/*! 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 85 A Hooke's law force \(-k x\) and... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified
(a) The function is one possible option; constant shift possible. (b) Stable position: \( x = F/k \). (c) Graph: Parabola from \(-5\) to \(5\). (d) No unstable position. (e) Max: \(4F/k\), Min: \(-2F/k\), Max speed: \(\sqrt{4F^2/km}\) at \(F/k\).

Step by step solution

01

Determine the Potential Energy Function

We know the force is the negative gradient of the potential energy, given by \( F(x) = - \frac{dU}{dx} \).Combining both forces, we have \(-kx + F = - \frac{dU}{dx} \). Integrating this equation with respect to \( x \), we find the potential energy function by separating the terms:\[U(x) = \int (-kx + F) \, dx = -\frac{k}{2}x^2 + Fx + C,\]where \( C \) is the integration constant. We can set \( C \) such that \( C = -\frac{F^2}{2k} \) to match the given expression, hence:\[U(x) = \frac{1}{2}kx^2 - Fx - \frac{F^2}{2k}.\]This is a possible function, but other functions differing by a constant are also valid since potentials are defined up to an arbitrary constant.
02

Find the Stable Equilibrium Position

To find the stable equilibrium position, we need the force to be zero:\(-kx + F = 0.\)Solving for \(x\), we get \[ x = \frac{F}{k}. \]This is the stable equilibrium position because the potential energy function is a parabolic function opening upwards.
03

Graph the Potential Energy Function

We want to graph \(U(x)\) in units of \(F^2/k \) versus \(x\) in units of \(F/k\) between \( -5F/k \) and \( 5F/k \). Let \( u = \frac{U(x)}{F^2/k} \) and \( x' = \frac{x}{F/k} \), so\[ U(x) = \frac{1}{2}\left(\frac{kF^2}{k^2} \cdot (x')^2\right) - \left( F \cdot \frac{F}{k} \cdot x' \right) - \frac{F^2}{2k},\]which simplifies to \[u(x') = \frac{1}{2}(x')^2 - x' - \frac{1}{2}.\]Now plot \( u(x') \) for \(-5 \leq x' \leq 5\).
04

Identify Unstable Equilibrium Positions

An unstable equilibrium occurs where the second derivative of \(U(x)\) is positive. Calculate \[ \frac{d^2U}{dx^2} = k. \]Since \(k > 0\), no position in this parabolic function corresponds to an unstable position because the curvature is never downward.
05

Determine Maximum and Minimum x-values

For total energy \( E = F^2/k \), the potential energy at the turning points satisfies\[ \frac{1}{2}kx^2 - Fx - \frac{F^2}{2k} = \frac{F^2}{k}.\]Simplifying yields:\[ \frac{1}{2}kx^2 - Fx - \frac{3F^2}{2k} = 0.\]This quadratic equation in \(x\) gives maximum and minimum \(x\) values upon solving:\[ x = \frac{F \pm \sqrt{9F^2}}{k} = \frac{F \pm 3F}{k}, \]resulting in \( x = \frac{4F}{k} \) or \( x = -\frac{2F}{k}. \)
06

Find Maximum Speed and Its Location

Maximum speed occurs where potential energy is minimum, i.e., kinetic energy is maximum.Using conservation of energy:\[ \frac{1}{2}mv_{max}^2 = E - U(x_{eq}),\]where \(x_{eq} = F/k\) is the equilibrium position and\(U(x_{eq}) = -F^2/(2k)\).Thus,\[ E - U(x_{eq}) = \frac{2F^2}{k},\]\[ \frac{1}{2}mv_{max}^2 = \frac{2F^2}{k},\]hence\[ v_{max} = \sqrt{\frac{4F^2}{km}}.\]The speed is maximum at \( x = F/k \), where potential energy is minimum.

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 91Ó°ÊÓ!

Key Concepts

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

Potential Energy Function
Potential Energy is an essential concept when it comes to understanding forces and motions. The potential energy function describes how the stored energy in a system changes as the system's configuration changes. For a Hooke's law spring and a constant force combination, the potential energy function can be found by integrating the overall force. The force equation is given by the sum of the forces, \(-kx + F = - \frac{dU}{dx} \), where \(k\) is the spring constant, and \(F\) is the constant force.

Upon integrating this force with respect to position \(x\), we obtain the potential energy function:\[U(x) = \frac{1}{2}kx^2 - Fx - \frac{F^2}{2k}.\]This function provides details on how potential energy changes for different positions \(x\). One important aspect to note is that potential energy functions are defined up to an arbitrary constant, meaning this isn't the only potential energy function possible, but rather a common form for clarity in calculations. Integration constants are adjusted to align the function with energy boundary conditions or specific scenarios.
Equilibrium Position
In physics, the equilibrium position is where the sum of forces on an object is zero, indicating a stable or balanced state. For systems governed by Hooke's Law and additional forces, equilibrium is where the system naturally settles without external disturbances.

