---
title: "A block of mass \\(m\\) is initially at rest on a frictionless horizontal surface. Starting at time \\(t = 0\\), a time-varying force \\(F(t) = F_0 \\sin\\left(\\dfrac{\\pi t}{T}\\right)\\) is exerted on the block in the positive x-direction for the time interval \\(0 \\le t \\le T\\), where \\(F_0\\) and \\(T\\) are positive constants. What is the change in velocity of the block during this time interval?"
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url: "https://nerd-notes.com/ubq/117580/"
date_modified: "2026-08-04T07:49:11+00:00"
---

# A block of mass \(m\) is initially at rest on a frictionless horizontal surface. Starting at time \(t = 0\), a time-varying force \(F(t) = F_0 \sin\left(\dfrac{\pi t}{T}\right)\) is exerted on the block in the positive x-direction for the time interval \(0 \le t \le T\), where \(F_0\) and \(T\) are positive constants. What is the change in velocity of the block during this time interval?

A block of mass \(m\) is initially at rest on a frictionless horizontal surface. Starting at time \(t = 0\), a time-varying force \(F(t) = F_0 \sin\left(\dfrac{\pi t}{T}\right)\) is exerted on the block in the positive x-direction for the time interval \(0 \le t \le T\), where \(F_0\) and \(T\) are positive constants. What is the change in velocity of the block during this time interval?

![A graph of force F versus time t. The horizontal axis is labeled time t, with tick marks at 0, T/2, and T. The vertical axis is labeled force F, with a tick mark at F_0. A smooth, symmetric curve starting at the origin (0, 0), rising to a peak of F_0 at t = T/2, and returning to 0 at t = T, representing one half-cycle of a sine wave labeled F(t) = F_0 sin(pi t / T). No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785829751-qpViOz.jpg)

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

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/117580/*
