Question

(a) Find the useful power output of an elevator motor that lifts a 2400-kg load a height of 35.0 m in 12.0 s, if it also increases the speed from rest to 4.0 m/s. Note that the total mass of the counterbalanced system is 10000 kg—so that only 2400 kg is raised in height, but the full 10000 kg is accelerated.

(b)What does it cost, if electricity is $0.09 per kW ⋅ h ?

Thank you

Answer 1

(a) Energy provided by motor(E) = Kinetic energy of mass + Potential energy of mass
E = (1/2)Mv² +mgH
m is 2400 kg which is lifted
M = 10,000 kg which is accelerated
v is final velocity = 4 m/s
H is height by which mass m is raised = 35 m
E = (1/2)*10000*4²] +[2400*9.81*35] = 904040 J
Power produced by the elevator = Energy Produced(E) /time
= 904040 / 12 = 75336.67 J/s
(b) Energy produced = 904040 J
Now the cost of electricity = 0.9 /kWh
1kW h = 1000*(3600) J = 36*10⁵ J
$0.09 =36*10⁵ J  
For 1 J = $ 0.09 /(36*10⁵)
For 904040 J = $ 904040 *[0.09 /(36*10⁵)]
= $ 0.226