The lift is operating at a steady speed of 10 km/h. Neglecting friction and air drag, and assuming that the average mass of each loaded chair is 250 kg, determine the power required to operate this ski lift. Also, estimate the power required to accelerate this ski lift in 7 s to its operating speed when it is first turned on.

Step-by-step explanation:

Power is needed for (1) acceleration and (2) lifting the loaded chairs. This two parts can be calculated separately and then added together.

(1) Power for acceleration:

The final speed of the lift is

V = (10 km/h) (1 h*1000 m60 sec*1 km) = 2.887 m/s.

Then the power needed is

Pa=12m (V2-V20) / Δt=12 (50*250 kg) (2.778 m/s) 2=9.6 kW.

(2) Power for lift

Assume that the acceleration is constant (i. e. power supply is constant), its value will be

a=ΔVΔt=2.778 m/s5 s=0.556 m/s2.

Then the vertical lift during acceleration will be

(12at2) * (2001000) = 1.36 m.

Hence, the power needed to increase the potential energy of the lift is

Pg=mgΔhΔt = (50*250 kg) (9.89 m/s2) (1.36 m) / (5 s) = 3.41 kW.

Then the total Power required is

Ptotal=Pa+Pg=9.6+34.1=43.7 kW.