---
title: "A particle of mass \\(m\\) moves along the positive \\(x\\)-axis under the influence of a single force given by \\(F(x) = \\dfrac{\\beta}{x}\\), where \\(\\beta\\) is a positive constant and \\(x > 0\\). The particle moves from position \\(x = x_0\\) to \\(x = 2x_0\\), and subsequently from \\(x = 2x_0\\) to \\(x = 4x_0\\). What is the ratio \\(\\dfrac{\\Delta K_2}{\\Delta K_1}\\), where \\(\\Delta K_1\\) is the change in the particle’s kinetic energy over the first displacement and \\(\\Delta K_2\\) is the change in its kinetic energy over the second displacement?"
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url: "https://nerd-notes.com/ubq/123974/"
date_modified: "2026-09-28T13:29:48+00:00"
---

# A particle of mass \(m\) moves along the positive \(x\)-axis under the influence of a single force given by \(F(x) = \dfrac{\beta}{x}\), where \(\beta\) is a positive constant and \(x > 0\). The particle moves from position \(x = x_0\) to \(x = 2x_0\), and subsequently from \(x = 2x_0\) to \(x = 4x_0\). What is the ratio \(\dfrac{\Delta K_2}{\Delta K_1}\), where \(\Delta K_1\) is the change in the particle’s kinetic energy over the first displacement and \(\Delta K_2\) is the change in its kinetic energy over the second displacement?

A particle of mass \(m\) moves along the positive \(x\)-axis under the influence of a single force given by \(F(x) = \dfrac{\beta}{x}\), where \(\beta\) is a positive constant and \(x > 0\). The particle moves from position \(x = x_0\) to \(x = 2x_0\), and subsequently from \(x = 2x_0\) to \(x = 4x_0\). What is the ratio \(\dfrac{\Delta K_2}{\Delta K_1}\), where \(\Delta K_1\) is the change in the particle's kinetic energy over the first displacement and \(\Delta K_2\) is the change in its kinetic energy over the second displacement?

- **A.** \(\dfrac{1}{4}\)
- **B.** \(\dfrac{1}{2}\)
- **C.** \(1\)
- **D.** \(2\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/123974/*