Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 18581 by Tinkutara last updated on 25/Jul/17

In moving a body of mass m up and down a rough incline plane of inclination θ, work done is (S is length of the planck, and μ is coefficient of friction).

$$\mathrm{In}\:\mathrm{moving}\:\mathrm{a}\:\mathrm{body}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{up}\:\mathrm{and} \\ $$$$\mathrm{down}\:\mathrm{a}\:\mathrm{rough}\:\mathrm{incline}\:\mathrm{plane}\:\mathrm{of}\:\mathrm{inclination} \\ $$$$\theta,\:\mathrm{work}\:\mathrm{done}\:\mathrm{is}\:\left(\mathrm{S}\:\mathrm{is}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{planck},\right. \\ $$$$\left.\mathrm{and}\:\mu\:\mathrm{is}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{friction}\right). \\ $$

Answered by ajfour last updated on 25/Jul/17

W=2μmgScos θ

$$\mathrm{W}=\mathrm{2}\mu\mathrm{mgScos}\:\theta \\ $$

Commented by Arnab Maiti last updated on 25/Jul/17

Force of friction =μmgcosθ Work done to take the body up against force of friction is =μmgcosθ×S Again when it is taken to down again work done against force and friction=μmgScosθ So total work done against force of friction, w_1 =2μmgScosθ The work done against gravitational force =0 , as final height= initial height The work done for change of position is olso 0, as final position =initial position. Hence net work done, W=w_1 ∴ W=2μmgScosθ

$$\mathrm{Force}\:\mathrm{of}\:\mathrm{friction}\:=\mu\mathrm{mgcos}\theta \\ $$$$\mathrm{Work}\:\mathrm{done}\:\mathrm{to}\:\mathrm{take}\:\mathrm{the}\:\mathrm{body} \\ $$$$\mathrm{up}\:\mathrm{against}\:\mathrm{force}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{is} \\ $$$$=\mu\mathrm{mgcos}\theta×\mathrm{S} \\ $$$$\mathrm{Again}\:\mathrm{when}\:\mathrm{it}\:\mathrm{is}\:\mathrm{taken}\:\mathrm{to}\:\mathrm{down} \\ $$$$\mathrm{again}\:\mathrm{work}\:\mathrm{done}\:\mathrm{against}\:\mathrm{force} \\ $$$$\mathrm{and}\:\mathrm{friction}=\mu\mathrm{mgScos}\theta \\ $$$$\mathrm{So}\:\mathrm{total}\:\mathrm{work}\:\mathrm{done}\:\mathrm{against} \\ $$$$\mathrm{force}\:\mathrm{of}\:\mathrm{friction},\:\mathrm{w}_{\mathrm{1}} =\mathrm{2}\mu\mathrm{mgScos}\theta \\ $$$$\mathrm{The}\:\mathrm{work}\:\mathrm{done}\:\mathrm{against}\:\mathrm{gravitational} \\ $$$$\mathrm{force}\:=\mathrm{0}\:,\:\mathrm{as}\:\mathrm{final}\:\mathrm{height}=\:\mathrm{initial} \\ $$$$\mathrm{height} \\ $$$$\mathrm{The}\:\mathrm{work}\:\mathrm{done}\:\mathrm{for}\:\mathrm{change}\:\mathrm{of} \\ $$$$\mathrm{position}\:\mathrm{is}\:\mathrm{olso}\:\mathrm{0},\:\mathrm{as}\:\mathrm{final}\:\mathrm{position} \\ $$$$=\mathrm{initial}\:\mathrm{position}. \\ $$$$\mathrm{Hence}\:\mathrm{net}\:\mathrm{work}\:\mathrm{done},\:\mathrm{W}=\mathrm{w}_{\mathrm{1}} \\ $$$$\therefore\:\mathrm{W}=\mathrm{2}\mu\mathrm{mgScos}\theta \\ $$

Commented by Tinkutara last updated on 25/Jul/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com