/*! 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 14 A fighter jet is launched from a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A fighter jet is launched from an aircraft carrier with the aid of its own engines and a steam-powered catapult. The thrust of its engines is \(2.3 \times 10^{5} \mathrm{~N}\). In being launched from rest it moves through a distance of \(87 \mathrm{~m}\) and has a kinetic energy of \(4.5 \times 10^{7} \mathrm{~J}\) at liftoff. What is the work done on the jet by the catapult?

Short Answer

Expert verified
The work done by the catapult is \(2.499 \times 10^{7} \, \mathrm{J}\).

Step by step solution

01

Determine Total Work Done

The total work done on the jet is equal to the change in its kinetic energy at liftoff. Since the jet starts from rest, the initial kinetic energy is 0 J. Therefore, the total work done is equal to the final kinetic energy of the jet at liftoff, which is given as \(4.5 \times 10^{7} \mathrm{~J}\).
02

Calculate Work Done by Engine

The work done by the jet's engines can be calculated using the formula \(W = F \times d\), where \(F\) is the force supplied by the engines and \(d\) is the distance over which the force is applied. Given that the thrust of the engines is \(2.3 \times 10^{5} \, \mathrm{N}\) and the distance is \(87 \, \mathrm{m}\), the work done by the engines is \(W = (2.3 \times 10^{5} \, \mathrm{N}) \times 87 \, \mathrm{m}\). Compute this to get \(W = 2.001 \times 10^{7} \, \mathrm{J}\).
03

Calculate Work Done by Catapult

The work done by the catapult is the remaining work needed to reach the total kinetic energy. Use the equation: \(W_{catapult} = W_{total} - W_{engine}\), where \(W_{total} = 4.5 \times 10^{7} \, \mathrm{J}\) and \(W_{engine} = 2.001 \times 10^{7} \, \mathrm{J}\). Substitute these values to calculate \(W_{catapult} = 4.5 \times 10^{7} \, \mathrm{J} - 2.001 \times 10^{7} \, \mathrm{J} = 2.499 \times 10^{7} \, \mathrm{J}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinetic Energy
Kinetic energy is a type of energy that an object possesses due to its motion. Imagine pushing a toy car across the floor. The faster it goes, the more kinetic energy it has. For an object that is moving, its kinetic energy can be calculated using the formula: \[ KE = \frac{1}{2} m v^2 \]*Where:*
  • \(m\) is the mass of the object
  • \(v\) is the velocity of the object
In the case of the fighter jet, its kinetic energy at liftoff is given as \(4.5 \times 10^7\, \mathrm{J}\). The jet starts from rest, which means its initial kinetic energy is \(0\, \mathrm{J}\). As the jet moves down the runway, its kinetic energy increases due to the force applied by its engines and the catapult. Understanding kinetic energy helps us see how energy transforms from one type to another. When an object speeds up, the work done on it gets converted into kinetic energy. It's like filling a balloon with air—the more you blow, the fuller it gets!
Work Done
"Work done" is a term used to describe the energy transferred when a force moves an object over a distance. Consider lifting a book off the ground: your muscles exert a force, and as the book rises, work is done on it. The formula for work done is:\[ W = F \times d \times \cos(\theta) \]*Where:*
  • \(W\) is the work done
  • \(F\) is the force applied
  • \(d\) is the distance moved in the direction of the force
  • \(\theta\) is the angle between the force and the displacement direction
In the problem of the fighter jet, the engines do work as they exert a thrust of \(2.3 \times 10^5\, \mathrm{N}\) over a distance of \(87\, \mathrm{m}\). The work done by the engines is calculated without angle consideration here (as both force and motion are in the same direction), amounting to \(2.001 \times 10^7\, \mathrm{J}\). By understanding work done, we can quantify how much energy is used to move objects, which is essential in understanding energy conversion processes. It provides insight into the efficiency of engines and other machinery.
Force and Motion
Force is a push or pull exerted on an object, and it can cause changes in the object's motion. This is a fundamental idea in physics. Imagine pushing a swing: the harder you push, the faster it moves. The relationship between force, mass, and acceleration is captured by Newton's Second Law, which states:\[ F = m \times a \]*Where:*
  • \(F\) is the force
  • \(m\) is the mass
  • \(a\) is the acceleration
In the context of the fighter jet, the engines produce a thrust—this is the force helping push the jet forward. As the engines generate this force, they accelerate the jet from rest, causing that increase in velocity and kinetic energy. Additionally, the catapult provides an extra force component to achieve the needed speed for liftoff. Understanding the interplay between force and motion allows us to predict and control how objects will move when different forces are applied. It's key to knowing how vehicles operate, from cars to airplanes, making it a foundational principle in engineering and physics. As a final note, remember that force alone doesn't cause motion, but rather it's the net force (or unbalanced force) that leads to acceleration or changes in velocity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(1200-\mathrm{kg}\) car is being driven up a \(5.0^{\circ}\) hill. The frictional force is directed opposite to the motion of the car and has a magnitude of \(f=524 \mathrm{~N}\). A force \(\overrightarrow{\mathrm{F}}\) is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight \(\overrightarrow{\mathrm{W}}\) and the normal force \(\overrightarrow{\mathrm{F}}_{\mathrm{N}}\) directed perpendicular to the road surface. The length of the road up the hill is \(290 \mathrm{~m}\). What should be the magnitude of \(\overrightarrow{\mathbf{F}}\), so that the net work done by all the forces acting on the car is \(+150 \mathrm{~kJ}\) ?

