---
title: "A particle moves counterclockwise once around a circular path of radius R centered at (R, 0) in the xy-plane. During this motion, the particle is subjected to a position-dependent force \\(\\vec{F} = -kx\\,\\hat{i}\\), where k is a positive constant. Which of the following correctly describes the net work done on the particle by \\(\\vec{F}\\) over one complete revolution, along with the correct justification?"
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url: "https://nerd-notes.com/ubq/120667/"
date_modified: "2026-08-23T04:42:33+00:00"
---

# A particle moves counterclockwise once around a circular path of radius R centered at (R, 0) in the xy-plane. During this motion, the particle is subjected to a position-dependent force \(\vec{F} = -kx\,\hat{i}\), where k is a positive constant. Which of the following correctly describes the net work done on the particle by \(\vec{F}\) over one complete revolution, along with the correct justification?

A particle moves counterclockwise once around a circular path of radius R centered at (R, 0) in the xy-plane. During this motion, the particle is subjected to a position-dependent force \(\vec{F} = -kx\,\hat{i}\), where k is a positive constant. Which of the following correctly describes the net work done on the particle by \(\vec{F}\) over one complete revolution, along with the correct justification?

![A Cartesian coordinate system with a horizontal x-axis and a vertical y-axis intersecting at the origin labeled O. A circle of radius R is shown in the xy-plane, centered at the point (R, 0) on the positive x-axis and tangent to the y-axis at the origin. An arrowhead on the upper arc of the circle points toward the upper-left, and an arrowhead on the lower arc points toward the lower-right, indicating a counterclockwise trajectory. At a representative point on the upper half of the circle, a horizontal arrow points to the left, labeled \vec{F}. A dashed vertical reference line extends to a tick mark labeled R on the x-axis, and the rightmost point of the circle on the x-axis is labeled 2R. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1787460153-B02hai.jpg)

- **A.** The net work is non-zero because the force always acts antiparallel to the particle's displacement along the path, resulting in continuous mechanical energy dissipation.
- **B.** The net work is zero because \(\vec{F}\) is a conservative force derived from a potential energy function, so the work done over any closed path is zero.
- **C.** The net work is zero because the force is perpendicular to the particle's velocity vector at every point along the circular path.
- **D.** The net work is non-zero because the magnitude of the force increases with distance from the y-axis, preventing the work done on the forward and return segments from cancelling.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120667/*