---
title: "A block of mass \\(m\\) is initially at rest at position \\(x = 0\\) on a frictionless horizontal surface. Beginning at time \\(t = 0\\), a time-dependent horizontal force \\(F(t) = F_0 \\sin(\\omega t)\\) is applied to the block in the positive \\(x\\)-direction, where \\(F_0\\) and \\(\\omega\\) are positive constants. Which of the following expressions represents the maximum speed \\(v_{\\text{max}}\\) attained by the block?"
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url: "https://nerd-notes.com/ubq/117573/"
date_modified: "2026-08-04T07:49:10+00:00"
---

# A block of mass \(m\) is initially at rest at position \(x = 0\) on a frictionless horizontal surface. Beginning at time \(t = 0\), a time-dependent horizontal force \(F(t) = F_0 \sin(\omega t)\) is applied to the block in the positive \(x\)-direction, where \(F_0\) and \(\omega\) are positive constants. Which of the following expressions represents the maximum speed \(v_{\text{max}}\) attained by the block?

A block of mass \(m\) is initially at rest at position \(x = 0\) on a frictionless horizontal surface. Beginning at time \(t = 0\), a time-dependent horizontal force \(F(t) = F_0 \sin(\omega t)\) is applied to the block in the positive \(x\)-direction, where \(F_0\) and \(\omega\) are positive constants. Which of the following expressions represents the maximum speed \(v_{\text{max}}\) attained by the block?

![A rectangular block of mass labeled m rests on a flat horizontal line representing a frictionless surface. A single horizontal force arrow labeled F(t) points to the right, attached to the right face of the block. Below the surface, a horizontal x-axis arrow points to the right. No other vectors, labels, or background lines appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-setup-1785829749-wk1kU9.jpg)

- **A.** \(\dfrac{2 F_0}{m\omega}\)
- **B.** \(\dfrac{F_0}{m\omega}\)
- **C.** \(\dfrac{F_0}{2 m\omega}\)
- **D.** \(\dfrac{\pi F_0}{m\omega}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/117573/*