/*! 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 48 Once ignited, a small rocket mot... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Once ignited, a small rocket motor on a spacecraft exerts a constant force of \(10 \mathrm{~N}\) for \(7.80 \mathrm{~s}\). During the burn, the rocket causes the \(100-\mathrm{kg}\) craft to accelerate uniformly. Determine that acceleration.

Short Answer

Expert verified
The acceleration is \(0.1\, \mathrm{m/s^2}\).

Step by step solution

01

Understand the Given Problem

We are given a constant force exerted by a rocket, a time duration for which this force is applied, and the mass of the spacecraft. We need to find the acceleration of the spacecraft.
02

Identify Relevant Formula

To find acceleration when a force is applied, we use Newton's second law: \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.
03

Rearrange the Formula for Acceleration

Rearrange the formula to solve for acceleration \( a \): \[ a = \frac{F}{m} \]
04

Plug in the Known Values

Substitute the given force \( F = 10 \mathrm{~N} \) and mass \( m = 100 \mathrm{~kg} \) into the formula: \[ a = \frac{10 \mathrm{~N}}{100 \mathrm{~kg}} \]
05

Calculate the Acceleration

Perform the division: \[ a = 0.1 \mathrm{~m/s^2} \]. Thus, the acceleration of the spacecraft is \( 0.1 \mathrm{~m/s^2} \).

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.

Force and Acceleration
Newton's Second Law of Motion is a fundamental concept in physics that explains how an object responds to forces acting upon it. When a force is applied to an object, it accelerates. The relationship is described by the formula: \( F = ma \), where \( F \) stands for force, \( m \) for mass, and \( a \) for acceleration.
This equation means that the force applied to an object is equal to the mass of that object multiplied by its acceleration. It helps us understand how changes in force or mass affect acceleration.
  • A stronger force on a constant mass results in a greater acceleration.
  • A heavier object requires more force to achieve the same acceleration as a lighter object.
Newton's Second Law is essential in predicting how objects will move, whether they're on Earth or in space.
Uniform Acceleration
Uniform acceleration refers to a constant rate of increase in velocity that an object experiences. When the acceleration does not change over time, it is uniform. In the context of rocket motion, when a constant force, such as from a small motor, is exerted, it produces uniform acceleration.
The significance of uniform acceleration lies in its simplicity, making calculations straightforward. Since velocity change is constant, it's predictable. In our rocket example, the spacecraft accelerates uniformly due to the steady force output of the motor. This means:
  • Velocity increases by the same amount every second as long as the force is constant.
  • The paths of objects with uniform acceleration are easy to calculate or forecast.
Recognizing uniform acceleration enables us to apply straightforward equations to determine speed and distance over time.
Mass and Weight
The terms 'mass' and 'weight' are often misunderstood as interchangeable, but they describe different properties. Mass refers to the amount of material in an object and is a scalar quantity, meaning it does not have a direction. It is measured in kilograms (kg) and remains constant regardless of location.
Weight, on the other hand, is a measure of the gravitational force acting on an object, a vector quantity with both magnitude and direction. On Earth, it changes with the strength of the gravitational field. Weight is calculated using \( W = mg \), where \( g \) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\) on Earth).
  • Weight can change depending on where the object is in the universe.
  • Mass is constant and fundamental to the object itself.
Understanding this distinction is crucial, especially in space contexts, where gravity's effects differ drastically from those on Earth.
Spacecraft Motion
Spacecraft motion depends heavily on the principles of force and acceleration. Unlike on Earth, where gravity and friction are significant factors in motion, space presents unique challenges and opportunities for motion with far fewer resistance factors.
In the vacuum of space, once a spacecraft has accelerated, it will continue to move at a constant velocity unless acted upon by another force. This inertia is due to the minimal friction and gravitational effects in space compared to those on Earth. Rockets use short bursts of force for acceleration, as seen in our exercise, during which a constant force results in uniform acceleration, helping to change speed and direction efficiently.
  • Inertia allows spacecraft to travel vast distances with minimal energy after initial acceleration.
  • Precision control of motion and direction requires understanding Newton's laws and applying them accurately in the absence of Earth's frictional forces.
Grasping these principles ensures successful navigation and operation of spacecraft in various missions across the cosmos.

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

An elevator starts from rest with a constant upward acceleration. It moves \(2.0 \mathrm{~m}\) in the first \(0.60 \mathrm{~s}\). A passenger in the elevator is holding a \(3.0-\mathrm{kg}\) package by a vertical string. What is the tension in the string during the accelerating process?

A car whose weight is \(F_{W}\) is on a ramp which makes an angle \(\theta\) to the horizontal. How large a perpendicular force must the ramp withstand if it is not to break under the car's weight? As rendered in Fig. \(3-6\), the car's weight is a force \(\overrightarrow{\mathbf{F}}_{W}\) that pulls straight down on the car. We take components of \(\overrightarrow{\mathbf{F}}\) along the incline and perpendicular to it. The ramp must balance the force component \(F_{W} \cos \theta\) if the car is not to crash through the ramp. In other words, the force exerted on the car by the ramp, upwardly perpendicular to the ramp, is \(F_{N}\) and \(F_{N}=F_{W} \cos \theta\).

A cord passing over a frictionless, massless pulley has a \(4.0\) -kg object tied to one end and a \(12-\mathrm{kg}\) object tied to the other. Compute the acceleration and the tension in the cord.

Imagine a planet having a mass twice that of Earth and a radius equal to \(1.414\) times that of Earth. Determine the acceleration due to gravity at its surface.

Compute the mass of the Earth, assuming it to be a sphere of radius \(6370 \mathrm{~km}\). Use the fact that \(\mathrm{g}=9.81 \mathrm{~m} / \mathrm{s}^{2}\) and give your answer to three significant figures. Let \(M_{E}\) be the mass of the Earth, and \(m\) the mass of an object. The weight of the object on the planet's surface is equal to \(m g .\) It is also equal to the gravitational force \(G\left(M_{E} m\right) R_{E}^{2}\), where \(R_{E}\) is the Earth's radius. Hence, $$ m g=G \frac{M_{E} m}{R_{E}^{2}} $$ from which \(\quad M_{E}=\frac{g R_{E}^{2}}{G}=\frac{\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\left(6.37 \times 10^{6} \mathrm{~m}\right)^{2}}{6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}}=5.97 \times 10^{24} \mathrm{~kg}\)

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.