Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A particle moves along the \(x\) axis under the influence of a conservative force that is described by $$\overrightarrow{\mathbf{F}}=-\alpha x e^{-\beta x^{2}} \hat{\mathbf{i}}$$ where \(\alpha\) and \(\beta\) are constants. Find the potential energy function \(U(x)\).

Short Answer

Expert verified
The potential energy function will be \(U(x) = -\alpha e^{-\beta x^{2}}\).

Step by step solution

01

Observe the Force Component

Notice that the force \(\overrightarrow{\mathbf{F}}\) is only along the x-axis and its expression is given by \(-\alpha x e^{-\beta x^{2}}\). The \(\hat{\mathbf{i}}\) indicates the direction of the force, which in this case is in the x-direction.
02

Implement the Relation Between Force and Potential Energy

Recall that the force is the negative gradient of potential energy. Hence, in terms of potential energy \(U(x)\), the force can be expressed as \(\overrightarrow{\mathbf{F}} = - \nabla{U(x)}\), where \(\nabla\) is the nabla operator symbolizing a vector differential. Since the force is acting only in the x-direction, this can be simplified to \(\overrightarrow{\mathbf{F}} = -\frac{d}{dx} U(x)\).
03

Solve for the Potential Energy Function

Now integrate the force equation with respect to \(x\) to solve for \(U(x)\). This will yield: \(U(x) = - \int -\alpha x e^{-\beta x^{2}} dx\). To solve this type of integral, make a substitution, let \(u = \beta x^{2}\). This will convert the integral into the form: \(U(x) = \alpha \int e^{-u} du\). This integral results in \(-\alpha e^{-u}\). Now, revert back to the x variable by substituting \(u\) with \(\beta x^{2}\). Hence, the potential energy function will be \(U(x) = -\alpha e^{-\beta x^{2}}\). Finally, normally a constant of integration would abound, but dealing with potential energy we can choose our zero level freely, and therefore we can set this constant to zero.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Understanding Conservative Forces
At the heart of many physical systems is the concept of a conservative force. These are forces that have the special property of conserving mechanical energy, regardless of the path taken by the object upon which they act. In other words, the work performed by a conservative force on an object moving between two points is the same, no matter what path is taken.

Examples of conservative forces include gravity and electrostatic forces. Notably, in the context of the provided exercise, knowing that a force is conservative allows us to deduce the existence of a potential energy function, which is a scalar field representing the potential energy per unit mass at every point in space. This potential energy function is especially significant because it is related to the force experienced by a particle at any position along its trajectory. It's crucial to recognize that the change in potential energy is negative to the work done by the conservative force.
The Nabla Operator in Physics
The nabla operator, denoted as abla, is a vector differential operator used extensively in physics. It enables the calculation of quantities like the gradient, divergence, and curl, which are central to the fields of classical mechanics, electromagnetism, and fluid dynamics.

In the simplest terms, the nabla operator is a symbolic representation that encapsulates the information of partial derivatives concerning the spatial variables. When applied to a scalar field like the potential energy function, it yields the gradient of that field, which tells us how the function changes over space. In the case of a one-dimensional problem like our textbook exercise, the nabla operator reduces to a simple derivative with respect to the single spatial variable.
Gradient in Physics
To gain deeper insights into conservative forces and potential energy, it's helpful to explore the gradient. In physics, the gradient represents the rate and direction of change in a scalar field. The gradient points in the direction of the greatest increase of the function, and its magnitude corresponds to the rate of increase.

In a physical sense, for a potential energy field, the gradient would indicate the direction in which the potential energy increases most rapidly. As seen in our exercise, the gradient also has a direct relationship with the force, given by abla U(x)abla U(x)—the force is the negative gradient of the potential energy function. This negative sign conveys that the force experienced by a particle will be in the direction that decreases its potential energy.
Integration in Physics
The process of integration in physics is a fundamental tool used to solve many problems, such as finding potential energies from forces. Integration can be thought of as summing infinitesimally small quantities to find a whole. It allows us to reconstruct a function (like potential energy) from its derivative (like force).

In our exercise, integration is employed to determine the potential energy function from the given force function. By integrating the negative of the force with respect to position, we are effectively accumulating the work done by the force over a certain distance, which corresponds to the change in potential energy. This integral gives us the potential energy function up to a constant, which can typically be set to zero at a point where the potential energy is arbitrarily defined to be zero.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

To disable ballistic missiles during the early boost phase of their flight, an "electromagnetic rail gun," to be carried in low-orbit Earth satellites, has been proposed. The gun might fire a \(2.38-\mathrm{kg}\) maneuverable projectile at \(10.0 \mathrm{~km} / \mathrm{s}\). The kinetic energy carried by the projectile is sufficient on impact to disable a missile even if it carries no explosive. (A weapon of this kind is a "kinetic energy" weapon.) The projectile is accelerated to muzzle speed by electromagnetic forces. Suppose instead that we wish to fire the projectile using a spring (a "spring" weapon). What must the force constant be in order to achieve the desired speed after compressing the spring \(1.47 \mathrm{~m} ?\)

In one dimension, the magnitude of the gravitational force of attraction between a particle of mass \(m_{1}\) and one of mass \(m_{2}\) is given by $$F_{x}(x)=G \frac{m_{1} m_{2}}{x^{2}}$$ where \(G\) is a constant and \(x\) is the distance between the particles. (a) What is the potential energy function \(U(x) ?\) Assume that \(U(x) \rightarrow 0\) as \(x \rightarrow \infty,(b)\) How much work is required to increase the separation of the particles from \(x=x_{1}\) to \(x=\) \(x_{1}+d ?\)

Each minute, \(73,800 \mathrm{~m}^{3}\) of water passes over a waterfall \(96.3 \mathrm{~m}\) high. Assuming that \(58.0 \%\) of the kinetic energy gained by the water in falling is converted to electrical energy by a hydroelectric generator, calculate the power output of the generator. (The density of water is \(1000 \mathrm{~kg} / \mathrm{m}^{3}\).)

Show that for the same initial speed \(v_{0}\) the speed \(v\) of a projectile will be the same at all points at the same elevation, regardless of the angle of projection. Ignore air drag.

A body is rolling horizontally without slipping with speed \(v\). It then rolls up a hill to a maximum height \(h\). If \(h=3 v^{2} / 4 g\) what might the body be?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free