/*! 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 29 The rocket sled shown below dece... [FREE SOLUTION] | 91Ó°ÊÓ

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The rocket sled shown below decelerates at a rate of \(196 \mathrm{m} / \mathrm{s}^{2}\). What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is \(2.10 \times 10^{3} \mathrm{kg}.\)

Short Answer

Expert verified
The force necessary to produce the deceleration of the rocket sled is \(-4.116 \times 10^5 \mathrm{N}\).

Step by step solution

01

Identify given variables

Before solving the problem, let's list down the given variables from the exercise: - Deceleration rate (a): \(- 196 \, \mathrm{m/s^2}\) - Mass of the system (m): \(2.10 \times 10^3 \, \mathrm{kg}\)
02

Use Newton's second law of motion to calculate the force

Newton's second law of motion states that the force acting on an object is equal to the product of its mass and acceleration. The formula is: \(F = m \cdot a\) We will plug in the given values into the formula to calculate the force (F).
03

Plug in the values and calculate the force

Replace the variables in the formula with the given values: \(F = (2.10 \times 10^3 \mathrm{kg}) \cdot (-196 \mathrm{m/s^2})\) Calculate the force: \(F = -4.116 \times 10^5 \mathrm{N}\)
04

Interpret the result

The force necessary to produce the deceleration of the rocket sled is \(-4.116 \times 10^5 \mathrm{N}\). The negative sign represents that the force acts in the opposite direction to the motion, which is consistent with the deceleration scenario.

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Key Concepts

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

Deceleration
Deceleration occurs when an object slows down; it is the opposite of acceleration. In physics, deceleration is simply described as negative acceleration. For the rocket sled, which originally moves at a certain speed, the deceleration rate is given as -196 m/s². The sign indicates the slowdown is in the opposite direction to the initial motion.
When understanding deceleration, it's important to consider the factors causing it, like friction or external forces. In this exercise, there is no power from the rocket, and the given deceleration is due to an external force applied to stop the sled.
  • Deceleration is essentially the rate of decrease in velocity.
  • It is calculated just like acceleration, with a change in velocity over time, but it is negative.
  • In this context, the rocket sled undergoes a deceleration, which means it is slowing down over time.
Being clear on what deceleration means helps in understanding how external forces interact with moving objects, which is crucial for calculating the needed stopping force.
Force Calculation
To calculate force, we use Newton's Second Law of Motion, which simplifies to the equation: \( F = m \cdot a \). This formula is crucial because it connects mass, acceleration, and force, showing how they influence one another in motion dynamics. When you know the mass and the rate of deceleration, as given in the exercise, you can determine the force required to bring the object to a stop. For the sled:
  • Mass \( m \) is the total weight of the system, \(2.10 \times 10^3\) kg.
  • Deceleration (acceleration) \( a \) is \(-196 m/s^2\).
The calculation with these values is straightforward:\[ F = 2.10 \times 10^3 \times (-196) \]From this, the force \( F \) is calculated to be \(-4.116 \times 10^5\) N. The negative sign indicates the direction of the force is opposite to the motion. Understanding this concept is fundamental because it allows you to predict how much effort is needed to change the motion of objects under different conditions.
Mass and Acceleration Relationship
The relationship between mass and acceleration is core to comprehending motion. According to Newton's Second Law, acceleration is produced when a force acts on a mass. The acceleration is directly proportional to the net force and inversely proportional to the mass of the object. In simpler terms:
  • The greater the mass of an object, the more force it needs to accelerate.
  • Conversely, for a fixed force, a larger mass will result in smaller acceleration.
For the rocket sled example:
  • Mass of the sled system \( m \) is \(2.10 \times 10^3\) kg, which indicates how heavy the system is.
  • The calculated acceleration (here deceleration) \( a \) is \(-196 \mathrm{m/s^2}\).
The relationship shows why such a large force is needed to decelerate massive objects. It illustrates how heavier objects resist changes in motion more due to their larger mass. Grasping this mass and acceleration connection is vital for predicting and understanding motion patterns.

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Most popular questions from this chapter

An object is acted on by three simultaneous forces: \(\overrightarrow{\mathbf{F}}_{1}=(-3.00 \hat{\mathbf{i}}+2.00 \hat{\mathbf{j}}) \mathrm{N}\) \(\overrightarrow{\mathbf{F}}_{2}=(6.00 \hat{\mathbf{i}}-4.00 \hat{\mathbf{j}}) \mathrm{N}, \quad\) and \(\overrightarrow{\mathbf{F}}_{3}=(2.00 \hat{\mathbf{i}}+5.00 \hat{\mathbf{j}}) \mathrm{N} .\) The object experiences acceleration of \(4.23 \mathrm{m} / \mathrm{s}^{2} .\) (a) Find the acceleration vector in terms of \(m\). (b) Find the mass of the object. (c) If the object begins from rest, find its speed after 5.00 s. (d) Find the components of the velocity of the object after 5.00 S.

What is the deceleration of the rocket sled if it comes to rest in \(1.10 \mathrm{s}\) from a speed of \(1000.0 \mathrm{km} / \mathrm{h}\) ? (Such deceleration caused one test subject to black out and have temporary blindness.)

Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of \(75.0 \mathrm{N},\) the second exerts a force of \(90.0 \mathrm{N},\) friction is \(12.0 \mathrm{N},\) and the mass of the third child plus wagon is \(23.0 \mathrm{kg}\). (a) What is the system of interest if the acceleration of the child in the wagon is to be calculated? (See the free-body diagram.) (b) Calculate the acceleration. (c) What would the acceleration be if friction were 15.0 N?

Suppose that you are viewing a soccer game from a helicopter above the playing field. Two soccer players simultaneously kick a stationary soccer ball on the flat field; the soccer ball has mass 0.420 kg. The first player kicks with force \(162 \mathrm{N}\) at \(9.0^{\circ}\) north of west. At the same instant, the second player kicks with force \(215 \mathrm{N}\) at \(15^{\circ}\) east of south. Find the acceleration of the ball in \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) form.

A 120-kg astronaut is riding in a rocket sled that is sliding along an inclined plane. The sled has a horizontal component of acceleration of \(5.0 \mathrm{m} / \mathrm{s}^{2}\) and a downward component of \(3.8 \mathrm{m} / \mathrm{s}^{2} .\) Calculate the magnitude of the force on the rider by the sled. (Hint: Remember that gravitational acceleration must be considered.)

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