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--- theorems:work-energy_theorem [2021/04/25 12:47]
cleonis Removed tab 'intuitive' the concrete mathematical derivation is the first step. Added why is it interesting section
+++ theorems:work-energy_theorem [2021/07/03 10:19]
cleonis [Concrete] More efficient derivation of the Work-Energy theorem
@@ Line 76: / Line 76: @@
 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 the following two relations will be used:
+In the course of the derivation acceleration $a$ is restated as follows:
-$$ ds = v \ dt  \qquad (2.1)  $$
-$$ dv = \frac{dv}{dt}dt  \qquad (2.2)   $$
+$$ a = \frac{dv}{dt} = \frac{dv}{ds} \frac{ds}{dt} = v \frac{dv}{ds}   \qquad (2.1) $$
 The integral for acceleration from a starting point $s_0$ to a final point $s$.
-$$ \int_{s_0}^s a \ ds \qquad (2.3)  $$
+$$ \int_{s_0}^s a \ ds \qquad (2.2)  $$
-Use (2.1) to change the differential from *ds* to *dt*. Since the differential is changed the limits change accordingly.
-$$ \int_{t_0}^t a \ v \ dt \qquad (2.4) $$
-Rearrange the order, and write the acceleration $a$ as $\tfrac{dv}{dt}$
-$$ \int_{t_0}^t v \ \frac{dv}{dt} \ dt  \qquad (2.5) $$
-Use (2.2) for a second change of differential, again the limits change accordingly.
-$$ \int_{v_0}^v v \ dv  \qquad (2.6)  $$
+Use (2.1) to change the differential from $ds$ to $dv$. Since the differential is changed the limits change accordingly.
-Putting everything together:
+$$ \int_{s_0}^s a \ ds = \int_{s_0}^s v \frac{dv}{ds} \ ds = \int_{v_0}^v v \ dv = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2  \qquad (2.3) $$
-$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2  \qquad (2.7) $$
+So we have:
+$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2  \qquad (2.4) $$
 Combining with $F=ma$ gives the work-energy theorem
-$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2  \qquad (2.8) $$
+$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2  \qquad (2.5) $$
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