/*! 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 94 Two \(20 \mathrm{~kg}\) spheres ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 94

Two \(20 \mathrm{~kg}\) spheres are fixed in place on a \(y\) axis, one at \(y=0.40 \mathrm{~m}\) and the other at \(y=-0.40 \mathrm{~m}\). A \(10 \mathrm{~kg}\) ball is then released from rest at a point on the \(x\) axis that is at a great distance (effectively infinite) from the spheres. If the only forces acting on the ball are the gravitational forces from the spheres, then when the ball reaches the \((x, y)\) point \((0.30 \mathrm{~m}, 0)\), what are (a) its kinetic energy and (b) the net force on it from the spheres, in unit- vector notation?

Short Answer

Expert verified
(a) The kinetic energy at (0.30 m, 0) is \( \frac{2G \times 200}{0.50} \). (b) The net force is in the x-direction, \( 2F_{1x} \hat{i} \).

Step by step solution

01

Calculate Potential Energy at the Starting Point

Assuming the initial potential energy is zero at a great distance (infinitely far away), we do not have to make any calculations for potential energy at infinity as it is our reference point.
02

Calculate Potential Energy at (x, y) = (0.30 m, 0)

To find the potential energy of the ball when it is at (x, y) = (0.30, 0), calculate the gravitational potential energy due to both spheres. The formula for the gravitational potential energy between two point masses is: \[ U = -\frac{G m_1 m_2}{r} \]Here, the gravitational constant \( G = 6.674 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2 \), and \( m_1 = 10 \,\text{kg} \) is the mass of the ball, and \( m_2 = 20 \,\text{kg} \) for each sphere.
03

Calculate the Distance to Each Sphere

For the upper sphere at (y = 0.40 m), the distance \( r_1 = \sqrt{(0.30)^2 + (0.40)^2} = 0.50 \text{ m} \).For the lower sphere at (y = -0.40 m), the distance \( r_2 = \sqrt{(0.30)^2 + (0.40)^2} = 0.50 \text{ m} \) as well.
04

Calculate Gravitational Potential Energies U1 and U2

Now, calculate the gravitational potential energy due to both spheres:For the upper sphere:\[ U_1 = -\frac{G \times 10 \times 20}{0.50} \]For the lower sphere:\[ U_2 = -\frac{G \times 10 \times 20}{0.50} \]
05

Compute Total Gravitational Potential Energy

The total potential energy \( U \) when the ball is at (0.30 m, 0) is given by:\[ U = U_1 + U_2 = -2 \times \frac{G \times 10 \times 20}{0.50} = -\frac{2G \times 200}{0.50} \]
06

Calculate Kinetic Energy at (0.30 m, 0)

From conservation of energy, initial potential energy plus initial kinetic energy equals final kinetic energy plus final potential energy.Since the initial kinetic energy at infinity is 0, and initial potential energy also is 0, all energy converts to kinetic:\[ K = -U = \frac{2G \times 200}{0.50} \]
07

Calculate Net Force at (0.30 m, 0)

The force on the ball due to one sphere is:\[ F_1 = \frac{G \times 10 \times 20}{(0.50)^2} \]The force due to the axis symmetry is symmetric in y-component, thus y-components cancel each other. Therefore, only x-components (direction towards each sphere) need to be considered.
08

Calculate X Components of Forces

Both x-components of force due to symmetry and equal forces:\[ F_{1x} = F_1 \times \cos{\theta} = F_1 \left(\frac{0.30}{0.50}\right) \]\[ F_{2x} = F_1 \left(\frac{0.30}{0.50}\right) \]Combine them:\[ F_{net, x} = 2F_{1x} \]
09

Solution to Forces in Unit Vectors

Since the y-components cancel, the net force is only in the x-direction. Thus, the net force in unit-vector form is:\[ \vec{F} = F_{net,x} \hat{i} \]

