---
title: "A block of mass \\(m\\) is released from rest at the top of a ramp of length \\(L\\) inclined at an angle \\(\\theta\\) above the horizontal. As the block slides down the ramp, a position-dependent retarding force of magnitude \\(f(x) = bx\\) acts on it, where \\(x\\) is the distance traveled along the incline from the top and \\(b\\) is a positive constant. What is the speed of the block when it reaches the bottom of the ramp?"
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url: "https://nerd-notes.com/ubq/117547/"
date_modified: "2026-08-04T07:48:59+00:00"
---

# A block of mass \(m\) is released from rest at the top of a ramp of length \(L\) inclined at an angle \(\theta\) above the horizontal. As the block slides down the ramp, a position-dependent retarding force of magnitude \(f(x) = bx\) acts on it, where \(x\) is the distance traveled along the incline from the top and \(b\) is a positive constant. What is the speed of the block when it reaches the bottom of the ramp?

A block of mass \(m\) is released from rest at the top of a ramp of length \(L\) inclined at an angle \(\theta\) above the horizontal. As the block slides down the ramp, a position-dependent retarding force of magnitude \(f(x) = bx\) acts on it, where \(x\) is the distance traveled along the incline from the top and \(b\) is a positive constant. What is the speed of the block when it reaches the bottom of the ramp?

- **A.** \(\sqrt{2gL\sin\theta - \dfrac{2bL^2}{m}}\)
- **B.** \(\sqrt{2gL\sin\theta + \dfrac{bL^2}{m}}\)
- **C.** \(\sqrt{2gL\sin\theta - \dfrac{bL^2}{2m}}\)
- **D.** \(\sqrt{2gL\sin\theta - \dfrac{bL^2}{m}}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/117547/*