/*! 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 38 In deep space, sphere \(A\) of m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In deep space, sphere \(A\) of mass \(20 \mathrm{~kg}\) is located at the origin of an \(x\) axis and sphere \(B\) of mass \(10 \mathrm{~kg}\) is located on the axis at \(x \equiv\) \(0.80 \mathrm{~m} .\) Sphere \(B\) is released from rest while sphere \(A\) is held at the origin. (a) What is the gravitational potential energy of the twosphere system just as \(B\) is released? (b) What is the kinetic energy of \(B\) when it has moved \(0.20 \mathrm{~m}\) toward \(A\) ?

Short Answer

Expert verified
(a) The initial potential energy is \(-1.67 \times 10^{-8} \, \text{J}\). (b) The kinetic energy of \( B \) is \(1.11 \times 10^{-8} \, \text{J}\).

Step by step solution

01

Calculate the Initial Gravitational Potential Energy

The gravitational potential energy (U) between two masses is given by the formula: \( U = -\frac{G m_1 m_2}{r} \), where \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \), \( m_1 = 20 \, \text{kg} \), \( m_2 = 10 \, \text{kg} \), and \( r = 0.80 \, \text{m} \). Substitute the values into the formula to find the initial potential energy:\[U_i = -\frac{6.674 \times 10^{-11} \times 20 \times 10}{0.80}\]Calculate this to find the initial gravitational potential energy.
02

Calculate the Gravitational Potential Energy After Moving

When sphere \( B \) has moved \(0.20 \, \text{m}\) towards \( A \), the new distance \( r \) between them is \( 0.80 - 0.20 = 0.60 \, \text{m} \). Use the potential energy formula: \[U_f = -\frac{G m_1 m_2}{r} = -\frac{6.674 \times 10^{-11} \times 20 \times 10}{0.60}\]Calculate \( U_f \) to find the potential energy after \( B \) has moved towards \( A \).
03

Determine the Change in Potential Energy

The change in gravitational potential energy \( \Delta U \) is given by:\[\Delta U = U_f - U_i\]Use the potential energy values calculated in Steps 1 and 2 to find \( \Delta U \).
04

Use Conservation of Energy to Find Kinetic Energy

Initially, all energy is stored as potential energy. As \( B \) moves towards \( A \), some of this potential energy is converted into kinetic energy. Since there are no non-conservative forces acting,\[\Delta U = K \text{ (change in potential energy equals kinetic energy gained)}\]Therefore, \( K = \Delta U \). Compute \( K \) using the change in potential energy calculated in Step 3.

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 91Ó°ÊÓ!

Key Concepts

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

Conservation of Energy
The principle of conservation of energy is a fundamental concept in physics. It tells us that energy in an isolated system cannot be created or destroyed. Instead, energy transforms from one form to another.
In the given exercise, the spheres conserve energy as they interact through gravitational forces since there are no external forces at play.
This means that the initial energy stored as gravitational potential energy between the two spheres is gradually converted into kinetic energy as sphere \(B\) moves towards sphere \(A\).

Some key points about energy conservation in this context include:
  • The total mechanical energy (sum of potential and kinetic energy) remains constant.
  • The decrease in gravitational potential energy equals the increase in kinetic energy.
  • Energy is merely transferred within the system, not lost.

Understanding this concept allows us to predict how one form of energy can transform into another, aiding in solving problems involving gravitational interactions.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. When an object moves, it transforms the potential energy it possesses into kinetic energy.
In this scenario, as sphere \(B\) starts moving towards sphere \(A\), it begins to accelerate due to gravitational forces.
This transformation is quantifiable using the formula for kinetic energy: \( K = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.

To calculate the kinetic energy gained by sphere \(B\):
  • First, determine the change in gravitational potential energy.
  • Use the principle of conservation of energy to equate this change to the kinetic energy gained.
  • This relationship holds because no external forces (e.g., friction) are acting on the system.

This concept is crucial for solving problems where objects in motion are influenced by forces like gravity, helping us quantify the energy transition from potential to kinetic.
Gravitational Force
Gravitational force is one of the fundamental forces of nature. It acts between two masses and is always attractive.
In the exercise, gravitational force is responsible for pulling the spheres towards each other.
The law of universal gravitation defines the gravitational force \( F \) between two masses as \( F = \frac{G m_1 m_2}{r^2} \), where \( m_1 \) and \( m_2 \) are the masses, \( r \) is the distance between their centers, and \( G \) is the gravitational constant.

Understanding gravitational force in this context involves:
  • Recognizing how it causes the spheres to accelerate towards one another.
  • Knowing that the force depends on the mass of the objects and their separation distance.
  • Seeing that changes in distance impact the strength of the gravitational interaction.

This force serves as the pivotal factor in determining the motion and interaction of masses in space, illustrating the continuous interplay between gravitational potential energy and kinetic energy.

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

At what altitude above Earth's surface would the gravitational acceleration be \(4.9 \mathrm{~m} / \mathrm{s}^{2} ?\)

Hunting a black hole. Observations of the light from a certain star indicate that it is part of a binary (two-star) system. This visible star has orbital speed \(v=270\) \(\mathrm{km} / \mathrm{s}\), orbital period \(T=1.70\) days, and approximate mass \(m_{1}=6 M_{s}\), where \(M_{s}\) is the Sun's mass, \(1.99 \times\) \(10^{50} \mathrm{~kg}\). Assume that the visible star and its companion star, which is dark and unseen, are both in circular orbits (Fig. \(13-47\) ). What integer multiple of \(M_{s}\) gives the approximate mass \(m_{2}\) of the dark star?

A very early, simple satellite consisted of an inflated spherical aluminum balloon \(30 \mathrm{~m}\) in diameter and of mass \(20 \mathrm{~kg}\). Suppose a meteor having a mass of \(7.0 \mathrm{~kg}\) passes within \(3.0\) \(\mathrm{m}\) of the surface of the satellite. What is the magnitude of the gravitational force on the meteor from the satellite at the closest approach?

(a) What linear speed must an Earth satellite have to be in a circular orbit at an altitude of \(160 \mathrm{~km}\) above Earth's surface? (b) What is the period of revolution?

Mile-high building. In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constructed. Ignoring Earth's rotation, find the change in your weight if you were to ride an elevator from the street level, where you weigh \(600 \mathrm{~N}\), to the top of the building.

See all solutions

Recommended explanations on Physics Textbooks

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.