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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?

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    $\begingroup$ Possible duplicate of Kinetic energy doesn't make sense $\endgroup$
    – JMac
    Commented May 1, 2017 at 17:00
  • $\begingroup$ This is a very complicated way to find the distance when you know the final and initial velocities. But if you do it right you find the same result. $\endgroup$
    – nasu
    Commented May 1, 2017 at 18:21
  • $\begingroup$ " the speed is $ at$". No, it's not. That's the change in velocity. $\endgroup$
    – Bill N
    Commented May 9, 2017 at 14:54

2 Answers 2

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I believe your problem is in the statement "the speed is a*t".

That's not true - the change in speed is $at$. Since you are starting out with initial speed $v$ and final speed $s$ (might I recommend that you change your notation... $s$ is really a terrible letter to use for velocity, and you are mixing the terms "velocity" and "speed" which are not the same thing), the thing you can say is

$$s = v + a\cdot t$$

See if that helps you make sense of it.

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Work done on an object is $F d = mad$. Let us assume constant acceleration so that $$ a = \frac{s^2-v^2}{2d}~. $$ where we used the equation $s^2-v^2=2ad$. Then, $$ w = mad = m \frac{s^2-v^2}{2d} d = \frac{1}{2} m s^2 - \frac{1}{2} mv^2 $$ which is the same as your earlier formula.

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