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27 August, 05:27

A ski lift has a one-way length of 1 km and a vertical rise of200m. The chairs are spaced 20 m apart, and each chair can seatthree people. The lift is operating at a steady spedd of 10 km/h. Neglecting friction and ari drag ans assuming that the average massof each loaded chair is 250 kg, determine the power required tooperate the ski lift. Also estimate the power required toaccelerate this ski lift in 5 s to its operating speed when it isfirst turned

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  1. 27 August, 07:30
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    a) 68.125 KW

    b) 43.04 KW

    Explanation:

    Distance = d = 1 km

    Height = h = 200 m

    Spacing between chairs = D = 20 m

    No. of people per chair = 3

    Speed = V = 10 km/h = 10000m/3600 s=2.8 m/s

    mass of chair = 250 kg

    Work to operate sky lift

    W = mgh

    Number of chairs any moment operational = N = 1 km/20=1000m/20=50

    So, total mass of chairs = 50 * 250 = 12500kg

    so, w = mgh=12500*9.8*200=2452500 j

    Power is rate of work - we need time for operation time of lift

    Time = t = distance/speedf = 1km / (10km/h) = 1km / (10 km/3600s) = 360s

    So, Power P = Work/time=w/t=12500/360=68125 j/s=68.125 KW

    Now calculating power for operating speed in 5 sec

    We calculate accelleration=a for 5 sec

    a = speed / time = V/52.8/5=0.28 m/sec2

    for vertical acceleration we calculate θ angle first;

    tanθ = height / distance = 200/1000 = 0.2

    ==> θ=11.34°

    Vertical acceleration = a₁=a sinθ = 0.28 * sin 11.34=0.10835 m/sec2

    to calculate height gained during startup use;

    S=vit+1/2at2

    here s=H

    vi=0m/s

    t=5 sec

    ==> H = 0+1/2a₁*t=0.5*0.10835 * 5²=0.5*0.10835*5*5=1.362 m

    Total Work = mgH+0.5*m*V²=12500 (9.8*1.362+0.5*2.8*2.8) = 215240.56 j

    Again power = work / time=215240.56/5=43048.112 W=43.04 KW
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