Question

An elevator can carry a maximum load of 1800 kg (elevator + passengers ) is moving up with a constant speed of 2 m/s. The frictional force opposing the motion is 400 N. Determine the minimum power delivered by

Answer 1

Given that,

Mass of elevator + passengers, $M=1800Kg$

Speed, $v=2m/s$

Friction force, $f=400N$

Power delivered, $P=force\times velocity.........\left(1\right)$

Force acting downward,

$F=mg+f$

$F=1800\times 10+4000$

$F=22000N$

Put the value of F in equation (1)

$P=22000\times 2$

$P=44000Watts$