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Old May 17th 2017, 04:56 AM   #1
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Energy Transformation

Greetings to all
This is my first post in this nice forum , my question is :

In any system , is it physically possible to have a situation where E - U < 0 ??

where ,
E : the total energy for this system .
U : the potential energy for this system .

Note : mathematically i have proved that it is impossible , but what about Physically ?

Thanks in advance and best regards.
Razi
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Old May 17th 2017, 05:56 AM   #2
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Originally Posted by Razi View Post
Greetings to all
This is my first post in this nice forum , my question is :

In any system , is it physically possible to have a situation where E - U < 0 ??

where ,
E : the total energy for this system .
U : the potential energy for this system .

Note : mathematically i have proved that it is impossible , but what about Physically ?

Thanks in advance and best regards.
Razi
This is going to depend in part on your definition of E and U, your sign convention and the timescale of the process.
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Old May 17th 2017, 06:09 AM   #3
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Smile

Dear Studiot

Thanks for replay , about the E it is the total mechanical energy for a conservative system which is equal to :

E = K + U
K : Kinetic Energy ,
U : Potential Energy ,

mathematically , E - U = K = (1/2)mv^2 always (+ve) >0 because (1/2) is +ve , m is scalar quantity and always +ve , v^2 is always +ve , so the kinetic energy , mathematically is > 0 .
Is there any case in the nature for E - U < 0 ?? ( Physically )? this is my question .

Best regards
Razi
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Old May 17th 2017, 07:21 AM   #4
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is the E the same as the E in E=m/(c)(c)?
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Old May 17th 2017, 08:43 AM   #5
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No , it is not energy equivalent to mass , it is just the total mechanical energy :-)
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Old May 17th 2017, 12:18 PM   #6
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I will have time to write a longer piece tomorrow but here is something to think about.

Thank you for the clarification.

I don't know where your data comes from but as I said things depend upon your definitions and sign conventions.

For instance gravitational potential energy is defined as


$\displaystyle U = - G\frac{{Mm}}{r}$


Notice the negative sign which is the convention that allows the potential energy to tend towards zero at very large distances.

As regards kinetic energy your formula is incomplete as is only takes account of the translational energy of a body. Rigid bodies or systems of particles may also have rotational kinetic energy.

The complete formula is


$\displaystyle KE = \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}$
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Old May 17th 2017, 12:40 PM   #7
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Thanks again for your replay and for your time

I am studying physics by using very famous book :
" Physics for scientists and engineers with modern physics , 5th ed. "
i have found this question in the question section of chapter 8 , page : 239 , question number 9 .

fortunately , till now i am just studying the linear motion , so the rotational motion will be in the next chapters so we can ignore the value of ( 1/2 ) Iw^2 right now .

Physically , i could not find ( as i know ) any case matched with this hypothesis so i have asked you :-) .

I will wait your answer , and thanks again so much
Best regards
Razi
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Old May 17th 2017, 12:44 PM   #8
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Thanks again for your replay and for your time

I am studying physics by using very famous book :
" Physics for scientists and engineers with modern physics , 5th ed. "
i have found this question in the question section of chapter 8 , page : 239 , question number 9 .

fortunately , till now i am just studying the linear motion , so the rotational motion will be in the next chapters so we can ignore the value of ( 1/2 ) Iw^2 right now .

Physically , i could not find ( as i know ) any case matched with this hypothesis so i have asked you :-) .

I will wait your answer , and thanks again so much
Best regards
Razi
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