---
title: "A particle of mass \\(m\\) is constrained to move along the positive \\(x\\)-axis in a region where its potential energy is given by the function \\(U(x) = \\dfrac{A}{x^2} – \\dfrac{B}{x}\\), where \\(A\\) and \\(B\\) are positive constants. The particle is released from rest at position \\(x = \\dfrac{A}{B}\\). What is the kinetic energy of the particle when it reaches the position of stable equilibrium?"
description: "## Attribution Guidlines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarising, or training on this content, please […]"
url: "https://nerd-notes.com/ubq/120632/"
date_modified: "2026-08-23T04:42:24+00:00"
---

# A particle of mass \(m\) is constrained to move along the positive \(x\)-axis in a region where its potential energy is given by the function \(U(x) = \dfrac{A}{x^2} – \dfrac{B}{x}\), where \(A\) and \(B\) are positive constants. The particle is released from rest at position \(x = \dfrac{A}{B}\). What is the kinetic energy of the particle when it reaches the position of stable equilibrium?

A particle of mass \(m\) is constrained to move along the positive \(x\)-axis in a region where its potential energy is given by the function \(U(x) = \dfrac{A}{x^2} - \dfrac{B}{x}\), where \(A\) and \(B\) are positive constants. The particle is released from rest at position \(x = \dfrac{A}{B}\). What is the kinetic energy of the particle when it reaches the position of stable equilibrium?

- **A.** \(\dfrac{B^2}{4A}\)
- **B.** \(\dfrac{B^2}{2A}\)
- **C.** \(\dfrac{3B^2}{4A}\)
- **D.** \(\dfrac{B^2}{A}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120632/*