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Problem Set 4 Solutions

Problem Set 4 Solutions - 1 Physics 200a PSIV SOLUTIONS 1(i...

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1 Physics 200a PSIV SOLUTIONS 1. (i) Using the dot product show that if A + B is perpendicular to A - B , then A = B .(ii) Use the dot product to relate vector components ( A x , A y ) in the standard frame to ( A x , A y ) measured in a frame rotated counterclockwise by an angle θ . Use the fact that components of a vector are just dot products with the corresponding basis vectors: A x = A · i etc. (i) ( A + B ) · ( A - B ) = A · A - B · B + A · B - B · A (1) = A 2 - B 2 (2) since A · B = B · A (3) So, in order for the dot product to be 0, A 2 must equal B 2 and A = B . (ii) A x = A · i = A · ( i cos θ + j sin θ ) = A x cos θ + A y sin θ (4) and so on. 2. If F ( x ) = Ax 3 / 2 find the work done by it in moving a body from x = 0 to x = A . Work done W = A 0 Ax 3 / 2 dx = Ax 5 / 2 5 / 2 A 0 = 2 5 A 7 / 2 . (5) 3. You have to choose between a 225-W fridge that costs $ 1150 and a 425-W model that costs $ 850. The first runs 11% of the time and the second runs 20% of the time. If electricity costs 9.5 cents/kWh when will you break even? (Use hours as the unit of time at first, go to days to report final answer.) Let us run these machines for t hours. Equating the cost of cheap machine C c = $850 + ( . 425 t kWh )( . 2)( . 095$ /kWh ) (6) to running the fancy one C f = $1150 + ( . 225 t kWh )( . 11)( . 095$ /kWh ) (7) we get t in hours. We convert to days and find 2180 days or about 6 years. 4. See the masses in Figure (1) which start out at rest. (i) Find the velocity of the 14 kg mass just before it hits the ground. (ii) Find the maximum height reached by the 8 kg (and don’t worry about hitting the pulley). (iii) Find the fraction of mechanical energy left when the system finally comes to rest. Initially we have the 14 kg with a PE of mgh = 14 * 9 . 8 * 5. Just before it hits the ground, it has a speed v , the other mass has a height 5 m and the same speed. Thus 14 * 9 . 8 * 5 = 8 * 9 . 8 * 5 + 1 2 v 2 (14 + 8) (8) which yields v = 5 . 17 m/s . (ii) The 8kg mass keeps going upwards till it comes to rest. The extra height it gets (above the 5m) is h = v 2 / 2 g = 1 . 36 m . (iii) At the point we reach in part (ii) we have no energy of any kind for the 14kg mass. The energy of the 8kg mass is best measured when it is at rest at highest point of 6 . 36m . Initially the only energy was the potential energy of the 14 kg mass at height 5m. Thus E final E initial = 8 * 6 . 36 * 9 . 8 14 * 5 * 9 . 8 = . 727 (9)
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2 so that roughly 27% of energy is lost.
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