#$&* course phy 201 Aug 1 4:10pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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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 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 = Fnet / 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 ------------------------------------------------ Self-critique Rating: ********************************************* 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: Knowing vf^2=v0^2+2a ds, we rearrange to get a ds=1/2(vf^2-v0^2). This shows us that it’s half the change in v^2, so (a ds) is proportional to the change in v^2. (1/2 in this case) Fnet=ma. So Fnet is proportional to a. (m in this case). Therefore, (Fnet ds) is proportional to (a ds), which I said is proportional to the change in v^2. Now we can say that (Fnet`ds) is proportional to the change in v^2. (1/2 in this case) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ Self-critique Rating: ------------------------------------------------ Self-critique Rating: ********************************************* Question: How do our experimental results confirm or cause us to reject this hypothesis? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You could tell if the results confirm or cause us to reject by simply graphing what you know and comparing it to (a ds=1/2*(change in v^2)). If (a ds) is proportional to the change in (vf^2) then the graph of (vf^2 vs. a ds) will more than likely be linear. You could always confirm this by just a simple linear graph of (vf^2 vs. ds). confidence 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): ok ------------------------------------------------ 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: 1 mile = 5280 ft, 1 hour = 3600 seconds, 1 inch = 2.54 cm, 1 foot = 12 inches. 1 mile * 5280 ft / mile = 5280 ft, 5280 ft = 5280 ft * 12 in/ft * 2.54 cm / in = 160934 cm 160934 cm * 1 m / (100 cm) = 1609.34 m 1609.34 m * 1 km / (1000 m) = 1.60934 km. 35 mi / hr = 35 mi / hr * (1.60934 km / 1 mi) = 56 (mi * km / (mi * hr) ) = 56 (mi / mi) * (km / hr) = 56 km / hr. 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. 35 mi/hr * ( 5280 ft / mi) * ( 1 hr / 3600 sec) = 53.33 ft/sec. 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): ok " Self-critique (if necessary): ------------------------------------------------ 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: 1 mile = 5280 ft, 1 hour = 3600 seconds, 1 inch = 2.54 cm, 1 foot = 12 inches. 1 mile * 5280 ft / mile = 5280 ft, 5280 ft = 5280 ft * 12 in/ft * 2.54 cm / in = 160934 cm 160934 cm * 1 m / (100 cm) = 1609.34 m 1609.34 m * 1 km / (1000 m) = 1.60934 km. 35 mi / hr = 35 mi / hr * (1.60934 km / 1 mi) = 56 (mi * km / (mi * hr) ) = 56 (mi / mi) * (km / hr) = 56 km / hr. 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. 35 mi/hr * ( 5280 ft / mi) * ( 1 hr / 3600 sec) = 53.33 ft/sec. 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): ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!