Problem 9 - Physics for Scientists and Engineers with Modern Physics

Question

(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 \(U(2.0 \mathrm{~m})=0 .\)

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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:\[dU(x) = \frac{k}{x^3} \, dx\].

Step 3: 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.

Step 4: 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 \( C \):\[0 = -\frac{k}{8} + C \implies C = \frac{k}{8}\].

Step 5: 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 \( x \).

Key Concepts

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.