theorems:work-energy_theorem
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theorems:work-energy_theorem [2021/07/03 10:19] cleonis [Concrete] More efficient derivation of the Work-Energy theorem |
theorems:work-energy_theorem [2021/07/18 10:57] (current) cleonis [Concrete] Reverted to the derivation with an intermediary stage, it is transparent whereas the version-with-less-steps wasn't |
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| The following derivation is more abstract. This derivation is from the outset designed to accomodate an arbitrary acceleration profile. | The following derivation is more abstract. This derivation is from the outset designed to accomodate an arbitrary acceleration profile. | ||
| - | In the course of the derivation acceleration $a$ is restated as follows: | + | The two relations that constitute the mathematics of the Work-Energy theorem: |
| + | $$ ds = v \ dt \qquad (2.1) $$ | ||
| - | $$ a = \frac{dv}{dt} = \frac{dv}{ds} \frac{ds}{dt} = v \frac{dv}{ds} \qquad (2.1) $$ | + | $$ dv = a \ dt \qquad (2.2) $$ |
| The integral for acceleration from a starting point $s_0$ to a final point $s$. | The integral for acceleration from a starting point $s_0$ to a final point $s$. | ||
| - | $$ \int_{s_0}^s a \ ds \qquad (2.2) $$ | + | $$ \int_{s_0}^s a \ ds \qquad (2.3) $$ |
| - | Use (2.1) to change the differential from $ds$ to $dv$. Since the differential is changed the limits change accordingly. | + | Intermediary step: change of the differential according to (2.1), with corresponding change of limits. |
| - | $$ \int_{s_0}^s a \ ds = \int_{s_0}^s v \frac{dv}{ds} \ ds = \int_{v_0}^v v \ dv = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (2.3) $$ | + | $$ \int_{t_0}^t a \ v \ dt \qquad (2.4) $$ |
| - | So we have: | + | Change the order: |
| - | $$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (2.4) $$ | + | $$ \int_{t_0}^t v \ a \ dt \qquad (2.5) $$ |
| - | Combining with $F=ma$ gives the work-energy theorem | + | Change of differential according to (2.2), with corresponding change of limits. |
| - | $$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad (2.5) $$ | + | $$ \int_{v_0}^v v \ dv \qquad (2.6) $$ |
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| + | Putting everything together: | ||
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| + | $$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (2.7) $$ | ||
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| + | Multiply both sides with $m$ and apply $F=ma$ to arrive at the work-energy theorem: | ||
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| + | $$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad (8) $$ | ||
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theorems/work-energy_theorem.1625300381.txt.gz · Last modified: 2021/07/03 10:19 by cleonis
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