Find the useful power output (in w) of an elevator motor that lifts a 2600 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 2600 kg is raised in height, but the full 10,000 kg is accelerated. what does it cost (in cents), if electricity is $0.0900 per kw · h?

P = 251,916.667 W

Cost = 2,267.25 cents

Explanation:

To solve this question we will use the Work Energy Theorem, which is

Where

= Change in Potential Energy

= Change in Kinetic Energy

Change in Potential Energy

Where

= Initial Potential Energy

= Final Potential Energy

= Mass of System = 10,000 kg

= Acceleration due to gravity = 9.81 m/s

= Initial Height = 0

= Final Height = 30 m

Inputting the values we get the answer for

Change in Kinetic Energy

Where

= Initial Kinetic Energy

= Final Kinetic Energy

= Mass of System = 10,000 kg

= Acceleration due to gravity = 9.81 m/s

= Initial Velocity = 0 m/2

= Final Velocity = 4 m/s

Inputting the values we get the answer for

Total Work

Inputting the values

a) Finding the useful Power Output

Where

= Power Output

= Work Done = 3,023,000J

= Time = 12s

Inputting the values

P = 251,916.667 W

b) Finding the Total Cost

Cost = $0.0900 x P/1000

Cost = $0.0900 x (251,916.667/1000)

Cost = $22.67 or 2,267.25 cents

d.7.5cm is your answer