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closet mathemetician
closet mathemetician is offline
#24
Dec10-09, 10:01 AM
P: 44
Here’s my version of the derivation of the kinetic energy formula (first post using latex, so go easy)

Kinetic energy is force times change in distance.

[tex]
KE = F \Delta x
[/tex]

Force is mass times acceleration, which is a change in velocity over time. We can write this as

[tex]
F= m \frac{\Delta v}{\Delta t}
[/tex]


Now the change in distance, [tex] \Delta x [/tex] can be expressed as velocity times time


[tex]
\Delta x = \bar{v} t
[/tex]


Notice that the velocity is is average velocity, where we have a constant change in velocity, or acceleration.

Average velocity can be written as the sum of the initial and final velocities divided by two.

[tex]
\bar v = \frac{v_0 + v_f}{2}
[/tex]


Now let's look at final velocity. We can write this as the initial velocity plus the change in velocity over time multiplied by time

[tex]
v_f = v_0 + \frac{\Delta v}{\Delta t} \Delta t
[/tex]

which can also be written as

[tex]
v_f = v_0 + at
[/tex]


Plugging this into the aveage velocity formula gives

[tex]
\bar v = \frac{v_0 + \left[ v_0 + \frac{\Delta v}{\Delta t} \Delta t \right]}{2}
[/tex]

which simplifies to

[tex]
\bar v = v_0 + \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t
[/tex]

Now, we plug this into the change in distance formula

[tex]
\Delta x = \left[v_0 + \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t \right] \Delta t
[/tex]

[tex]
= v_0 \Delta t + \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t^2
[/tex]

Now, let's assume the beginning velocity is zero. Then we take the product of the above and force


[tex]

F\Delta x = \left( m\frac{\Delta v}{\Delta t} \right) \left( \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t \Delta t \right)
[/tex]


Collecting terms we arrive at


[tex]
KE = m \frac{1}{2} \Delta v \Delta v
[/tex]

Which of course is

[tex]
KE = \frac{1}{2} mv^2
[/tex]