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