/*! 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 23 The X-15 is air-launched from un... [FREE SOLUTION] | 91Ó°ÊÓ

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The X-15 is air-launched from under the wing of a B-52 mother ship. Immediately after launch, the pilot starts the XLR-99 rocket engine, which provides 57,000 Ib of thrust. For the first moments, the X-15 accelerates in horizontal flight. The gross weight of the airplane at the start is \(34,000 \mathrm{lb}\). Calculate the initial acceleration of the airplane.

Short Answer

Expert verified
The initial acceleration of the airplane is approximately 1.676 times the acceleration due to gravity.

Step by step solution

01

Identify the Forces Involved

The only significant force acting on the X-15 horizontally immediately after launch is the thrust from the rocket engine. The thrust provided by the XLR-99 rocket engine is 57,000 lb.
02

Apply Newton's Second Law

Newton's Second Law states that the acceleration of an object is the net force acting on it divided by its mass. In terms of weight and force in pounds, this can be simplified to \( a = \frac{F}{m} \), where \( a \) is acceleration, \( F \) is the thrust (57,000 lb), and \( m \) is the weight of the airplane (34,000 lb).
03

Calculate the Initial Acceleration

Using the formula \( a = \frac{F}{m} \), substitute the known values: \( a = \frac{57,000}{34,000} \). Simplify this expression to find the acceleration.
04

Simplifying the Expression

Divide 57,000 by 34,000: \( a = \frac{57,000}{34,000} = 1.676 \). The units cancel since both force and weight are given in pounds, leaving acceleration in terms of \( \text{g}_{ ext{Earth}} \) (acceleration due to gravity), but essentially unitless as a ratio.

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Key Concepts

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

Rocket Thrust
In the context of rockets, thrust is the force that propels the rocket forward. It can be understood as the reaction force described by Newton's Third Law of Motion, which states that every action has an equal and opposite reaction. When a rocket engine fires, it expels gas particles at high speed in one direction, and the rocket moves in the opposite direction. This is what creates the thrust. The XLR-99 rocket engine used in the X-15 generates a thrust of 57,000 pounds. This means the engine is continuously pushing the rocket with a force equivalent to 57,000 pounds in a specific direction. Thrust is a crucial aspect because without this force, the rocket wouldn't be able to move or change velocity. To explore further:
  • Thrust must overcome the opposing forces such as drag and the inertia of the rocket.
  • The amount of thrust can be varied during flight to control acceleration.
  • A higher thrust means a greater capacity to accelerate or lift heavier payloads.
Understanding thrust is essential for figuring out how rockets move through different phases of their trajectory.
Initial Acceleration
Initial acceleration refers to the rate at which an object begins to speed up just after it's set in motion. It is particularly important in scenarios like rocket launches because it sets the pace for the rest of the journey. For the X-15 immediately after launch, calculating initial acceleration helps us understand how quickly the plane will increase its speed from rest.Using Newton's Second Law, we calculate initial acceleration as:\[ a = \frac{F}{m} \]where:
  • \( F \) is the force (thrust) of 57,000 lb generated by the rocket engine
  • \( m \) is the mass, or more accurately in this context, the weight of the X-15 at 34,000 lb
Substituting these values gives us the formula expressing the initial acceleration:\[ a = \frac{57,000}{34,000} = 1.676 \]This result indicates that initially, the X-15 experiences an acceleration of 1.676 times Earth's gravitational acceleration, which propels the aircraft forward.
Force and Mass Relationship
The force and mass relationship is a core element of Newton's Second Law, which establishes that the force applied to an object results in acceleration proportional to the force itself and inversely proportional to the object's mass. In simpler terms, the greater the force applied to an object, the more it accelerates. Conversely, the more massive an object, the less it accelerates for a given force.Mathematically, this is expressed with the equation:\[ a = \frac{F}{m} \]This equation highlights:
  • Force \( F \) is directly proportional to acceleration \( a \). If force increases, so does acceleration.
  • Mass \( m \) is inversely proportional to acceleration. As mass increases, acceleration decreases for the same force.
In the exercise, the X-15's initial acceleration is the result of dividing the rocket's thrust by its mass. Because the X-15 has a significant weight (or mass), the rocket needs a powerful engine to achieve noticeable acceleration.This relationship between force, mass, and acceleration is fundamental to mastering problems in mechanics and is extensively used in various engineering fields, especially in aerospace, to design efficient rockets and vehicles that can achieve desired performances.

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Most popular questions from this chapter

In the four-stroke, reciprocating, internal combustion engine that powers most automobiles as well as most small general aviation aircraft, combustion of the fuel-air mixture takes place in the volume between the top of the piston and the top of the cylinder. (Reciprocating engines are discussed in Ch. 9.) The gas mixture is ignited when the piston is essentially at the end of the compression stroke (called top dead center), when the gas is compressed to a relatively high pressure and is squeezed into the smallest volume that exists between the top of the piston and the top of the cylinder. Combustion takes place rapidly before the piston has much time to start down on the power stroke. Hence, the volume of the gas during combustion stays constant; that is, the combustion process is at constant volume. Consider the case where the gas density and temperature at the instant combustion begins are \(11.3 \mathrm{~kg} / \mathrm{m}^{3}\) and \(625 \mathrm{~K}\), respectively. At the end of the constant-volume combustion process, the gas temperature is \(4000 \mathrm{~K}\). Calculate the gas pressure at the end of the constant-volume combustion. Assume that the specific gas constant for the fuel-air mixture is the same as that for pure air.

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