Expression of power

Jun 2019
10
0
Power is defined by P=W/t where W is work done and t is the time.
If force F is constant, then power can also be P=Fv where v is the velocity, which sort of can be derived from P=W/t=Fs/t=Fv. Since F=ma, also P=mav..........(1)
However, if work done is turned into kinetic energy, E=(1/2)mv^2, then P=E/t=(1/2)(mv^2)/t. If that's the case, then v/t=a gives P=(1/2)mav.........(2)
So why is that (1) and (2) not the same?
 

topsquark

Forum Staff
Apr 2008
2,926
608
On the dance floor, baby!
The problem winds up being due to Calculus.

Here it is:
\(\displaystyle P = \dfrac{dW}{dt} = \dfrac{d}{dt} \left ( \dfrac{1}{2} m v^2 \right ) = \dfrac{1}{2} m \left ( \dfrac{d}{dt} v^2 \right )\)

When you take the derivative you get an extra factor of 2 in there:
\(\displaystyle P = \dfrac{1}{2} m \cdot \left ( 2v \dfrac{dv}{dt} \right ) = mav\)

There should be a non-Calculus derivation for this but I can't think of it off the top of my head.

-Dan
 
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