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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 -kg craft to accelerate uniformly. Determine that acceleration.

Short Answer

Expert verified
The acceleration of the spacecraft is \(0.1 \text{ m/s}^2\).

Step by step solution

01

Identify known values

We are given the following values in the problem: the force exerted by the rocket motor, \( F = 10 \text{ N} \), the time duration of the force application, \( t = 7.80 \text{ s} \), and the mass of the spacecraft, \( m = 100 \text{ kg} \).
02

Recall Newton's Second Law

Newton's Second Law of Motion tells us that force equals mass times acceleration: \( F = ma \). This formula will help us find the acceleration when the force and mass are known.
03

Solve for acceleration

Rearrange the equation from Step 2 to solve for acceleration \( a \):\[ a = \frac{F}{m} \]Now, substitute the given values into the equation:\[ a = \frac{10 \text{ N}}{100 \text{ kg}} \]
04

Calculate the acceleration

Carry out the division in the equation from Step 3:\[ a = 0.1 \text{ m/s}^2 \]This value represents the constant acceleration of the spacecraft.

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Key Concepts

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

Force and Acceleration
When we talk about force and acceleration, we delve into one of the core concepts of physics, governed by Newton's Second Law of Motion. This law fundamentally states that the force acting on an object is equal to the mass of that object multiplied by the acceleration produced. In mathematical terms, this is expressed as:
  • \( F = ma \)
Force and acceleration share a direct relationship. That means when a force is applied to an object, it will accelerate in the direction of the force. The magnitude of this acceleration depends on both the amount of force applied and the mass of the object.
Think of pushing a toy car versus pushing a real car. The toy car will accelerate much more easily with a small force due to its low mass. Conversely, a real car will need a far greater force to achieve the same acceleration.
This principle is vital in understanding how spacecraft move, as well as in countless other real-world applications.
Uniform Acceleration
Uniform acceleration refers to a consistent change in velocity over time. This means that every second, the object's speed increases (or decreases) by the same amount. In the given exercise, the spacecraft's acceleration remains constant while the force is applied.
This is an ideal situation often used in physics problems to simplify calculations. In everyday terms, if a car accelerates uniformly, it would be like pushing down the gas pedal steadily, increasing speed smoothly without jerking or sudden changes.
Uniform acceleration allows us to easily calculate changes in speed and position over time using simple equations. This kind of motion helps in predicting future positions or velocities, making it extremely useful in planning space missions or even in designing amusement park rides.
In the exercise, the constant force provides a uniform acceleration to the spacecraft, ensuring smooth and predictable movement.
Mass and Force Relationship
The relationship between mass and force is a key aspect of Newton's Second Law. They are directly connected to how an object accelerates. According to the law:
  • The larger the mass of an object, the more force required to change its motion.
  • If the same force is applied to two objects of different masses, the object with the smaller mass will experience a greater acceleration.
This explains why the 100 kg spacecraft in the exercise, despite the force of 10 N, has a relatively small acceleration.
The principle helps in understanding phenomena like why different vehicles need different amounts of power to accelerate or stop, and it's crucial in designing systems like rockets, where getting the balance between mass and force right is essential for effective motion.
Calculating Acceleration
To calculate acceleration, you can use the formula derived from Newton's Second Law. The rearranged equation to find acceleration given force and mass is:
\[ a = \frac{F}{m} \]This formula allows you to determine how quickly an object will speed up or slow down when a known force is applied. In the exercise, we used this formula to find the spacecraft's acceleration.
Simply plug the force of 10 N and the mass of 100 kg into the equation:
\[ a = \frac{10 \text{ N}}{100 \text{ kg}} = 0.1 \text{ m/s}^2 \]This tells us that the spacecraft accelerates at a rate of 0.1 meters per second squared, indicating a steady increase in velocity due to the engine's thrust. Understanding how to calculate acceleration is crucial for predicting how objects will move under different forces, an essential skill in many fields of engineering and science.

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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 \(3.0\) -kg package by a vertical string. What is the tension in the string during the accelerating process?

An unknown force acting on a \(50.0\) -g body floating in space produces a constant acceleration of \(20.0 \mathrm{~cm} / \mathrm{s}^{2}\). If the same force is now made to act on a different body, also in space, producing a constant acceleration of \(40.0 \mathrm{~cm} / \mathrm{s}^{2}\), what is the mass of that body?

Having hauled it to the top of a tilted driveway, a child is holding a wagon from rolling back down. The driveway is inclined at \(20^{\circ}\) to the horizontal. If the wagon weighs \(150 \mathrm{~N}\), with what force must the child pull on the handle if the handle is parallel to and pointing up the incline?

A constant force acts on a \(5.0\) kg object and reduces its velocity from \(7.0 \mathrm{~m} / \mathrm{s}\) to \(3.0 \mathrm{~m} / \mathrm{s}\) in a time of \(3.0\) s. Determine the force. We must first find the acceleration of the object, which is constant because the force is constant. Taking the direction of motion as positive, from Chapter 2 $$ a=\frac{v_{f}-v_{i}}{t}=\frac{-4.0 \mathrm{~m} / \mathrm{s}}{3.0 \mathrm{~s}}=-1.33 \mathrm{~m} / \mathrm{s}^{2} $$ Use \(F=m a\) with \(m=5.0 \mathrm{~kg}\) : $$ F=(5.0 \mathrm{~kg})\left(-1.33 \mathrm{~m} / \mathrm{s}^{2}\right)=-6.7 \mathrm{~N} $$ The minus sign indicates that the force is a retarding force, directed opposite to the motion.

A horizontal cable pulls a 200 -kg cart along a horizontal track. The tension in the cable is \(500 \mathrm{~N}\). Starting from rest, \((a)\) How long will it take the cart to reach a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) ? (b) How far will it have gone?

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