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Conservation of energy Suppose an object with mass \(m\) moves in a region \(R\) in a conservative force field given by \(\mathbf{F}=-\nabla \varphi,\) where \(\varphi\) is a potential function in a region \(R .\) The motion of the object is governed by Newton's Second Law of Motion, \(\mathbf{F}=m \mathbf{a},\) where a is the acceleration. Suppose the object moves from point \(A\) to point \(B\) in \(R\) a. Show that the equation of motion is \(m \frac{d \mathbf{v}}{d t}=-\nabla \varphi\) b. Show that \(\frac{d \mathbf{v}}{d t} \cdot \mathbf{v}=\frac{1}{2} \frac{d}{d t}(\mathbf{v} \cdot \mathbf{v})\) c. Take the dot product of both sides of the equation in part (a) with \(\mathbf{v}(t)=\mathbf{r}^{\prime}(t)\) and integrate along a curve between \(A\) and B. Use part (b) and the fact that \(\mathbf{F}\) is conservative to show that the total energy (kinetic plus potential) \(\frac{1}{2} m|\mathbf{v}|^{2}+\varphi\) is the same at \(A\) and \(B\). Conclude that because \(A\) and \(B\) are arbitrary, energy is conserved in \(R\).

Short Answer

Expert verified
Question: Prove that the total energy of an object with mass \(m\) moving through a conservative force field governed by a potential function \(\varphi\) remains constant within region \(R\). Answer: Using Newton's second law of motion, the equation of motion, and deriving a relationship between velocity and acceleration, we showed that the total energy (kinetic plus potential) remains constant within region \(R\). This is because the left side of the equation represents the change in kinetic energy and the right side represents the change in potential energy. Since the force field is conservative, the total energy at any two points in region \(R\) is the same, proving that energy is conserved in the region.

Step by step solution

01

Part a: Equation of motion

Given the conservative force field by \(\mathbf{F} = -\nabla \varphi\), we will use Newton's second law of motion, \(m\mathbf{a} = \mathbf{F}\), to derive the equation of motion. Here, \(\mathbf{a}\) represents acceleration, which can be represented as the time derivative of the velocity vector, \(\mathbf{v}\). So we have: \(m \frac{d \mathbf{v}}{d t} = -\nabla \varphi\)
02

Part b: Relationship between velocity and acceleration

In this step, we will show that \(\frac{d \mathbf{v}}{d t} \cdot \mathbf{v}=\frac{1}{2} \frac{d}{d t}(\mathbf{v} \cdot \mathbf{v})\). First, we differentiate the dot product of velocity with itself with respect to time: \(\frac{d}{d t}(\mathbf{v} \cdot \mathbf{v}) = \frac{d}{d t}(v_x^2 + v_y^2 + v_z^2)\) Since the derivative of the sum is the sum of derivatives: \(\frac{d}{d t}(\mathbf{v} \cdot \mathbf{v}) = 2v_x \frac{dv_x}{dt} + 2v_y \frac{dv_y}{dt} + 2v_z \frac{dv_z}{dt}\) Factor out the 2: \(\frac{1}{2} \frac{d}{d t}(\mathbf{v} \cdot \mathbf{v}) = v_x \frac{dv_x}{dt} + v_y \frac{dv_y}{dt} + v_z \frac{dv_z}{dt}\) Now, we can express this as: \(\frac{1}{2} \frac{d}{d t}(\mathbf{v} \cdot \mathbf{v}) = \mathbf{v} \cdot \frac{d \mathbf{v}}{d t}\)
03

c. Energy conservation

Take the dot product of both sides of the equation of motion derived in part a with \(\mathbf{v}(t)=\mathbf{r}^{\prime}(t)\): \(m (\frac{d \mathbf{v}}{d t} \cdot \mathbf{v}) = -(\nabla\varphi \cdot \mathbf{v})\) From part b, we have \(\frac{d \mathbf{v}}{d t} \cdot \mathbf{v}=\frac{1}{2} \frac{d}{d t}(\mathbf{v} \cdot \mathbf{v})\). Substitute it into the equation: \(m (\frac{1}{2} \frac{d}{d t}(\mathbf{v} \cdot \mathbf{v})) = -(\nabla\varphi \cdot \mathbf{v})\) Rearrange the equation and integrate both sides with respect to time over the curve between points \(A\) and \(B\): \(\frac{1}{2}m \int_A^B \frac{d}{dt} (\mathbf{v} \cdot \mathbf{v}) dt = -\int_A^B (\nabla\varphi \cdot \mathbf{v}) dt\) The left side of the equation represents the change in kinetic energy, while the right side represents the change in potential energy. Since \(\mathbf{F}\) is conservative, the integral on the right is path-independent and can be rewritten as \(\varphi(A) - \varphi(B)\). Thus, we have: \(\frac{1}{2} m (|\mathbf{v(B)}|^2 - |\mathbf{v(A)}|^2) = \varphi(A) - \varphi(B)\) Rearrange to get: \(\frac{1}{2}m |\mathbf{v(A)}|^2 + \varphi(A) = \frac{1}{2}m |\mathbf{v(B)}|^2 + \varphi(B)\) This equation tells us that the total energy (kinetic plus potential) at points \(A\) and \(B\) is the same. Since points \(A\) and \(B\) are arbitrary within region \(R\), we can conclude that energy is conserved in \(R\).

