#$&* course PHY 121 6/26 2This just sent without any identifying information so I am re-sending it. 010. `query 10
.............................................
Given Solution: 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. ** STUDENT QUESTION: I wasn’t sure what equation to use to find KE the second way. What does Kef stand for? INSTRUCTOR RESPONSE: In general f stands for 'final' and 0 for 'initial'. Just as v0 and vf stand for initial and final velocities, we'll use KEf and KE0 to stand for initial and final kinetic energies. STUDENT QUESTION: Ok I know the other equation now but I still don’t really understand it. How come you multiply each velocity by 0.5? I don’t really understand the second equation KE = 1/2 m v^2 INSTRUCTOR RESPONSE On one level, KE = 1/2 m v^2 is simply a formula you have to know. It isn't hard to derive that formula, as you'll see soon, and the 1/2 arises naturally enough. A synopsis of the derivation: If force F_net is applied to mass m through displacement `ds then: a = F_net / m, and vf^2 = v0^2 + 2 a `ds It's not difficult to rearrange the result of these two equations to get F_net * `ds = 1/2 m vf^2 - 1/2 m v0^2. You'll see the details soon, but that's where the formula KE = 1/2 m v^2 comes from; the 1/2 or 0.5 is part of the solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There is nothing I just don't understand. It's a matter of assimilating this new information. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `q (This question applies primarily to General College Physics students and University Physics students, though Principles of Physics students are encouraged, if they wish, to answer the question). In terms of the equations of motion why do we expect that a * `ds is proportional to the change in v^2, and why do we then expect that the change in v^2 is proportional to Fnet `ds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Acceleration is measured in m/s^2 times. If you multiply that times distance, you would get m^2/s^2. Squaring velocity would also give m^2/s^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: In a nutshell: since vf^2 = v0^2 + 2 a `ds, a `ds = 1/2 (vf^2 - v0^2), so a `ds is proportional to the change in v^2 since F_net = m a, F_net * `ds = m a * `ds so F_net * `ds is proportional to a * `ds Thus F_net `ds is proportional to a * `ds, which in turn is proportional to the change in v^2. Thus F_net `ds is proportional to the change in v^2. More detail: It's very important in physics to be able to think in terms of proportionality. To say that y is proportional to x is to say that for some k, y = k x. That is, y is a constant multiple of x. To say that a * `ds is proportional to the change in v^2 is to say that for some k, a * `ds = k * ( change in v^2)--i.e., that a * `ds is a constant multiple of the change in v^2. In terms of the equations of motion, we know that vf^2 = v0^2 + 2 a `ds so a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2. So a `ds is a constant multiple (1/2) of the change in v^2. Formally we have a `ds = k ( change in v^2) for the specific k value k = 1/2. Now since Fnet = m a, we conclude that Fnet * `ds = m a * `ds and since a `ds = k * ( change in v^2) for the specific k value k = 1/2, we substitute for a * `ds to get Fnet `ds = m * k * (change in v^2), for k = 1/2. Now m and k are constants, so m * k is constant. We can therefore revise our value of k, so that it becomes m * 1/2 or m / 2 With this revised value of k we have Fnet * `ds = k * (change in v^2), where now k has the value m / 2. That is, we don't expect Fnet * `ds to be proportional to the change in velocity v, but to the change in the square v^2 of the velocity. STUDENT COMMENT: I am still a bit confused. Going through the entire process I see how these values correlate but on my own I am not coming up with the correct solution. I am getting lost after we discover the a `ds is a constant multiple of (1/2) the change in v^2. Is it that I should simply substitute the k into the equation? Or am I missing something else? INSTRUCTOR RESPONSE: The short answer is that by the fourth equation of uniformly accelerated motion, a `ds = 1/2 (vf^2 - v0^2), which is half the change in v^2, so that a `ds is proportional to the change in v^2. (The proportionality constant between a `ds and change in v^2 is the constant number 1/2). F_net = m a, where m is the mass of the object. So F_net is proportional to a. (The proportionality constant between F_net and a is the constant mass m). Thus F_net `ds is proportional to a `ds, which we have seen is proportional to the change in v^2. The conclusion is the F_net `ds is proportional to the change in v^2. (The proportionality constant between F_net `ds and change in v^2 is 1/2 m.) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I'm working on processing this. ------------------------------------------------ Self-critique Rating: 2 ------------------------------------------------ Self-critique Rating: ********************************************* Question: How do our experimental results confirm or cause us to reject this hypothesis? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Is this the hypothesis for the last question? The more force, the more acceleration. