/*! 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 15 A small \(8.00 \mathrm{~kg}\) ro... [FREE SOLUTION] | 91Ó°ÊÓ

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A small \(8.00 \mathrm{~kg}\) rocket burn fuel that exerts a time-varying upward force on the rocket (assume constant mass) as the rocket moves upward from the launch pad. This force obeys the equation \(F=A+B t^{2} .\) Measurements show that at \(t=0,\) the force is \(100.0 \mathrm{~N}\) and at the end of the first \(2.00 \mathrm{~s}\), it is \(150.0 \mathrm{~N}\). (a) Find the constants \(A\) and \(B\), including their SI units. (b) Find the net force on this rocket and its acceleration (i) the instant after the fuel ignites and (ii) \(3.00 \mathrm{~s}\) after the fuel ignites. (c) Suppose that you were using this rocket in outer space, far from all gravity. What would its acceleration be 3.00 s after fuel ignition?

Short Answer

Expert verified
The constants \(A\) and \(B\) are \(100N\) and \(12.5N/s^{2}\) respectively. The net force and acceleration the instant after the fuel ignites are \(78N\) and \(9.75m/s²\) respectively. Three seconds after ignition, these values are \(112.6N\) and \(14.07m/s²\) respectively. In the absence of gravity, the acceleration three seconds after ignition would be \(23.4m/s²\).

Step by step solution

01

Find the Constants A and B

From the problem, we know that when \(t=0\), the force \(F=100N\). This enables us to write the following equation from \(F=A+Bt^{2}\): \(A=100N\) since the right term of the equation vanishes to zero. For \(t=2s\) and \(F=150N\), substitute these values into the equation to get: \(150 = 100 + 4B\). Solving for \(B\) hence gives \(B=12.5N/s^{2}\). Hence, \(A=100N\) and \(B=12.5N/s^{2}\).
02

Calculate Net Force the Instant After the Fuel Ignites

The instant after the fuel ignites is effectively \(t=0\). Substituting these values into the equation \(F=A+Bt^{2}\), we get \(F = 100N - mg\), where \(m\) is the mass of the rocket and \(g\) is acceleration due to gravity (9.8 m/s²). Substituting \(m=8kg\) and \(g=9.8 m/s²\) into the equation gives \(F = 100N - (8 kg)(9.8 m/s²) = 78N\).
03

Calculate Acceleration the Instant After the Fuel Ignites

Apply Newton's second law \(F = ma\) to solve for acceleration \(a\). Substituting known values gives \(78N = 8kg * a\). Solving for \(a\) hence gives \(a = 9.75 m/s²\).
04

Calculate Net Force 3s After the Fuel Ignites

Substitute \(t=3s\) into the equation \(F=A+Bt^{2}\), resulting in \(F = 100N +12.5N/s² * (3s)^2 = 187.5N\). Then, the net force is \(F - mg = 187.5N - (8kg)(9.8m/s²) = 112.6 N\).
05

Calculate Acceleration 3s After the Fuel Ignites

Use Newton's second law \(F = ma\) to solve for acceleration \(a\). Substituting known values gives \(112.6N = 8kg * a\). Solving for \(a\) gives \(a = 14.07 m/s²\).
06

Calculate Acceleration in Outer Space 3s After Fuel Ignition

In outer space, far from gravity, the net force on the rocket would be similar to the gross force exerted by the fuel. Hence, with \(F=187.5N\), the acceleration \(a\) can be calculated with Newton's second law as \(187.5N = 8kg * a\). Solving for \(a\) gives \(a = 23.4 m/s²\) in the absence of gravity.

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental principle in physics that explains how forces affect motion. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
Translated into a formula, this law is written as: \[ F = ma \] where:
  • \( F \) is the net force applied to the object, measured in newtons (N).
  • \( m \) is the mass of the rocket, in kilograms (kg), which remains constant.
  • \( a \) is the acceleration produced, in meters per second squared \((m/s²)\).
In our rocket problem, Newton's Second Law helps us understand how the varying force applied by the rocket fuel affects its motion. As different forces act on the rocket at different times, the resulting acceleration changes, which we calculate using this second law.
Forces and Motion
Forces and motion are the heart of dynamics in physics, describing how things move and why they change their state of motion.
Force is any interaction that, when unopposed, changes the motion of an object. Forces can cause an object to start moving, stop moving, or change direction.
In the case of the rocket, the force exerted by its fuel plays a crucial role. The motion is not just defined by fuel thrust, but also by gravity's impact. For example, when the rocket ignites initially, the net force is a result of subtracting the gravitational force from the thrust force. The formula for net force immediately after ignition (\( t=0 \)) becomes:\[ F_{ ext{net}} = A - mg \]where:
  • \( A \) is the initial force through rocket propulsion.
  • \( m \) is the rocket's mass.
  • \( g \) is the acceleration due to gravity, about \(9.8 \, m/s²\).
Understanding forces in motion helps calculate how these forces culminate into the rocket's ascent moment after moment.
Rocket Propulsion
Rocket propulsion is a process that allows rockets to move by expelling mass in one direction to generate a force in the opposite direction, following Newton's Third Law.
In this scenario, the rocket uses a time-varying force, which means that the amount of thrust changes over time, represented by the equation: \[ F = A + Bt^2 \] Here:
  • \( A \) is the force at initialization.
  • \( B \) describes how the force increases over time, adjusted by \( t^2 \), the square of time.
The varying force enables rockets to adjust their acceleration at different phases of flight, providing flexibility in how the rocket moves through space.
It's essential for propulsion to be carefully calculated to ensure the rocket can travel the intended path, overcoming gravity and, eventually, operating differently when in vacuum conditions, such as space.
Time-varying Forces
Time-varying forces are forces that change in magnitude or direction as time progresses. This is highly relevant in rocket applications, as the output force is not static.
In our example, the time-varying force equation is given by: \[ F = A + B t^2 \] Where:
  • \( A \) is the initial thrust provided by the rocket's engines.
  • \( B \) quantifies how much the force increases as time progresses.
These forces allow the rocket to gradually increase its thrust, typically necessary for lifting off and reaching desired altitudes.
When considering time-varying forces, understanding the time component allows engineers and scientists to predict and control the motion precisely, as seen in the problem when calculating the net forces at 0, 2, and 3 seconds.

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Most popular questions from this chapter

Estimate the mass in kilograms and the weight in pounds of a typical sumo wrestler. How do your estimates for the wrestler compare to your estimates of the average mass and weight of the students in your physics class? Do a web search if necessary to help make the estimates. In your solution list what values you assume for the quantities you use in making your estimates.

A loaded elevator with very worn cables has a total mass of \(2200 \mathrm{~kg},\) and the cables can withstand a maximum tension of \(28,000 \mathrm{~N}\) (a) Draw the free-body force diagram for the elevator. In terms of the forces on your diagram, what is the net force on the elevator? Apply Newton's second law to the elevator and find the maximum upward acceleration for the elevator if the cables are not to break. (b) What would be the answer to part (a) if the elevator were on the moon, where \(g=1.62 \mathrm{~m} / \mathrm{s}^{2} ?\)

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