/*! 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] | 91Ó°ÊÓ

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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 is \( 4G \times 20 \times 10 \). (b) The net force is zero.

Step by step solution

01

Understanding Gravitational Potential Energy at Infinite Distance

Initially, the 10 kg ball is released from a point with effectively infinite distance from the two spheres. At this point, the gravitational potential energy due to both spheres is effectively zero because gravitational forces diminish with distance. Thus, the initial potential energy, \( U_0 \), is zero.
02

Calculating Potential Energy at Point (0.30, 0)

When the ball reaches point \((0.30 \mathrm{~m}, 0)\), we calculate the potential energy due to each sphere. The gravitational potential energy between two masses is given by \( U = -\frac{G m_1 m_2}{r} \), where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the respective masses, and \( r \) is the distance between them.
03

Distance to Each Sphere

Calculate the distances between the ball at \((0.30 \mathrm{~m}, 0)\) and each of the spheres. For both spheres:- Distance to sphere at \((0, 0.40)\): \[ r_1 = \sqrt{(0.30 - 0)^2 + (0 - 0.40)^2} = \sqrt{0.30^2 + 0.40^2} = 0.50 \mathrm{~m} \]- Distance to sphere at \((0, -0.40)\): \[ r_2 = \sqrt{(0.30 - 0)^2 + (0 + 0.40)^2} = \sqrt{0.30^2 + 0.40^2} = 0.50 \mathrm{~m} \]
04

Calculate Potential Energy at Point (0.30, 0)

Calculate the potential energy using the gravitational potential energy formula for each sphere:- Energy due to the first sphere: \[ U_1 = -\frac{G \cdot 20 \cdot 10}{0.50} \]- Energy due to the second sphere: \[ U_2 = -\frac{G \cdot 20 \cdot 10}{0.50} \]Combine these to get the total potential energy at this point:\[ U = U_1 + U_2 = -2\left( \frac{G \cdot 20 \cdot 10}{0.50} \right) \]
05

Compute Change in Potential and Kinetic Energy

Initial potential energy was zero, and final potential energy is \( U = -2\left( \frac{G \cdot 20 \cdot 10}{0.50} \right) \). The ball's change in potential energy is equal to its change in kinetic energy due to energy conservation:\[ \Delta KE = - \Delta U = 2\left( \frac{G \cdot 20 \cdot 10}{0.50} \right) \]
06

Calculate Net Gravitational Force

Each sphere creates a gravitational force on the ball in the \( y \)-direction. The force from each sphere is calculated by the formula:- Force due to a single sphere: \[ F = \frac{G \cdot 20 \cdot 10}{0.50^2} \]Since the setup is symmetric with one mass above and one below the \( x \)-axis, the net force in the \( y \)-direction will be zero. Thus, the forces exerted by the spheres cancel out along the \( y \) axis, resulting in no net force along that direction.
07

Formulate the Answer

The kinetic energy of the ball at the moment it reaches \((0.30 \mathrm{~m}, 0)\) is \( KE = 4G \times 20 \times 10 \). The net force exerted on the ball by the two spheres is zero as forces cancel out symmetrically.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object has due to its motion. When a ball is at rest, like the 10 kg ball at a point on the x-axis, its kinetic energy is zero. However, as it starts to move, this energy increases. Kinetic energy can be calculated using the formula:
  • \( KE = \frac{1}{2}mv^2 \)
where \( m \) is the mass of the object and \( v \) is its velocity.
This formula shows that kinetic energy is directly proportional to the mass and the square of the velocity. In our exercise, energy conservation plays a role. As the ball moves from a point of zero potential energy at infinite distance to the point (0.30 m, 0), it gains kinetic energy equal to the potential energy lost. This change is due to gravitational forces acting on it.
So the greater the velocity or mass, the higher the kinetic energy acquired, as depicted by the calculation at the final point, resulting in the ball's kinetic energy being significant towards the end of its path.
Gravitational Force
Gravitational force is a force of attraction that exists between any two masses. This force is what pulls objects towards one another. The strength of the gravitational force between two objects depends on their masses and the distance between them.
The gravitational force can be calculated with the formula:
  • \( F = \frac{Gm_1m_2}{r^2} \)
where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between them.
In the given exercise, the 10 kg ball is under the influence of gravitational forces from two spheres, each weighing 20 kg, which are located symmetrically about the x-axis. These forces act in opposite directions along the y-axis. Therefore, their net result in the y-direction cancels out. However, they contribute to the ball's kinetic energy through potential energy changes.
Energy Conservation
Energy conservation is a principle stating that energy cannot be created or destroyed; it can only be transformed from one form to another. In physics, this means that the total energy of an isolated system remains constant over time.
For our exercise, when the ball starts moving towards the point (0.30 m, 0), energy conservation dictates that the potential energy lost due to the gravitational forces from the spheres is transformed into kinetic energy. Initially, at an infinite distance, the potential energy is zero. As it approaches the two spheres, it converts this potential energy into kinetic energy while maintaining total energy.
This principle helps us understand that any potential energy reduction in the system must equivalently increase the kinetic energy of the ball, ensuring the total energy remains constant throughout its motion.
Newton's Law of Universal Gravitation
Newton's Law of Universal Gravitation explains how every point mass attracts every other point mass in the universe with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The mathematical expression for this is:
  • \( F = \frac{Gm_1m_2}{r^2} \)
where \( F \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between the centers of their masses.
This universal law applies to all objects with mass. In our exercise, it helps us calculate the gravitational forces exerted by the two 20 kg spheres on the 10 kg ball. Despite their symmetrical placement, they exert equal and opposite forces on the ball along the y-axis leading to a net zero force. This understanding is crucial in predicting motion under gravitational influence, allowing us to determine resultant forces and energy transformations that take place.

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Most popular questions from this chapter

In a double-star system, two stars of mass \(3.0 \times 10^{30} \mathrm{~kg}\) each rotate about the system's center of mass at radius \(1.0 \times 10^{11} \mathrm{~m}\). (a) What is their common angular speed? (b) If a meteoroid passes through the system's center of mass perpendicular to their orbital plane, what minimum speed must it have at the center of mass if it is to escape to "infinity" from the two-star system?

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