Question

Find the useful power output (in W) of an elevator motor that lifts a 2500 kg load a height of 30.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10,000 kg—so that only 2500 kg is raised in height, but the full 10,000 kg is accelerated.

watts

What does it cost (in cents), if electricity is $0.0900 per kW · h?

cents

Answer 1

Given Data

mass, m = .2500 kg

speed , v = 4.00 m/s.

time, t = 12.0 s,

height , h =30.0 m

mass , M = 10,000 kg

Work Done, W = ΔKE+ΔPE
= 1/2Mv^2+mgh

                      = 1/2*10000*4^2 + 2500*9.8*30

                      = 815000

P = W/t

   = 815000 / 12

   = 67916.67 W

1 kWh = (1*10^3) * 3600 watt

Hence cost of 67916.67 W will be = { 67916.67 W / [(1*10^3) * 3600 ] } * 9 cents

                                                  = 0.1692