Let's assume that an object has KE expressed by velocity $v$ and mass $m$.  If work $w$ has been done to this object such that its new speed (same direction) is $s$. There are 2 ways(I know) to calculate $w$. The first is  $ \frac {1}{2} ms^2 -  \frac {1}{2} mv^2$. The other way is (assuming you know distance and acceleration) to use $w=mda $ then $d= \frac {1}{2}at^2 + vt  $   $\frac{2d}{a}= t^2 + \frac {2v}{a}t$  $\frac {2d}{a} +\frac {v^2}{a^2} = t^2 + \frac {2v}{a}t + \frac {v^2}{a^2}$   $\frac {2d}{a} +\frac {v^2}{a^2} = (t+\frac {v}{a})^2$  $\sqrt {\frac {2d}{a} +\frac {v^2}{a^2}} = t + \frac{v}{a}$   $\sqrt {\frac {2da+v^2}{a^2}}= t + \frac{v}{a}$  $\frac {\sqrt {2da+v^2}}{a} = t + \frac{v}{a}$   $\frac {\sqrt {2da+v^2}-v}{a} = t $   and  the speed is a*t which is equal to   ${\sqrt {2da+v^2}-v}$  remember that   $ da = \frac {w}{m}$  $ s = \sqrt {\frac {2w}{m}+v^2}-v$  $ (s+v)^2 = \frac{2w}{m} +v^2$  $ \frac{1}{2}m((s+v)^2-v^2) = w $  simplify and it will be  $ w= \frac {1}{2} ms^2 + \frac{1}{2} mvs$  which isn't quite the other formula. Why is it that my formula doesn't work?
Kinetic energy is simple, right?! from the equation it seems as though if you know the speed and mass you'll know the work. Let's try to prove it. let's assume we know the acceleration and distance for now. $$w=mda$$ So the distance is $$d= 1/2 *at^2$$ Then  $$t = \sqrt{2d/a}$$ So $a*t$, which is the speed, is  $$a*\sqrt{2d/a} = a* \sqrt{2d}/\sqrt{a} = \sqrt{2da} = \sqrt{2w/m}$$ So regardless of the distance or acceleration, if you know the work and the mass, you can calculate the speed which is  $$v=\sqrt{2w/m}$$  What if you know the speed and the mass? You can easily calculate the work. $$v = \sqrt{2w/m}$$ Then  $$v^2 = 2w/m$$  That would mean  $$m*1/2*v^2=W $$ Rearrange it and  $$W= 1/2 *m *v^2$$ would be the equation.  What if you have an initial speed and a new speed and want calculate the amount of work done after an initial speed to get the new speed? You'll just subtract the total work from the initial speeds work.  $$1/2mv^2 - 1/2ms^2$$  where s is  the initial speed and v is the new speed. However if you do it the other way which is  $$d= 1/2at^2 + vt$$ and solve it the answer would be $$1/2ms^2  + 1/2msv$$ Why aren't they equal?
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