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A-man-pushes-a-box-of-40kg-up-an-incline-plane-of-15-if-the-man-applies-a-horizontal-force-of-200N-and-the-box-moves-up-the-plane-a-distance-of-20m-at-a-constant-velocity-and-the-coefficient-of-frict




Question Number 28644 by NECx last updated on 28/Jan/18
A man pushes a box of 40kg up  an incline plane of 15°,if the man  applies a horizontal force of  200N and the box moves up the  plane a distance of 20m at a  constant velocity and the  coefficient of friction is 0.10,  find   a)workdone by the man on the  box.  b)workdone against friction.
$${A}\:{man}\:{pushes}\:{a}\:{box}\:{of}\:\mathrm{40}{kg}\:{up} \\ $$$${an}\:{incline}\:{plane}\:{of}\:\mathrm{15}°,{if}\:{the}\:{man} \\ $$$${applies}\:{a}\:{horizontal}\:{force}\:{of} \\ $$$$\mathrm{200}{N}\:{and}\:{the}\:{box}\:{moves}\:{up}\:{the} \\ $$$${plane}\:{a}\:{distance}\:{of}\:\mathrm{20}{m}\:{at}\:{a} \\ $$$${constant}\:{velocity}\:{and}\:{the} \\ $$$${coefficient}\:{of}\:{friction}\:{is}\:\mathrm{0}.\mathrm{10}, \\ $$$${find}\: \\ $$$$\left.{a}\right){workdone}\:{by}\:{the}\:{man}\:{on}\:{the} \\ $$$${box}. \\ $$$$\left.{b}\right){workdone}\:{against}\:{friction}. \\ $$
Commented by NECx last updated on 29/Jan/18
please help
$${please}\:{help} \\ $$
Commented by Tinkutara last updated on 29/Jan/18
Work done by man on box=Fcos θ×20  =200cos 15°×20=1000(√2)((√3)+1)  f=μN=μ(Fsin θ+mgcos θ)  =0.1(200sin 15°+40gcos 15°)
$${Work}\:{done}\:{by}\:{man}\:{on}\:{box}={F}\mathrm{cos}\:\theta×\mathrm{20} \\ $$$$=\mathrm{200cos}\:\mathrm{15}°×\mathrm{20}=\mathrm{1000}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right) \\ $$$${f}=\mu{N}=\mu\left({F}\mathrm{sin}\:\theta+{mg}\mathrm{cos}\:\theta\right) \\ $$$$=\mathrm{0}.\mathrm{1}\left(\mathrm{200sin}\:\mathrm{15}°+\mathrm{40}{g}\mathrm{cos}\:\mathrm{15}°\right) \\ $$
Commented by Tinkutara last updated on 29/Jan/18
Commented by NECx last updated on 30/Jan/18
thank you so much
$${thank}\:{you}\:{so}\:{much} \\ $$

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