course Phy 231 ??????Q???l?\??assignment #010010. `query 10
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10:46:36 Query introductory problem set 3 #'s 7-12 Describe two ways to find the KE gain of an object of known mass under the influence of a known force acting for a given time, one way based on finding the distance the object moves and the other on the change in the velocity of the object, and explain why both approaches reach the same conclusion.
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RESPONSE --> One way to find KE involves finding 'ds, by using the equation: 'ds = (F/2m)*'dt^2, which can then be translated into: KE = .5F^2 / m * 'dt^2, by substituting some of the relationships explained in the problem set flow charts. Another way to find KE, is from velocity, where KE = 0.5 * m * v^2 The KE can be calculated from the initial velocity point (we'll call it KE1) and then again from the change in velocity (we'll call KE2). The KE gain can then be calculated from the difference between KE 2 and KE1. We should reach the same conclusion either way because the root equations used to reach both final answers are the same. For example, in both equations we use the a = F / m confidence assessment: 2
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10:53:28 ** First way: KE change is equal to the work done by the net force, which is net force * displacement, or Fnet * `ds. Second way: KE change is also equal to Kef - KE0 = .5 m vf^2 - .5 m v0^2. **
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RESPONSE --> OK I believe I was incorrect on the first part. So the work done by Fnet, is equal to Fnet * 'ds. I will commit this to my notes. I have also noted the equation for change in KE, which is equal to: 0.5 m * vf^2 - 0.5 m * v0^2 self critique assessment: 2
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10:53:52 General College Physics and Principles of Physics: prob 2.04 convert 35 mi/hr to km/hr, m/s and ft/s.
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RESPONSE --> OK - different class confidence assessment: 3
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10:54:06 We need a conversions between miles and meters, km and ft, and we also need conversions between hours and seconds. We know that 1 mile is 5280 ft, and 1 hour is 3600 seconds. We also know that 1 inch is 2.54 cm, and of course 1 foot is 12 inches. 1 mile is therefore 1 mile * 5280 ft / mile = 5280 ft, 5280 ft = 5280 ft * 12 in/ft * 2.54 cm / in = 160934 cm, which is the same as 160934 cm * 1 m / (100 cm) = 1609.34 m, which in turn is the same as 1609.34 m * 1 km / (1000 m) = 1.60934 km. Thus 35 mi / hr = 35 mi / hr * (1.60934 km / 1 mi) = 56 (mi * km / (mi * hr) ) = 56 (mi / mi) * (km / hr) = 56 km / hr. We can in turn convert this result to m / s: 56 km/hr * (1000 m / km) * (1 hr / 3600 sec) = 15.6 (km * m * hr) / (hr * km * sec) = 15.6 (km / km) * (hr / hr) * (m / s) = 15.6 m/s. The original 35 mi/hr can be converted directly to ft / sec: 35 mi/hr * ( 5280 ft / mi) * ( 1 hr / 3600 sec) = 53.33 ft/sec.
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RESPONSE --> OK - not my class self critique assessment: 3
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10:55:44 Gen phy and prin phy prob 2.16: sports car 95 km/h stops in 6.2 s; find acceleration
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RESPONSE --> OK - not my class confidence assessment: 3
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10:55:50 ** 95 km/hr = 95,000 m / (3600 sec) = 26.3 m/s. So change in velocity is `dv = 0 m/s - 26.3 m/s = -26.3 m/s. Average acceleration is aAve = `dv / `dt = -26.3 m/s / (6.2 s) = -4. m/s. So the time to come to a stop is `dt = `ds / vAve = 50 m / (12.5 m/s) = 4 s. Acceleration is rate of velocity change = change in velocity / change in clock time = -25 m/s / (4 s) = -4.2 m/s^2. Extension: One 'g' is the acceleration of gravity, 9.8 m/s^2. So the given acceleration is -4.2m/s^2 / [ (9.8 m/s^2) / 'g' ] = -.43 'g'.
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RESPONSE --> OK - not my class self critique assessment: 3
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12:09:14 univ phy 2.66 train 25m/s 200 m behind 15 m/s train, accel at -.1 m/s^2. Will the trains collide and if so where? Describe your graph.
