law of conservation

Question

Find the useful power output (in W) of an elevator motor that lifts a 2600 kg load a height of 45.0 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 2600 kg is raised in height, but the full 10,000 kg is accelerated. (need a number.)

(a) 

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

(b)

Answer 1

(a) Useful power P = mgh/t = 2600kg·9.8m/s²·45m/12s = 95550 W

(b) Total power N = P + Mv²/(2t) = 95550W + 10000kg·16m²/s²/(24s) = 102216.67 W

Then Total work A = Nt = 102216.67W·12s = 0.341 kWh; Total cost C = 0.341kWh·$0.0900/kWh = $0.031