Physics Energy Question

Question

An elevator has mass of 1000 kg and carries a load of 800 kg. A constant frictional force of 4000N retards its motion upward.

a) Calculate the minimum power delivered by the motor to lift the elevator at a constant speed v=3.0m/s.

b) Determine an expression for the power delivered by the motor when providing an upward acceleration of 2.0m/s².

Answer 1

(a) As speed is constant, net force on the elevator must be zero. Forces downwards are 1000g + 800g because of the mass and 4000 because of friction. Taking g=9.81 m/s/s gives total force downwards = 21658 N

In 1 s elevator moves 3m upwards

Work done = Force x distance = 21658 x 3 = 64974 J

Power = work / time = 64974 / 1 = 64974 W or 64.974 kW

(b) If acceleration upwards is 2 m/s/s, from F=ma net force upwards = 1800 x 2 = 3600 N

So total force upwards = 21658 + 3600 = 25258 N

Work done in 1 s = 25258 x 3 = 75774 J

So power = work / time = 75774 / 1 = 75774 W or 75.774 kW