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Two bodies are interacting by a conservative force. Show that the mechanical energy of an isolated system consisting of two bodies interacting with a conservative force is conserved. (Hint: Start by using Newton's third law and the definition of work to find the work done on each body by the conservative force.)

Short Answer

Expert verified
The mechanical energy conservation for an isolated system with two bodies interacting by a conservative force can be demonstrated by using Newton's third law and the definition of work. Calculating the work done on each body due to the conservative force and applying Newton's third law, we find the total work done on both bodies. By relating this to the change in potential energy and establishing the conservation of mechanical energy, we can conclude that the mechanical energy of such a system is conserved. The final conservation of energy equation is \(\Delta E = W_T + (-W_T) = 0\), showing that the change in mechanical energy, \(\Delta E\), is zero.

Step by step solution

01

Define mechanical energy and conservative force

Mechanical energy is defined as the sum of kinetic energy (K) and potential energy (U) of the system, represented as E = K + U. A conservative force is a force that does not change the overall mechanical energy of the system when it does work on the system. Work done by a conservative force is path-independent and can be determined from the difference in potential energy between the initial and final states of the system. In this case, Body 1 will experience a force F12 due to Body 2, and Body 2 will experience a force F21 due to Body 1. According to Newton's third law, F12 = -F21.
02

Calculate the work done on each body by the conservative force

Let's consider the work done on both bodies due to the conservative force. We can use the definition of work to find the work done on each body: Work done W on a body is given by: \(W = \vec{F} \cdot \vec{d}\), where \(\vec{F}\) is the force vector and \(\vec{d}\) is the displacement vector. For Body 1, the work done by the conservative force is \(W_{12} = \vec{F}_{12} \cdot \vec{d}_{12}\), and for Body 2, the work done by the conservative force is \(W_{21} = \vec{F}_{21} \cdot \vec{d}_{21}\), where \(\vec{F}_{12}\) and \(\vec{F}_{21}\) are the conservative forces on Body 1 and Body 2, respectively, and \(\vec{d}_{12}\) and \(\vec{d}_{21}\) are the displacement vectors of Body 1 and Body 2, respectively.
03

Apply Newton's third law and find the total work done

By Newton's third law, the forces on both bodies are equal in magnitude and opposite in direction, which means \(F12 = -F21 \Rightarrow \vec{F}_{12} = -\vec{F}_{21}\). Now, we will find the total work done on both bodies, which is the sum of the work done on Body 1 and Body 2. Total work done \(W_T = W_{12} + W_{21}\) Since work is the dot product of force and displacement, we have: \(W_T = (\vec{F}_{12} \cdot \vec{d}_{12}) + (\vec{F}_{21} \cdot \vec{d}_{21})\) Using Newton's third law, replace \(\vec{F}_{21}\) with \(-\vec{F}_{12}\) in the equation: \(W_T = (\vec{F}_{12} \cdot \vec{d}_{12}) - (\vec{F}_{12} \cdot \vec{d}_{21})\) Factor out the force \(\vec{F}_{12}\): \(W_T = \vec{F}_{12}\cdot(\vec{d}_{12} - \vec{d}_{21})\) Since both bodies belong to an isolated system, their displacements are related by: \(\vec{d}_{12} = -\vec{d}_{21}\) Therefore, we can simplify the total work done as: \(W_T = \vec{F}_{12} \cdot (2\vec{d}_{12})\), which simplifies to: \(W_T = 2\vec{F}_{12} \cdot \vec{d}_{12}\)
04

Establish the conservation of mechanical energy

For the conservative force, the work done is related to the change in potential energy, \(\Delta U\), of the system as: \(W_T = -\Delta U\) Combining the equations for \(W_T, \Delta U\), and the conservative force: \(\Delta U = -2\vec{F}_{12} \cdot \vec{d}_{12}\) Now recall that mechanical energy is conserved for an isolated system under the influence of a conservative force. Since total mechanical energy, E, is the sum of kinetic and potential energy, we can write the conservation of energy as: \(\Delta E = \Delta K + \Delta U = 0\) From this, it follows that: \(\Delta K = -\Delta U\) Considering the work-energy theorem, we have: \(\Delta K = W_T\) Therefore, \(\Delta K = -\Delta U = -2\vec{F}_{12} \cdot \vec{d}_{12}\), which means: \(\Delta E = W_T + (-W_T) = 0\) Since the change in mechanical energy, \(\Delta E\), is zero, we can conclude that the mechanical energy of an isolated system consisting of two bodies interacting with a conservative force is conserved.

