/*! 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 89 A rocket blasts off from rest an... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 89

A rocket blasts off from rest and attains a speed of \(45 \mathrm{~m} / \mathrm{s}\) in \(15 \mathrm{~s}\). An astronaut has a mass of \(57 \mathrm{~kg}\). What is the astronaut's apparent weight during takeoff?

Short Answer

Expert verified
The astronaut's apparent weight during takeoff is 729.6 N.

Step by step solution

01

Define Apparent Weight

The apparent weight of the astronaut is the normal force acting on the astronaut. During takeoff, this is the sum of the gravitational force and the force due to the rocket's acceleration.
02

Calculate Gravitational Force

The gravitational force (weight) acting on the astronaut can be calculated using the formula: \[ F_g = m imes g \] where \( m = 57 \text{ kg} \) is the mass of the astronaut and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. Plugging in the values: \[ F_g = 57 imes 9.8 = 558.6 \text{ N} \].
03

Calculate Rocket's Acceleration

The acceleration of the rocket can be calculated using the formula: \[ a = \frac{v}{t} \] where \( v = 45 \text{ m/s} \) is the final velocity and \( t = 15 \text{ s} \) is the time taken. Plugging in the values: \[ a = \frac{45}{15} = 3 \text{ m/s}^2 \].
04

Calculate Force Due to Rocket's Acceleration

The force experienced by the astronaut due to the rocket's acceleration is given by: \[ F_a = m imes a \] where \( m = 57 \text{ kg} \) is the mass of the astronaut and \( a = 3 \text{ m/s}^2 \) is the acceleration. Plugging in the values: \[ F_a = 57 \times 3 = 171 \text{ N} \].
05

Calculate Apparent Weight

The apparent weight of the astronaut is the sum of the gravitational force and the force due to the rocket's acceleration: \[ F_{apparent} = F_g + F_a \] \[ F_{apparent} = 558.6 + 171 = 729.6 \text{ N} \].

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.

Gravitational Force
Gravitational force is a fundamental concept that affects all objects with mass. It is the force that attracts two bodies towards each other. In the context of our exercise, the Earth exerts a gravitational force on the astronaut.

To find the gravitational force, we use the equation:
  • Formula: \( F_g = m \times g \)
  • \( m \) is the mass of the object (57 kg for the astronaut)
  • \( g \) is the acceleration due to Earth's gravity, approximately 9.8 m/s虏
By substituting these values, we compute the gravitational force: \[ F_g = 57 \times 9.8 = 558.6 \text{ N} \]

This value represents the weight of the astronaut when standing still at the Earth's surface.
Rocket Acceleration
Rocket acceleration is essential for understanding changes in motion during liftoff. It describes how a rocket鈥檚 velocity changes with time as it rises off the Earth's surface.

We calculate this acceleration using the formula:
  • Formula: \( a = \frac{v}{t} \)
  • \( v \) is the final velocity reached (45 m/s)
  • \( t \) is the time taken to reach this velocity (15 seconds)
Plug these values into the formula: \[ a = \frac{45}{15} = 3 \text{ m/s}^2 \]

This tells us how fast the speed of the rocket increases per second. Rocket acceleration creates additional force on the astronaut beyond what's felt from gravity.
Normal Force
Normal force is the support force exerted upon an object in contact with another stable object. For an astronaut in a rising rocket, this force is felt as a reaction to the rocket's acceleration.

It's crucial for calculating the apparent weight of the astronaut, as it includes both the gravitational force and the force due to the acceleration of the rocket.

Given:
  • Gravitational Force: 558.6 N
  • Force from Rocket Acceleration: 171 N
The apparent weight, experienced as normal force, is: \[ F_{apparent} = F_g + F_a \] \[ F_{apparent} = 558.6 + 171 = 729.6 \text{ N} \]

This calculated force is what the astronaut feels as their weight inside the accelerating rocket.
Newton's Laws of Motion
Newton's Laws of Motion are critical in understanding how forces affect the motion of objects. The concepts behind these laws help us analyze situations like a rocket's takeoff.

**First Law (Law of Inertia):**
  • An object at rest stays at rest, and an object in motion stays in motion unless acted upon by an external force.
  • The astronaut inside the rocket would remain at rest unless the rocket provides a force to propel them upwards.


**Second Law (Law of Acceleration):**
  • The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Represented by \( F = m \times a \).
  • It shows us that the acceleration of the rocket, and thus the force felt by the astronaut, changes directly based on the forces applied.


**Third Law (Action-Reaction Law):**
  • For every action, there is an equal and opposite reaction.
  • The rocket pushes downwards upon the Earth's surface; in reaction, the Earth pushes the rocket upwards, causing the rise and subsequent forces experienced by the astronaut.


Understanding these laws aids in grasping why the apparent weight changes during a rocket's launch.

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 350 -kg sailboat has an acceleration of \(0.62 \mathrm{~m} / \mathrm{s}^{2}\) at an angle of \(64^{\circ}\) north of east. Finc the magnitude and direction of the net force that acts on the sailboat.

(a) Calculate the magnitude of the gravitational force exerted on a \(425-\mathrm{kg}\) satellite that is a distance of two earth radii from the center of the earth. (b) What is the magnitude of the gravitational force exerted on the earth by the satellite? (c) Determine the magnitude of the satellite's acceleration. (d) What is the magnitude of the earth's acceleration? A space traveler weighs \(540 \mathrm{~N}\) on earth. What will the traveler weigh on another planet whose radius is three times that of earth and whose mass is twice that of earth?

A person is attempting to push a refrigerator across a room. He exerts a horizontal force on the refrigerator, but it does not move. (a) What other horizontal force must be acting on the refrigerator? (b) How are the magnitude and direction of this force related to the force that the person exerts? (c) Suppose that the person applies the force such that it is the largest possible force before the refrigerator begins to move. What factors determine the magnitude of this force? Problem Consult Multiple-Concept Example 9 to explore a model for solving this problem. A person pushes on a 57 -kg refrigerator with a horizontal force of \(-267 \mathrm{~N}\); the minus sign indicates that the force is directed along the \(-x\) direction. The coefficient of static friction is \(0.65 .\) (a) If the refrigerator does not move, what is the magnitude and direction of the static frictional force that the floor exerts on the refrigerator? (b) What is the magnitude of the largest pushing force that can be applied to the refrigerator before it just begins to move?

A toboggan slides down a hill and has a constant velocity. The angle of the hill is \(8.00^{\circ}\) with respect to the horizontal. What is the coefficient of kinetic friction between the surface of the hill and the toboggan?

A student is skateboarding down a ram p that is \(6.0 \mathrm{~m}\) long and inclined at \(18^{\circ}\) with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is \(2.6 \mathrm{~m} / \mathrm{s}\). Neglect friction and find the speed at the bottom of the ramp.
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Electromagnetism

Read Explanation

Physical Quantities And Units

Read Explanation

Kinematics Physics

Read Explanation

Particle Model of Matter

Read Explanation

Rotational Dynamics

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.