PRACTICE IT Use the worked example above to help you solve this problem. A 1.18

Question

PRACTICE IT Use the worked example above to help you solve this problem. A 1.18 x 103-kg elevator car carries a maximum load of 8.90 x 102 kg. A constant friction force of 4.06 x 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 73.0 P = | 97.9 hp EXERCISE HINTS: GETTING STARTED II'M STUCK 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.) kw hp

Explanation / Answer

1] Total opponent force, Fo = (1.18*10^3 + 8.9* 10^2)*9.8 + 4.06*10^3 = 24346 N

P = Fo*V/1000 = 24346*3/1000 3*5.830 = 73 kw

P = 73 kw*1.34 hp/kw = 97.82 hp

2] Opponent force Fo = (-1.18*10^3 - 8.9*10^2)*9.8 + 4.06*10^3 = 16226 N

P = Fo*V/1000 = 16226*3/1000 = 48.68 kw

P = 48.68 kw* 1.34 hp/kw = 48.64 hp

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