/*! 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. 46 A particle of mass m moving alon... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A particle of mass m moving along the x-axis has velocity vx=v0sin(Ï€³æ/2L). How much work is done on the particle as it moves (a) from x=0to x=Land

(b) from x=0tox=2L?

Short Answer

Expert verified

a). mv024

b).-mv024

Step by step solution

01

Step 1. Given Information

Mass of particle=m

Velocity of the particle vx=vosin(Ï€³æ/2L)

02

Step 2. Part a). Work done on the particle

When it moves from x=0to x=L. For that the we will proceed as

W=∫abFdxW=∫abmadxax=vxdvxdxdvxdx=v0Ï€2LcosÏ€³æ2L

Now,

W=∫abmadx=∫0Lmv02Ï€2LsinÏ€³æ2LcosÏ€³æ2Ldx=mv02Ï€4L∫0L2sinÏ€³æ2LcosÏ€³æ2Ldx=mv02Ï€4L∫0LsinÏ€³æLdx

Using identity: sin2θ=2sinθcosθ

=-mv0214cosÏ€³¢L-cos0=mv024

03

Step 3. Part b). Work done on the particle

We need to determine work done on the particle when it formx=0to x=2L

Now,

W=∫abmadx=∫02Lmv02Ï€2LsinÏ€³æ2LcosÏ€³æ2Ldx=mv02Ï€4L∫02L2sinÏ€³æ2LcosÏ€³æ2Ldx=mv02Ï€4L∫0LsinÏ€³æLdx=-mv02cosÏ€³æL2L=-mv024cos2Ï€³¢L-cos0=-mv024

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.