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In-moving-a-body-of-mass-m-up-and-down-a-rough-incline-plane-of-inclination-work-done-is-S-is-length-of-the-planck-and-is-coefficient-of-friction-

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

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