Use the worked example above to help you solve this problem. A 1.09 103-kg eleva
ID: 1445459 • Letter: U
Question
Use the worked example above to help you solve this problem. A 1.09 103-kg elevator car carries a maximum load of 7.70 102 kg. A constant friction force of 4.06 103 N retards its motion upward, as shown in the figure. What minimum power, in kilowatts and horsepower, must the motor deliver to lift the fully loaded elevator at a constant speed of 3.00 m/s? P = kW P = hp
Use the values from PRACTICE IT to help you work this exercise. Suppose the same elevator car with the same load descends at 3.00 m/s. What minimum power is required? (Here, the motor removes energy from the elevator by not allowing it to fall freely.) P = kW P = hp
Explanation / Answer
elavator is moving with constant speed, hence
a = 0
using Fnet= ma
F - mg - f = 0
F = mg + f = (1.09 x 10^3 + 7.70 x 10^2) x 9.8 + (4.06 x 10^3)
F = 22288 N
Power = F. v = 22288 x 3 =66864 W = 66.86 kW
in hp = 66.86 x 1.34 = 89.66 hp
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when going down,
F + f - mg = 0
F = (1.09 x 10^3 + 7.70 x 10^2) x 9.8 - (4.06 x 10^3)
F = 14168 N
Power = F. v = 14168 x 3 = 42504 W = 42.50 kW
in hp = 41.50 x 1.34 = 57 hp
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