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2 UG=0J- A) UG B) Us C) KT D) these are all zero 1 1) At the point labeled "0" in the figure, which of the following energy terms are nonzero? Select all that apply. A) UG B) Us C) KT 0 2) At the point labeled "I" in the figure, which of the following energy terms are nonzero? Select all that apply.

2 UG=0J- A) UG B) Us C) KT D) these are all zero 1 1) At the point labeled "0" in the figure, which of the following energy terms are nonzero? Select all that apply. A) UG B) Us C) KT 0 2) At the point labeled "I" in the figure, which of the following energy terms are nonzero? Select all that apply.

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Can you please do 1, 2, 7 and 8. THANK YOU

2. 1 UG = 0J- 1) At the point labeled "O" in the figure, which of the following energy terms are nonzero? Select all that apply. A) UG B) Us C) K- D) these are all zero 2) At the point labeled “I" in the figure, which of the following energy terms are nonzero? Select all that apply. A) UG B) Us C) KT D) these are all zero
Transcribed Image Text:2. 1 UG = 0J- 1) At the point labeled "O" in the figure, which of the following energy terms are nonzero? Select all that apply. A) UG B) Us C) K- D) these are all zero 2) At the point labeled “I" in the figure, which of the following energy terms are nonzero? Select all that apply. A) UG B) Us C) KT D) these are all zero
7) Given the chosen coordinate system, calculate the height of the pendulum bob at point "0." Assume all five digits in the experimentally determined speed are significant. A) 5.3450 6 cm B) 5.350 51 cm C) 5.605 36 cm D) 5.61108 cm E) 5.920 31 cm 8) This calculation represents an idealization - that there is no friction. As we saw in the data analysis, friction did exist in this experiment although its affect on the result of this previous calculation would have been negligible and the specific analysis approximately negated it anyway. Even so, given that friction really did remove some energy from the system before the first pass through the photogate, which of the following is true about the real height ho? A) There would need to have been more gravitational potential energy in the system at point "0" than this calculation assumes there was, since the work done by nonconservative forces reduces the amount of energy in the system. Therefore, since g and m do not change, ho would need to be greater than calculated in order to account for both the energy that remains in the system as observed at point "1" and the energy that's wasted as nonconservative work from point "0" to point "1." B) The height calculated is larger than would have been true because friction takes energy out of the system and since m and g are fixed, the only thing that can be different is h. H has to go down to make up for the work done by friction.
Transcribed Image Text:7) Given the chosen coordinate system, calculate the height of the pendulum bob at point "0." Assume all five digits in the experimentally determined speed are significant. A) 5.3450 6 cm B) 5.350 51 cm C) 5.605 36 cm D) 5.61108 cm E) 5.920 31 cm 8) This calculation represents an idealization - that there is no friction. As we saw in the data analysis, friction did exist in this experiment although its affect on the result of this previous calculation would have been negligible and the specific analysis approximately negated it anyway. Even so, given that friction really did remove some energy from the system before the first pass through the photogate, which of the following is true about the real height ho? A) There would need to have been more gravitational potential energy in the system at point "0" than this calculation assumes there was, since the work done by nonconservative forces reduces the amount of energy in the system. Therefore, since g and m do not change, ho would need to be greater than calculated in order to account for both the energy that remains in the system as observed at point "1" and the energy that's wasted as nonconservative work from point "0" to point "1." B) The height calculated is larger than would have been true because friction takes energy out of the system and since m and g are fixed, the only thing that can be different is h. H has to go down to make up for the work done by friction.
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