PRACTICE IT A 1.02 103-kg elevator car carries a maximum load of 7.40 102 kg. A

Question

PRACTICE IT A 1.02 103-kg elevator car carries a maximum load of 7.40 102 kg. A constant friction force of 3.96 103 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
EXERCISE 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

A)F = mg + friction
F= (1.02* 10 ^ 3 + 7.40 * 10 ^ 2)* 9.8 + 3.96 * 10 ^ 3 =21208 N

power = F * V = F * 3 =21208*3=63624 W=6.36 *10^4kW

1kW=1.34hp=====>power=6.36*10^4*1.34hp=8.52*10^4 hp

B)When falling the force of friction removes energy so the motor has less work to do.
In the formula for Force you would SUBTRACT the friction instead of adding it.
To get
P=((1.02* 10 ^ 3 + 7.40 * 10 ^ 2)* 9.8 - 3.96 * 10 ^ 3)*3 =39864W=3.98*10^4kW

1kW=1.34hp=====>power=3.98*10^4*1.34hp=5.33*10^4 hp

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