A 650-kilogram elevator starts from rest. It moves upwards for 3 seconds with constant acceleration until it reaches its cruising speed of 1.75 m/s. During this three-second interval, the elevator motor must provide power to move the elevator. Calculate the average power provided by the motor during this time interval? Show your work. *Hint*: Two forces act on the elevator. One is weight pointing downwards, and the other is the force of motor pointing upwards. Calculate work done by the motor.

The elevator, powered by a motor, is pulled upwards with a constant acceleration, starting from rest. The elevator accelerates for a period of 3 seconds, until it reaches a constant velocity of 1.75 m/s. The mass of this elevator is 650 kg, so the gravitational force acting on the elevator downwards is, Fg=mgg=9.8 m/s2Fg=650×9.8Fg=6370 N The motor has to do some work against this weight of the elevator, in order to pull it upwards. So the motor also pulls on the elevator with a certain applied force, upwards. Thus, there will be a net force acting on the elevator along its direction of motion, due to the combination of these two forces, given as Fnet=F-Fg The negative sign indicates that the two forces act opposite to each otherThe net force acting on the elevator is given as Fnet=maa is the acceleration of the elevator This constant acceleration of the elevator can be found as a=vf-vi∆tvf is the final velocity after time ∆tvi is the initial velocity Since the elevator starts from rest, vi=0 Thus, the acceleration of the elevator is a=vf∆ta=1.753a=0.58 m/s2 Thus, the net force acting on the elevator is Fnet=maFnet=650×0.58Fnet=379.167 NFrom this net force, the force applied by the elevator can be found as Fnet=F-FgF=Fnet+FgF=379.167+6370F=6749.167 N This is the force applied by the elevator, which is only applied for a duration of 3 seconds. In this time of 3 seconds, the elevator moves with a constant acceleration of 0.58. So the distance moved by the elevator in these 3 seconds can be found using kinematical equations as d=vit+12at2t=3 svi=0a=0.58 m/s2d=0+12×0.58×32d=2.61 m The work done by the motor in moving the elevator to this distance is given as W=FdW=6749.167×2.61W=17615.326 JFrom the above found work, the power delivered by the motor can be found. Power is defined as the work done per unit time. Thus, Power, P=WtP=17615.3263P=5871.7753 Watt This is the power provided by the motorAnswer: Power delivered by the motor is 5871.7753 W

Answered: A 650-kilogram elevator starts from…