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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

Calculate the gradient \(\nabla f\) of the following functions, \(f(x, y, z):(\mathbf{a}) f=\ln (r),\) (b) \(f=r^{n}\), (c) \(f=g(r),\) where \(r=\sqrt{x^{2}+y^{2}+z^{2}}\) and \(g(r)\) is some unspecified function of \(r .\) [Hint: Use the chain rule.]

(a) The force exerted by a one-dimensional spring, fixed at one end, is \(F=-k x,\) where \(x\) is the displacement of the other end from its equilibrium position. Assuming that this force is conservative (which it is) show that the corresponding potential energy is \(U=\frac{1}{2} k x^{2},\) if we choose \(U\) to be zero at the equilibrium position. (b) Suppose that this spring is hung vertically from the ceiling with a mass \(m\) suspended from the other end and constrained to move in the vertical direction only. Find the extension \(x_{\mathrm{o}}\) of the new equilibrium position with the suspended mass. Show that the total potential energy (spring plus gravity) has the same form \(\frac{1}{2} k y^{2}\) if we use the coordinate \(y\) equal to the displacement measured from the new equilibrium position at \(x=x_{\mathrm{o}}\) (and redefine our reference point so that \(U=0\) at \(y=0\) ).

Near to the point where I am standing on the surface of Planet \(X\), the gravitational force on a mass \(m\) is vertically down but has magnitude \(m \gamma y^{2}\) where \(\gamma\) is a constant and \(y\) is the mass's height above the horizontal ground. (a) Find the work done by gravity on a mass \(m\) moving from \(\mathbf{r}_{1}\) to \(\mathbf{r}_{2}\), and use your answer to show that gravity on Planet \(X,\) although most unusual, is still conservative. Find the corresponding potential energy. (b) Still on the same planet, I thread a bead on a curved, frictionless, rigid wire, which extends from ground level to a height \(h\) above the ground. Show clearly in a picture the forces on the bead when it is somewhere on the wire. (Just name the forces so it's clear what they are; don't worry about their magnitude.) Which of the forces are conservative and which are not? (c) If I release the bead from rest at a height \(h\), how fast will it be going when it reaches the ground?

If a particle's potential energy is \(U(\mathbf{r})=k\left(x^{2}+y^{2}+z^{2}\right),\) where \(k\) is a constant, what is the force on the particle?

For a system of \(N\) particles subject to a uniform gravitational field g acting vertically down, prove that the total gravitational potential energy is the same as if all the mass were concentrated at the center of mass of the system; that is, \(U=\sum_{\alpha} U_{\alpha}=M g Y\) where \(M=\sum m_{\alpha}\) is the total mass and \(\mathbf{R}=(X, Y, Z)\) is the position of the \(\mathrm{CM},\) with the \(y\) coordinate measured vertically up. [Hint: We know from Problem 4.5 that \(\left.U_{\alpha}=m_{\alpha} g y_{\alpha} .\right]\)

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