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The explanation for this result: On a ramp with fixed slope the acceleration is constant so a `ds is simply proportional to `ds specifically a `ds = k * `ds for k = a. In the preceding question we saw why a * `ds = k * (change in v^2), with k = 1/2. In our experiment the object always accelerated from rest. So the change in v^2 for each trial would be from 0 to vf^2. the change would therefore be just change in v^2 = vf^2 - 0^2 = vf^2. Thus if a `ds is proportional to the change in vf^2, our graph of vf^2 vs. a `ds should be linear. The slope of this graph would just be our value of k in the proportionality a * `ds = k * (change in v^2), where as we have seen k = 1/2 We wouldn't even need to determine the actual value of the acceleration a. To confirm the hypothesis all we need is a linear graph of vf^2 vs. `ds. (we could of course use that slope with our proportionality to determine a, if desired) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Still working on this. ------------------------------------------------ Self-critique Rating: 2 ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qGeneral College Physics and Principles of Physics: convert 35 mi/hr to km/hr, m/s and ft/s. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 35 miles/hour * 5280 feet/mile * 12 inches/foot * 2.54 cm/inch * 1 meter/100 cm * 1 km/1000 m = 5,643704/100,000 = 56.33 km/hour 35 miles/hour * 5280 feet/mile * 12 inches/foot * 2.54 cm/inch * 1 meter/100 cm * 1 hour/60 minutes * 1 minute/60 seconds = 5,632,704/360,000 = 15.65 m/s 35 miles/hour * 5280 feet/mile * 1 hour/60 minutes * 1 minute/60 seconds = 184,800/3600= 51.33 ft/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aWe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got the first two fine and continue to get 51.33 instead of 53.33 for the last one. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: Openstax problem 2.16: A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. Assuming uniform acceleration, what are its acceleration and the distance covered? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The change in velocity is 0 m/s to 30 m/s, which is v0-vf = 30 m/s - 0 m/s = 30 m/s. If you divide the change in velocity by the time, you will get 30 m/s / 7 s = 4.29 m/s^2 for the acceleration. For the distance covered, you start by finding the average velocity (0 m/s + 30 m/s)/2 = 15 m/s. Multiply the 15 m/s times the 7 seconds, for the distance of 105 meters. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: It seems unlikely that the significant figures in this problem are realistic. What are the chances that an animal's speed and the corresponding time interval are both measured to 3-significant-figure accuracy (which might be realistic, though with different parts of the animal's body at any instant having different forward speeds the velocity measurement would be very challenging), and both yield such round numbers (1 in 100 chance of each means 1 in 10 000 chance this would occur). However if we accept the significant figures specified for this problem, the result can be obtained as follows: Acceleration is rate of change of velocity with respect to clock time, so that acceleration = (change in velocity) / (change in clock time) = 30.0 meters / (7.00 seconds) = 4.29 meters / second^2. Note that 30 / 7 = 4.28571428571... , but with 3-significant-figure information we can only be confident of our 3-significant-figure rounding of this result. The displacement of the cheetah will be `ds = vAve * `dt = (0 + 30.0 m/s) / 2 * 7.00 s = 105 meters. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qGen phy and prin phy prob 2.16: car accelerates uniformly from rest to 95 km/h in 6.2 s; find acceleration YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `dv / `dt = acceleration 95 km/hr / 6.2 s * 1 hr/60 min * 1 min/60 sec * 1000 m/km = 26.39 m/s divided by 6.2 seconds = 4.26 m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** 95 km/hr = 95,000 m / (3600 sec) = 26.3 m/s. So change in velocity is `dv = 26.3 m/s = 0 m/s = 26.3 m/s. Average acceleration is aAve = `dv / `dt = 26.3 m/s / (6.2 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'. STUDENT QUESTION: How did we know that the final velocity was 0? INSTRUCTOR RESPONSE: The final velocity was 0 because the car came to rest. Summary of what we were given: Initial velocity is 95 km/hr, or 26.3 m/s. Final velocity is 0, since the car came to rest. The velocity makes this change in a time interval of 6.2 seconds. We can easily reason out the result using the definition of acceleration: The acceleration is the rate at which velocity changes with respect to clock time, which by the definition of rate is (change in velocity) / (change in clock time) The change in velocity from the initial 0 m/s to the final 26.3 m/s is 26.3 m/s, so acceleration = change in velocity / change in clock time = 26.3 m/s / (6.2 s) = 4.2 m/s^2. We could also have used the equations of uniformly accelerated motion, with vf = 26.3 m/s, v0 = 0 and `dt = 6.2 seconds. However in this case it is important to understand that the definition of acceleration can be applied directly, with no need of the equations. (solution using equations: 2d equation is vf = v0 + a `dt, which includes our three known quantities; solving for a we get a = (vf - v0) / `dt = (26.3 m/s - 0 m/s) / (6.2 s) = 4.2 m/s^2.) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My graph had one line going across at 25 m/s and the other going down from 15 m/s to 0. My math looked like this: The slower train that is accelerating at a negative rate is going to go from 15 m/s to 0 m/s. The change in velocity will be -15 m/s. If you divide that by the acceleration, you will get that it will take 15 seconds for the train to come to a complete stop. In 15 seconds, the train traveling at 25 m/s will go 375 meters. However, it is 200 meters behind the train that is now sitting on the tracks. This means that the trains will collide at 575 meters down the track from time point zero. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): It looks like I misinterpreted the question and thought that the first, slower train was slowing down, while the faster train was maintaining its speed. ------------------------------------------------ Self-critique Rating: 2" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!