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Key Concepts

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

Newton's Third Law
Newton's third law of motion is essential to our understanding of forces and interactions between objects. It states that for every action, there is an equal and opposite reaction. This means whenever one object exerts a force on a second object, the second object simultaneously exerts a force equal in magnitude and opposite in direction on the first object.

To illustrate, consider two skaters pushing against each other on an ice rink. As they exert force on one another, both skaters move in opposite directions with forces that are equal in size but directed oppositely. This concept is the cornerstone for analyzing the interaction of bodies in a system, and it lays the foundation for the conservation of mechanical energy in systems where only internal forces act, like the two bodies mentioned in the exercise interacting with a conservative force.
Conservative Force
A conservative force is fundamental to energy conservation. The characteristic of a conservative force is that the work it does on an object moving between two points is independent of the path taken. This is crucial because it implies that the total mechanical energy (kinetic plus potential) of an object influenced solely by conservative forces remains constant as the object moves.

Gravitational and elastic spring forces are examples of conservative forces. If a ball is thrown into the air, gravity will do work on it during its ascent and descent, yet the total mechanical energy at any point in its trajectory will remain the same (discounting air resistance). Understanding conservative forces explains why, in the given problem, the work done by such forces on the two-body system relates directly to changes in potential energy, hence reinforcing the concept of energy conservation within an isolated system.
Work-Energy Theorem
The work-energy theorem is a powerful concept within physics that connects the work done by forces to the change in kinetic energy of an object. It states that the net work done by forces on an object is equal to the change in that object's kinetic energy. Mathematically, it can be expressed as: \[ \text{Work} = \text{Change in Kinetic Energy} \[\Delta K = W\]\]

For instance, when a person pushes a stationary block, the work done by the person leads to an increase in the block's kinetic energy, causing it to move. Conversely, when a block sliding on a rough surface comes to a halt, the work done by friction (a non-conservative force) is equal to the reduction in the block's kinetic energy. This principle helps us predict the new velocities of objects after forces have done work on them and is directly applied in the exercise solution to relate the conservative force’s work to changes in kinetic energy.
Isolation of Systems in Physics
The concept of an isolated system in physics is crucial when studying conservation laws. An isolated system is one where no energy or matter is exchanged with the surroundings. This isolation creates a closed environment in which we can confidently apply conservation laws, such as the conservation of mechanical energy.

In the context of the problem given, comparing two bodies as an isolated system and assuming that there are no external forces means that the total mechanical energy within the system is conserved. The assumption of no external work allows us to consider only the internal interactions, making the analysis and application of energy conservation feasible. In real-world scenarios, perfectly isolated systems don't exist, as there are always some external interactions, be it friction with the environment or gravitational effects from other bodies. However, assuming a system is isolated is a practical abstraction that simplifies calculations and helps us understand the fundamental principles governing physical interactions.

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Most popular questions from this chapter

The force exerted by a diving board is conservative, provided the intemal friction is negligible. Assuming friction is negligible, describe changes in the potential energy of a diving board as a swimmer drives from it, starting just before the swimmer steps on the board until just after his feet leave it.

The massless spring of a spring gun has a force constant \(k=12 \mathrm{N} / \mathrm{cm} .\) When the gun is aimed vertically, a \(15-\mathrm{g}\) projectile is shot to a height of \(5.0 \mathrm{m}\) above the end of the expanded spring. (See below.) How much was the spring compressed initially?

Shown below is a box of mass \(m_{1}\) that sits on a frictionless incline at an angle above the horizontal \(\theta=30^{\circ} .\) This box is connected by a relatively massless string, over a frictionless pulley, and finally connected to a box at rest over the ledge, labeled \(m_{2} .\) If \(m_{1}\) and \(m_{2}\) are a height \(h\) above the ground and \(m_{2}>>m_{1}:\) (a) What is the initial gravitational potential energy of the system? (b) What is the final kinetic energy of the system?

Can a non-conservative force increase the mechanical energy of the system?

A projectile of mass 2 kg is fired with a speed of 20 \(\mathrm{m} / \mathrm{s}\) at an angle of \(30^{\circ}\) with respect to the horizontal. (a) Calculate the initial total energy of the projectile given that the reference point of zero gravitational potential energy at the launch position. (b) Calculate the kinetic energy at the highest vertical position of the projectile. (c) Calculate the gravitational potential energy at the highest vertical position. (d) Calculate the maximum height that the projectile reaches. Compare this result by solving the same problem using your knowledge of projectile motion.

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