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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 [2021/07/18 10:57]
cleonis [Concrete] Reverted to the derivation with an intermediary stage, it is transparent whereas the version-with-less-steps wasn't
@@ 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:
+The two relations that constitute the mathematics of the Work-Energy theorem:
-$$ ds = v \ dt  \qquad (2.1)  $$
+$$ ds = v \ dt \qquad (2.1) $$
+$$ dv = a \ dt \qquad (2.2) $$
-$$ dv = \frac{dv}{dt}dt  \qquad (2.2)   $$
 The integral for acceleration from a starting point $s_0$ to a final point $s$.
@@ Line 86: / Line 87: @@
 $$ \int_{s_0}^s a \ ds \qquad (2.3)  $$
-Use (2.1) to change the differential from *ds* to *dt*. Since the differential is changed the limits change accordingly.
+Intermediary step: change of the differential according to (2.1), with corresponding change of limits.
 $$ \int_{t_0}^t a \ v \ dt \qquad (2.4) $$
-Rearrange the order, and write the acceleration $a$ as $\tfrac{dv}{dt}$
+Change the order:
-$$ \int_{t_0}^t v \ \frac{dv}{dt} \ dt  \qquad (2.5) $$
+$$ \int_{t_0}^t v \ a \ dt \qquad (2.5) $$
-Use (2.2) for a second change of differential, again the limits change accordingly.
+Change of differential according to (2.2), with corresponding change of limits.
-$$ \int_{v_0}^v v \ dv  \qquad (2.6)  $$
+$$ \int_{v_0}^v v \ dv \qquad (2.6) $$
 Putting everything together:
-$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2  \qquad (2.7) $$
+$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (2.7) $$
-Combining with $F=ma$ gives the work-energy theorem
+Multiply both sides with $m$ and apply $F=ma$ to arrive at the work-energy theorem:
-$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2  \qquad (2.8) $$
+$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad (8) $$
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