How much energy is saved per airplane

Click For Summary
SUMMARY

The discussion focuses on calculating energy savings for an airline by removing 100 kg of paint from each airplane. For part (a), the gravitational potential energy saved by not lifting the paint to an altitude of 26,000 m is calculated using the formula Ug = mgh, resulting in a significant reduction in energy expenditure. In part (b), the kinetic energy savings from not accelerating the paint to a cruising speed of 270 m/s is determined using the kinetic energy formula K = 1/2 mv². The total energy savings combines both potential and kinetic energy reductions.

PREREQUISITES
  • Understanding of gravitational potential energy (Ug = mgh)
  • Knowledge of kinetic energy formula (K = 1/2 mv²)
  • Basic principles of physics related to energy conservation
  • Familiarity with units of mass, height, and speed in physics calculations
NEXT STEPS
  • Calculate gravitational potential energy for different altitudes using Ug = mgh
  • Explore kinetic energy calculations for varying masses and speeds using K = 1/2 mv²
  • Research energy efficiency strategies in aviation
  • Investigate the impact of weight reduction on fuel consumption in aircraft
USEFUL FOR

Aerospace engineers, airline executives, physics students, and anyone interested in optimizing fuel efficiency in aviation through energy savings calculations.

cstout
Messages
37
Reaction score
0

Homework Statement



An airline executive decides to economize by reducing the energy, and thus the amount of fuel, required for long distance flights. He orders the ground crew to remove the paint from the outer surface of each plane. The paint removed from a single plane has a mass of approximately 100 kg. (a) If the airplane cruises at an elevation of 26,000 m, how much energy is saved in not having to lift the paint to that altitude? (b) How much energy is saved per airplane by not having to move that amount of paint from rest to a cruising speed of 270 m/s?


Homework Equations



K=1/2mv2

W= Change in Kinetic Energy


The Attempt at a Solution


Part A - I don't know what to do

Part B.
K =(1/2)(100kg)(270)^2 ----but this is wrong
 
Physics news on Phys.org
cstout said:

Homework Statement



An airline executive decides to economize by reducing the energy, and thus the amount of fuel, required for long distance flights. He orders the ground crew to remove the paint from the outer surface of each plane. The paint removed from a single plane has a mass of approximately 100 kg. (a) If the airplane cruises at an elevation of 26,000 m, how much energy is saved in not having to lift the paint to that altitude? (b) How much energy is saved per airplane by not having to move that amount of paint from rest to a cruising speed of 270 m/s?


Homework Equations



K=1/2mv2

W= Change in Kinetic Energy


The Attempt at a Solution


Part A - I don't know what to do

Part B.
K =(1/2)(100kg)(270)^2 ----but this is wrong
I think in A they are asking for the savings in work done against gravity...that is, the savings in Potential Energy (PE). For B, I believe they are looking for the total energy savings with both PE and KE savings summed together.
 
in part a you must determine the amount of energy required to lift 100kg to an altitude of 26,000 m. Remember gravitational potential energy is given by Ug=mgh
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
9
Views
2K