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| 183_notes:examples:energy_in_a_spring-mass_system [2014/10/28 12:49] – pwirving | 183_notes:examples:energy_in_a_spring-mass_system [2014/10/28 13:07] (current) – pwirving | ||
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| Surroundings: | Surroundings: | ||
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| + | $E_{f} = E_{i} + W$ The energy principle. | ||
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| + | $K = \dfrac{1}{2}mv^2$ | ||
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| + | $U_{spring} = \dfrac{1}{2}k_{s}s^2$ | ||
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| $E_{f} = E_{i} + W$ | $E_{f} = E_{i} + W$ | ||
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| + | The initial energy and the final energy can be expressed as the addition of kinetic and potential parts. | ||
| $K_{f} + U_{f} = K_{i} + U_{i} + W$ | $K_{f} + U_{f} = K_{i} + U_{i} + W$ | ||
| - | $K_{f}$ and $W$ cancel out. | + | We know that no work is done by the surroundings because the point where the wall force is applied does not move. |
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| + | We also know that when the stretch is maximum, the block is momentarily at rest (the turning point). | ||
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| + | With this information we know that $K_{f}$ and $W$ are zero. | ||
| $0 + \dfrac{1}{2}k_{s}s^2_{f} = \dfrac{1}{2}mv^2_{i} + \dfrac{1}{2}k_{s}s^2_{i}$ | $0 + \dfrac{1}{2}k_{s}s^2_{f} = \dfrac{1}{2}mv^2_{i} + \dfrac{1}{2}k_{s}s^2_{i}$ | ||
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| + | The amplitude corresponds to the stretch $s_{f}$. We want to rearrange the equations to isolate $s_{f}$. | ||
| $\dfrac{1}{2}k_{s}s^2_{f} = \dfrac{1}{2}(0.2 kg)(0.5 m/s)^2 + \dfrac{1}{2}(12 N/m)(0.03 m)^2$ | $\dfrac{1}{2}k_{s}s^2_{f} = \dfrac{1}{2}(0.2 kg)(0.5 m/s)^2 + \dfrac{1}{2}(12 N/m)(0.03 m)^2$ | ||
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| + | We can further simplify the equation by substituting values in for the variables. | ||
| $\dfrac{1}{2}k_{s}s^2_{f} = 0.03 J$ | $\dfrac{1}{2}k_{s}s^2_{f} = 0.03 J$ | ||
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| + | Solve for $s_{f}$. | ||
| $s_{f} = \sqrt {2(0.03 J)/(12 N/m)} = 0.07 m$ | $s_{f} = \sqrt {2(0.03 J)/(12 N/m)} = 0.07 m$ | ||
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| b: | b: | ||
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| + | $v_{max}$ is maximum when its final state is when $s_{f} = 0$ which is when the block has only kinetic energy. | ||
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| + | The total initial energy when there is only kinetic energy is equal to the total final energy when there is only potential energy. | ||
| $K_f + 0 = 0.03 J$ | $K_f + 0 = 0.03 J$ | ||
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| + | Substitute in $\dfrac{1}{2}mv^2$ | ||
| $\dfrac{1}{2}mv^2_{f} = 0.03 J$ | $\dfrac{1}{2}mv^2_{f} = 0.03 J$ | ||
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| + | Isolate $v_{max}$ to solve for the maximum velocity. | ||
| $v_{max} = \sqrt {2(0.03 J)/(0.2 kg)} = 0.55 m/s$ | $v_{max} = \sqrt {2(0.03 J)/(0.2 kg)} = 0.55 m/s$ | ||