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Key Concepts

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

Conservative Force Field
A conservative force field is one where the work done by or against the force depends solely on the starting and ending points, and not on the path taken. This means that if you move an object in a loop within such a field, the net work done is zero. In mathematical terms, a force field \( \mathbf{F} \) is conservative if it can be expressed as the negative gradient of a scalar potential function \( \varphi \), so \( \mathbf{F} = -abla \varphi \).
Key characteristics of conservative force fields include:
  • Path independence: The work done only depends on the initial and final positions.
  • Existence of potential energy: You can define a potential energy function \( \varphi \), linked to how \( \mathbf{F} \) does work.
  • Reversible processes: All energy processes in a conservative field can be reversed without a loss of energy.
Conservative fields are foundational for understanding energy conservation. They ensure that when an object moves from one point to another in the field, energy is neither lost nor gained but simply transformed between kinetic and potential forms. This allows for easier calculations and predictions in physics problems.
Newton's Second Law of Motion
Newton's Second Law of Motion is a cornerstone of classical mechanics. It states that the force acting on an object is equal to the mass of that object multiplied by its acceleration: \( \mathbf{F} = m \mathbf{a} \). This simple equation encapsulates how objects change their motion when subjected to forces.
Here's a breakdown:
  • Force (\( \mathbf{F} \)): A vector quantity that represents a push or pull on an object. It can cause a change in the object's velocity.
  • Mass (\( m \)): A scalar quantity representing the amount of matter in an object, contributing to its "inertia," or resistance to changes in motion.
  • Acceleration (\( \mathbf{a} \)): The rate of change of velocity of an object. It's also a vector quantity that indicates how quickly an object speeds up, slows down, or changes direction.
When applying this law in a conservative force field, we can relate it to energy conservation. If an object is influenced by such a field, the forces corresponding to the gradient of the potential tell us how the energy dynamics play out along the object's path. It helps connect the dots between forces, motion, and energy transformations.
Kinetic and Potential Energy
Kinetic and potential energy are two fundamental forms of energy in physics, central to understanding energy conservation.
- **Kinetic Energy** is the energy of motion. For an object with mass \( m \) and velocity \( \mathbf{v} \), the kinetic energy \( KE \) is given by \(KE = \frac{1}{2} m |\mathbf{v}|^2\). This energy increases with higher speeds and more mass.
- **Potential Energy** is the stored energy of position. In a conservative force field, it can be represented by the potential function \( \varphi \). For instance, gravitational potential energy near the Earth's surface can be calculated by \( \varphi = mgh \), where \( h \) is the height above a reference point.
The interplay between these energies is what drives many physical phenomena. For a system in a conservative field, as an object moves, energy constantly shifts between kinetic and potential forms. One goes up while the other goes down, but their sum – the total mechanical energy – remains constant if no external work is performed.
The relationship between kinetic and potential energy in such systems serves as a perfect demonstration of the conservation of energy, highlighting how energy can be transformed but never destroyed or created from nothing.

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Most popular questions from this chapter

The Navier-Stokes equation is the fundamental equation of fluid dynamics that models the flow in everything from bathtubs to oceans. In one of its many forms (incompressible, viscous flow), the equation is $$\rho\left(\frac{\partial \mathbf{V}}{\partial t}+(\mathbf{V} \cdot \nabla) \mathbf{v}\right)=-\nabla p+\mu(\nabla \cdot \nabla) \mathbf{V}$$ In this notation, \(\mathbf{V}=\langle u, v, w\rangle\) is the three-dimensional velocity field, \(p\) is the (scalar) pressure, \(\rho\) is the constant density of the fluid, and \(\mu\) is the constant viscosity. Write out the three component equations of this vector equation. (See Exercise 40 for an interpretation of the operations.)

Gravitational potential The gravitational force between two point masses \(M\) and \(m\) is $$\mathbf{F}=G M m \frac{\mathbf{r}}{|\mathbf{r}|^{3}}=G M m \frac{\langle x, y, z\rangle}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}}$$ where \(G\) is the gravitational constant. a. Verify that this force field is conservative on any region excluding the origin. b. Find a potential function \(\varphi\) for this force field such that \(\mathbf{F}=-\nabla \varphi\). c. Suppose the object with mass \(m\) is moved from a point \(A\) to a point \(B\), where \(A\) is a distance \(r_{1}\) from \(M,\) and \(B\) is a distance \(r_{2}\) from \(M .\) Show that the work done in moving the object is \(G M m\left(\frac{1}{r_{2}}-\frac{1}{r_{1}}\right)\). d. Does the work depend on the path between \(A\) and \(B ?\) Explain.

Let \(S\) be the cylinder \(x^{2}+y^{2}=a^{2},\) for \(-L \leq z \leq L\) a. Find the outward flux of the field \(\mathbf{F}=\langle x, y, 0\rangle\) across \(S\) b. Find the outward flux of the field \(\mathbf{F}=\frac{\langle x, y, 0\rangle}{\left(x^{2}+y^{2}\right)^{p / 2}}=\frac{\mathbf{r}}{|\mathbf{r}|^{p}}\) across \(S\), where \(|\mathbf{r}|\) is the distance from the \(z\) -axis and \(p\) is a real number. c. In part (b), for what values of \(p\) is the outward flux finite as \(a \rightarrow \infty(\text { with } L\) fixed)? d. In part (b), for what values of \(p\) is the outward flux finite as \(L \rightarrow \infty\) (with \(a\) fixed)?

Alternative construction of potential functions Use the procedure in Exercise 71 to construct potential functions for the following fields. $$\quad \mathbf{F}=\langle x, y\rangle$$

Flux integrals Assume the vector field \(\mathbf{F}=\langle f, g\rangle\) is source free (zero divergence) with stream function \(\psi\). Let \(C\) be any smooth simple curve from \(A\) to the distinct point \(B\). Show that the flux integral \(\int_{\mathcal{C}} \mathbf{F} \cdot \mathbf{n} d s\) is independent of path; that is, \(\int_{\mathcal{C}} \mathbf{F} \cdot \mathbf{n} d s=\psi(\mathcal{B})-\psi(A)\)

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