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Chapter 7: Problem 49

(a) Derive an expression for the potential energy of an object subject to a force \(F_{x}=a x-b x^{3},\) where \(a=5 \mathrm{N} / \mathrm{m}\) and \(b=2 \mathrm{N} / \mathrm{m}^{3},\) taking \(U=0\) at \(x=0 .\) (b) Graph the potentialenergy curve for \(x>0\) and use it to find the turning points for an object whose total energy is \(-1 \mathrm{J}\)

Short Answer

Expert verified
The potential energy of the object is \(-2.5x^2 + 0.5x^4\). The turning points can be identified by solving \(-1 = -2.5x^2 + 0.5x^4\).

Step by step solution

01

Derive the Potential Energy

The potential energy \(U\) is defined as the negative integral of the force over distance, namely \(U = -\int F_x dx\). We substitute \(F_x = 5x - 2x^3\) and integrate with respect to \(x\) from 0 to \(x\), which is the particular solution for this case. Thus, the integral can be set up as \(U = -\int_{0}^{x} (5x - 2x^3) dx\).
02

Calculate the Integral

Calculating the definite integral, \(U = -[2.5x^2 - 0.5x^4]_{0}^{x}\) which simplifies to \(U = -2.5x^2 + 0.5x^4\).
03

Graph the Potential Energy

Graph the function \(U = -2.5x^2 + 0.5x^4\) for \(x > 0\).
04

Identify the Turning Points

The turning points are where the total energy equals the potential energy. Equate the total energy to the potential energy and solve for \(x\). The equation is \(-1 = -2.5x^2 + 0.5x^4\). Solving this will give the turning points.

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Key Concepts

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

Deriving Potential Energy
Understanding the derivation of potential energy is key for grasping how forces affect the energy stored within a system. In the example problem, the force applied is dependent on the position, described by the equation \(F_{x} = ax - bx^3\). To derive the potential energy (\(U\)) from this force, we used the principle that potential energy is the negative integral of the force over the distance. The process began with setting up the integral \(U = -\int F_x dx\) and then inserting our specific force function.
Force and Potential Energy Relationship
The relationship between force and potential energy is a cornerstone in mechanics. The force here is a restoring force, proportional to the position and modified by a cubic term which represents the nonlinear characteristics of the force as the object moves. Through integration, we found that the work done by this force translates into a change in potential energy. We expressed this relationship mathematically, with potential energy being the integral of the force. It's important to remember, potential energy is a scalar quantity, meaning it has no direction, just magnitude.
Graphing Potential Energy Curves
To visualize how potential energy varies with position, graphing the derived potential energy equation is quite useful.
Graphing \(U = -2.5x^2 + 0.5x^4\) helps us see how the energy landscape changes. It allows students to visualize the concept of a potential well, where the system's equilibrium points and stability can be analyzed. A graph makes it easier to comprehend the effects of the non-linear term \(-bx^3\) on the overall potential energy, where the curve bends differently compared to a simple parabola.
Calculating Turning Points in Potential Energy
Turning points in potential energy are where the physical system can change its direction of motion. These points occur when the derivative of the potential energy with respect to position is zero. For our exercise, we calculated the turning points by setting the total energy equal to the potential energy and solving for \(x\). The total energy included both kinetic and potential energy, but at the turning points, it's all potential. This calculation demonstrated when the object would stop moving (at the maxima) and where it would have maximum speed (at the minima).

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Most popular questions from this chapter

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