An elevator can carry a maximum load of 1800 kg (elevator +passengers ) is moving up with a constant speed of `2ms^(-1)` .The frictional force opposing the motion is 4000 N . Determine the minimum power delivered by the motor to the elevator in watss as well as in horse power .

Explanation

The downward force on the elevator is ,
`F = mg +f " where " f = ` friction force
` = 1800 xx10+ 4000`
` = 18000 +4000`
` = 22000 N `
The motor supply enough power to balance this force .
Hence Power `P = Fv`
` = 22000 xx2 `
` = 44000 W `
` = (44000)/(746) hp `
`= 58.98 ` hp
` :. P approx = 59 ` hp .