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Chapter 5: Problem 92

Compute the initial upward acceleration of a rocket of mass \(1.3 \times 10^{4} \mathrm{~kg}\) if the initial upward force produced by its engine (the thrust) is \(2.6 \times 10^{5} \mathrm{~N}\). Do not neglect the gravitational force on the rocket.

Short Answer

Expert verified
The initial upward acceleration of the rocket is \(10.2 \mathrm{~m/s^2}\).

Step by step solution

01

Identify the Forces Acting on the Rocket

The rocket is subjected to two forces: the upward thrust force from the engine, \( F_{thrust} = 2.6 \times 10^5 \mathrm{~N} \), and the downward gravitational force, \( F_{gravity} = m \cdot g \), where \(m = 1.3 \times 10^4 \mathrm{~kg}\) is the mass of the rocket, and \( g = 9.8 \mathrm{~m/s^2} \) is the acceleration due to gravity.
02

Calculate the Gravitational Force

We calculate the gravitational force using the formula \( F_{gravity} = m \cdot g \).Substitute the values: \[ F_{gravity} = 1.3 \times 10^4 \times 9.8 = 1.274 \times 10^5 \mathrm{~N} \].
03

Calculate the Net Force on the Rocket

The net force \( F_{net} \) acting on the rocket is the difference between the thrust force and the gravitational force: \[ F_{net} = F_{thrust} - F_{gravity} \]. Substitute the known values: \[ F_{net} = 2.6 \times 10^5 - 1.274 \times 10^5 = 1.326 \times 10^5 \mathrm{~N} \].
04

Calculate the Upward Acceleration

Use Newton's second law \( F_{net} = m \cdot a \) to find the acceleration \( a \). Rearrange the equation for \( a \): \[ a = \frac{F_{net}}{m} \].Substitute the known values: \[ a = \frac{1.326 \times 10^5}{1.3 \times 10^4} = 10.2 \mathrm{~m/s^2} \].

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Key Concepts

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

Rocket Propulsion
Rocket propulsion is the process that allows rockets to move through space. This movement is due to the action-reaction principle described by Newton's Third Law of Motion. When a rocket engine fires, it expels gas out of its engine at high speed in the opposite direction of the intended movement. This expulsion of gas produces a force, called thrust, which propels the rocket forward.
Rockets are powered by engines that burn propellants to create hot gases. These gases are pushed out of the engine's nozzle to produce thrust.
  • The nozzle is specially designed to increase the speed of these gases, which increases the thrust force.
  • The greater the rate at which the gas is expelled, the larger the thrust force and the faster the rocket accelerates.
Efficient rocket propulsion is essential, especially in overcoming forces like gravity, which tries to pull the rocket back to the ground as it is launching.
Gravitational Force
Gravitational force is the force due to gravity that pulls objects towards the center of the Earth. This force exists everywhere on Earth and even in space, being one of the four fundamental forces of nature.
On Earth, this force is calculated using the formula \( F_{gravity} = m \cdot g \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity, typically \( 9.8 \mathrm{~m/s^2} \).
For rockets:
  • This force opposes the upward motion created by the thrust force.
  • The rocket must overcome this gravitational force to ascend into space effectively.
The gravitational force is essential in determining the net force that acts on the rocket, affecting its acceleration.
Thrust Force
Thrust force is the force generated by the rocket engines to propel the rocket upward. It is essential in moving the rocket against gravitational forces and atmospheric drag.
Thrust is generally created by the rapid expulsion of high-pressure gases:
  • In the case of rockets, this occurs when the rocket engines combust fuel and oxidizer.
  • The magnitude of thrust force depends on the velocity at which the gases are ejected, as well as the mass of the gases.
This force must be greater than the gravitational force for the rocket to lift off and must be constantly managed to handle different phases of flight, from liftoff to reaching orbit. Newton's Second Law explains this interaction with: \( F_{net} = m \cdot a \), showing how the net force, which includes thrust and gravity, affects the rocket's acceleration.

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Most popular questions from this chapter

A motorcycle and \(60.0 \mathrm{~kg}\) rider accelerate at \(3.0 \mathrm{~m} / \mathrm{s}^{2}\) up a ramp inclined \(10^{\circ}\) above the horizontal. What are the magnitudes of (a) the net force on the rider and (b) the force on the rider from the motorcycle?

Three forces act on a particle that moves with unchanging velocity \(\vec{v}=(2 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(7 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} .\) Two of the forces are \(\vec{F}_{1}=(2 \mathrm{~N}) \hat{\mathrm{i}}+\) \((3 N) \hat{j}+(-2 N) \hat{k}\) and \(\vec{F}_{2}=(-5 N) \hat{i}+(8 N) \hat{j}+(-2 N) \hat{k} .\) What is the third force?

A nucleus that captures a stray neutron must bring the neutron to a stop within the diameter of the nucleus by means of the strong force. That force, which "glues" the nucleus together, is approximately zero outside the nucleus. Suppose that a stray neutron with an initial speed of \(1.4 \times 10^{7} \mathrm{~m} / \mathrm{s}\) is just barely captured by a nucleus with diameter \(d=1.0 \times 10^{-14} \mathrm{~m}\). Assuming the strong force on the neutron is constant, find the magnitude of that force. The neutron's mass is \(1.67 \times 10^{-27} \mathrm{~kg}\).

A block is projected up a frictionless inclined plane with initial speed \(v_{0}=3.50 \mathrm{~m} / \mathrm{s} .\) The angle of incline is \(\theta=32.0^{\circ} .(\mathrm{a})\) How far up the plane does the block go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?

A shot putter launches a \(7.260 \mathrm{~kg}\) shot by pushing it along a straight line of length \(1.650 \mathrm{~m}\) and at an angle of \(34.10^{\circ}\) from the horizontal, accelerating the shot to the launch speed from its initial speed of \(2.500 \mathrm{~m} / \mathrm{s}\) (which is due to the athlete's preliminary motion). The shot leaves the hand at a height of \(2.110\) \(\mathrm{m}\) and at an angle of \(34.10^{\circ}\), and it lands at a horizontal distance of \(15.90 \mathrm{~m}\). What is the magnitude of the athlete's average force on the shot during the acceleration phase? (Hint: Treat the motion during the acceleration phase as though it were along a ramp at the given angle.)
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