/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 CALE An object moving in the \(x... [FREE SOLUTION] | 91Ó°ÊÓ

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CALE An object moving in the \(x y\) -plane is acted on by a conservative force described by the potential-energy function \(U(x, y)=\alpha\left(1 / x^{2}+1 / y^{2}\right),\) where \(\alpha\) is a positive constant. Derive an expression for the force expressed in terms of the unit vectors \(\hat{i}\) and \(\hat{J} .\)

Short Answer

Expert verified
\(\vec{F}(x, y) = \frac{2\alpha}{x^3} \hat{i} + \frac{2\alpha}{y^3} \hat{j} \).

Step by step solution

01

Understand Potential Energy and Force Relation

To find the force associated with the potential energy function, we use the relation between force and potential energy, which is given by \[ \vec{F} = -abla U \].This means that the force is the negative gradient of the potential energy function \(U(x, y)\).
02

Find the Gradient of the Potential Energy

The gradient of \(U(x, y)\) is the vector of partial derivatives of \(U\) with respect to \(x\) and \(y\). So, we compute: \[ \frac{\partial U}{\partial x} = \alpha \left( -2 \cdot \frac{1}{x^3} \right) = -\frac{2\alpha}{x^3} \] \[ \frac{\partial U}{\partial y} = \alpha \left( -2 \cdot \frac{1}{y^3} \right) = -\frac{2\alpha}{y^3} \].
03

Apply the Negative Sign to the Gradient

Taking the negative of the gradient computed in the previous step, the force \(\vec{F}\) is given by: \[ \vec{F}(x, y) = -\left( -\frac{2\alpha}{x^3} \hat{i} - \frac{2\alpha}{y^3} \hat{j} \right) \] which simplifies to: \[ \vec{F}(x, y) = \frac{2\alpha}{x^3} \hat{i} + \frac{2\alpha}{y^3} \hat{j} \].
04

Write the Final Expression for Force

The final expression for the force vector in terms of unit vectors \(\hat{i}\) and \(\hat{j}\) is: \[ \vec{F}(x, y) = \frac{2\alpha}{x^3} \hat{i} + \frac{2\alpha}{y^3} \hat{j} \].

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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 the stored energy of an object due to its position, arrangement, or state. In physics, the potential energy associated with a conservative force can be defined by a potential energy function. For the problem at hand, the potential energy function is given by \( U(x, y) = \alpha \left( \frac{1}{x^2} + \frac{1}{y^2} \right) \), where \( \alpha \) is a positive constant.
Understanding potential energy is central because it helps us determine how a conservative force can perform work on an object, without relying on any external factors.

Consider these points when studying potential energy:
  • It depends on the system's configuration, such as the position of particles.
  • It is related to forces that restore energy, like gravitational and electrostatic forces.
  • It's often used to derive forces when its gradient is considered.
In this exercise, potential energy allows the calculation of force via its derivatives, contributing to an understanding of the work done by the conservative force.
Gradient
The gradient is a concept from vector calculus that describes how a multi-variable function changes at a given point. When dealing with potential energy, the gradient indicates the direction and rate of the steepest increase in potential energy.
The mathematical representation of the gradient is the vector of partial derivatives. For a potential energy function \( U(x, y) \), its gradient \( abla U \) is given by:
\[ abla U = \left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y} \right) \]
Computing this gives us the information needed to understand how force is derived from potential energy. It shows us in which direction the potential energy changes fastest.

Key aspects of a gradient include:
  • It points towards the maximum rate of change of the function.
  • In this problem, the gradient helps find the force vector by differentiating the potential energy function with respect to space coordinates.
  • The negative gradient is used to find the force because force acts in the direction of decreasing potential energy.
The completeness of this concept allows us to effectively use the potential energy to calculate forces acting on an object.
Force Vector
A force vector is a mathematical representation that describes both the magnitude and direction of a force acting on an object. In a conservative system, the force vector can be directly derived from the potential energy function through the concept of the gradient.
In this exercise, the force vector \( \vec{F} \) is shown to be:
\[ \vec{F}(x, y) = - abla U(x, y) = \frac{2\alpha}{x^3} \hat{i} + \frac{2\alpha}{y^3} \hat{j} \]
Here, \( \hat{i} \) and \( \hat{j} \) are unit vectors in the \( x \)- and \( y \)-directions, respectively.

Important points about force vectors include:
  • They provide essential information about both the direction and strength of the force.
  • The negative sign in \( -abla U \) ensures the force relates correctly to the decreasing potential energy.
  • They help in illustrating how an object would move under the influence of this force.
Thus, understanding how the force vector is derived from potential energy is crucial for comprehending the interactions in a conservative force field.
Partial Derivatives
Partial derivatives are tools used in calculus to measure how a function changes as its variables change. For a function of two variables like \( U(x, y) \), each partial derivative represents how \( U \) changes with respect to one variable, while holding the other constant.
In this problem, the partial derivatives are computed as follows:
\[ \frac{\partial U}{\partial x} = -\frac{2\alpha}{x^3} \]
\[ \frac{\partial U}{\partial y} = -\frac{2\alpha}{y^3} \]
These expressions help describe the gradient of the potential energy function.

Key points on partial derivatives include:
  • They are fundamental in finding how functions change directionally in relation to their individual variables.
  • They enable us to understand how potential energy varies in different planes, crucial for deriving forces.
  • They provide the necessary components for constructing the gradient, which is used in the force calculation.
Through partial derivatives, we unlock the ability to assess each component's role in the overall function's behavior, essential for solving physics problems involving multiple variables.

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Most popular questions from this chapter

While a roofer is working on a roof that slants at \(36^{\circ}\) above the horizontal, he accidentally nudges his \(85.0-\mathrm{N}\) toolbox, causing it to start sliding downward, starting from rest. If it starts 4.25 \(\mathrm{m}\) from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 \(\mathrm{N} ?\)

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