To find the stable equilibrium position for the given potential energy function, we set the derivative (force) to zero: \(-kx + F = 0\). Solving this equation yields \[x = \frac{F}{k}.\]This is the point where the system reaches stability because any small movement away from this point will be counteracted by the spring force attempting to restore balance. In this scenario, the stable equilibrium position corresponds to a minimum in the potential energy curve, which resembles a parabola opening upwards.
Force Analysis
Understanding the combined effects of forces helps to predict the behavior of objects within a system. With Hooke's Law, the force on an object is proportional to its displacement and acts in the opposite direction. This is represented as \(-kx\) for a spring force. Additionally, a constant conservative force \(F\) in the positive direction adds to the overall force.

Analyzing these forces helps derive potential energy functions and solve equilibrium equations. When combining both forces without external influences, they create a predictable pattern within the system governed by their sum, \( -kx + F \). This approach is fundamental for predicting the dynamics and stability in physics problems, demonstrating how diverse forces interact through direct addition to exert influence on a system's motion or static condition.
Kinetic Energy
Kinetic Energy is the energy of motion. When an object moves, it possesses kinetic energy proportional to its mass and the square of its velocity. The expression for kinetic energy is \(\frac{1}{2} mv^2\), where \(m\) is mass and \(v\) is the velocity.

In the context of Hooke’s Law forces and potential energy, kinetic energy becomes maximum when potential energy is at its minimum due to energy conservation. Given total energy \(E\) is constant, any drop in potential energy results in increased kinetic energy:\[\frac{1}{2}mv_{max}^2 = E - U(x_{eq}).\]By finding the maximum speed at \(x = \frac{F}{k}\), where potential energy reaches its lowest, the motion dynamics can be better understood. This allows for prediction of speeds and motion trajectories under specific initial energy conditions.
Conservation of Energy
The principle of Conservation of Energy states that the total energy in an isolated system remains constant, though it can change forms. This principle is crucial when analyzing systems like those described by Hooke's Law, where potential energy and kinetic energy exchange roles during motion.

In a system influenced by conservative forces, potential and kinetic energy sum up to a constant total energy \(E\). Determining the range and behavior of an object within an energy-conserving system entails ensuring that the potential energy function aligns with total energy:\[\frac{1}{2}kx^2 - Fx - \frac{F^2}{2k} = E.\]Through these relationships, we can find conditions for maximum and minimum positions (turning points) and determine motion characteristics like speed and direction at different points within the system. Energy is neither created nor destroyed; it is merely transferred between energy states for informative predictions and analyses.

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

In an experiment, one of the forces exerted on a proton is \(\overrightarrow{\boldsymbol{F}}=-\alpha x^{2} \hat{i},\) where \(\alpha=12 \mathrm{N} / \mathrm{m}^{2} .\) (a) How much work does \(\overrightarrow{\boldsymbol{F}}\) do when the proton moves along the straight- line path from the point \((0.10 \mathrm{m}, 0)\) to the point \((0.10 \mathrm{m}, 0.40 \mathrm{m}) ?(\mathrm{b})\) Along the straightline path from the point \((0.10 \mathrm{m}, 0)\) to the point \((0.30 \mathrm{m}, 0) ?\) (c) Along the straight-line path from the point \((0.30 \mathrm{m}, 0)\) to the point \((0.10 \mathrm{m}, 0) ?(\mathrm{d})\) Is the force \(\overrightarrow{\boldsymbol{F}}\) conservative? Explain. If \(\overrightarrow{\boldsymbol{F}}\) is conservative, what is the potential energy function for it? Let \(U=0\) when \(x=0\) .

A cutting tool under microprocessor control has several forces acting on it. One force is \(\overrightarrow{\boldsymbol{F}}=-\alpha x y^{2} \hat{\boldsymbol{j}},\) a force in the negative \(y\) -direction whose magnitude depends on the position of the tool. The constant is \(\alpha=2.50 \mathrm{N} / \mathrm{m}^{3} .\) Consider the displacement of the tool from the origin to the point \(x=3.00 \mathrm{m}, y=3.00 \mathrm{m} .\) (a) Calculate the work done on the tool by \(\vec{F}\) if this displacement is along the straight line \(y=x\) that connects, these two points. (b) Calculate the work done on the tool by \(\vec{F}\) if the tool is first moved out along the \(x\) -axis to the point \(x=3.00 \mathrm{m}, y=0\) and then moved parallel to the \(y\) -axis to the point \(x=3.00 \mathrm{m}\) , \(y=3.00 \mathrm{m}\) . (c) Compare the work done by \(\vec{F}\) along these two paths. Is \(\vec{F}\) conservative or nonconservative? Explain.

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?

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?

You are testing a new amusement park roller coaster with an empty car with mass 120 \(\mathrm{kg}\) . One part of the track is a vertical loop with radius 12.0 \(\mathrm{m}\) . At the bottom of the loop (point \(A\) ) the car has speed \(25.0 \mathrm{m} / \mathrm{s},\) and at the top of the loop (point \(B )\) it has speed 8.0 \(\mathrm{m} / \mathrm{s}\) . As the car rolls from point \(A\) to point \(B,\) how much work is done by friction?

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.