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I'm 12 so please don't laugh at my math.

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 it's 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$ Now, I know what you are thinking, what other way do you need? Anyway, the other way is to(assuming you know distance and acceleration)  $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?

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?

I'm 12 so please don't laugh at my math.

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 it's 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$ Now, I know what you are thinking, what other way do you need? Anyway, the other way is to(assuming you know distance and acceleration) $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?

I'm 12 so please don't laugh at my math.

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 it's 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$ Now, I know what you are thinking, what other way do you need? Anyway, the other way is to(assuming you know distance and acceleration) $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?