Question

(a) Find the useful power output (in W) of an elevator motor that lifts a 2700 kg load a height of 25.0 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 2700 kg is raised in height, but the full 10,000 kg is accelerated.

W

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

cents

Answer 1

Power = P = ?

mass = m = 2700 kg

height h = 25 m

time = t= 12sec

initial speed = 0

final speed = 4 m/s

total mass = M =10000 kg

here are two components of power.

1st power required to raise . P1 = mgh/t = 2700*9.8*25 /12 = 55125 W

2nd power required for acceleration. =M*a*h/t = M*(v-u)/t * h/t

here a = acceleration

sp P2= 6944.44 W

so total power will be = 62069.44 W = 6.20 * 10⁴W = 62 kW

b) 12 sec = 0.0033 hrs

total cost = $0.0900 per kW · h * 62 * 0.0033 = 1.84 cents