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Which of the following forces is conservative? (a) \(\mathbf{F}=k(x, 2 y, 3 z)\) where \(k\) is a constant. (b) \(\mathbf{F}=k(y, x, 0) .\) (c) \(\mathbf{F}=k(-y, x, 0)\). For those which are conservative, find the corresponding potential energy \(U,\) and verify by direct differentiation that \(\mathbf{F}=-\nabla U\).

Short Answer

Expert verified
Force (a) is conservative with potential energy \( U = -kx^2/2 - ky^2 - 3kz^2/2 \).

Step by step solution

01

Understand Conservative Forces

A force is conservative if the work done by the force around any closed path is zero or, equivalently if it can be expressed as the gradient of a scalar potential function.
02

Check for Zero Curl

For a force \( \mathbf{F} = (F_x, F_y, F_z) \) to be conservative, its curl should be zero, i.e., \( abla \times \mathbf{F} = 0 \). Calculate the curl for each force.
03

Calculate Curl for Each Force (a)

For \( \mathbf{F} = k(x, 2y, 3z) \), the curl is \( abla \times \mathbf{F} = (0, 0, 0) \), which is zero. Thus, force (a) is conservative.
04

Calculate Curl for Force (b)

For \( \mathbf{F} = k(y, x, 0) \), the curl is \( abla \times \mathbf{F} = (0, 0, -k) \), which is non-zero. Thus, force (b) is not conservative.
05

Calculate Curl for Force (c)

For \( \mathbf{F} = k(-y, x, 0) \), the curl is \( abla \times \mathbf{F} = (0, 0, 2k) \), which is non-zero. Thus, force (c) is not conservative.
06

Find Potential Energy for Force (a)

Since force (a) is conservative, we can find potential energy \( U \) by integrating: \[ U = - \int{kx dx} - \int{2ky dy} - \int{3kz dz} = -kx^2/2 - ky^2 - 3kz^2/2 \]
07

Verify Force Equation

To verify, differentiate the potential energy function: \( abla U = (kx, 2ky, 3kz) \). Thus, \( \mathbf{F} = -abla U \) confirms our force \( \mathbf{F} = k(x, 2y, 3z) \) is correctly represented by its potential energy.

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Key Concepts

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

Potential Energy
Potential energy is an important concept in physics, representing the energy stored in a system due to the position or configuration of its components. This energy can be transformed into kinetic energy or other forms of energy, allowing a system to do work. In the context of conservative forces, the potential energy is related to how forces act within the system.

When a force is conservative, there exists a potential energy function, often denoted as \( U \), such that the force can be derived as the negative gradient of this potential energy. This is expressed mathematically as \( \mathbf{F} = -abla U \).

For example, in our problem exercise, the force \( \mathbf{F} = k(x, 2y, 3z) \) is conservative, meaning we can find its potential energy by integrating the components of the force. The potential energy function obtained, \( U = -\frac{kx^2}{2} - ky^2 - \frac{3kz^2}{2} \), shows how the energy is distributed with respect to \( x \), \( y \), and \( z \) in this 3-dimensional system.

By understanding potential energy, we can predict how movement or changes in configuration can affect a system's behavior, making it a fundamental concept in physics.
Gradient
The gradient is a vector operation that describes the rate and direction of change in a scalar field. Think of a hill: the gradient at any point on the hill points in the direction of the steepest ascent and its magnitude tells you how steep the hill is.

For a scalar field \( U(x, y, z) \), the gradient is denoted by \( abla U \) and results in a vector \((\frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z})\). It indicates how the potential energy changes with position in the space.

In the exercise, when we computed \( abla U \) for the potential energy function \( U = -\frac{kx^2}{2} - ky^2 - \frac{3kz^2}{2} \), we found that \( abla U = (kx, 2ky, 3kz) \). This confirms the conservative force \( \mathbf{F} = -abla U \).

Understanding gradients is crucial, as they give insight into how forces will act on particles, pushing them along the path of decreasing potential energy in a conservative field.
Curl
Curl is a vector operation used to determine the rotation or 'twist' of a vector field in three-dimensional space. When you find the curl of a vector field, you get another vector that indicates the axis of rotation and the magnitude of the rotation in that field.

For a force field to be conservative, its curl must be zero, \( abla \times \mathbf{F} = 0 \). This implies that the path taken between two points does not affect the work done, characteristic of conservative fields.

In the exercise, we calculated the curl for several force fields to determine their nature. The force \( \mathbf{F} = k(x, 2y, 3z) \) showed a curl of zero, confirming it as conservative. However, forces \( \mathbf{F} = k(y, x, 0) \) and \( \mathbf{F} = k(-y, x, 0) \) had non-zero curls, making them non-conservative.

Knowing about curl helps us understand the behavior of force fields. It's essential in physics when determining whether certain forces will conserve mechanical energy and how they will influence particle motion. Understanding curl aids in visualizing how fields behave in space, especially in electromagnetism and fluid dynamics.

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Most popular questions from this chapter

Consider a small frictionless puck perched at the top of a fixed sphere of radius \(R\). If the puck is given a tiny nudge so that it begins to slide down, through what vertical height will it descend before it leaves the surface of the sphere? [Hint: Use conservation of energy to find the puck's speed as a function of its height, then use Newton's second law to find the normal force of the sphere on the puck. At what value of this normal force does the puck leave the sphere?]

A mass \(m\) is in a uniform gravitational field, which exerts the usual force \(F=m g\) vertically down, but with \(g\) varying with time, \(g=g(t) .\) Choosing axes with \(y\) measured vertically up and defining \(U=m g y\) as usual, show that \(\mathbf{F}=-\nabla U\) as usual, but, by differentiating \(E=\frac{1}{2} m v^{2}+U\) with respect to \(t,\) show that \(E\) is not conserved.

A mass \(m\) moves in a circular orbit (centered ón the origin) in the field of an attractive central force with potential energy \(U=k r^{n} .\) Prove the virial theorem that \(T=n U / 2\).

(a) Consider an electron (charge \(-e\) and mass \(m\) ) in a circular orbit of radius \(r\) around a fixed proton (charge \(+e\) ). Remembering that the inward Coulomb force \(k e^{2} / r^{2}\) is what gives the electron its centripetal acceleration, prove that the electron's KE is equal to \(-\frac{1}{2}\) times its \(\mathrm{PE}\); that is, \(T=-\frac{1}{2} U\) and hence \(E=\frac{1}{2} U\). (This result is a consequence of the so-called virial theorem. See Problem 4.41.) Now consider the following inelastic collision of an electron with a hydrogen atom: Electron number 1 is in a circular orbit of radius \(r\) around a fixed proton. (This is the hydrogen atom.) Electron 2 approaches from afar with kinetic energy \(T_{2} .\) When the second electron hits the atom, the first electron is knocked free, and the second is captured in a circular orbit of radius \(r^{\prime} .\) (b) Write down an expression for the total energy of the three-particle system in general. (Your answer should contain five terms, three PEs but only two KEs, since the proton is considered fixed.) (c) Identify the values of all five terms and the total energy \(E\) long before the collision occurs, and again long after it is all over. What is the KE of the outgoing electron 1 once it is far away? Give your answers in terms of the variables \(T_{2}, r,\) and \(r^{\prime}\).

By writing a \(\cdot \mathbf{b}\) in terms of components prove that the product rule for differentiation applies to the dot product of two vectors; that is, \(\frac{d}{d t}(\mathbf{a} \cdot \mathbf{b})=\frac{d \mathbf{a}}{d t} \cdot \mathbf{b}+\mathbf{a} \cdot \frac{d \mathbf{b}}{d t}\).

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