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June 25, 2019, 09:59:46 pm

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#### Mao

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##### Deriving Ek
« on: January 04, 2008, 09:22:51 am »
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I was just wondering the other day, many of the VCE physics formulas are derived from very simple equations, e.g. the 5 formulas dealing with constant acceleration and etc...

however no textbooks i've come across to have shown the workings to $E_k=\frac{1}{2} \cdot m \cdot v^2$
obviously you do not need that for the course, but just out of curiosity, where is this formula derived from? if at all??

i tried substituting stuff to $w=f \cdot d$, but most of it ended nowhere...
this was what i got...

$w=f \cdot d$

given that $v^2=u^2+2 \cdot a \cdot d \Rightarrow d= \frac{v^2-u^2}{2 \cdot a}$

therefore $w=E_k=m \cdot a \cdot \left( \frac{v^2-u^2}{2 \cdot a} \right)$

$\Rightarrow E_k=\frac{1}{2} \cdot m \cdot \left( v^2-u^2 \right)$ given that $a \neq 0$

and i'm stuck....

Thanks!!
« Last Edit: January 04, 2008, 09:25:06 am by Obsolete Chaos »
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#### Collin Li

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##### Re: Deriving Ek
« Reply #1 on: January 04, 2008, 09:31:50 am »
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http://en.wikipedia.org/wiki/Kinetic_energy#Derivation_and_definition

This is the correct way to do it. To fix your equation, you have to consider that Work is an input of energy, so the $(v^2-u^2)$ shows a change in velocity, proportional to the change in work. Kinetic energy sets a reference point: $u=0 \implies E_k=0$.

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##### Re: Deriving Ek
« Reply #2 on: January 04, 2008, 10:07:42 am »
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To perhaps most accurately define work done you'd need to use line integrals, however think about this:

$\mbox{Work Done} = \int_{x=a}^{x=b} \mbox{F dx}$

$= \int_{x=a}^{x=b} \mbox{ma dx}$

$= \int_{v=a}^{v=v_0} \mbox{mv dv}$ (Since a = dv/dt = dv/dx * dx/dt = dv/dx * v)

$= \frac{1}{2} mv_0^2$ (We don't worry about v=a, since we want total KE, we want the work done from rest!).
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#### Mao

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