/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 What is the average power output... [FREE SOLUTION] | 91Ó°ÊÓ

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What is the average power output of an elevator that lifts \(885 \mathrm{~kg}\) a vertical height of \(32.0 \mathrm{~m}\) in \(11.0 \mathrm{~s} ?\)

Short Answer

Expert verified
The average power output is approximately 25,239.27 Watts or 25.24 kW.

Step by step solution

01

Understand the Problem

The problem asks for the average power output of an elevator. We're given the mass of the object lifted, the height it is lifted to, and the time taken for the lift. Our goal is to determine the average power output using these values.
02

Identify the Relevant Formula

To find the average power, we use the formula for power related to work done over time. The formula is:\[ P = \frac{W}{t} \]where \(P\) is the power, \(W\) is the work done, and \(t\) is the time taken. We also need the formula for work done against gravity, \(W = mgh\), where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the height.
03

Calculate the Work Done

We substitute the given values into the work formula. Using the gravitational acceleration \(g = 9.8\, \text{m/s}^2\), we calculate:\[ W = mgh = 885 \times 9.8 \times 32.0 = 277,632 \text{ Joules} \]
04

Calculate the Average Power

Now that we have the work done, we use the power formula to find the average power:\[ P = \frac{W}{t} = \frac{277,632}{11.0} = 25,239.27 \text{ Watts} \]
05

Consider the Answer

The average power is calculated as approximately \(25,239.27 \text{ Watts}\), or \(25.24 \text{ kW}\) since \(1 \text{ kW} = 1000 \text{ Watts}\). This is a large power output, typical for an elevator lifting significant weight quickly.

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Key Concepts

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

Work-Energy Principle
In physics, the Work-Energy Principle is a powerful tool to understand how energy is transferred in various systems. It states that the work done on an object is equal to the change in its kinetic energy. This fundamental principle can be observed in numerous scenarios, especially when forces act to move objects over distances.
For instance, when an elevator lifts a heavy object, work is done against gravitational forces. This results in a change in potential energy rather than kinetic energy, although the principle still applies. The amount of work done is calculated by multiplying the force applied (which equals the object's weight) by the distance moved, adhering to the formula:
  • Work = Force × Distance
In our solution, the force applied was the force of gravity acting on the mass of the object, calculated using the expression:
  • \( W = mgh \)
This indicates the work done against gravity to lift an object to a certain height.
Overall, the Work-Energy Principle helps determine how motion or changes in position affect the energy state of a system.
Gravitational Potential Energy
Gravitational potential energy is a form of energy related to the height of an object in a gravity field. When an object is lifted vertically against the force of gravity, work is done and its gravitational potential energy increases.
It's calculated using the formula:
  • \( Ug = mgh \)
where
  • \( m \) is the mass,
  • \( g \) is the gravitational acceleration (~9.8 m/s² on Earth),
  • \( h \) is the height above the ground.
In the context of the elevator problem, the gravitational potential energy signifies the energy stored when lifting the weight.
This energy can be thought of as potential energy waiting to do work, such as when the elevator descends and the potential energy converts back to kinetic energy or another energy form. It's a vital concept in understanding how energy transformation and conservation happen in lifting scenarios.
Average Power Calculation
Average power calculation involves determining how much work is done or energy is transformed over a period of time. Power is essentially the rate of doing work or transferring energy. When calculating average power, it’s crucial to consider the entire duration over which energy transformation occurs.
The formula for calculating average power is:
  • \( P = \frac{W}{t} \)
where
  • \( W \) is the work done (measured in Joules), and
  • \( t \) is the time (measured in seconds) over which the work is done.
In our elevator scenario, we found the work done using gravitational potential energy (\( mgh \)), and then divided by time to find power.
The result, expressed in watts, tells us the rate at which energy is used or work is done over the lifting duration. Thus, average power provides insight into how efficient a system is at performing work over time. Understanding this concept is essential when examining motors, engines, or any systems where work or energy transfer is involved over a specific period.

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Most popular questions from this chapter

A 65 -kg hiker climbs to the top of a 4200 -m-high mountain. The climb is made in 5.0 h starting at an elevation of 2800 m. Calculate \((a)\) the work done by the hiker against gravity, \((b)\) the average power output in watts and in horsepower, and \((c)\) assuming the body is 15\(\%\) efficient, what rate of energy input was required.

A \(65-\mathrm{kg}\) hiker climbs to the top of a 4200 -m-high mountain. The climb is made in \(5.0 \mathrm{~h}\) starting at an elevation of \(2800 \mathrm{~m} .\) Calculate \((a)\) the work done by the hiker against gravity, (b) the average power output in watts and in horsepower, and (c) assuming the body is \(15 \%\) efficient, what rate of energy input was required.

A ball is attached to a horizontal cord of length \(\ell\) whose other end is fixed, Fig. \(8-42 .(a)\) If the ball is released, what will be its speed at the lowest point of its path? \((b)\) A peg is located a distance \(h\) directly below the point of attachment of the cord. If \(h=0.80 \ell,\) what will be the speed of the ball when it reaches the top of its circular path about the peg?

The Lunar Module could make a safe landing if its vertical velocity at impact is \(3.0 \mathrm{~m} / \mathrm{s}\) or less. Suppose that you want to determine the greatest height \(h\) at which the pilot could shut off the engine if the velocity of the lander relative to the surface is (a) zero; (b) \(2.0 \mathrm{~m} / \mathrm{s}\) downward; (c) \(2.0 \mathrm{~m} / \mathrm{s}\) upward. Use conservation of energy to determine \(h\) in each case. The acceleration due to gravity at the surface of the Moon is \(1.62 \mathrm{~m} / \mathrm{s}^{2}\)

(II) An outboard motor for a boat is rated at 55 hp. If it can move a particular boat at a steady speed of \(35 \mathrm{~km} / \mathrm{h}\), what is the total force resisting the motion of the boat?

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