Every textbook writes KE = ½mv² and moves on. Dimensional analysis tells me the m and the v² have to be there, but the ½ looks like it was put in to make something cancel later.
Is there a derivation where the ½ falls out on its own instead of being chosen? I have done one calculus course, so an integral is fine. I just want to see the step it comes from.
Accepted Answer
It falls out of one integral. Nothing is chosen.
W = ∫ F dx, and F = ma = m dv/dt, so W = ∫ m (dv/dt) dx.
The whole trick is rewriting dx. Since dx = v dt:
W = ∫ m (dv/dt) v dt = ∫ m v dv
That integral is over velocity instead of position, and it is elementary:
∫ from u to v of m v dv = ½mv² − ½mu²
So the ½ is just the ½ from ∫v dv = v²/2, the same one you get integrating any linear function.
Worth flagging the argument people reach for first, because it does not actually work.
The tempting one: it starts at 0 and ends at v, so the average speed is v/2, and W = F·d hands you the ½.
From rest at constant acceleration the average really is v/2, so you land on the right number. But the time average of v is (u+v)/2 only when the acceleration is constant, and ½mv² holds for any force at all. It is a mnemonic that happens to work in the easy case, not a derivation.
Without calculus: v² = u² + 2as gives as = (v² − u²)/2. Multiply by m and the left side is F·s, the work done. The ½ comes from the 2 in that equation, which came from the ½at² you already know.
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