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Chapter 7: Problem 46

The loaded cab of an elevator has a mass of \(5.0 \times 10^{3} \mathrm{~kg}\) and moves \(210 \mathrm{~m}\) up the shaft in \(23 \mathrm{~s}\) at constant speed. At what average rate does the force from the cable do work on the cab?

Short Answer

Expert verified
The average power is approximately 447.4 kW.

Step by step solution

01

Understand the Problem

We need to find the average rate at which work is done by the force of the cable as it lifts the elevator cab. This is equivalent to finding the power exerted by the force.
02

Identify Key Variables

Identify the key data given: the mass of the cab, \( m = 5.0 \times 10^3 \mathrm{~kg} \); the height it moves, \( h = 210 \mathrm{~m} \); the time duration, \( t = 23 \mathrm{~s} \).
03

Recall the Power Formula

Power \( P \) is the rate at which work \( W \) is done, which can be calculated by the formula \( P = \frac{W}{t} \).
04

Calculate Work Done

The work done by the force lifting the elevator is equal to the gravitational potential energy gained, calculated by \( W = mgh \), where \( g \) is gravitational acceleration, \( 9.8 \mathrm{~m/s^2} \). Substitute the values: \( W = 5.0 \times 10^3 \times 9.8 \times 210 \).
05

Calculate the Total Work

Compute the total work done: \( W = 5.0 \times 10^3 \times 9.8 \times 210 = 1.029 \times 10^7 \mathrm{~J} \).
06

Substitute into the Power Formula

Use the power formula: \( P = \frac{1.029 \times 10^7}{23} \). Calculate \( P \).
07

Compute the Average Power

Perform the division: \( P = 4.474 \times 10^5 \mathrm{~W} \) or \( 447.4 \mathrm{~kW} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power
Power is a fundamental concept in physics that describes how quickly work is done. It is defined as the rate of doing work or transferring energy. Mathematically, we use the formula \( P = \frac{W}{t} \) where \( P \) represents power, \( W \) is the work done, and \( t \) stands for time.
In the context of the elevator problem, power is how fast the cable does work to lift the elevator cab. Here, power is calculated based on how quickly energy in the form of gravitational potential energy is transferred to the cab.
  • The faster the elevator cab is lifted, the more power is required.
  • The unit of power is the watt (\( \ ext{W} \)), which is equivalent to one joule per second.
This helps you understand not just how much energy is consumed, but how efficiently that energy is used over time.
Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses because of its position in a gravitational field. It relates to the work done by the gravitational force when an object is moved. The formula used to calculate it is \( W = mgh \), where:
  • \( m \) is the mass
  • \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \) on Earth)
  • \( h \) is the height
In the elevator problem, as the cab is lifted to a height of 210 meters, its gravitational potential energy increases. This increase corresponds to the work done by the force (in this case, by the cable) to lift the cab against the gravitational pull. Understanding this concept is crucial because it directly explains how energy is stored and transferred when lifting an object at constant speed.
Kinematics
Kinematics is the study of motion without considering its causes. In the context of the elevator, even though it moves at a constant speed, kinematics helps us understand the displacement and timing during motion.
The elevator moves 210 meters up in 23 seconds, and it's important because it provides the duration over which work is performed. This information is critical for calculating power since we need the total time while the elevator is in motion.
  • Kinematics doesn't directly calculate energy or force, but it describes the motion essential for these calculations.
  • Understanding timing and displacement is necessary for finding average speed and, subsequently, the rate at which work is done.
This concept shows how a basic understanding of time and motion is necessary for navigating more complex concepts like power and energy.
Mechanics
Mechanics is the branch of physics dealing with the motion of objects and the forces acting upon them. It combines elements of kinematics, dynamics, and energy concepts like work and power in solving real-world problems.
In the elevator scenario:
  • Mechanics helps determine how forces interact to produce motion.
  • It ensures the balance of forces, allowing the elevator to move at a constant speed.
  • Here, when the force from the cable equals the gravitational force, the cab moves smoothly without acceleration.
By integrating these principles, mechanics provides a complete picture that allows calculations of energy transfer and efficiency, as seen in this problem dealing with the elevator cab. This understanding is essential for solving complex problems involving multiple force interactions and various energies.

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Most popular questions from this chapter

A \(0.35 \mathrm{~kg}\) ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring \((k=450 \mathrm{~N} / \mathrm{m})\) whose other end is fixed. The ladle has a kinetic energy of \(10 \mathrm{~J}\) as it passes through its equilibrium position (the point at which the spring force is zero). (a) At what rate is the spring doing work on the ladle as the ladle passes through its equilibrium position? (b) At what rate is the spring doing work on the ladle when the spring is compressed \(0.10 \mathrm{~m}\) and the ladle is moving away from the equilibrium position?

(a) In 1975 the roof of Montreal's Velodrome, with a weight of \(360 \mathrm{kN}\), was lifted by \(10 \mathrm{~cm}\) so that it could be centered. How much work was done on the roof by the forces making the lift? (b) In 1960 a Tampa, Florida, mother reportedly raised one end of a car that had fallen onto her son when a jack failed. If her panic lift effectively raised \(4000 \mathrm{~N}\) (about \(\frac{1}{4}\) of the car's weight) by \(5.0 \mathrm{~cm}\), how much work did her force do on the car?

A skier is pulled by a towrope up a frictionless ski slope that makes an angle of \(12^{\circ}\) with the horizontal. The rope moves parallel to the slope with a constant speed of \(1.0 \mathrm{~m} / \mathrm{s}\). The force of the rope does \(880 \mathrm{~J}\) of work on the skier as the skier moves a distance of \(7.0\) \(\mathrm{m}\) up the incline. (a) If the rope moved with a constant speed of \(2.0\) \(\mathrm{m} / \mathrm{s}\), how much work would the force of the rope do on the skier as the skier moved a distance of \(8.0 \mathrm{~m}\) up the incline? At what rate is the force of the rope doing work on the skier when the rope moves with a speed of (b) \(1.0 \mathrm{~m} / \mathrm{s}\) and (c) \(2.0 \mathrm{~m} / \mathrm{s}\) ?

A machine carries a \(4.0 \mathrm{~kg}\) package from an initial position of \(\vec{d}_{i}=(0.50 \mathrm{~m}) \hat{\mathrm{i}}+(0.75 \mathrm{~m}) \hat{\mathrm{j}}+(0.20 \mathrm{~m}) \hat{\mathrm{k}}\) at \(t=0\) to a final position of \(\vec{d}_{f}=(7.50 \mathrm{~m}) \hat{\mathrm{i}}+(12.0 \mathrm{~m}) \hat{\mathrm{j}}+(7.20 \mathrm{~m}) \hat{\mathrm{k}}\) at \(t=12 \mathrm{~s}\). The constant force applied by the machine on the package is \(\vec{F}=(2.00 \mathrm{~N}) \hat{\mathrm{i}}+(4.00 \mathrm{~N}) \hat{\mathrm{j}}+(6.00 \mathrm{~N}) \hat{\mathrm{k}}\). For that displacement, find (a) the work done on the package by the machine's force and (b) the average power of the machine's force on the package.

Only one force is acting on a \(2.8 \mathrm{~kg}\) particle-like object whose position is given by \(x=4.0 t-5.0 t^{2}+2.0 t^{3}\), with \(x\) in meters and \(t\) in seconds. What is the work done by the force from \(t=0 \mathrm{~s}\) to \(t=6.0 \mathrm{~s}\) ?
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