How much energy is saved per airplane

AI Thread Summary
Removing paint from an airplane can save significant energy by reducing the weight that needs to be lifted and accelerated. For part A, the energy savings from lifting 100 kg to an altitude of 26,000 m can be calculated using the gravitational potential energy formula, Ug = mgh. In part B, the kinetic energy savings from not having to accelerate that weight 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. This approach highlights the importance of weight reduction in improving fuel efficiency for airlines.
cstout
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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
 
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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
 
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