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Chapter 6: Problem 22

The energy height, \(H\), of an aircraft of mass \(m\) at altitude \(h\) and with speed \(v\) is defined as its total energy (with the zero of the potential energy taken at ground level) divided by its weight. Thus, the energy height is a quantity with units of length. a) Derive an expression for the energy height, \(H\), in terms of the quantities \(m, h\), and \(v\). b) A Boeing 747 jet with mass \(3.5 \cdot 10^{5} \mathrm{~kg}\) is cruising in level flight at \(250.0 \mathrm{~m} / \mathrm{s}\) at an altitude of \(10.0 \mathrm{~km} .\) Calculate the value of its energy height. Note: The energy height is the maximum altitude an aircraft can reach by "zooming" (pulling into a vertical climb without changing the engine thrust). This maneuver is not recommended for a 747 , however.

Short Answer

Expert verified
Question: Calculate the energy height (maximum altitude) of a Boeing 747 jet with a mass of 3.5 x 10^5 kg, flying at an altitude of 10.0 x 10^3 m, and a speed of 250.0 m/s. Answer: To calculate the energy height, follow the steps below: 1. Calculate potential energy (PE): PE = mgh = (3.5 x 10^5)(9.81)(10 x 10^3) 2. Calculate kinetic energy (KE): KE = (1/2)mv^2 = (1/2)(3.5 x 10^5)(250)^2 3. Calculate total energy (TE): TE = PE + KE 4. Calculate the energy height (H) using the derived expression and the weight of the aircraft (mg): H = TE/mg Following these steps, you can find the energy height (maximum altitude) of the Boeing 747 jet in meters.

Step by step solution

01

Calculate potential energy

The gravitational potential energy (PE) of an object at a given height is calculated by the formula: \[PE = mgh\] where \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is the altitude.
02

Calculate kinetic energy

The kinetic energy (KE) of an object is calculated using the formula: \[KE = \frac{1}{2}mv^2\] where \(m\) is mass and \(v\) is the speed.
03

Determine total energy

The total energy (TE) of the aircraft is the sum of its potential energy and kinetic energy: \[TE = PE + KE\]
04

Derive expression for energy height

To derive the expression for energy height in terms of the quantities \(m, h,\) and \(v\), divide the total energy by the aircraft's weight (\(mg\)): \[H = \frac{TE}{mg}\]
05

Calculate energy height for Boeing 747 jet

Given the mass, altitude, and speed for the Boeing 747 jet, we can use the formulas from Steps 1 to 4 to calculate its energy height: \[m = 3.5 \cdot 10^{5} \mathrm{~kg}\] \[h = 10.0 \cdot 10^{3} \mathrm{~m}\] \[v = 250.0 \mathrm{~m/s}\] First, calculate the potential energy: \[PE = (3.5 \times 10^{5})(9.81)(10 \times 10^3)\] Next, calculate the kinetic energy: \[KE = \frac{1}{2}(3.5 \times 10^{5}) (250)^2\] Find the total energy by adding PE and KE: \[TE = PE + KE\] Find the energy height H using the derived expression, and the weight of the aircraft (\(mg\)): \[H = \frac{TE}{mg}\] This will give us the energy height (maximum altitude) of the Boeing 747 jet in meters.

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Key Concepts

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

Gravitational Potential Energy
Imagine an object lifted above the ground鈥攊t鈥檚 being held against the pull of gravity. Gravitational potential energy (\textbf{PE}) represents the work done to elevate that object and is a form of stored energy. The higher the object, the greater the potential energy.

When dealing with problems in physics, we calculate this energy with the formula: \[PE = mgh\], where:
  • \(m\) is the mass of the object in kilograms.
  • \(g\) is the acceleration due to gravity (\(9.81 \text{ m/s}^2\) on Earth).
  • \(h\) is the height or altitude above the reference point, often ground level, in meters.

The concept of gravitational potential energy is crucial when considering the energy profile of an aircraft in flight. For the aircraft, this energy is indicative of its capability to do work by virtue of its altitude.

Kinetic Energy
As your textbook teaches, energy comes in many forms, and when we discuss movement, we talk about kinetic energy (\textbf{KE}). Kinetic energy is the energy of motion鈥攅very moving object has kinetic energy.

The formula to determine the kinetic energy of an object is: \[KE = \frac{1}{2}mv^2\]Here,
  • \(m\) stands for the mass of the object (how heavy it is), and
  • \(v\) is the velocity or speed of the object (how fast it鈥檚 moving).

In the case of aircraft, the faster they're moving, or the more massive they are, the more kinetic energy they possess. This kinetic energy contributes to the aircraft's ability to continue in motion and overcome resistive forces.

Total Mechanical Energy
Let's combine what we've learned: an aircraft, like any object, can possess both potential and kinetic energy simultaneously. Total mechanical energy (\textbf{TE}) is simply the sum of an object鈥檚 potential and kinetic energies.

To obtain the total mechanical energy, we use the equation: \[TE = PE + KE\].

For an aircraft, TE reflects its overall energy state, accounting for both its altitude and speed. Understanding the concept of total mechanical energy is essential because it reveals the total energy resources the aircraft can convert from one form to another while preserving energy conservation laws鈥攕ignificant when considering maneuvers or flight dynamics.

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Most popular questions from this chapter

A high jumper approaches the bar at \(9.0 \mathrm{~m} / \mathrm{s}\). What is the highest altitude the jumper can reach, if he does not use any additional push off the ground and is moving at \(7.0 \mathrm{~m} / \mathrm{s}\) as he goes over the bar?

Two masses are connected by a light string that goes over a light, frictionless pulley, as shown in the figure. The 10.0 -kg mass is released and falls through a vertical distance of \(1.00 \mathrm{~m}\) before hitting the ground. Use conservation of mechanical energy to determine: a) how fast the 5.00 -kg mass is moving just before the 10.0 -kg mass hits the ground; and b) the maximum height attained by the 5.00 -kg mass.

A 5.00 -kg ball of clay is thrown downward from a height of \(3.00 \mathrm{~m}\) with a speed of \(5.00 \mathrm{~m} / \mathrm{s}\) onto a spring with \(k=\) \(1600 . \mathrm{N} / \mathrm{m} .\) The clay compresses the spring a certain maximum amount before momentarily stopping a) Find the maximum compression of the spring. b) Find the total work done on the clay during the spring's compression.

A 0.500 -kg mass is attached to a horizontal spring with \(k=100 . \mathrm{N} / \mathrm{m}\). The mass slides across a frictionless surface. The spring is stretched \(25.0 \mathrm{~cm}\) from equilibrium, and then the mass is released from rest. a) Find the mechanical energy of the system. b) Find the speed of the mass when it has moved \(5.00 \mathrm{~cm}\). c) Find the maximum speed of the mass.

A basketball of mass \(0.624 \mathrm{~kg}\) is shot from a vertical height of \(1.2 \mathrm{~m}\) and at a speed of \(20.0 \mathrm{~m} / \mathrm{s}\). After reaching its maximum height, the ball moves into the hoop on its downward path, at \(3.05 \mathrm{~m}\) above the ground. Using the principle of energy conservation, determine how fast the ball is moving just before it enters the hoop.
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