# Expression of power

#### Leonson

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
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

benit13 and Leonson