
#1
Apr1609, 10:12 PM

P: 4

Hi,
I am working on a derivation of of the power equation. It is very simple, I would like to know if it is correct. I started with the equation for mechanical work over a changing path, and changing force. This is from Wikipedia, and it agrees with my text. W = [tex]\int[/tex] F _{º} ds where: _{º} is the symbol for dot product F is the force vector; and s is the position vector. Then to make it power I just put (1/delta t) in front of it to make: P = [tex]\frac{1}{\Delta t}[/tex] [tex]\int[/tex] F _{º} ds Let me know if this checks out. Thanks! 



#2
Apr1709, 01:34 AM

Emeritus
Sci Advisor
PF Gold
P: 6,238

Yes, that is correct, in as much as you want to calculate the average power that the force did during the motion through the specified path. The reason is that you calculated the total work done through the motion, and divided that by the total time it took to complete that motion.
If you want to have the instantaneous power, then you have to limit yourself to an infinitesimal amount of path: [tex] P_{inst} = \frac{1}{\delta t} \int_{\delta s} F . ds = \frac{F . \delta s}{\delta t} = F . \frac{\delta s}{\delta t} = F . v [/tex] So the instantaneous power done by a force on a moving point is given by the dot product of the force and the velocity of that point. 



#3
Apr1709, 07:07 PM

P: 4

Thanks, I was going for average power.



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