|Mar3-13, 04:04 PM||#1|
The "h" in mgh.. Can you change it?
Okay I still can't get my head around the idea of gravitational potential energy. I think that having questions answered may help my understanding.
Cats usually always land on their feet, so I'll use one for my example.
1) I hold the cat ten meters off the ground I'm standing on. h is 10 meters, right?
2) I place a boulder beneath the cat, which takes up five meters. The cat is five meters above that. The h for his gravitational potential energy is half, right? He can only fall five meters now? Or is it still ten meters?
3) I put the cat back where he was, ten meters up, but now I dig a hole, straight down into the ground, ten meters deep. If I release the cat he'll fall twenty meters. Did my digging that hole give the cat additional potential energy, without even moving the cat?
I'm probably misunderstanding something. Thank you for any answers and answers may prompt more questions.
|Mar3-13, 04:10 PM||#2|
You would be considering the change in potential energy or the potential energy relative to a reference point.
So in the first case, this is 10 m from the surface on which you are standing, so the change in PE is mgΔh = mg(10-0).
In the second case, relative to the ground, the cat is still Δh= 10 m. Relative to the surface of the boulder Δh = 10-5 = 5 so the energy relative to the boulder is halved.
Basically, you need to consider the change in energy.
|Mar3-13, 06:53 PM||#3|
You are free to set your h=0 line wherever you want. It is only changes in PE that are important here. For your second example the change in PE will be the same if you say that the cat falls from h=10 to h=5 or if you say that it falls from h=5 to h=0. Or you could even have it fall from h=0 to h=-5. All of those will give you the same answer.
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