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GPE and KE...please help argh 
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#1
May404, 10:42 AM

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Hi I am just finishing my AS physics coursework and just as i thought i understood it someone tells me something else. When you pick up an object i understand that it gains GPE and the amount of this depends on the height it is picked up to, hence GPE = mgh. But when you drop the object what happens to the energy? Does it all suddenly convert to KE, or when it is falling does the GPE gradually turn to KE?
In my coursework i calculated GPE and KE, assuming they should be roughly the same values and they were but then some1 said to me that they should be inversely proportional to each other. Can some1 please help me thankyou. meawinner 


#2
May404, 11:24 AM

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just think about conservation of energy. when the object is falling, the height changes, by virtue of energy conservation, the object gains kinetic energy.



#3
May404, 12:35 PM

P: 11

If you lift a ball to some heigth h, you obtain a potential energy due to gravity that is equal to mgh. When the ball drops, the energy is gradually converted to KE, not all at once. To emphasize this point, imagine dropping the ball from heigth h. A friend catches the ball at height h/2. Since we already know the formula for potential energy due to gravity, our new potential would be equal to mgh/2. The ball still maintains an energy potential that was not converted to kinectic. All the potential energy is converted to kinetic energy at a time (only at the instantaneous moment right before impact) when h=0 after the point of release.



#4
May404, 12:51 PM

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GPE and KE...please help argh
When we have energy conservation, it means that the SUM of kinetic energy and potential energy remains equal at all times. Those who told you that the kinetic energy should be inversely proportional to the potential energy are totally wrong; they haven't learnt the elementary distinction between a sum and a product. 


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