/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 24 A 150 g particle at x = 0 is m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 150 g particle at x=0 is moving at 2.00 m/s in the +x-direction. As it moves, it experiences a force given by Fx=(0.250N)sin(x/2.00m). What is the particle’s speed when it reaches x=3.14 m?

Short Answer

Expert verified

Velocity of the particle at given xis 3.26m/s

Step by step solution

01

Step 1. Given Information

Mass of pariclem=150g=0.15kg

Initial Velocity vi=2m/s

Initial position xi=0m

Final position xf=3.14m

Variable force actingFx=(0.250N)sin(x/2m)

Final velocityvf=?

02

Step 2. Find the velocity of a particle

We have given variable force acting on a particle and we need to find the velocity at given x. We can do so by calculating work done by the force for moving particles from x=0to x=3.14m and then by equating total work done as a change in kinetic energy

W=∫xixfFxdx=∫03.140.250sinx2dx=-0.50cosx2=0.5J

Now,

W=∆KW=12mvf2-12mvi212mvf2=W+12mvi2=0.5+12×0.15×22=0.8vf2=2×0.80.15=10.67vf=3.26m/s

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.