Question Number 25308 by NECx last updated on 08/Dec/17
$${A}\:{man}\:{pushes}\:{a}\:{box}\:{of}\:\mathrm{40}{kg}\:{up}\:{an} \\ $$$${incline}\:{of}\:\mathrm{15}°.{If}\:{the}\:{man}\:{applied} \\ $$$${a}\:{horizontal}\:{force}\:\mathrm{200}{N}\:{and}\:{the} \\ $$$${box}\:{moves}\:{up}\:{the}\:{plane}\:{a}\:{distance} \\ $$$${of}\:\mathrm{20}{m}\:{at}\:{a}\:{constant}\:{velocity}\:{and} \\ $$$${the}\:{coefficient}\:{of}\:{friction}\:{is}\:\mathrm{0}.\mathrm{10}, \\ $$$${find} \\ $$$$\left.{a}\right){workdone}\:{by}\:{the}\:{man}\:{on}\:{the} \\ $$$${box} \\ $$$$\left.{b}\right){workdone}\:{against}\:{friction}. \\ $$
Answered by jota+ last updated on 08/Dec/17
$$\theta=\mathrm{15}\:\:\:\:{F}=\mathrm{200}\:\mathrm{N}\:\:\:{s}=\mathrm{20}\:\mathrm{m} \\ $$$${m}=\mathrm{40}\:{kg}\:\:\:\mu=\mathrm{0}.\mathrm{10} \\ $$$$ \\ $$$${F}_{{s}} ={F}\mathrm{cos}\:\theta−{mg}\mathrm{sin}\:\theta−\mu{N}. \\ $$$$\:\:\:\:\:\:{N}={mg}\mathrm{cos}\:\theta+{F}\mathrm{sin}\:\theta \\ $$$$\left.{a}\right)\:{W}={F}_{{s}} {s} \\ $$$$\left.{b}\right)\:{W}=−\mu{Ns} \\ $$