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- Thread starter Haftred
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a = v^2 - v

ΣWork = ΣF * d = mad = 0.5m(v^2 - v

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HallsofIvy

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Quantities, mv^2, 35mv^2, are also conserved!

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Originally posted by PrudensOptimus

ΣWork = ΣF * d = mad = 0.5m(v^2 - v_{0}^2) = ke

So ummmm ... This is relative kinetic energy not average kinetic energy?

That is, relative to the kinetic energy of the object when v

Interesting derivation. Thanks.

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energy is the ability to do work.

work is force times distance.

kinetic energy is that energy a body has due to its motion.

so lets see:

[tex]

W=\int_A^B \mathbf{F}\cdot d\mathbf{s}

[/tex]

but by Newton

[tex]

\mathbf{F}=m\mathbf{a}

[/tex]

so

[tex]

W=\int_A^Bm\mathbf{a}\cdot d\mathbf{s} = m\int_{t_{A}}^{t_{B}}\mathbf{a}\cdot\mathbf{v}dt\\

= \frac{m}{2}\int_{t_{A}}^{t_{B}}\frac{d(v^2)}{dt}\! dt = \frac{1}{2}mv_B^2 - \frac{1}{2}mv_A^2

[/tex]

so you see that the work done is equal to the change in kinetic energy between the two points, and this is a derivation of the formula for that kinetic energy.

work is force times distance.

kinetic energy is that energy a body has due to its motion.

so lets see:

[tex]

W=\int_A^B \mathbf{F}\cdot d\mathbf{s}

[/tex]

but by Newton

[tex]

\mathbf{F}=m\mathbf{a}

[/tex]

so

[tex]

W=\int_A^Bm\mathbf{a}\cdot d\mathbf{s} = m\int_{t_{A}}^{t_{B}}\mathbf{a}\cdot\mathbf{v}dt\\

= \frac{m}{2}\int_{t_{A}}^{t_{B}}\frac{d(v^2)}{dt}\! dt = \frac{1}{2}mv_B^2 - \frac{1}{2}mv_A^2

[/tex]

so you see that the work done is equal to the change in kinetic energy between the two points, and this is a derivation of the formula for that kinetic energy.

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