Question

What is the power required of a dc motor to lift a load of 3 kg accross an inclined plane of 10 degrees from the horizontal plane? (The inclined plane is frictionless).

Answer 1

The power required depends on the speed of lifting.

Equation of power is given by,

$P=F×v$

Where $P$ is the power in $W$, $F$ is the force in $N$ and $v$ is the velocity in $ms^{-1}$

The force required for lifting a body up a frictionless inclined plane of $\theta$ degrees from the horizontal plane is given by,

$F=mgsin\theta\\ F=3kg×9.81ms^{-2}×sin(10\degree)\\ F=3kg×9.81ms^{-2}×0.1736\\ F=5.11N$

The power required is therefore,

$P=(5.11×v)W$

Assuming a lifting speed of $0.1ms^{-2},$

The power required is,

$P=(5.11×0.1)W\\ P=0.51W$