/*! 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 An object moving in the \(x y\) ... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 36

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 \(\hat{y}\) .

Short Answer

Expert verified
The force is \( \vec{F} = 2\alpha \frac{1}{x^3} \hat{i} + 2\alpha \frac{1}{y^3} \hat{j} \).

Step by step solution

01

Understand the Relationship between Force and Potential Energy

In classical mechanics, the force associated with a potential energy function can be determined using the gradient of the potential energy. Specifically, the force is given by the negative gradient of the potential energy function. The formula for calculating this is: \[ \vec{F} = -abla U \]where \( abla U \) is the gradient of \( U(x,y) \).
02

Calculate the Gradient of the Potential Energy Function

The potential energy function is given by \( U(x, y) = \alpha \left( \frac{1}{x^2} + \frac{1}{y^2} \right) \). We need to calculate the gradient by taking partial derivatives:\[ \frac{\partial U}{\partial x} = \alpha \frac{d}{dx}\left( \frac{1}{x^2} \right) = -2\alpha \frac{1}{x^3} \]\[ \frac{\partial U}{\partial y} = \alpha \frac{d}{dy}\left( \frac{1}{y^2} \right) = -2\alpha \frac{1}{y^3} \]
03

Express the Gradient as a Vector

The gradient of \( U \), denoted as \( abla U \), can be expressed as:\[ abla U = \left( \frac{\partial U}{\partial x} \right) \hat{i} + \left( \frac{\partial U}{\partial y} \right) \hat{j} = \left( -2\alpha \frac{1}{x^3} \right) \hat{i} + \left( -2\alpha \frac{1}{y^3} \right) \hat{j} \]
04

Compute the Force Using the Negative Gradient

According to Step 1, the force \( \vec{F} \) can be determined as the negative of the gradient of \( U(x,y) \):\[ \vec{F} = -abla U = 2\alpha \frac{1}{x^3} \hat{i} + 2\alpha \frac{1}{y^3} \hat{j} \]This expression represents the force in terms of the unit vectors \( \hat{i} \) and \( \hat{j} \).

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Key Concepts

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

Conservative Force
Conservative forces are unique because they are path-independent. This means the work done by a conservative force when moving an object from one point to another does not depend on the path taken. The only thing that matters is the initial and final positions.

Some common examples of conservative forces include gravity and the electrostatic force.

The key feature of a conservative force is that it can be derived from a potential energy function, denoted as \( U \). This characteristic allows for energy conservation within a system, making it easier to analyze mechanical problems.

When dealing with a conservative force, the total mechanical energy (kinetic plus potential) of a system remains constant, provided no non-conservative forces (like friction) are acting on it. This property simplifies the task of predicting how the system will behave over time.
Gradient of a Function
The gradient of a function is a vector that points in the direction of the greatest rate of increase of the function. It's like a little arrow showing which way to go to climb uphill fastest on a 3D landscape.

In mathematical terms, the gradient is a vector of partial derivatives. For a function of two variables such as \( U(x, y) \), the gradient is given by:
  • \( abla U = \left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y} \right) \)


This represents how the function \( U \) changes with respect to each variable. For example, \( \frac{\partial U}{\partial x} \) tells us how \( U \) changes as \( x \) changes, holding \( y \) constant. Similarly, \( \frac{\partial U}{\partial y} \) shows the change with \( y \), holding \( x \) constant.

Understanding gradients is crucial because they help us understand how a function behaves and how to use this information, such as determining the direction and magnitude of forces in physics.
Classical Mechanics
Classical mechanics is the branch of physics that deals with the motion of objects and the forces acting upon them. It is based on principles laid down by Isaac Newton, so it's often called Newtonian mechanics. This area of physics covers the motion of everything, from spinning wheels to orbiting planets.

Key concepts in classical mechanics include:
  • Newton's Laws of Motion: These laws describe how forces affect motion.
  • Work and Energy: The relationship between the force applied and the movement it causes.
  • Conservation Laws: These include the conservation of energy and momentum, which are central to predicting physical systems' outcomes.


In problems involving potential energy and forces, classical mechanics uses the concept of energy conservation to make solving problems easier and more intuitive by converting a problem into mathematical language.

Despite its age, classical mechanics remains a powerful tool in science and engineering, offering essential insights for everything from the design of vehicles and structures to the exploration of space.

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Most popular questions from this chapter

Up and Down the Hill. A \(28-\mathrm{kg}\) rock approaches the foot of a hill with a speed of 15 \(\mathrm{m} / \mathrm{s}\) . This hill slopes upward at a constant angle of \(40.0^{\circ}\) above the horizontal. The coefficients of static and kinetic friction between the hill and the rock are 0.75 and 0.20 , respectively. (a) Use energy conservation to find the maximum height above the foot of the hill reached by the rock. (b) Will the rock remain at rest at its highest point, or will it slide back down the hill? (c) If the rock does slide back down, find its speed when it returns to the bottom of the hill.

A Hooke's law force \(-k x\) and a constant conservative force \(F\) in the \(+x\) -direction act on an atomic ion. (a) Show that a possible potential-energy function for this combination of forces is \(U(x)=\frac{1}{2} k x^{2}-F x-F^{2} / 2 k\) . Is this the only possible function? Explain. (b) Find the stable equilibrium position. (c) Graph \(U(x)\) (in units of \(F^{2} / k )\) versus \(x\) (in units of \(F / k )\) for values of \(x\) between \(-5 F / k\) and 5\(F / k\) . (d) Are there any unstable equilibrium positions? (e) If the total energy is \(E=F^{2} | k,\) what are the maximum and minimum valnes of \(x\) that the ion reaches in its motion? If the ion has mass \(m,\) find its maximum speed if the total energy is \(E=F^{2} / k .\) For what value of \(x\) is the speed maximum?

An ideal spring of negligible mass is 12.00 \(\mathrm{cm}\) long when nothing is attached to it. When you hang a \(3.15-\mathrm{kg}\) weight from it, you measure its length to be 13.40 \(\mathrm{cm}\) . If you wanted to store 10.0 \(\mathrm{J}\) of potential energy in this spring, what would be its total length? Assume that it continues to obey Hooke's law.

You are asked to design a spring that will give a \(1160-\mathrm{kg}\) satellite a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) relative to an orbiting space shuttle. Your spring is to give the satellite a maximum acceleration of 5.00 \(\mathrm{g}\) . The spring's mass, the recoil kinetic energy of the shuttle, and changes in gravitational potential energy will be negligible. (a) What must the force constant of the spring be? (b) What distance must the spring be compressed?

A certain spring is found not to obey Hooke's law; it exerts a restoring force \(F_{x}(x)=-\alpha x-\beta x^{2}\) if it is stretched or compressed, where \(\alpha=60.0 \mathrm{N} / \mathrm{m}\) and \(\beta=18.0 \mathrm{N} / \mathrm{m}^{2} .\) The mass of the spring is negligible. (a) Calculate the potentinl-energy function \(U(x)\) for this spring. Let \(U=0\) when \(x=0\) (b) An object with mass 0.900 \(\mathrm{kg}\) on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 \(\mathrm{m}\) to the right (the \(+x\) -direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 \(\mathrm{m}\) to the right of the \(x=0\) equilibrium position?
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