How Is the Potential Energy of a Skydiver-Earth System Reducing?

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The potential energy of a skydiver-Earth system reduces as the skydiver falls, with the rate of change calculated using the formula d(PE)/dt = mg(dh/dt). Given a 60 kg skydiver at a terminal speed of 50 m/s, the potential energy decreases at a rate of -29,430 Joules per second. This energy loss occurs because the height (h) of the skydiver decreases as they fall. The term "Earth system reducing" refers to the combined potential energy of the skydiver and the Earth, emphasizing their relationship rather than treating them as separate entities. Understanding this relationship clarifies the calculation of potential energy changes during free fall.
Crashdowngurl
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A 60 Kg skydiver is falling at a constant terminal speed of 50 m/s. At what rate is the potential energy of the skydiver - Earth system reducing? What happens to this energy?
Potential energy = m x g x h
Where
m = mass of the body
g = acceleration due to gravity
h = height of the object





I don't understand what it's asking in this problem and in all the reading I've done there has been some sort of height or distance measurement involved for this type of question dealing with potential energy. The Earth system reducing is throwing me and I am not sure at all where to begin. I do know that formula is this though:


Potential energy = m x g x h
Where
m = mass of the body
g = acceleration due to gravity
h = height of the object
 
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Take the derivative of the potential energy wrt time.

Does that ring any bells?
 
Potential energy = m x g x h
Where
m = mass of the body
g = acceleration due to gravity
h = height of the object

Yes, that's true. And you are asked "At what rate is the potential energy of the skydiver - Earth system reducing?". "earth system reducing" is not a separate object! It's simply asking how fast the potential energy of the skydiver is changing. Okay, m is not changing- the mass remains 60kg. g is not changing in this example (it would if the change in height were really large but I don't think that's intended here- no "space diving"!) so the only thing changing is h.

If you can use calculus, you could think: Rate of change is the derivative with respect to time so d(PE)/dt= mg dh/dt. dh/dt is just the rate of change of height- we are told that that is -50 m/s. The potential energy is changing at a rate of -50*m*g= -50*60*9.81= -29430 Joules per second.

If you haven't take calculus, since the speed is a constant, you can just pick some arbitrary height to start with: let's say at some instant the skidiver is h= 1000 m above the earth. His potential energy (relative to the earth- that's why it is a "skydiver-earth system") is 60*9.81*1000= 588600 Joules. One second later, he is 50 m lower: h= 950 m and the potential energy is 60*9.81*950= 559170 Joules. In one second, he has lost potential energy 588600- 559170=29430 as before.

If
 
Originally posted by Crashdowngurl
At what rate is the potential energy of the skydiver - Earth system reducing? ...

I don't understand what it's asking in this problem and in all the reading I've done there has been some sort of height or distance measurement involved for this type of question dealing with potential energy. The Earth system reducing is throwing me and I am not sure at all where to begin.
Halls discussed this a bit, but one thing that may be throwing you off is just the way that the question is asked. First realize that the potential energy is a property of the "sky diver"+"earth" system, not just the skydiver. Often you'll see questions like "What's the change in potential energy of an object when it is raised a height h?". These questions are sloppily worded! The "potential energy" is not "in" the object, it is a relationship between two objects. Of course, since so many books (and instructors) speak sloppily, when someone does ask it correctly it can throw you off.
 
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