Average Power Output for 260000 kg Jet at 9200m Altitude

Click For Summary
The discussion centers on calculating the average power output of a 260,000 kg jet after it reaches a cruising altitude of 9,200 meters and a speed of 235 m/s. The initial kinetic energy is zero, and the final kinetic energy is calculated as 7,179,250,000 Joules. However, the calculation neglects the change in potential energy, which is crucial for determining the total work done. Participants suggest incorporating both the change in kinetic energy and potential energy into the work equation to find the correct power output. The conversation emphasizes the importance of considering all forms of energy in such calculations.
spockjones20
Messages
23
Reaction score
1

Homework Statement



620 seconds after starting its take-off, a 260000 kg jet reaches its cruising altitude of 9200 meters and flies with a speed of 235 m/s. Neglecting friction, what has been the average power output of the engines?

Homework Equations



ΔKE = Work
Power = Work/time

The Attempt at a Solution



Initial KE = 0
Final KE = .5*260000*235^2 = 7179250000
Power = 7179250000/620 = 11579435.48 Watts

Can someone see an issue with what I am trying to do here? I tried to type in this answer but apparently it is wrong.
 
Physics news on Phys.org
spockjones20 said:

Homework Statement



620 seconds after starting its take-off, a 260000 kg jet reaches its cruising altitude of 9200 meters and flies with a speed of 235 m/s. Neglecting friction, what has been the average power output of the engines?

Homework Equations



ΔKE = Work
Power = Work/time

The Attempt at a Solution



Initial KE = 0
Final KE = .5*260000*235^2 = 7179250000
Power = 7179250000/620 = 11579435.48 Watts

Can someone see an issue with what I am trying to do here? I tried to type in this answer but apparently it is wrong.

I haven't checked your math, but I can see you left out the change in potential energy.
 
Ok. Would that go into the work equation?
 
spockjones20 said:
Ok. Would that go into the work equation?

I believe so.

Try this: ΔKE + ΔPE = Work

and see what happens.
 

Thread 'Magnitude of buoyant force in fluids of different densities'

The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?
View full post »

Similar threads

How many crates can you load in a minute with different average power outputs?
  • · Replies 1 ·
Replies
1
Views
2K
Calculate Avg Power Output for 75kg Person Walking 100ft in 3min
  • · Replies 9 ·
Replies
9
Views
2K
Power output of a hydroelectric plant
  • · Replies 4 ·
Replies
4
Views
3K
Power output by basketball player when jumping
  • · Replies 7 ·
Replies
7
Views
3K
How Is the Average Power Output Calculated for an Airplane Propeller?
  • · Replies 1 ·
Replies
1
Views
1K
Determining power output at different time intervals
  • · Replies 4 ·
Replies
4
Views
2K
Calculating Pump Power and Fountain Height from Water Ejection Speed
  • · Replies 20 ·
Replies
20
Views
4K
What is the Average Power Output of a Human Heart Pump?
  • · Replies 4 ·
Replies
4
Views
4K
Why Doesn't My Power Calculation Match Real-World Scenarios?
  • · Replies 4 ·
Replies
4
Views
3K
What is the average power of a boy
  • · Replies 3 ·
Replies
3
Views
1K