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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

A particle of mass \(m_{1}\) and speed \(v_{1}\) collides with a second particle of mass \(m_{2}\) at rest. If the collision is perfectly inelastic (the two particles lock together and move off as one) what fraction of the kinetic energy is lost in the collision? Comment on your answer for the cases that \(m_{1} \ll m_{2}\) and that \(m_{2} \ll m_{1}\).

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?

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, \(\frac{d}{d t}(\mathbf{a} \cdot \mathbf{b})=\frac{d \mathbf{a}}{d t} \cdot \mathbf{b}+\mathbf{a} \cdot \frac{d \mathbf{b}}{d t}\).

In one dimension, it is obvious that a force obeying Hooke's law is conservative (since \(F=-k x\) depends only on the position \(x,\) and this is sufficient to guarantee that \(F\) is conservative in one dimension). Consider instead a spring that obeys Hooke's law and has one end fixed at the origin, but whose other end is free to move in all three dimensions. (The spring could be fastened to a point in the ceiling and be supporting a bouncing mass \(m\) at its other end, for instance.) Write down the force \(\mathbf{F}(\mathbf{r})\) exerted by the spring in terms of its length \(r\) and its equilibrium length \(r_{\mathrm{o}} .\) Prove that this force is conservative. [Hints: Is the force central? Assume that the spring does not bend.]

An interesting one-dimensional system is the simple pendulum, consisting of a point mass \(m\), fixed to the end of a massless rod (length \(l\) ), whose other end is pivoted from the ceiling to let it swing freely in a vertical plane, as shown in Figure \(4.26 .\) The pendulum's position can be specified by its angle \(\phi\) from the equilibrium position. (It could equally be specified by its distance \(s\) from equilibrium \(-\) indeed \(s=l \phi-\) but the angle is a little more convenient.) (a) Prove that the pendulum's potential energy (measured from the equilibrium level) is \(U(\phi)=m g l(1-\cos \phi)\). Write down the total energy \(E\) as a function of \(\phi\) and \(\dot{\phi}\). (b) Show that by differentiating your expression for \(E\) with respect to \(t\) you can get the equation of motion for \(\phi\) and that the equation of motion is just the familiar \(\Gamma=I \alpha\) (where \(\Gamma\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration \(\ddot{\phi}\) ). (c) Assuming that the angle \(\phi\) remains small throughout the motion, solve for \(\phi(t)\) and show that the motion is periodic with period \(\tau_{\mathrm{o}}=2 \pi \sqrt{l / g}\).

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