/*! 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 18 A plane weighing \(220 \mathrm{k... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 18

A plane weighing \(220 \mathrm{kN} \quad(25 \text { tons })\) lands on an aircraft carrier. The plane is moving horizontally at $67 \mathrm{m} / \mathrm{s}(150 \mathrm{mi} / \mathrm{h})$ when its tailhook grabs hold of the arresting cables. The cables bring the plane to a stop in a distance of $84 \mathrm{m}$ (a) How much work is done on the plane by the arresting cables? (b) What is the force (assumed constant) exerted on the plane by the cables? (Both answers will be underestimates, since the plane lands with the engines full throttle forward; in case the tailhook fails to grab hold of the cables, the pilot must be ready for immediate takeoff.)

Short Answer

Expert verified
Question: Calculate the work done by the arresting cables and the constant force exerted on the airplane. Solution: 1. Convert the plane's weight to mass: \(m = \frac{220 \, kN}{9.81 \, m/s^2} = 22416.31 \, kg\) 2. Calculate the initial kinetic energy (KE_initial): KE_initial = \(\frac{1}{2}mv^2 = \frac{1}{2} \times 22416.31 \, kg \times (67 \, m/s)^2 = 50,786,772.79 \, J\) 3. Apply the work-energy theorem: W = KE_final - KE_initial = \(0 - 50,786,772.79 \, J = -50,786,772.79 \, J\) 4. Calculate the force exerted by the arresting cables: F = \(-\frac{W}{d} = -\frac{-50,786,772.79 \, J}{84 \, m} = 605,080.87 \, N\) Answer: The work done by the arresting cables is approximately \(-50,786,772.79 \, J\) and the constant force exerted on the airplane by the cables is approximately \(605,080.87 \, N\).

Step by step solution

01

(Understanding the Given Information)

(First, let's recall the information given in the problem: The plane's initial weight is \(220 \, kN\), its initial velocity is \(67 \, m/s\), and it comes to a stop over a distance of \(84 \, m\). We'll use these values in our calculations.)
02

(Calculate the Initial and Final Kinetic Energies)

(We need to find out the initial and final kinetic energies of the plane. Remember, kinetic energy is given by the formula KE = \(\frac{1}{2}mv^2\). The initial kinetic energy (KE_initial) of the plane is given by \(\frac{1}{2}M(v_1)^2\) where \(m\) is the mass of the plane \((\frac{220 \, kN}{9.81 \, m/s^2})\), and \(v_1\) is the initial velocity of the plane, \(67 \, m/s\). Since the plane comes to a stop, its final kinetic energy (KE_final) is \(0.\))
03

(Applying Work-Energy Theorem)

(Now, we will apply the work-energy theorem which states that the work done on an object is equal to the change in its kinetic energy. Using the formula, W = \(\Delta\)KE = KE_final - KE_initial, we can find the work done by the arresting cables on the plane.)
04

(Calculating the Work Done)

(Plug in the values of initial and final kinetic energies into the work-energy theorem formula. W = \(0 - \frac{1}{2}M(v_1)^2\). Solve for W to get the work done by the arresting cables.)
05

(Finding the Force Exerted by the Arresting Cables)

(We can find the constant force exerted on the plane by the cables using the definition of work done: \(W = Fd \cos\theta\), where \(F\) is the force, \(d\) is the stopping distance \(84 \, m\), and \(\theta\) is the angle between the force and displacement (in this case, \(\theta = 180^{\circ}\) because the force acts opposite to the displacement). Since \(W\) is negative and the cosine of \(180^{\circ}\) is \(-1\), we can rearrange this equation as \(F = -\frac{W}{d}\). Plug in the values of \(W\) and \(d\) and solve for the force exerted by the arresting cables.) Now, let's compute the values for the work done by the arresting cables and the force exerted by them using the information and formulas we derived above.

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 91影视!

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

Sean climbs a tower that is \(82.3 \mathrm{m}\) high to make a jump with a parachute. The mass of Sean plus the parachute is \(68.0 \mathrm{kg} .\) If \(U=0\) at ground level, what is the potential energy of Sean and the parachute at the top of the tower?
Sam pushes a \(10.0-\mathrm{kg}\) sack of bread flour on a frictionless horizontal surface with a constant horizontal force of \(2.0 \mathrm{N}\) starting from rest. (a) What is the kinetic energy of the sack after Sam has pushed it a distance of \(35 \mathrm{cm} ?\) (b) What is the speed of the sack after Sam has pushed it a distance of \(35 \mathrm{cm} ?\)
The number of kilocalories per day required by a person resting under standard conditions is called the basal metabolic rate (BMR). (a) To generate 1 kcal, Jermaine's body needs approximately 0.010 mol of oxygen. If Jermaine's net intake of oxygen through breathing is 0.015 mol/min while he is resting, what is his BMR in kcal/day? (b) If Jermaine fasts for 24 h, how many pounds of fat does he lose? Assume that only fat is consumed. Each gram of fat consumed generates \(9.3 \mathrm{kcal}\)
Hilda holds a gardening book of weight \(10 \mathrm{N}\) at a height of $1.0 \mathrm{m}\( above her patio for \)50 \mathrm{s}$. How much work does she do on the book during that 50 s?
A car moving at \(30 \mathrm{mi} / \mathrm{h}\) is stopped by jamming on the brakes and locking the wheels. The car skids 50 ft before coming to rest. How far would the car skid if it were initially moving at $60 \mathrm{mi} / \mathrm{h} ?$ [Hint: You will not have to do any unit conversions if you set up the problem as a proportion. \(]\)
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Famous Physicists

Read Explanation

Rotational Dynamics

Read Explanation

Wave Optics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Space Physics

Read Explanation
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.