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bogdanovich [222]

3 years ago

12

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 incr

eases 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?

1 answer:

Annette [7]3 years ago

4 0

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

$W = dP + dK\\$

Where

$dP$ = Change in Potential Energy

$dK$ = Change in Kinetic Energy

Change in Potential Energy

$P_{i} = mgh_{i}\\ P_{f} = mgh_{f}$

Where

$P_{i}$ = Initial Potential Energy

$P_{f}$ = Final Potential Energy

$m$ = Mass of System = 10,000 kg

$g$ = Acceleration due to gravity = 9.81 m/s

$h_{i}$ = Initial Height = 0

$h_{f}$ = Final Height = 30 m

Inputting the values we get the answer for $dP$

$dP = P_{f} - P_{i}\\dP= mgh_{f} - mgh_{i}\\ dP= 10000(9.81)(30) - 0\\ dP= 2943000$

Change in Kinetic Energy

$K_{i} = \frac{1}{2} mv_{i} ^2\\ K_{f} = \frac{1}{2} mv_{f} ^2$

Where

$K_{i}$ = Initial Kinetic Energy

$K_{f}$ = Final Kinetic Energy

$m$ = Mass of System = 10,000 kg

$g$ = Acceleration due to gravity = 9.81 m/s

$v_{i}$ = Initial Velocity = 0 m/2

$v_{f}$ = Final Velocity = 4 m/s

Inputting the values we get the answer for $dK$

$dK = K_{f} - K_{i}\\ dK = \frac{1}{2} mv_{f} ^2 - \frac{1}{2} mv_{i} ^2\\ dK = \frac{1}{2} (10000)(4)^2 - 0 \\ dK = 80000$

Total Work

$W = dP + dK\\$

Inputting the values

$W = 2943000 + 80000$

$W = 3,023,000$

a) Finding the useful Power Output

$P = \frac{W}{t}$

Where

$P$ = Power Output

$W$ = Work Done = 3,023,000J

$t$ = Time = 12s

Inputting the values

$P = \frac{3,023,000}{12}\\ P = 251,916.667$

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

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