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RESPONSE --> From simple reasoning, we know that the passenger train is slowing by -0.1 m/s every second, so it will take 10 seconds to decrease the velocity by 1 m/s. Reasoning further, we know that since the freight train is travelling at 15 m/s and the pass. train is travelling at 25 m/s (a difference of 10 m/s), it will take 100 seconds for the pass. train to reach the same velocity as the freight train. We then can deduce that if the trains have not collided within this timeframe, they will not collide at all, since they would then be travelling at the same rate. Therefore, we can use 'dt = 100s for both trains and see what the 'ds equals to. If the 'ds of the pass. train is greater than the 'ds of the freight then know they will have collided. If not, then they will not have collided. Pass. train: 'ds = vAve * 'dt 'ds = (25 m/s + 15 m/s) / 2 * 100s 'ds = 20 m/s * 100 s = 2000 m Freight train: 'ds = vAVe * 'dt 'ds = 15 m/s * 100 s = 1500 m + 200 m = 1700 m So yes, they will collide. Where? The trains will collide at the point where 'ds is the same for both trains over a certain 'dt. I graphed the 'ds vs. 'dt on a graph for both trains and found the point where they intersected. The intersection occurred near the 22.5 s = 'dt mark, which corresponds to about the 537.5 m mark. confidence assessment: 2
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12:17:45 ** If we assume the passenger train is at position x = 0 at clock time t = 0 we conclude that the position function is x(t) = x0 + v0 t + .5 a t^2; in this case a = -.1 m/s&2 and x0 was chosen to be 0 so we have x(t) = 25 m/s * t + .5 * (-.1m/s^2) * t^2 = 25 m/s * t - .05 m/s^2 * t^2. To distinguish the two trains we'll rename this function x1(t) so that x1(t) = 25 m/s * t - .05 m/s^2 * t^2. At t = 0 the freight train, which does not change speed so has acceleration 0 and constant velocity 15 m/s, is 200 m ahead of the passenger train, so the position function for the freight train is x2(t) = 200 m + 15 m/s * t . The positions will be equal if x1 = x2, which will occur at any clock time t which solves the equation 25 t - .05 t^2 = 200 + 15 t(units are suppressed here but we see from the units of the original functions that solutions t will be in seconds). Rearranging the equation we have -.05 t^2 + 10 t - 200 = 0. The quadratic formula tells us that solutions are t = [ - 10 +- sqrt( 10^2 - 4 * (-.05) * (-200) ) ] / ( 2 * .05 ) Simplifying we get solutions t = 22.54 and t = 177.46. At t = 22.54 seconds the trains will collide. Had the trains been traveling on parallel tracks this would be the instant at which the first train overtakes the second. t = 177.46 sec would be the instant at which the second train again pulled ahead of the slowing first train. However since the trains are on the same track, the accelerations of both trains will presumably change at the instant of collision and the t = 177.46 sec solution will not apply. GOOD STUDENT SOLUTION: for the two trains to colide, the 25 m/s train must have a greater velocity than the 15 m/s train. So I can use Vf = V0 + a('dt). 15 = 25 + (-.1)('dt) -10 = -.('dt) 'dt = 100 so unless the displacement of the 25 m/s train is greater than the 15 m/s train in 100 s, their will be no colision. 'ds = 15 m/s(100) + 200 m 'ds = 1700 m 'ds = 25 m/s(100) + .5(-.1)(100^2) = 2000 m. The trains collide. **
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RESPONSE --> OK I was close... Interesting solution using the quadratic equation. At one point, I actually did have the following: v0 'dt - 0.5 a 'dt^2 = 'ds = vAve 'dt, which can be simplified to: 25 m/s 'dt - 0.05 m/s^2 'dt^2 = 15 m/s 'dt however, I left out the 200 m part, so I ended up having to just use trial and error. Had I added this in, I would have been able to construct the quadratic equation of: dt^2 - 200 'dt - 4000 = 0 self critique assessment: 2
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