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A block of mass 20 kg is moved with constant velocity along an inclined plane of inclination 37° with help of a force of constant power 50 W. If the coefficient of kinetic friction between block and surface is 0.25, then what fraction of power is used against gravity?

Fatima , 7 Years ago

Grade 12th pass

2 Answers

Eshan

From equilibrium of forces along the incline,
$mgsin\theta+\mu mg cos\theta=F$
$\implies F=160N$
Power= $Fv$ =constant= $50W$
$\implies v=\dfrac{5}{16}m/s$
Power used against gravity= $\vec{F_g}.\vec{v}=-mgsin\theta v=37.5W$
Hence fraction of power used= $75%$ %

Last Activity: 7 Years ago

Gitanjali Rout

From equilibrium of forces along the incline,
mgsin\theta+\mu mg cos\theta=F
\implies F=160N
Power=Fv=constant=50W
\implies v=\dfrac{5}{16}m/s
Power used against gravity=\vec{F_g}.\vec{v}=-mgsin\theta v=37.5W
Hence fraction of power used=75%%

Last Activity: 7 Years ago

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