Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum  1Likes
May 17th 2017, 04:56 AM

#1  Junior Member
Join Date: Apr 2017 Location: Turkey
Posts: 15
 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

 
May 17th 2017, 05:56 AM

#2  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 480

Originally Posted by Razi 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.

 
May 17th 2017, 06:09 AM

#3  Junior Member
Join Date: Apr 2017 Location: Turkey
Posts: 15

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

 
May 17th 2017, 07:21 AM

#4  Junior Member
Join Date: May 2017
Posts: 9

is the E the same as the E in E=m/(c)(c)?

 
May 17th 2017, 08:43 AM

#5  Junior Member
Join Date: Apr 2017 Location: Turkey
Posts: 15

No , it is not energy equivalent to mass , it is just the total mechanical energy :)

 
May 17th 2017, 12:18 PM

#6  Senior Member
Join Date: Apr 2015 Location: Somerset, England
Posts: 480

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
$\displaystyle KE = \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}$

 
May 17th 2017, 12:40 PM

#7  Junior Member
Join Date: Apr 2017 Location: Turkey
Posts: 15

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

 
May 17th 2017, 12:44 PM

#8  Junior Member
Join Date: Apr 2017 Location: Turkey
Posts: 15

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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