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 Potential Energy
Gravitational potential energy is a form of energy an object possesses due to its position in a gravitational field. In the context of our exercise, we focus on the gravitational potential energy between the two spheres and the ball. This energy is calculated using the formula:\[ U = -\frac{G m_1 m_2}{r} \] where:
  • \( G \) is the gravitational constant, \( 6.674 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2 \).
  • \( m_1 \) and \( m_2 \) are the masses of the objects interacting gravitationally鈥攊n this case, the ball and one of the spheres.
  • \( r \) is the distance between the centers of the two masses.
Gravitational potential energy is negative because the force is attractive, indicating that work must be done to separate the objects to an infinite distance apart. As the ball moves into the influence of the spheres, it loses gravitational potential energy which transforms into kinetic energy.
Conservation of Energy
The law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In the problem, we have potential energy converting into kinetic energy. Initially, the ball is at rest far away, making both its potential and kinetic energy zero. As the ball approaches the influence of the spheres:
  • The potential energy decreases due to the attractive gravitational force between the masses.
  • This lost potential energy becomes the ball's kinetic energy.
The kinetic energy when the ball reaches the position \((0.30 \, \text{m}, 0)\) is equal to the loss in potential energy. Thus, the ball鈥檚 increased speed corresponds to the energy gained from moving closer to the spheres.
Unit Vector Notation
Unit vector notation is a way to express vectors in coordinate form, typically with \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) representing the x, y, and z axes, respectively. A vector's direction and magnitude are critical in physics for accurately describing motion or force.For the exercise:
  • The net force on the ball needs to be expressed in terms of unit vector notation.
  • Because there is symmetry in the problem, the forces' y-components cancel out.
  • Thus, the remaining x-component of force is what we express as \( \vec{F} = F_{net,x} \hat{i} \).
This simplifies analyzing the ball's motion as it allows us to focus on one dimension without losing information on the force's direction.
Newton's Law of Universal Gravitation
Newton's Law of Universal Gravitation explains the attractive force between two masses. The force can be computed with the formula:\[ F = \frac{G m_1 m_2}{r^2} \] where:
  • \( F \) is the gravitational force between the masses.
  • \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2 \).
  • \( m_1 \) and \( m_2 \) are the masses involved (for our problem, the spheres and the ball).
  • \( r \) is the distance between the centers of the two masses.
This law is universal since it applies to all masses and distances. In the problem, two gravitational forces act on the ball due to each sphere. These forces are vectors, meaning they have both direction and magnitude. The exercise requires analyzing the x-components of these forces since the y-components cancel out, ultimately giving the net force responsible for the ball's acceleration.

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

We watch two identical astronomical bodies \(A\) and \(B\), each of mass \(m\), fall toward each other from rest because of the gravitational force on each from the other. Their initial center-to-center separation is \(R_{i} .\) Assume that we are in an inertial reference frame that is stationary with respect to the center of mass of this twobody system. Use the principle of conservation of mechanical energy \(\left(K_{f}+U_{f}=K_{i}+U_{i}\right)\) to find the following when the centerto-center separation is \(0.5 R_{i}:\) (a) the total kinetic energy of the system, (b) the kinetic energy of each body, (c) the speed of each body relative to us, and (d) the speed of body \(B\) relative to body \(A\). Next assume that we are in a reference frame attached to body \(A\) (we ride on the body). Now we see body \(B\) fall from rest toward us. From this reference frame, again use \(K_{f}+U_{f}=K_{i}+U_{i}\) to find the following when the center-to-center separation is \(0.5 R_{i}:(\mathrm{e})\) the kinetic energy of body \(B\) and (f) the speed of body \(B\) relative to body \(A\). (g) Why are the answers to (d) and (f) different? Which answer is correct?

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

(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?

The Sun's center is at one focus of Earth's orbit. How far from this focus is the other focus, (a) in meters and (b) in terms of the solar radius, \(6.96 \times 10^{8} \mathrm{~m}\) ? The eccentricity is \(0.0167\), and the semimajor axis is \(1.50 \times 10^{11} \mathrm{~m}\).

The Sun, which is \(2.2 \times 10^{20} \mathrm{~m}\) from the center of the Milky Way galaxy, revolves around that center once every \(2.5 \times 10^{8}\) years. Assuming each star in the Galaxy has a mass equal to the Sun's mass of \(2.0 \times 10^{30} \mathrm{~kg}\), the stars are distributed uniformly in a sphere about the galactic center, and the Sun is at the edge of that sphere, estimate the number of stars in the Galaxy.
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Turning Points in Physics

Read Explanation

Electricity

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Rotational Dynamics

Read Explanation

Further Mechanics and Thermal Physics

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.