Problem 14 - An Introduction to Mechanics

Question

A particle of mass \(m\) moves in one dimension along the positive \(x\) axis. It is acted on by a constant force directed toward the origin with magnitude \(B\), and an inverse-square law repulsive force with magnitude \(A / x^{2}\) ( \(a\) ) Find the potential energy function \(U(x)\). (b) Sketch the energy diagram for the system when the maximum kinetic energy is \(K_{0}=\frac{1}{2} m v_{0}^{2}\). (c) Find the equilibrium position, \(x_{0}\).

Step-by-Step Solution

Answer

(a) \( U(x) = -Bx + \frac{A}{x} \). (b) Energy diagram shows potential \( U(x) \) and kinetic energy \( K_0 \). (c) Equilibrium at \( x_0 = \sqrt{-A/B} \) if real.

Step 1: Determine the potential energy function for the constant force

The potential energy function due to a constant force directed toward the origin is given by the integral of the force with respect to displacement. Thus, for the constant force \( B \), the potential energy is \( U_1(x) = -B x \). The negative sign indicates that the force is toward the origin.

Step 2: Determine the potential energy function for the inverse-square law force

The inverse-square force is repulsive and its potential energy function is found by integrating the force \( F = -\frac{A}{x^2} \). Therefore, the potential energy is given by \( U_2(x) = -\int -\frac{A}{x^2} \,dx = \frac{A}{x} + C \), where \( C \) is an arbitrary constant (often set to zero for convenience).

Step 3: Combine potential energy functions to find total potential energy

The total potential energy function, \( U(x) \), is the sum of the potential energies from both forces: \( U(x) = U_1(x) + U_2(x) = -Bx + \frac{A}{x} \).

Step 4: Sketch the energy diagram for the system

To sketch the energy diagram, note that the potential energy \( U(x) = -Bx + \frac{A}{x} \) will depict a curve starting with high potential energy for small \( x \) (due to \( \frac{A}{x} \)) and decreasing as \( x \) increases (dominated by \(-Bx\)). The maximum kinetic energy \( K_0 = \frac{1}{2} m v_0^2 \) is available to the particle, and we can plot a horizontal line at \( K_0 \) to show the total energy (potential + kinetic).

Step 5: Find the equilibrium position, \( x_0 \)

The equilibrium position occurs where the force on the particle is zero. Since force is the negative gradient of potential energy, set \( \frac{dU}{dx} = 0 \). Therefore, \( \frac{d}{dx}(-Bx + \frac{A}{x}) = B + \frac{A}{x^2} = 0 \). Solving for \( x \), we have \[ B = -\frac{A}{x^2} \]. Rearranging gives \( x^2 = -\frac{A}{B} \). Thus, \( x_0 = \sqrt{-\frac{A}{B}} \). However, if \( A = 0 \), the equilibrium is at the origin. This analysis holds if \( A \) and \( B \) are such that \( A/B < 0 \), ensuring a real positive solution.

Key Concepts

Equilibrium Position

Understanding the equilibrium position is essential in physics, especially in problems involving forces. An equilibrium position is a point where the net force acting on a particle is zero. Here, a particle is under the influence of two forces: a constant force moving it towards the origin and an inverse-square repulsive force away from the origin. At equilibrium, these forces perfectly balance each other out.

• If the net force is zero, the particle doesn't move, indicating that it is in equilibrium.
• In mathematical terms, this is where the derivative of the potential energy with respect to position equals zero, \( \frac{dU}{dx} = 0 \).

For our specific case, the forces lead us to the potential energy equation \( U(x) = -Bx + \frac{A}{x} \). To find the equilibrium position \( x_0 \), we solve \( B + \frac{A}{x^2} = 0 \). After simplifying, this helps derive \( x_0 = \sqrt{-\frac{A}{B}} \), provided the conditions allow for a real positive solution.

Inverse-Square Law

The inverse-square law is a powerful principle in physics. It describes any physical law stating that a specified quantity is inversely proportional to the square of the distance from its source. In the context of the problem, this inverse-square force acts repulsively on the particle.

• The force can be formulated as \( F = -\frac{A}{x^2} \), where \( A \) represents the strength of the force and \( x \) is the distance from the source.
• This indicates that as the particle moves further from the origin, the force becomes weaker.

The inverse-square law is pivotal because it reveals how fields like gravity and electromagnetism operate over distances. It influences the potential energy function, forming one of the core factors in our overall potential energy \( U(x) = -Bx + \frac{A}{x} \). This impacts how we find equilibrium positions as well.

Constant Force

A constant force is one that remains unchanged in magnitude and direction over time. For this particular exercise, the constant force is directed towards the origin, with a magnitude of \( B \). These forces are vital as they simplify many real-world scenarios allowing us to predict movement accurately.

• The potential energy functional form due to a constant force is given by \( U = -B x \), highlighting how the potential energy changes linearly with displacement.
• It represents a linear field, contrasting with the more complex behaviors seen in non-linear forces.

In this problem, the presence of both the constant force towards the origin and the inverse-square repulsive force makes the force calculation a bit more intricate. But understanding that one of these forces remains uniform lays the groundwork for exploring the system's entire potential energy landscape, allowing us to predict behavior when combined with other forces like the inverse-square force.