Question

An elevator car has a mass of 800 kg and is carrying passengers having a combined mass of 500 kg. As the elevator ascends, it encounters a constant frictional force of 3600 N. (a) What must be the minimum power delivered by the motor to lift the elevator car at a constant speed of 3.00 m/s? (b) What power must the motor deliver at the instant its speed is 3.00 m/s if it is designed to provide an upward acceleration of 1.00 m/s??

Answer 1

a) To move with the constant speed the motor should deliver the force that is equal to the total weight of the elevator with people plus the friction force (in other words, the force that pulls the elevator up should be equal to the forces that pull it down):

$$F_{motor} = F_{friction} + F_{weight}$$

where $F_{friction} = 3600N$, and

$$F_{weight} = mg$$

where $m = 800kg + 500kg = 1300kg$ is the total mass, and $g = 9.8m/s^2$ is the gravitational acceleration.

Then the minimum power will be:

$$P = F_{motor}v =( F_{friction} + mg)v$$

where $v = 3m/s$ is the speed of the lift. Thus, obtain:

$$P = (3600N + 1300kg\cdot 9.8m/s^2)\cdot 3m/s = 49020\space W$$

b) To move with the acceleration, the motor should provide the following force (according to the second Newton's law):

$$F_{motor} = F_{friction} + F_{weight} + ma$$

where $a = 1m/s^2$ is the desired acceleration. Thus, the power will be:

$$P = F_{motor}v =(F_{friction} +mg + ma)v = (F_{friction} +m(g + a))v \\ P = (3600N + 1300kg\cdot (9.8m/s^2 + 1 m/s^2))\cdot 3m/s = 52920\space W$$

Answer. a) 4920 W, b) 52920 W.