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phy131studiof15:lectures:finalp2sol [2015/12/02 09:46] mdawber [Question 4] |
phy131studiof15:lectures:finalp2sol [2015/12/02 09:49] mdawber [Question 6] |
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===== Question 5 ===== | ===== Question 5 ===== | ||

+ | {{f11finalq5fig.png}} | ||

- | a) $kx_{0}=mg$ | + | A spring of spring constant $k=10 \,\mathrm{N/m}$ is hanging from the ceiling. A 500 g wooden Santa is attached to the spring changing it's equilibrium position by $x_{0}$. In all of the following questions you may neglect the mass of the spring. |

+ | | ||

+ | a) (5 points) Find $x_{0}$. | ||

+ | | ||

+ | $kx_{0}=mg$ | ||

$x_{0}=\frac{0.5\times9.8}{10}=0.49\mathrm{m}$ | $x_{0}=\frac{0.5\times9.8}{10}=0.49\mathrm{m}$ | ||

- | b) $\Delta PE=\int_{x_{0}}^0 F_{Net}\,dx=\int_{x_{0}}^0(kx-mg)\,dx=-\frac{1}{2}kx_{0}^{2}+mgx_{0}=-\frac{1}{2}\times10\times0.49^{2}+0.5\times9.8\times0.49=1.2\mathrm{J}$ | + | b) (5 points) The Santa is pushed back up so the spring has the same length as it had before the Santa was attached. How much potential energy has been added to the system? |

+ | | ||

+ | $\Delta PE=\int_{x_{0}}^0 F_{Net}\,dx=\int_{x_{0}}^0(kx-mg)\,dx=-\frac{1}{2}kx_{0}^{2}+mgx_{0}=-\frac{1}{2}\times10\times0.49^{2}+0.5\times9.8\times0.49=1.2\mathrm{J}$ | ||

If we took the spring in the other direction then | If we took the spring in the other direction then | ||

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So you can see that there is the same amount of energy stored in either direction, which is important for simple harmonic motion to occur. | So you can see that there is the same amount of energy stored in either direction, which is important for simple harmonic motion to occur. | ||

- | c) $f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}=\frac{1}{2\pi}\sqrt{\frac{10}{0.5}}=0.71\mathrm{Hz}$ | + | c) (5 points) The Santa is then released and executes simple harmonic motion.What is the frequency of the simple harmonic motion? |

- | d) $v_{max}=\omega A= 2\pi\times0.71\times0.49=2.19\mathrm{m/s}$ occurs at displacement $x=0\mathrm{m}$ from $x_{0}$ | + | $f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}=\frac{1}{2\pi}\sqrt{\frac{10}{0.5}}=0.71\mathrm{Hz}$ |

- | e) $a_{max}=\omega^{2} A=(2\pi\times0.71)^{2}\times0.49=9.8\mathrm{m/s^{2}}$ occurs at displacements $x=-0.49\mathrm{m},x=0.49\mathrm{m}$ from $x_{0}$ | + | d) (5 points) What is the magnitude of the maximum velocity during simple harmonic motion? Give the position of the point or points at which this occurs in terms of displacement from $x_{0}$. |

+ | | ||

+ | $v_{max}=\omega A= 2\pi\times0.71\times0.49=2.19\mathrm{m/s}$ occurs at displacement $x=0\mathrm{m}$ from $x_{0}$ | ||

+ | | ||

+ | e) (5 points) What is the magnitude of the maximum acceleration during simple harmonic motion? Give the position of the point or points at which this occurs in terms of displacement from $x_{0}$. | ||

+ | | ||

+ | $a_{max}=\omega^{2} A=(2\pi\times0.71)^{2}\times0.49=9.8\mathrm{m/s^{2}}$ occurs at displacements $x=-0.49\mathrm{m},x=0.49\mathrm{m}$ from $x_{0}$ | ||

===== Question 6 ===== | ===== Question 6 ===== | ||

- | a) For an adiabatic process | + | {f11finalq7fig.png}} |

+ | | ||

+ | Santa slides down a chimney which has a volume of 10\,m$^{3}$. The chimney is sealed at the bottom by a floor, and sealed at the top by Santa's belly. We can treat the air in the chimney as an ideal diatomic gas. (Note: Santa's suit has very good thermal insulation!) | ||

+ | | ||

+ | a) (5 points) When Santa slides down the chimney from top to bottom he reduces the volume of the air by a factor of 5 (ie. $V_{final}=\frac{V_{initial}}{5}=2\,\mathrm{m^{3}}$). If the initial pressure is standard atmospheric pressure what is the final pressure. (Hint: No heat is added to the air in this process and we can approximate it is quasistatic.) | ||

+ | | ||

+ | For an adiabatic process | ||

$P_{i}V_{i}^{\gamma}=P_{f}V_{f}^{\gamma}$ where in this case $\gamma=\frac{7}{5}$ | $P_{i}V_{i}^{\gamma}=P_{f}V_{f}^{\gamma}$ where in this case $\gamma=\frac{7}{5}$ | ||

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$P_{f}=P_{i}(\frac{V_{i}}{V_{f}})^{\gamma}=101.3\mathrm{kPa}\times{5}^{7/5}=964.2\mathrm{kPa}$ | $P_{f}=P_{i}(\frac{V_{i}}{V_{f}})^{\gamma}=101.3\mathrm{kPa}\times{5}^{7/5}=964.2\mathrm{kPa}$ | ||

- | b) As the gas is ideal | + | b) (5 points) If the initial temperature of the air in the chimney was $2\mathrm{^{o}C}$ what is the final temperature of the air in the chimney? |

+ | | ||

+ | As the gas is ideal | ||

$P_{i}V_{i}=nRT_{i}$ | $P_{i}V_{i}=nRT_{i}$ | ||

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$T_{f}=\frac{964.2}{101.3}\times\frac{1}{5}\times275=523.5\mathrm{K}$ | $T_{f}=\frac{964.2}{101.3}\times\frac{1}{5}\times275=523.5\mathrm{K}$ | ||

- | c) For an isothermal process | + | c) (5 points) To get Santa out of the chimney a fire is lit under it which adds heat to the air, so that it expands isothermally. How much work is done by the gas as it expands isothermally back to it's original volume. |

+ | | ||

+ | For an isothermal process | ||

$W=nRT\ln{\frac{V_{f}}{V_{i}}}$ | $W=nRT\ln{\frac{V_{f}}{V_{i}}}$ | ||

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$W=964.2\times10^{3}\times 2\ln(5)=3103\mathrm{kJ}$ | $W=964.2\times10^{3}\times 2\ln(5)=3103\mathrm{kJ}$ | ||

- | d) For an isothermal process | + | d) (5 points) What is the pressure in the chimney when Santa has returned to his original position? |

+ | | ||

+ | For an isothermal process | ||

$P_{i}V_{i}=P_{f}V_{f}$ | $P_{i}V_{i}=P_{f}V_{f}$ | ||

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$P_{f}=\frac{V_{f}}{V_{i}}P_{i}=\frac{964.2}{5}=192.8\mathrm{kPa}$ | $P_{f}=\frac{V_{f}}{V_{i}}P_{i}=\frac{964.2}{5}=192.8\mathrm{kPa}$ | ||

- | e) Red is first process, green is second | + | e) (5 points) Sketch a pressure-volume diagram which shows both processes, making sure to label each process and the direction in which they occur. |

+ | | ||

+ | Red is first process, green is second | ||

{{finalf11q7solfig.png}} | {{finalf11q7solfig.png}} | ||