#$&* course Phy 201 010. `query 10
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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 wasnt 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 dont really understand it. How come you multiply each velocity by 0.5? I dont 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, andvf^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): Ok. I forgot the KE0 in the second equation but i'm guessing you only need that if the initial velocity is something other than zero. ------------------------------------------------ Self-critique Rating: 2
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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 don't understand how you rearranged the equation vf^2 = v0^2 + 2 a `ds, to get, a `ds = 1/2 (vf^2 - v0^2). ------------------------------------------------ Self-critique Rating:
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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): Didn't do the lab. ------------------------------------------------ Self-critique Rating: ------------------------------------------------ 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: 35mi/hr * 1.6 = 56km/hr = 56000m/hr = 933.33m/min = 15.5m/s 51.33ft/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 think your ft/sec is off ------------------------------------------------ Self-critique Rating:
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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): the question asks to find the acceleration of a car from rest to 95km/h not 95km/h to rest. were we supposed to find the deceleration of the car as well? ------------------------------------------------ Self-critique Rating:
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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): the question asks to find the acceleration of a car from rest to 95km/h not 95km/h to rest. were we supposed to find the deceleration of the car as well? ------------------------------------------------ Self-critique Rating:
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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): the question asks to find the acceleration of a car from rest to 95km/h not 95km/h to rest. were we supposed to find the deceleration of the car as well? ------------------------------------------------ Self-critique Rating: