#$&* course Phy 201 query 10010. `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 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):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* 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: a *’ds ~ vf^2 - v0^2 ~ Fnet * ‘ds vf^s= v0^2 +2a’ds —> a * ‘ds= (vf^2 - v0^2)/ 2 Then we need to incorporate ‘ds into the Fnet equation: Fnet=m*a (multiply both sides by ‘ds) ‘ds * Fnet=m*a*’ds *******I’m not sure how to get rid of mass in this equation. 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): Is it because the mass is a constant that we can discount it to make Fnet * ‘ds proportional to m*a*’ds?
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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): ------------------------------------------------ Self-critique 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): We have different results for the last unit conversion, I redid my math, but maybe I’m still doing something wrong. ------------------------------------------------ Self-critique Rating:3
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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:OK ********************************************* 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: 95km/hr (1000m/km)=95000m/hr (1hr/3600s) = 26.39m/s So the change in velocity is 26.39m/s - 0m/s= 26.39m/s Then we can find the average acceleration by dividing the change in velocity by the change in time: aAve=‘dv/‘dt =26.39m/s/6.2 =4.26m/s^2 confidence 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):OK ------------------------------------------------ Self-critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!