# Power as force times velocity and change in KE

I was noodling over problems and coming up with one for my students. The concept of power as force times velocity shows up on the Regents exam and I was thinking of a baseball.

Specifically, a baseball mass is 0.142 kg. A pitcher winds up and hurls it at 31 m/s. I assume it takes 0.1 s for the pitcher to release the ball, so he is accelerating it while this is happening. I realize that the ball isn't accelerating in a straight line, but I am going to assume that the acceleration is linear for simplicity's sake.

So if I use the force x velocity method I get this:

$$a = \frac{31 m/s}{0.1s} = 310 m/s^2$$ $$F = ma = (0.142kg)(310m/s^2)=44.02 N$$ $$Fv = (44.02N)(31 m/s) = 1362.62 N\cdot m/s=1362.62\space J/s=1362.62W$$

Which seems all well and good. But let's say I did this with the ∆KE, using ∆KE/∆t

$$KE=\frac{1}{2}mv^2 = \frac{1}{2}(0.142kg)(31m/s)^2=(0.071kg)(961m^2/s^2)=68.231J$$ $$68.231J/0.1s = 682.31 W$$

Now, intuitively I would think I would get the same answer and I thought maybe $$Fv$$ only applies to average velocity. In which case I am fine I just need to remember to divide by 2, but I was looking up the same kind of problem in HS physics texts and they don't seem to mention that. But that could just be an omission because of many simplifying assumptions, or I just missed it.

So what did I do wrong here? I suspect I am being simply stupid, missing some obvious point.

ETA: whenever I do calculations with gravitational potential energy everything comes out as it should -- divide the PE by the time it takes to lift something and you get power, and if I assume that one is lifting h meters high multiplied by force (mg) then everything works out. Now in the baseball case the baseball is accelerating, rather than being lifted at a constant $$v$$, which might be the source of the glitch here.

Thanks!

$$P=\vec F \cdot \vec v$$ is the instantaneous power. As $$v$$ changes so does $$P$$. In contrast $$\Delta KE/\Delta t$$ gives average power.