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 minimumpowerdeliveredby the motor to theelevator in watss as well as in horse power .

An elevator can carry a maximum load of 1800 kg (elevator +passengers

1.	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 minimumpowerdeliveredby the motor to theelevator in watss as well as in horse power .
Answer» Solution :The downward force on the elevatoris , `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 .