The drawing shows two frictionless inclines that begin at ground level \((h=0 \mathrm{~m})\) and slope upward at the same angle \(\theta .\) One track is longer than the other, however. Identical blocks are projected up each track with the same initial speed \(v_{0}\). On the longer track the block slides upward until it reaches a maximum height \(H\) above the ground. On the shorter track the block slides upward, flies off the end of the track at a height \(H_{1}\) above the ground, and then follows the familiar parabolic trajectory of projectile motion. At the highest point of this trajectory, the block is a height \(H_{2}\) above the end of the track. The initial total mechanical energy of each block is the same and is all kinetic energy. (a) When the block on the longer track reaches its maximum height, is its final total mechanical energy all kinetic energy, all potential energy, or some of each? Explain. (b)When the block on the shorter track reaches the top of its trajectory after leaving the track, is its final total mechanical energy all kinetic energy, all potential energy, or some of each? Justify your answer. (c) Which is the greater height above the gound: \(H\) or \(H_{1}+H_{2}\) ? Why? The initial speed of each block is \(v_{0}=7.00 \mathrm{~m} / \mathrm{s},\) and each incline slopes upward at an angle of \(\theta=50.0^{\circ} .\) The block on the shorter track leaves the track at a height of \(H_{1}=1.25 \mathrm{~m}\) above the ground. Find (a) the height \(H\) for the block on the longer track and (b) the total height \(H_{1}+H_{2}\) for the block on the shorter track. Verify that your answers are consistent with your answers to the Concept Questions.

A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates. The crest of the second hill is circular, with a radius of \(r=36 \mathrm{~m} .\) Neglect friction and air resistance. What must be the height \(h\) of the first hill so that the skier just loses contact with the snow at the crest of the second hill?

The brakes of a truck cause it to slow down by applying a retarding force of \(3.0 \times 10^{3} \mathrm{~N}\) to the truck over a distance of \(850 \mathrm{~m}\). What is the work done by this force on the truck? Is the work positive or negative? Why?

Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the drive force is shut off, the snowmobile coasts to a halt. (a) During the coasting phase, what is the net external force acting on the snowmobile? (b) How does the work \(W\) done by the net external force depend on the magnitude and direction of the net force and on the displacement of the snowmobile? (c) What is the net force acting on the snowmobile during the constantvelocity phase? Explain. (d) The work-energy theorem is given by \(W=\frac{1}{2} m v_{\mathrm{f}}^{2}-\frac{1}{2} m v_{0}^{2}\) (Equation 6.3 ), where \(W\) is the work done by the net external force acting on the snowmobile, \(m\) is its mass, and \(v_{\mathrm{f}}\) and \(v_{0}\) are, respectively, its final and initial speeds. Suppose the net external force, \(m,\) and \(v_{\mathrm{f}}\) and \(v_{0}\) are known. By using only the work-energy theorem, can one determine the time \(t\) it takes for the snowmobile to coast to a halt? The snowmobile and its rider have a mass of \(136 \mathrm{~kg}\). Under the influence of a drive force of \(205 \mathrm{~N}\), it is moving at a constant velocity, whose magnitude is \(5.50 \mathrm{~m} / \mathrm{s}\). Its drive force is shut off. Find (a) the distance in which the snowmobile coasts to a halt and (b) the time required to do so.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.