theorems:work-energy_theorem
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theorems:work-energy_theorem [2021/04/24 22:54] cleonis Capital letters in the title |
theorems:work-energy_theorem [2021/04/24 23:28] cleonis [Concrete] Fixed some yptos, added a sentence on release of potential energy linear with distance |
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| This exemplifies that it is very useful to have an equation that expresses force acting over distance. | This exemplifies that it is very useful to have an equation that expresses force acting over distance. | ||
| - | In the case of the paddle wheel setup used by Joule the friction limits the rate at which the weight descends. Other experiments corroborate that the amount of potential energy release on descent does not depend on the velocity of descent. If an object falls without friction the velocity increases linear with time. That corroborates that for free fall also the potential energy release is released linear with height. | + | In the case of the paddle wheel setup used by Joule the friction limits the rate at which the weight descends. Other experiments corroborate that the amount of potential energy release on descent does not depend on the velocity of descent. If an object falls without friction the velocity increases linear with time. That corroborates that for free fall also the potential energy release is linear with height. |
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| (1.1) is the derivative of (1.2), in that sense the two are not independent, but it's still possible to take advantage of the fact that there are two equations there. | (1.1) is the derivative of (1.2), in that sense the two are not independent, but it's still possible to take advantage of the fact that there are two equations there. | ||
| - | If an object is subject to uniform acceleration a over a distancs $(s - s_0)$ then how much change of velocity will that cause? | + | If an object is subject to uniform acceleration $a$ over a distancs $(s - s_0)$ then how much change of velocity will that cause? |
| So, we want to start with acceleration and distance traveled as a given, and obtain velocity. That means we need to work towards an expression that contains position, acceleration and velocity, but not time. | So, we want to start with acceleration and distance traveled as a given, and obtain velocity. That means we need to work towards an expression that contains position, acceleration and velocity, but not time. | ||
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| $$ s = s_0 + v_0\frac{(v-v_0)}{a} + \tfrac{1}{2} a \frac{{(v-v_0)}^2}{a^2} \qquad (1.3) $$ | $$ s = s_0 + v_0\frac{(v-v_0)}{a} + \tfrac{1}{2} a \frac{{(v-v_0)}^2}{a^2} \qquad (1.3) $$ | ||
| - | We mulitply everything out, and move terms for position and acceleration to the left. | + | We multiply everything out, and move terms for position and acceleration to the left. |
| $$ a(s-s_0) = vv_0 - {v_0}^2 + \tfrac{1}{2}v^2 - vv_0 + \tfrac{1}{2} {v_0}^2 \qquad (1.4) $$ | $$ a(s-s_0) = vv_0 - {v_0}^2 + \tfrac{1}{2}v^2 - vv_0 + \tfrac{1}{2} {v_0}^2 \qquad (1.4) $$ | ||
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| Repeating the three equations that are sufficient to define the Work-Energy theorem: | Repeating the three equations that are sufficient to define the Work-Energy theorem: | ||
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| $$ v = v_0 + at \qquad (1.1) $$ | $$ v = v_0 + at \qquad (1.1) $$ | ||
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| $$ F=ma $$ | $$ F=ma $$ | ||
| - | + | $F=ma$ and the work-energy theorem are very closely connected indeed. By taking the derivative with respect to position $F=ma$ is recovered immediately. | |
| - | $F=ma$ and the work-energy theorem are very closely connected indeed. | + | |
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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. | ||
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| In the course of the derivation the following two relations will be used: | In the course of the derivation the following two relations will be used: | ||
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| $$ dv = \frac{dv}{dt}dt \qquad (2.2) $$ | $$ dv = \frac{dv}{dt}dt \qquad (2.2) $$ | ||
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| 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 $s$. | ||
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| $$ \int_{s_0}^s a \ ds \qquad (2.3) $$ | $$ \int_{s_0}^s a \ ds \qquad (2.3) $$ | ||
| - | Use (11) to change the differential from *ds* to *dt*. Since the differential is changed the limits change accordingly. | + | Use (2.1) to change the differential from *ds* to *dt*. Since the differential is changed the limits change accordingly. |
| $$ \int_{t_0}^t a \ v \ dt \qquad (2.4) $$ | $$ \int_{t_0}^t a \ v \ dt \qquad (2.4) $$ | ||
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| $$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad (2.8) $$ | $$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad (2.8) $$ | ||
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| This form of the work-energy theorem emphasizes that in order to define a potential energy an integral must be defined. | This form of the work-energy theorem emphasizes that in order to define a potential energy an integral must be defined. | ||
theorems/work-energy_theorem.txt · Last modified: 2021/07/18 10:57 by cleonis
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