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Ch 09: Work and Kinetic Energy
Chapter 9, Problem 9

(a) How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m? (b) How much power must the motor supply to do this in 50 s at constant speed?

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Work

Work is defined as the product of force and displacement in the direction of the force. In the context of lifting an elevator, the work done by the motor can be calculated using the formula W = F × d, where W is work, F is the force (equal to the weight of the elevator), and d is the height lifted. The weight can be determined by multiplying the mass of the elevator by the acceleration due to gravity (approximately 9.81 m/s²).
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Power

Power is the rate at which work is done or energy is transferred over time. It is calculated using the formula P = W/t, where P is power, W is work, and t is the time taken to do the work. In this scenario, to find the power required by the motor to lift the elevator in a specified time, one must first calculate the work done and then divide it by the time taken (50 seconds).
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Constant Speed

When an object moves at constant speed, it means that its velocity remains unchanged over time, implying that the net force acting on it is zero. In the case of the elevator, if it is lifted at constant speed, the upward force exerted by the motor must equal the downward gravitational force acting on the elevator. This condition simplifies the calculations for work and power, as the motor's force directly counteracts the weight of the elevator.
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