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

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

Answered by ajfour last updated on 25/Jul/17

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

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

$$\mathrm{Force}\mathrm{of}\mathrm{friction}=\mu \mathrm{mgcos}\theta$$$$\mathrm{Work}\mathrm{done}\mathrm{to}\mathrm{take}\mathrm{the}\mathrm{body}$$$$\mathrm{up}\mathrm{against}\mathrm{force}\mathrm{of}\mathrm{friction}\mathrm{is}$$$$=\mu \mathrm{mgcos}\theta \times \mathrm{S}$$$$\mathrm{Again}\mathrm{when}\mathrm{it}\mathrm{is}\mathrm{taken}\mathrm{to}\mathrm{down}$$$$\mathrm{again}\mathrm{work}\mathrm{done}\mathrm{against}\mathrm{force}$$$$\mathrm{and}\mathrm{friction}=\mu \mathrm{mgScos}\theta$$$$\mathrm{So}\mathrm{total}\mathrm{work}\mathrm{done}\mathrm{against}$$$$\mathrm{force}\mathrm{of}\mathrm{friction},{\mathrm{w}}_1=2\mu \mathrm{mgScos}\theta$$$$\mathrm{The}\mathrm{work}\mathrm{done}\mathrm{against}\mathrm{gravitational}$$$$\mathrm{force}=0,\mathrm{as}\mathrm{final}\mathrm{height}=\mathrm{initial}$$$$\mathrm{height}$$$$\mathrm{The}\mathrm{work}\mathrm{done}\mathrm{for}\mathrm{change}\mathrm{of}$$$$\mathrm{position}\mathrm{is}\mathrm{olso}0,\mathrm{as}\mathrm{final}\mathrm{position}$$$$=\mathrm{initial}\mathrm{position}.$$$$\mathrm{Hence}\mathrm{net}\mathrm{work}\mathrm{done},\mathrm{W}={\mathrm{w}}_1$$$$\therefore \mathrm{W}=2\mu \mathrm{mgScos}\theta$$

Commented by Tinkutara last updated on 25/Jul/17

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

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