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A-man-pushes-a-box-of-40kg-up-an-incline-of-15-If-the-man-applied-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-




Question Number 25308 by NECx last updated on 08/Dec/17
A man pushes a box of 40kg up an  incline of 15°.If the man applied  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 0.10,  find  a)workdone by the man on the  box  b)workdone against friction.
$${A}\:{man}\:{pushes}\:{a}\:{box}\:{of}\:\mathrm{40}{kg}\:{up}\:{an} \\ $$$${incline}\:{of}\:\mathrm{15}°.{If}\:{the}\:{man}\:{applied} \\ $$$${a}\:{horizontal}\:{force}\:\mathrm{200}{N}\:{and}\:{the} \\ $$$${box}\:{moves}\:{up}\:{the}\:{plane}\:{a}\:{distance} \\ $$$${of}\:\mathrm{20}{m}\:{at}\:{a}\:{constant}\:{velocity}\:{and} \\ $$$${the}\:{coefficient}\:{of}\:{friction}\:{is}\:\mathrm{0}.\mathrm{10}, \\ $$$${find} \\ $$$$\left.{a}\right){workdone}\:{by}\:{the}\:{man}\:{on}\:{the} \\ $$$${box} \\ $$$$\left.{b}\right){workdone}\:{against}\:{friction}. \\ $$
Answered by jota+ last updated on 08/Dec/17
θ=15    F=200 N   s=20 m  m=40 kg   μ=0.10    F_s =Fcos θ−mgsin θ−μN.        N=mgcos θ+Fsin θ  a) W=F_s s  b) W=−μNs
$$\theta=\mathrm{15}\:\:\:\:{F}=\mathrm{200}\:\mathrm{N}\:\:\:{s}=\mathrm{20}\:\mathrm{m} \\ $$$${m}=\mathrm{40}\:{kg}\:\:\:\mu=\mathrm{0}.\mathrm{10} \\ $$$$ \\ $$$${F}_{{s}} ={F}\mathrm{cos}\:\theta−{mg}\mathrm{sin}\:\theta−\mu{N}. \\ $$$$\:\:\:\:\:\:{N}={mg}\mathrm{cos}\:\theta+{F}\mathrm{sin}\:\theta \\ $$$$\left.{a}\right)\:{W}={F}_{{s}} {s} \\ $$$$\left.{b}\right)\:{W}=−\mu{Ns} \\ $$

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