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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:
  
 +$$ ds = v \ dt \qquad (2.1) $$
  
-$$ a = \frac{dv}{dt} = \frac{dv}{ds} \frac{ds}{dt} = v \frac{dv}{ds} ​  \qquad (2.1) $$+$$ dv = \ dt \qquad (2.2) $$
  
  
 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.
  
-$$ \int_{s_0}^a \ ds = \int_{s_0}^s ​v \frac{dv}{ds} \ ds = \int_{v_0}^v v \ dv = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2  ​\qquad (2.3) $$+$$ \int_{t_0}^a \ v \ dt \qquad (2.4) $$
  
-So we have:+Change the order:
  
-$$ \int_{s_0}^a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2  ​\qquad (2.4) $$+$$ \int_{t_0}^t v \ a \ dt \qquad (2.5) $$
  
-Combining ​with $F=ma$ gives the work-energy theorem+Change of differential according to (2.2), ​with corresponding change of limits.
  
-$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 ​ \qquad (2.5) $$+$$ \int_{v_0}^v v \ dv \qquad (2.6) $$ 
 + 
 +Putting everything together: 
 + 
 +$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (2.7) $$ 
 + 
 + 
 +Multiply both sides with $m$ and apply $F=ma$ to arrive at the work-energy theorem: 
 + 
 +$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad (8) $$
  
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theorems/work-energy_theorem.txt · Last modified: 2021/07/18 10:57 by cleonis