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Chapter 5: Problem 60

Aircraft Carrier Takeoff On an aircraft carrier, a jet can be catapulted from 0 to \(250 \mathrm{~km} / \mathrm{h}\) in \(2.00 \mathrm{~s}\) (see Figure \(5.20\) ). If the average force exerted by the catapult is \(9.35 \times 10^{5} \mathrm{~N}\), what is the mass of the jet? Figure \(5.20\) A jet takes off from the flight deck of an aircraft carrier.

Short Answer

Expert verified
The mass of the jet is approximately 5374 kg.

Step by step solution

01

Understand the Problem

We need to find the mass of the jet. We're given that the jet is accelerated from 0 to 250 km/h in 2 seconds, and the force exerted by the catapult is \(9.35 \times 10^5 \text{ N}\).
02

Convert Velocity

Convert the velocity of the jet from kilometers per hour to meters per second, since 1 km/h = 0.27778 m/s. Thus, 250 km/h converts to \(250 \times 0.27778 = 69.444 \text{ m/s}\).
03

Apply Newton's Second Law

According to Newton's Second Law, \( F = ma \), where \( F \) is force, \( m \) is mass, and \( a \) is acceleration. We will need to find acceleration \( a \).
04

Calculate Acceleration

Acceleration is the change in velocity divided by time: \(2 imes 10^5 ext{ kg} imes 34.72 ext{ m/s}^2 \) (by converting 34.72 m/s to m/s^2, which is not necessary since we already calculated force as acceleration) Finally, we have the jet's mass as: \( m = \frac{9.35 \times 10^{5} \text{ N}}{1.74 \times 10^2 ext{ m/s}^2} \approx 5373.6 ext{ kg} \).
05

Finalize the Solution

The mass of the jet is found to be approximately 5374 kg. This was calculated by rearranging the formula for force to solve for mass and using the given acceleration and force.

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Key Concepts

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

Force Calculation
Understanding how to calculate force is crucial in applying Newton's Second Law. According to this law, force (\( F \) ) is the product of mass (\( m \) ) and acceleration (\( a \) ). This can be represented by the equation: \[ F = ma \] In our exercise, the force exerted by the catapult on the jet is known: \( 9.35 \times 10^5 \text{ N} \). Here's the process to calculate force using Newton's Second Law:
  • Identify the variables given in the problem.
  • Use the equation \( F = ma \) to relate force, mass, and acceleration.
  • In this case, solve for mass once acceleration is determined.
This principle is fundamental in physics and helps us understand how force affects the motion of objects. Once you have either force or mass and acceleration, you can determine the missing variable.
Velocity Conversion
Before you can calculate acceleration, it's important to ensure all your units are consistent. This means converting velocities from kilometers per hour to meters per second, which is a standard unit of velocity in physics. Here's how to do it:
  • Know that 1 kilometer per hour is equal to 0.27778 meters per second.
  • Multiply the given velocity in km/h by this conversion factor.
  • For our jet, \( 250 \text{ km/h} \times 0.27778 \ = 69.444 \text{ m/s} \).
By converting the velocity, you ensure that your calculations for acceleration and subsequent mathematical operations are accurate. This step eliminates unit mismatches and keeps equations in a coherent format, which is crucial for precision in physics.
Acceleration Determination
Identifying acceleration is a key aspect of solving physics problems, especially when applying Newton's Second Law. Acceleration (\( a \) ) is defined as the change in velocity over time. In formula terms, this is expressed as:\[ a = \frac{\Delta v}{\Delta t} \] Let's break this down with our example:
  • Determine the change in velocity (\( \Delta v \) ). Here, it's from \( 0 \text{ m/s} \) to \( 69.444 \text{ m/s} \).
  • The time (\( \Delta t \) ) taken for this change is \( 2 \text{ s} \).
  • Plug these values into the formula: \[ a = \frac{69.444 \text{ m/s} - 0 \text{ m/s}}{2 \text{ s}} = 34.722 \text{ m/s}^2 \].
Accurate determination of acceleration allows you to link it with force and mass, effectively enabling you to resolve dynamics problems using Newton's framework.

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Most popular questions from this chapter

An object moves with no friction or air resistance. Initially, its kinetic energy is \(10 \mathrm{~J}\), and its gravitational potential energy is \(20 \mathrm{~J}\). What is its kinetic energy when its potential energy has decreased to 15 J? What is its potential energy when its kinetic energy has decreased to 5 J?

Your favorite uncle makes this statement: "A force that is always perpendicular to the velocity of a particle does no work on the particle." Is this statement true or false? If it is true, state why. If it is false, give a counterexample.

Predict \& Explain The work required to accelerate a car from 0 to \(50 \mathrm{~km} / \mathrm{h}\) is \(W\). (a) Is the work required to accelerate the car from \(50 \mathrm{~km} / \mathrm{h}\) to \(150 \mathrm{~km} / \mathrm{h}\) equal to \(2 W, 3 W, 8 W\), or \(9 W ?\) (b) Choose the best explanation from among the following: A. The work to accelerate the car depends on the speed squared. B. The final speed is three times the speed that was produced by the work \(W\). C. The increase in speed from \(50 \mathrm{~km} / \mathrm{h}\) to \(150 \mathrm{~km} / \mathrm{h}\) is twice the increase in speed from 0 to \(50 \mathrm{~km} / \mathrm{h}\).

Calculate A 0.14-kg pinecone is \(16 \mathrm{~m}\) above the ground. What is its gravitational potential energy?

A crow drops a \(0.11-\mathrm{kg}\) clam onto a rocky beach from a height of \(9.8 \mathrm{~m}\). What is the kinetic energy of the clam when it is \(5.0 \mathrm{~m}\) above the ground? What is its speed at that point?
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