Answer:
P = 251,916.667 W
Cost = 2,267.25 cents
Explanation:
To solve this question we will use the Work Energy Theorem, which is
[tex]W = dP + dK\\[/tex]
Where
[tex]dP[/tex] = Change in Potential Energy
[tex]dK[/tex] = Change in Kinetic Energy
Change in Potential Energy
[tex]P_{i} = mgh_{i}\\ P_{f} = mgh_{f}[/tex]
Where
[tex]P_{i}[/tex] = Initial Potential Energy
[tex]P_{f}[/tex] = Final Potential Energy
[tex]m[/tex] = Mass of System = 10,000 kg
[tex]g[/tex] = Acceleration due to gravity = 9.81 m/s
[tex]h_{i}[/tex] = Initial Height = 0
[tex]h_{f}[/tex] = Final Height = 30 m
Inputting the values we get the answer for [tex]dP[/tex]
[tex]dP = P_{f} - P_{i}\\dP= mgh_{f} - mgh_{i}\\ dP= 10000(9.81)(30) - 0\\ dP= 2943000[/tex]
Change in Kinetic Energy
[tex]K_{i} = \frac{1}{2} mv_{i} ^2\\ K_{f} = \frac{1}{2} mv_{f} ^2[/tex]
Where
[tex]K_{i}[/tex] = Initial Kinetic Energy
[tex]K_{f}[/tex] = Final Kinetic Energy
[tex]m[/tex] = Mass of System = 10,000 kg
[tex]g[/tex] = Acceleration due to gravity = 9.81 m/s
[tex]v_{i}[/tex] = Initial Velocity = 0 m/2
[tex]v_{f}[/tex] = Final Velocity = 4 m/s
Inputting the values we get the answer for [tex]dK[/tex]
[tex]dK = K_{f} - K_{i}\\ dK = \frac{1}{2} mv_{f} ^2 - \frac{1}{2} mv_{i} ^2\\ dK = \frac{1}{2} (10000)(4)^2 - 0 \\ dK = 80000[/tex]
Total Work
[tex]W = dP + dK\\[/tex]
Inputting the values
[tex]W = 2943000 + 80000[/tex]
[tex]W = 3,023,000[/tex]
a) Finding the useful Power Output
[tex]P = \frac{W}{t}[/tex]
Where
[tex]P[/tex] = Power Output
[tex]W[/tex] = Work Done = 3,023,000J
[tex]t[/tex] = Time = 12s
Inputting the values
[tex]P = \frac{3,023,000}{12}\\ P = 251,916.667[/tex]
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
