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Chapter 8: Problem 9

(II) A particle is constrained to move in one dimension along the \(x\) axis and is acted upon by a force given by \(\overrightarrow{\mathbf{F}}(x)=-\frac{k}{x^{3}} \hat{\mathbf{i}}\) where \(k\) is a constant with units appropriate to the SI system. Find the potential energy function \(U(x)\), if \(U\) is arbitrarily defined to be zero at \(x=2.0 \mathrm{~m},\) so that \(U(2.0 \mathrm{~m})=0 .\)

Short Answer

Expert verified
The potential energy function is \( U(x) = -\frac{k}{2x^2} + \frac{k}{8} \).

Step by step solution

01

Understand the relationship between force and potential energy

The force acting on a particle is related to the potential energy function via the equation \( \overrightarrow{\mathbf{F}}(x) = - \frac{dU(x)}{dx} \hat{\mathbf{i}} \). Given \( \overrightarrow{\mathbf{F}}(x) = -\frac{k}{x^3} \hat{\mathbf{i}} \), we can find \( U(x) \) by integrating the force equation.
02

Set up the integral

To find the potential energy \( U(x) \), integrate the force function with respect to \( x \):\[-\frac{dU(x)}{dx} = -\frac{k}{x^3}\].Integrating with respect to \( x \) gives:\[dU(x) = \frac{k}{x^3} \, dx\].
03

Integrate to find the potential energy

Integrate the expression from Step 2:\[U(x) = \int \frac{k}{x^3} \, dx = k \int x^{-3} \, dx\].This yields:\[U(x) = k \left(-\frac{1}{2x^2}\right) + C\],where \( C \) is the integration constant.
04

Apply the boundary condition to find the constant

We know \( U(2.0 \, \text{m}) = 0 \). Substitute \( x = 2.0 \, \text{m} \) into the expression for \( U(x) \):\[0 = k \left(-\frac{1}{2(2.0)^2}\right) + C\].Solving for \( C \):\[0 = -\frac{k}{8} + C \implies C = \frac{k}{8}\].
05

Write the final expression for the potential energy

Substitute the found constant \( C \) back into the expression for \( U(x) \):\[U(x) = -\frac{k}{2x^2} + \frac{k}{8}\].This is the expression for the potential energy as a function of \( x \).

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Key Concepts

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

Force
Force is a fundamental concept in physics that describes an interaction between objects that can cause a change in motion. In this exercise, the force acting on the particle is given by \( \overrightarrow{\mathbf{F}}(x) = -\frac{k}{x^{3}} \hat{\mathbf{i}} \). This equation represents a force that varies with the position \(x\) and is inversely proportional to \(x^3\).
The negative sign signifies that the force is attractive, pulling the particle back along the \(x\) axis. This is similar to how gravitational force works, pulling objects towards a center or point.
Understanding the direction and magnitude of force is essential because it directly relates to how a particle's potential energy will change as it moves along its path. By determining the potential energy function, we can predict how the particle's energy changes and understand the forces it experiences during movement.
Integration
Integration is a mathematical tool that allows us to find quantities like potential energy from rates of change, such as force. In our exercise, we find the function for potential energy \(U(x)\) by integrating the given force function.
A key relationship in this context is given by \(-\frac{dU(x)}{dx} = \overrightarrow{\mathbf{F}}(x)\), meaning that force is the negative derivative of potential energy with respect to position.
\[U(x) = \int \frac{k}{x^3} \, dx = k \int x^{-3} \, dx\]
Performing the integration returns \(-\frac{k}{2x^2} + C\), where \(C\) is the constant of integration. Integration effectively "builds" the energy function from the force, capturing the underlying physical process.
Boundary Conditions
When solving for the potential energy function, boundary conditions are crucial. They provide specific information that helps determine constants from integration that aren't directly observable.
In our problem, we are given a boundary condition: \(U(2.0 \, \text{m}) = 0\). This means that at the position of \(x = 2.0 \, \text{m}\), the potential energy is zero.
We substitute this condition into our integrated expression for potential energy: \[ 0 = -\frac{k}{8} + C\]
Solving this equation for \(C\) gives \(C = \frac{k}{8}\).
With this value of \(C\), the potential energy function becomes \[ U(x) = -\frac{k}{2x^2} + \frac{k}{8}\]
Boundary conditions ensure that the potential energy accurately reflects the scenario described and anchor the math to physical reality.

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Most popular questions from this chapter

(II) A 1200 -kg car rolling on a horizontal surface has speed \(v=75 \mathrm{~km} / \mathrm{h}\) when it strikes a horizontal coiled spring and is brought to rest in a distance of \(2.2 \mathrm{~m}\). What is the spring stiffness constant of the spring?

(III) The potential energy of the two atoms in a diatomic (two-atom) molecule can be written $$ U(r)=-\frac{a}{r^{6}}+\frac{b}{r^{12}} $$ where \(r\) is the distance between the two atoms and \(a\) and \(b\) are positive constants. (a) At what values of \(r\) is \(U(r)\) a minimum? A maximum? (b) At what values of \(r\) is \(U(r)=0 ?\) (c) Plot \(U(r)\) as a function of \(r\) from \(r=0\) to \(r\) at a value large enough for all the features in \((a)\) and \((b)\) to show. ( \(d\) ) Describe the motion of one atom with respect to the second atom when \(E<0,\) and when \(E>0 .(e)\) Let \(F\) be the force one atom exerts on the other. For what values of \(r\) is \(F>0, F<0, F=0 ?\) (f) Determine \(F\) as a function of \(r\).

(1I) A 56 -kg skier starts from rest at the top of a 1200 -m- long trail which drops a total of 230 m from top to bottom. At the bottom, the skier is moving 11.0 \(\mathrm{m} / \mathrm{s} .\) How much. energy was dissignated by friction?

(II) A 75 -kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a \(23^{\circ}\) hill. The skier is pulled a distance \(x=220 \mathrm{~m}\) along the incline and it takes 2.0 min to reach the top of the hill. If the coefficient of kinetic friction between the snow and skis is \(\mu_{\mathrm{k}}=0.10,\) what horsepower engine is required if 30 such skiers (max) are on the rope at one time?

(II) A pump lifts \(21.0 \mathrm{~kg}\) of water per minute through a height of \(3.50 \mathrm{~m}\). What minimum output rating (watts) must the pump motor have?
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