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183_notes:system_choice [2014/10/29 10:11] – caballero | 183_notes:system_choice [2021/04/15 17:17] – [Choosing a System Matters] stumptyl | ||
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+ | Section 7.7, 7.8 and 7.9 in Matter and Interactions (4th edition) | ||
+ | |||
===== Choosing a System Matters ===== | ===== Choosing a System Matters ===== | ||
- | Earlier, [[183_notes: | + | Earlier, [[183_notes: |
+ | |||
+ | ==== Lifting a box === | ||
+ | |||
+ | A person is lifting a box (of mass $m$) from the ground to a height $h$. At that height, the box is moving with an upward speed $v$. To lift the box, we assume the person applies a force $F$ upward. Let's consider three different choices of system to demonstrate what can be learned from each choice. | ||
+ | |||
+ | === System 1: Box, Earth, and Person === | ||
+ | |||
+ | While lifting the box, the person' | ||
+ | |||
+ | * System: Box, Earth, and Person; Surroundings: | ||
+ | * Initial State: Box on ground with no velocity; Box at height $h$ with velocity $v$ | ||
+ | |||
+ | $$\Delta E_{sys} = W_{surr}$$ | ||
+ | $$\Delta K_{box} + \Delta U_{grav} + \Delta E_{person} = 0$$ | ||
+ | $$\dfrac{1}{2}mv^2 + mgh + \Delta E_{person} = 0$$ | ||
+ | $$\Delta E_{person} = -\left(\dfrac{1}{2}mv^2 + mgh \right)$$ | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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 === | ||
+ | |||
+ | 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: | ||
+ | * Initial State: Box on ground with no velocity; Box at height $h$ with velocity $v$ | ||
+ | |||
+ | $$\Delta E_{sys} = W_{surr}$$ | ||
+ | $$\Delta K_{box} = W_{person} + W_{Earth}$$ | ||
+ | $$\dfrac{1}{2}mv^2 = Fh - mgh$$ | ||
+ | $$\dfrac{1}{2}mv^2 = \left(F-mg\right)h$$ | ||
- | ==== A falling ball === | + | 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. |
- | [{{ 183_notes: | + | Notice also that because the person is outside the system, we no longer consider the internal energy change |
- | Consider a ball initially at rest that begins to fall towards the Earth. | ||
- | In the first case, you choose the ball to be the system. The kinetic energy of the ball increases because the work that is done by the gravitational force due to Earth (which is in the surroundings) does positive work on the ball. The displacement of the ball is down and so is the gravitational force. | + | === System 3: Box and Earth === |
- | For case 1 with the ball as the system, | + | Finally, this system |
- | | + | |
- | | + | |
- | - $\Delta K_{ball} > 0$ because | + | |
- | In the second case, you choose the ball and the Earth to be the system. The kinetic energy of the ball still increases, but now there' | + | $$\Delta E_{sys} = W_{surr}$$ |
+ | $$\Delta K_{box} + \Delta U_{grav} = W_{person}$$ | ||
+ | $$\dfrac{1}{2}mv^2 + mgh = Fh$$ | ||
- | For case 1 with the ball as the system, | + | 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' |
- | - System: Ball + Earth; Surroundings: | + | ==== The Choice of System |
- | - Initial state: Ball at rest; Final state: Ball moving | + | |
- | - $\Delta K_{ball} > 0$ but why? $W_{surr} | + | |
- | This puzzle will be resolved by considering energy associated with the interactions | + | As you read, different choices of system result in different conclusions that you can make about the motion |