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| 183_notes:system_choice [2014/10/29 16:51] – caballero | 183_notes:system_choice [2018/05/29 21:46] – hallstein | ||
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| + | Section 7.7, 7.8 and 7.9 in Matter and Interactions (4th edition) | ||
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| ===== Choosing a System Matters ===== | ===== Choosing a System Matters ===== | ||
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| $$\dfrac{1}{2}mv^2 + mgh + \Delta E_{person} = 0$$ | $$\dfrac{1}{2}mv^2 + mgh + \Delta E_{person} = 0$$ | ||
| $$\Delta E_{person} = -\left(\dfrac{1}{2}mv^2 + mgh \right)$$ | $$\Delta E_{person} = -\left(\dfrac{1}{2}mv^2 + mgh \right)$$ | ||
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| + | Here we find that the person experiences an internal energy change that is negative, which makes sense because they are expending energy to lift the box. This system is the only one in which we can find this energy change because it is the only one that includes the person. | ||
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| + | Notice that in this system, the person does no work on the box because the box and person are in the system. Instead, we are accounting for the change in energy of the person. | ||
| === System 2: Box only === | === System 2: Box only === | ||
| - | Because the system is only a single particle, there' | + | Because the system is only a single particle, there can be no potential energy. There is work done by both the person and the Earth, but they have opposite signs because the applied force and the gravitational force are in different directions. |
| * System: Box; Surroundings: | * System: Box; Surroundings: | ||
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| $$\dfrac{1}{2}mv^2 = Fh - mgh$$ | $$\dfrac{1}{2}mv^2 = Fh - mgh$$ | ||
| $$\dfrac{1}{2}mv^2 = \left(F-mg\right)h$$ | $$\dfrac{1}{2}mv^2 = \left(F-mg\right)h$$ | ||
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| + | Here we find that the person had to exert a force that was larger than the weight of the box because the kinetic energy change is positive and thus the work done must be positive. We could not conclude that from an analysis of System 1; it required a different system. | ||
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| + | Notice also that because the person is outside the system, we no longer consider the internal energy change of the person, but the work the person does on the box while lifting it. | ||
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| $$\Delta K_{box} + \Delta U_{grav} = W_{person}$$ | $$\Delta K_{box} + \Delta U_{grav} = W_{person}$$ | ||
| $$\dfrac{1}{2}mv^2 + mgh = Fh$$ | $$\dfrac{1}{2}mv^2 + mgh = Fh$$ | ||
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| + | Here we find that a similar result as in System 2, but also explicitly note the the person does no work on the Earth because the displacement of the Earth as result of the person' | ||
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| + | ==== The Choice of System Determines Energy Accounting ==== | ||
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| + | As you read, different choices of system result in different conclusions that you can make about the motion of the system. How energy is accounted for is different in different systems (e.g. whether to consider internal energy changes or work done). Choosing different systems in which to analyze the energy can help draw more complete conclusions and explanations about the motion of systems. | ||