Problem 34 - University Physics with Modern Physics

Question

A particle is moving along the \(x\) -axis subject to the potential energy function \(U(x)=1 / x+x^{2}+x-1\) a) Express the force felt by the particle as a function of \(x\). b) Plot this force and the potential energy function. c) Determine the net force on the particle at the coordinate \(x=2.00 \mathrm{~m}\)

Step-by-Step Solution

Answer

Question: Determine the force acting on a particle at x=2.00 m, given that its potential energy function is U(x) = 1/x + x^2 + x - 1. Answer: The net force acting on the particle at x = 2.00 m is 4.25 N.

Step 1: (Part a: Derive force function F(x))

We know that the force acting on a particle due to potential energy is given by: F(x) = -dU(x)/dx Given, U(x) = 1/x + x^2 + x - 1 We will differentiate U(x) with respect to x to find the force function, F(x). F(x) = - d(1/x + x^2 + x - 1)/dx

Step 2: (Differentiate U(x) with respect to x)

Differentiating each term of U(x) with respect to x, we get: -d(1/x)/dx = - (-1/x^2) = 1/x^2 -d(x^2)/dx = -2x -d(x)/dx = -1 -d(-1)/dx = 0 So, F(x) = - (1/x^2 - 2x - 1)

Step 3: (Part b: Plot F(x) and U(x))

To plot F(x) = - (1/x^2 - 2x - 1) and U(x) = 1/x + x^2 + x - 1, you can use graphing software or online tools such as Desmos, Wolfram Alpha, or GeoGebra. Enter the given functions as mentioned and adjust the range of x values as needed to see the behavior of the functions.

Step 4: (Part c: Find net force at x=2.00 m)

Using the derived force function F(x) = - (1/x^2 - 2x - 1), we can determine the net force on the particle at x=2.00 m: F(2) = - (1/(2^2) - 2(2) - 1) F(2) = - (1/4 - 4 - 1) F(2) = - (-4.25) F(2) = 4.25 N The net force acting on the particle at x = 2.00 m is 4.25 N.

Key Concepts

Potential Energy

Potential energy refers to the stored energy in an object due to its position or state. In classical mechanics, the potential energy is often associated with the position of an object in a force field, such as gravitational or electric fields.
For the exercise, we focused on the potential energy function of a particle moving along the x-axis:

• Given function: \[ U(x) = \frac{1}{x} + x^2 + x - 1 \]
• Each term of this function contributes differently to the potential energy based on the particle's position along the x-axis.

Understanding potential energy helps in calculating the force acting on a particle, as energy conversion often involves forces. The potential well or peak in the graph of the function can hint at stable or unstable equilibrium positions for the particle.

Force Calculation

The relationship between force and potential energy is a key concept in classical mechanics. Force is related to the negative gradient of potential energy. In one dimension, this simplifies to:

• \[ F(x) = -\frac{dU(x)}{dx} \]

This expression shows that the direction of the force is opposite to the incline of potential energy. For our given potential energy:

• Original function: \[ U(x) = \frac{1}{x} + x^2 + x - 1 \]
• Derived force function: \[ F(x) = -\left( \frac{1}{x^2} - 2x - 1 \right) \]

Calculating force from potential energy involves finding the derivative of each term with respect to x, then summing them while considering the negative sign. This approach ensures we capture how changes in x affect the energy (and thus the force) experienced by the particle.

Differentiation

Differentiation is a mathematical process that calculates the rate of change of a function at any point. It is crucial in determining how potential energy varies with position (x), which in turn affects the force calculation.

• Differentiation of each term of \( U(x) \) is necessary to find \( F(x) \).
• For instance:
  • The derivative of the term \( \frac{1}{x} \) with respect to x is \(-\frac{1}{x^2} \).
  • The term \( x^2 \) becomes \( 2x \) after differentiation.
  • \( x \) simplifies to 1, and constant terms like -1 drop out as their derivative is zero.

By applying these derivatives and the negative slope rule, you arrive at the function to calculate force from potential energy. Differentiation provides a precise tool for understanding how small changes in position impact energy and subsequent forces.

Graphing Functions

Graphing functions is a practical way to visualize the behavior and relationships of mathematical equations. In this exercise, the objective was to plot both the potential energy function \( U(x) \) and the derived force function \( F(x) \).

• Visualizing \( U(x) = \frac{1}{x} + x^2 + x - 1 \) allows you to see potential peaks and troughs, which indicate where the particle may gain or lose speed.
• Plotting the derived force function \( F(x) = -\left( \frac{1}{x^2} - 2x - 1 \right) \) shows how force varies with position, offering insight into how the particle is pushed or pulled along the x-axis.

Using tools like Desmos or GeoGebra can easily help visualize these functions by correctly setting the range for x and interpreting the resulting graphs. This method not only aids in understanding the current state of the system but also in predicting future movement under the given potential energy landscape.