/*! 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 6 A rocket with a lift-off mass \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 6

A rocket with a lift-off mass \(20000 \mathrm{~kg}\) is blasted upwards with an initial acceleration of \(5.0 \mathrm{~m} / \mathrm{s}^{2}\). Calculate the initial thrust (force) of the blast. [NCERT] (a) \(3 \times 10^{5} \mathrm{~N}\) (b) \(2 \times 10^{8} \mathrm{~N}\) (c) \(4 \times 10^{5} \mathrm{~N}\) (d) \(5 \times 10^{5} \mathrm{~N}\)

Short Answer

Expert verified
The initial thrust of the blast is approximately \( 3 \times 10^5 \) N (option a).

Step by step solution

01

Identify the Known Quantities

The lift-off mass of the rocket is given as \( m = 20,000 \) kg and the initial acceleration is \( a = 5.0 \text{ m/s}^2 \).
02

Determine the Gravitational Force

Calculate the gravitational force acting on the rocket using the formula \( F_g = m \, g \), where \( g = 9.81 \text{ m/s}^2 \) is the acceleration due to gravity. This yields \( F_g = 20,000 \, \times 9.81 = 196,200 \text{ N} \).
03

Apply Newton's Second Law

The net force required for the rocket's acceleration can be found using \( F_{net} = m \, a \). Thus, \( F_{net} = 20,000 \times 5.0 = 100,000 \text{ N} \).
04

Calculate the Total Thrust

The total thrust of the rocket must overcome both the gravitational force and provide the additional force necessary for the initial acceleration, so it is \( F_{thrust} = F_g + F_{net} \). Therefore, \( F_{thrust} = 196,200 + 100,000 = 296,200 \text{ N} \).
05

Compare with Given Options

Rounding \( 296,200 \text{ N} \) to the nearest significant figure matches option (a) \( 3 \times 10^5 \text{ N} \).

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 91影视!

Key Concepts

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

Newton's Second Law
Newton's Second Law of Motion is essential for understanding how forces affect motion. This law states:
  • A body's acceleration is directly proportional to the net force acting on it.
  • This acceleration inversely depends on the body's mass.
  • Expressed mathematically, the law is given by the formula: \( F = m \cdot a \)
In the rocket problem, this principle helps determine the net force needed to achieve the given initial acceleration. Here, the net force helps the mass of the rocket accelerate upwards at 5.0 m/s虏. By applying Newton's Second Law, you identify how additional forces, such as gravitational pull, interact to influence the total thrust required.
Gravitational Force
Gravitational force is the attraction an object experiences towards the center of the Earth.
  • It acts in the downward direction on all mass affected by Earth's gravity.
  • The formula to calculate this force is \( F_g = m \cdot g \), where \( m \) is the mass, and \( g \) is the gravitational acceleration.
  • On Earth, \( g \) is approximately \( 9.81 \text{ m/s}^2 \).
In the context of a rocket, gravitational force resists its upward movement. It is vital to calculate this force, as it must be overcome by the rocket's engines to achieve lift-off. For our 20,000 kg rocket, the gravitational force is quite significant, coming in at 196,200 N.
Initial Acceleration
Initial acceleration is the acceleration a rocket needs to achieve at the very start of its motion. Achieving a high initial acceleration ensures the rocket overcomes inertia and gravitational forces effectively.
  • In this problem, the initial acceleration given is 5.0 m/s虏.
  • This acceleration helps the rocket to start its journey upwards immediately upon launch.
  • The net force producing this acceleration must be calculated separately alongside the gravitational force, as both affect the total thrust needed.
Thus, achieving the target initial acceleration is crucial for the success of the mission, as it indicates sufficient power has been generated to lift the rocket.
Net Force
Net force is the total force acting on an object when all individual forces are combined.
  • It determines the acceleration of an object according to Newton's Second Law.
  • To find net force, you multiply the rocket's mass by its acceleration (\( F_{net} = m \cdot a \)).
  • For our rocket, this calculation yields a net force of 100,000 N, needing to act to achieve the desired acceleration.
Net force is the result of subtracting opposing forces such as gravity from the total thrust force provided by the engines. Therefore, it plays a critical role in understanding how much more power the rocket needs to reach the intended acceleration.
Mass of Rocket
The mass of the rocket is a fundamental component in calculating forces acting on it.
  • In this example, the mass is given as 20,000 kg.
  • The mass influences both the force needed to achieve acceleration and the gravitational force.
  • Consideration of mass ensures accurate calculations of thrust and motion dynamics.
Mass is an unwavering constant that impacts both the required magnitude of thrust and the energies employed. It's essential to factor in when designing engines and estimating the power needed for mission success. Understanding the mass lets engineers anticipate how much fuel and thrust are necessary for lift-off.

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

A block of mass \(2 \mathrm{~kg}\) is placed on the floor. The coefficient of static friction is \(0.4\). If a force of \(2.8 \mathrm{~N}\) is applied on the block parallel to floor, the force of friction between the block and floor (Taking \(\left.g=10 \mathrm{~ms}^{-2}\right)\) is (a) \(2.8 \mathrm{~N}\) (b) \(8 \mathrm{~N}\) (c) \(2 \mathrm{~N}\) (d) zero

A cricket player lowers his hands while catching a ball, because (a) ball is heavy (b) ball is coming with high speed (c) it increases the time of impact and reduces the impact of force on his hands (d) Nonc of the above

Two masses of \(3 \mathrm{~kg}\) and \(5 \mathrm{~kg}\) are suspended from the ends of an unstreatchable massless cord passing over a frictionless pulley. When the masses are released, the pressure on the pulley is (a) \(2 \mathrm{~kg} \mathrm{f}\) (b) \(7.5 \mathrm{~kg}\) f (c) \(8 \mathrm{~kg} \mathrm{f}\) (d) \(15 \mathrm{kgf}\)

A chain lies on a rough horizontal table. It starts sliding when one-fourth of its length hangs over the edge of the table. The coefficient of static friction between the chain and the surface of the table is (a) \(\frac{1}{2}\) (b) \(\frac{1}{3}\) (c) \(\frac{1}{4}\) (d) \(\frac{1}{5}\)

A \(1000 \mathrm{~kg}\) lift is supported by a cable that can support \(2000 \mathrm{~kg}\). The shortest distance in which the lift can be stopped when it is descending with a speed of \(2.5 \mathrm{~ms}^{-1}\) is [Take \(\left.g=10 \mathrm{~ms}^{-2}\right]\) (a) \(1 \mathrm{~m}\) (b) \(2 \mathrm{~m}\) (c) \(\frac{5}{32} \mathrm{~m}\) (d) \(\frac{5}{16} \mathrm{~m}\)
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Oscillations

Read Explanation

Thermodynamics

Read Explanation

Engineering Physics

Read Explanation

Famous Physicists

Read Explanation

Mechanics and Materials

Read Explanation
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.