/*! 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 75 An \(1865-\mathrm{kg}\) airplane... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 75

An \(1865-\mathrm{kg}\) airplane starts at rest on an airport runway at sea level. (a) What is the change in mechanical energy of the airplane if it climbs to a cruising altitude of \(2420 \mathrm{m}\) and maintains a constant speed of \(96.5 \mathrm{m} / \mathrm{s}\) ? (b) What cruising speed would the plane need at this altitude if its increase in kinetic energy is to be equal to its increase in potential energy?

Short Answer

Expert verified
(a) The change in mechanical energy is 44,318,350 J. (b) The required cruising speed is approximately 153.42 m/s.

Step by step solution

01

Determine Potential Energy Change

To find the increase in the potential energy of the airplane, use the formula for gravitational potential energy: \(\Delta U = mgh\), where \(m\) is the mass of the airplane \(1865\, \text{kg}\), \(g\) is the acceleration due to gravity \(9.8\, \text{m/s}^2\), and \(h\) is the height \(2420\, \text{m}\).\[\Delta U = 1865 \times 9.8 \times 2420\]
02

Calculate Potential Energy Value

Now compute the potential energy:\[\Delta U = 1865 \times 9.8 \times 2420 = 44,318,350 \, \text{Joules}\]
03

Determine Kinetic Energy at Constant Speed

The kinetic energy of the airplane when it's flying at the constant speed of \(96.5 \, \text{m/s}\) is calculated using the formula for kinetic energy: \(K = \frac{1}{2}mv^2\).\[ K = \frac{1}{2} \times 1865 \times (96.5)^2\]
04

Calculate Kinetic Energy Value

Compute the kinetic energy:\[K = \frac{1}{2} \times 1865 \times 96.5^2 = 8,648,631.25 \, \text{Joules}\]
05

Determine Total Mechanical Energy Change for Part (a)

The total change in mechanical energy is the sum of the change in potential energy and the initial kinetic energy (since it starts from rest, the initial kinetic energy is 0). Therefore:\[\Delta E = \Delta U + 0 = 44,318,350 \, \text{J}\]
06

Set Condition for Equal Energy Increases for Part (b)

To find the cruising speed where the increase in kinetic energy equals the increase in potential energy, set \(\Delta K = \Delta U\). From Step 1, we know \(\Delta U = 44,318,350 \, \text{J}\).Solve for \(v\) with \[\frac{1}{2}mv^2 = 44,318,350\]
07

Calculate the Required Cruising Speed

Rearrange to solve for \(v\):\[ v^2 = \frac{2 \times 44,318,350}{1865} \]\[ v = \sqrt{\frac{2 \times 44,318,350}{1865}} \]Compute \(v\).\[ v \approx 153.42 \, \text{m/s} \]

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Key Concepts

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

Potential Energy
Potential energy is a form of energy that an object possesses due to its position or condition. Think about holding a ball above the ground. It has the potential to fall because of its height. The higher the ball, the more potential energy it has. More formally, potential energy is often calculated using the formula:\[ U = mgh \]where:
  • m is the mass of the object in kilograms.
  • g is the acceleration due to gravity, approximately \(9.8 \, \text{m/s}^2\).
  • h is the height above the reference point in meters.
For the airplane in our exercise, this potential energy changes as it climbs to a new altitude. This is because the airplane has gained height, thereby increasing its potential to do work due to its elevated position.
Kinetic Energy
Kinetic energy is the energy of motion. Any object that is moving, from a speeding car to the flowing water in a river, has kinetic energy. This type of energy depends on the mass of the object and its velocity. The formula to calculate kinetic energy is:\[ K = \frac{1}{2}mv^2 \]Here:
  • m is the mass of the object, in our case, the airplane.
  • v is the velocity of the object in meters per second (\text{m/s}).
As the airplane accelerates to its cruising speed, its kinetic energy increases. This is because both its mass and velocity factor into the energy calculation. In our scenario, even if the airplane maintains a constant speed, the kinetic energy associated with that speed remains significant.
Gravitational Potential Energy
Gravitational potential energy is a specific type of potential energy that relates to an object's position relative to the Earth. It's especially important when considering objects at different heights. To determine gravitational potential energy, we use the same formula for potential energy:\[ U = mgh \]Gravitational potential energy increases with height because the force of gravity acting on the object (here, the airplane) works over a longer distance. This energy can be converted into kinetic energy if the airplane descends without any other force acting on it, highlighting the interplay between these types of mechanical energy. The exercise demonstrates how gravitational potential energy plays a crucial role when an airplane climbs to a specific altitude.

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Most popular questions from this chapter

You ride your bicycle down a hill, maintaining a constant speed the entire time. (a) As you ride, does the gravitational potential energy of the you- bike-Earth system increase, decrease, or stay the same? Explain. (b) Does the kinetic energy of you and your bike increase, decrease, or stay the same? Explain. (c) Does the mechanical energy of the you-bike-Earth system increase, decrease, or stay the same? Explain.

As an Acapulco cliff diver drops to the water from a height of \(46 \mathrm{m},\) his gravitational potential energy decreases by \(25,000 \mathrm{J}\) What is the diver's weight in new tons?

An 8.70 -kg block slides with an initial speed of \(1.56 \mathrm{m} / \mathrm{s}\) up a ramp inclined at an angle of \(28.4^{\circ}\) with the horizontal. The coefficient of kinetic friction between the block and the ramp is \(0.62 .\) Use energy conservation to find the distance the block slides before coming to rest.

Predict/Explain When a ball of mass \(m\) is dropped from rest from a height \(h,\) its kinetic energy just before landing is \(K\). Now, suppose a second ball of mass \(4 m\) is dropped from rest from a height \(h / 4 .\) (a) Just before ball 2 lands, is its kinetic energy \(4 K, 2 K, K, K / 2,\) or \(K / 4 ?\) (b) Choose the best explanation from among the following: I. The two balls have the same initial energy. II. The more massive ball will have the greater kinetic energy. III. The reduced drop height results in a reduced kinetic energy.

Jeff of the Jungle swings on a \(7.6-\mathrm{m}\) vine that initially makes an angle of \(37^{\circ}\) with the vertical. If Jeff starts at rest and has a mass of \(78 \mathrm{kg}\), what is the tension in the vine at the lowest point of the swing?
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