EXAMPLE 5.12Power Delivered by an Elevator Motor (a) The motor exerts an upward

Question

EXAMPLE 5.12Power Delivered by an Elevator Motor
(a) The motor exerts an upward force [T with arrow] on the elevator. A frictional force [f with arrow] and the force of gravity M [g with arrow] act downward. (b) The free-body diagram for the elevator car.
GOAL Apply the force-times-velocity definition of power.

PROBLEM A 1.00 103-kg elevator carries a maximum load of 8.00 102 kg. A constant frictional force of 4.00 103 N retards its motion upward, as in the figure. What minimum power, in kilowatts and in horsepower, must the motor deliver to lift the fully loaded elevator at a constant speed of 3.00 m/s?

STRATEGY To solve this problem, we need to determine the force the elevator car's motor must deliver through the force of tension in the cable, [T with arrow] . Substituting this force together with the given speed v into P = Fv gives the desired power. The tension in the cable, T, can be found with Newton's second law.

P = ___ hp

P=___KW

Use the values from above 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

EXAMPLE 5.12Power Delivered by an Elevator Motor (a) The motor exerts an upward force [T with arrow] on the elevator. A frictional force [f with arrow] and the force of gravity M [g with arrow] act downward. (b) The free-body diagram for the elevator car. GOAL Apply the force-times-velocity definition of power. PROBLEM A 1.00 103-kg elevator carries a maximum load of 8.00 102 kg. A constant frictional force of 4.00 103 N retards its motion upward, as in the figure. What minimum power, in kilowatts and in horsepower, must the motor deliver to lift the fully loaded elevator at a constant speed of 3.00 m/s? STRATEGY To solve this problem, we need to determine the force the elevator car's motor must deliver through the force of tension in the cable, [T with arrow] . Substituting this force together with the given speed v into P = Fv gives the desired power. The tension in the cable, T, can be found with Newton's second law. A 1.20 x 10^3 - kg elevator car carries a maximum load of 7.40 x 10^2 kg . A constant friction force of 3.94 x 10^3 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? Use the values from above 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.)

Explanation / Answer

A.

When the elevator is going up T = mg + f

power delivered by T = TV = (mg + f) V = (1800 x 9.8 + 4000) (3) = 64920 W = 64.9 kW

since 1 HP = 746 W, 64920 W = 87 HP

B.

power delivered by T = TV = (mg + f) V = (1940 x 9.8 + 3940) (3) = 68856 W = 68.9 kW

since 1 HP = 746 W, 68856 W = 92.3 HP

C.

When the elevator is going down T + f = mg

T = 1940 x 9.8 - 3940 = 15072 N

power delivered by T = TV = 45216 W = 45.2 kW

since 1 HP = 746 W, 45216 W = 60.6 HP

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