Use the worked example above to help you solve this problem. A 1.03 103-kg eleva

Question

Use the worked example above to help you solve this problem. A 1.03 103-kg elevator car carries a maximum load of 8.60 102 kg. A constant friction force of 4.02 103 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

Part 2 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

Here

mass of elevator , m = 1.03 *10^3 Kg

F = 4.02 *10^3 N

mass of load ,ml = 8.6 *10^2 Kg = 860 Kg

speed , v = 3 m/s

Power = (F + (m + ml)g) * v

Power = (4.02 *10^3 + (1.03 *10^3 + 860) * 9.80) * 3

Power = 67626 W

power delivered is 67626 W = 67.6 kW

Power = 67626/(846) W

Power = 80 hp

-----------------------------------
for the elevator moving downwards

Power = ((m + ml)g - F) * v

Power = (- 4.02 *10^3 + (1.03 *10^3 + 860) * 9.80) * 3

Power = 43506 W

the power of engine is 43506 W = 43.6 kW

Power delivered = power delivered in W/(846 W/hp)

Power delivered = 43506 W/(846 W/hp)

Power delivered = 51.8 hp

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