Conservation of Energy airplane pilot

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Homework Help Overview

The discussion revolves around the conservation of energy principles applied to a scenario involving an airplane pilot who fell from a height of 400 m and landed in a snowbank. Participants are exploring the work done by various forces during the fall and subsequent stop, including gravity, air resistance, and the snow's impact.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to estimate the work done by the snow in bringing the pilot to rest and are questioning whether this is solely the work done by gravity or includes other factors like air resistance. There are discussions about using potential and kinetic energy to calculate work done and the average force exerted by the snow.

Discussion Status

Several participants have offered insights into the relationships between work, energy, and forces involved in the scenario. There is an ongoing exploration of how to calculate the work done by the snow and air resistance, with some participants suggesting that the work done by air resistance can be derived from the difference between potential energy and kinetic energy at terminal velocity. Multiple interpretations of the equations and terms involved are being discussed.

Contextual Notes

Participants are working under the assumption that the pilot's mass and terminal velocity are known, and there is a focus on the energy transformations occurring during the fall and landing. The depth of the crater and the height from which the pilot fell are also points of consideration in the calculations.

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An airplane pilot fell 400 m after jumping without his parachute opening. He landed in a snowbank, creating a crater 2 m deep, but survived with only minor injuries. Assume that the pilot's mass was 50 kg and his terminal velocity was 65 m/s.

(a) Estimate the work done by the snow in bringing him to rest.

Is it just work done by gravity, which is mgh, h being 402 m?

(b) Estimate the average force exerted on him by the snow to stop him.

No idea...

(c) Estimate the work done on him by air resistance as he fell.

No clue...
 
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Work-energy theorem: The net work done on an object is equal to the change in kinetic energy of the object.

Assume that the snow exerted the only force on the pilot as he came to a stop. Since he was at a terminal velocity, there's no need to use kinematics.

b)
Once you find the work done by the snow, use the standard equation for work to find the force

c)
at terminal velocity, air resistance equals weight.
 
physicsss said:
An airplane pilot fell 400 m after jumping without his parachute opening. He landed in a snowbank, creating a crater 2 m deep, but survived with only minor injuries. Assume that the pilot's mass was 50 kg and his terminal velocity was 65 m/s.

(a) Estimate the work done by the snow in bringing him to rest.


Is it just work done by gravity, which is mgh, h being 402 m?

No, there is also work done by air resistance- because he has a maximum velocity of 65 m/s. That's the speed with which he hits the snow so the work the snow does is equal to his kinetic energy at that speed.

(b) Estimate the average force exerted on him by the snow to stop him.

No idea...

You know the work done by the snow from (a) and you know that work is "force times distance"...

(c) Estimate the work done on him by air resistance as he fell.

No clue...

NOW use your idea from (a). His potential energy is what you said there- the total work done to bring him to a stop is equal to that. You calculated the work done by the snow in (a). The work done by the air resistance is the difference between his initial potential energy and the work done by the snow.

(Rember than the work done by the snow was the same as his kinetic energy at his terminal speed. That is, the work done by the air is the initial potential energy less the kinetic energy at terminal speed.)
 
Is a) W=mgh+1/2*mv^2 , where h = height of crater and v = terminal velocity?
 
physicsss said:
Is a) W=mgh+1/2*mv^2 , where h = height of crater and v = terminal velocity?
Yes, if by h you mean the depth of the crater (2 m).
 
For part c, is it mgh, where h is 400m, minus what I got for a) ??
 
physicsss said:
For part c, is it mgh, where h is 400m, minus what I got for a) ??
Almost. When it just hits the ground it has a KE that is less than its original PE (measured from the ground). The "missing" energy is that lost to air resistance.
 
mgh, where h is 402m, minus what I got for a) ??
 
wait... is Work done by friction=mgh-1/2*mv^2-mgd, where h is 400m and v is 65m/s, and d=2.0m?
 
Last edited:
  • #10
physicsss said:
mgh, where h is 402m, minus what I got for a) ??
That would work.
 
  • #11
physicsss said:
wait... is Work done by friction=mgh-1/2*mv^2-mgd, where h is 400m and v is 65m/s, and d=2.0m?
I would drop the mgd term. Assume that air resistance just acts while the object is falling, but not after it hits the snow.
 

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