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Question Number 34703 by NECx last updated on 10/May/18

A man pushes a box of 40kg up an  incline of 15°,if the man applies  a horizontal force 200N and the  box moves up the plane a distance  of 20m at a constant velocity and  the coefficient of friction is 0.1,  find   a)the workdone by the man on the  box  b)workdone against friction

$${A}\:{man}\:{pushes}\:{a}\:{box}\:{of}\:\mathrm{40}{kg}\:{up}\:{an} \\ $$$${incline}\:{of}\:\mathrm{15}°,{if}\:{the}\:{man}\:{applies} \\ $$$${a}\:{horizontal}\:{force}\:\mathrm{200}{N}\:{and}\:{the} \\ $$$${box}\:{moves}\:{up}\:{the}\:{plane}\:{a}\:{distance} \\ $$$${of}\:\mathrm{20}{m}\:{at}\:{a}\:{constant}\:{velocity}\:{and} \\ $$$${the}\:{coefficient}\:{of}\:{friction}\:{is}\:\mathrm{0}.\mathrm{1}, \\ $$$${find}\: \\ $$$$\left.{a}\right){the}\:{workdone}\:{by}\:{the}\:{man}\:{on}\:{the} \\ $$$${box} \\ $$$$\left.{b}\right){workdone}\:{against}\:{friction} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 10/May/18

work done by the man is  (mgsinθ+μmgcosθ)×s  =(40×9.8×sin15^o +0.1×40×9.8×cos15^o )×20  b)work done against friction  =μmgcosθ×s

$${work}\:{done}\:{by}\:{the}\:{man}\:{is} \\ $$$$\left({mgsin}\theta+\mu{mgcos}\theta\right)×{s} \\ $$$$=\left(\mathrm{40}×\mathrm{9}.\mathrm{8}×{sin}\mathrm{15}^{{o}} +\mathrm{0}.\mathrm{1}×\mathrm{40}×\mathrm{9}.\mathrm{8}×{cos}\mathrm{15}^{{o}} \right)×\mathrm{20} \\ $$$$\left.{b}\right){work}\:{done}\:{against}\:{friction} \\ $$$$=\mu{mgcos}\theta×{s} \\ $$

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