WEBVTT physics/ap-physics-1-2/fullerton
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Hello everyone and welcome back to Educator.com.
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I am Dan Fullerton and in this mini-lesson, we are going to go through page 1 of the APlusPhysics worksheet on springs.
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You can find the link to the worksheet down below the video.
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With that, let us get started.
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Question 1 -- In an experiment, a student applied various forces to a spring and measured the springs corresponding elongation.
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The table below shows his data.
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Question 1 says on the grid at right, plot the data points for force versus elongation, so I have a point here at (0,0).
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I have a point of elongation of 0.3, 1.3 and 1 N, at point (6,7) and 3 N somewhere right around there.
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At 4 at an elongation of 1; we have 4 N; at an elongation of 1.3, we have 5 N; and at an elongation of 1.5 m, we have 6 N.
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There is my general look at the data.
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Number 2 says draw our best fit line.
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To draw our best fit line, you are looking to have roughly the same amount of points above and below the line.
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I would recommend using as straight edge, so it should look something kind of like that.
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Number 3 -- Using your graph, calculate the spring constant of the spring showing all work.
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Well, the way we get the spring constant from a force versus an elongation graph -- if F = kx, then k = F/x and we are going to get that from this graph by taking the slope, where our slope is going to be our rise/run.
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What I am going to do is I am going to pick a couple of points on the line that are pretty easy to use and as we actually have 0 and our 1.5 m line at 6 and 0 and that looks to fit our best fit line pretty well...
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...I can actually use those easy points in this case, where my rise is going to be -- I go from 0 to 6 N, so we have risen 6 N and our run is 1.5 m, which is going to give me a spring constant of 4 N/m.
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Moving on to Number 4 -- A 10 N force is required to hold a stretched spring 0.2 m from its rest position.
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What is the potential energy stored in the stretched spring?
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Well, what we know is potential energy is 1/2 kx², however, we do not have (k) yet, but we know that we have a force of 10 N when it is stretched out by 0.2 m.
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If k = F/x, what I can do over here is I can replace (k) with F/x, so potential energy is 1/2 k (F/x) × x².
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I can divide (x) and the square, therefore, potential energy will be 1/2 F × (x)...
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...which will be 1/2, our force is 10 N, and our (x) is 0.2 m, so 10 × 0.2 = 2 and 1/2 of that is going to give me 1 N-m or 1 J -- Answer 1.
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Number 5 -- A 5 N force causes a spring to stretch 0.2 m.
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What is the potential energy stored in the spring?
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Well, we have a force of 5 N. We have a stretch of 0.2 m.
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Let us try this one a different way.
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Let us find the spring constant first, which is F/x, or 5 N/0.2 m, which should be 25 N/m.
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All right, so to find its potential energy, that is 1/2 kx² or 1/2 × 25 N/m × x², which will be 0.2 m² and we are just going to check this out quick...
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...25 × 0.5 × 0.2² will give us a value of 0.5 J -- Answer Number 2.
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Number 6 -- The spring of a toy car is wound by pushing the car backward with an average force of 15 N through a distance of 1/2 a meter.
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How much elastic potential energy is stored in the car's spring during this process?
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Well, the way we could do this is we have to do work on it in order to put energy in the spring, so the work that we do -- well, is going to be our force times our displacement, which is going to be 15 N × 0.5 m (displacement) or 7.5 J.
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The energy that we did in order to move that back must be what is now stored in the spring, so that is the elastic potential energy, 7.5 J.
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Number 7 -- The graph represents the relationship between the force applied to each of two springs (A) and (B) and their elongations.
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What physical quantity is represented by the slope of each line?
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The slope of the force versus elongation graph, remember, is the spring constant.
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One last problem here. Number 8 -- Same graph, same situation -- If a 1 kg mass is suspended from each spring and if each mass is at rest, how does the potential energy stored in spring (A) compare to the potential energy stored in spring (B)?
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Well, let us write our equation -- potential energy is 1/2 kx², but remember k = F/x, so potential energy is also 1/2 F/x × x².
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We can divide out that (x) to say that potential energy is 1/2 Fx.
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Now we have 1 kg mass suspended from each of these.
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If each mass is at rest, how does the potential energy stored in spring (A) compare to the potential energy stored in spring (B)?
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As I look at this, we need to figure out really what (x) is going to be.
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Let us figure out (x), so that will be F/k, so I am going to replace that equation again, so potential energy is 1/2 × k and my (x) is going to be F/k ², so that will be 1/2 kF²/k².
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So I could also write this as -- divide out the (k) and that will be F²/2 k.
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The force on each should be the same because we have the same gravitational force on the mass, but the one that has the larger spring constant is going to have a smaller potential energy because (k) is in the denominator, so in this case (A) has the bigger spring constant, so it is going to have less potential energy.
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(A) has less potential energy. That was a kind of an around about way to get to the answer on that one.
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Hopefully that gets you a great start on springs.
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If this did not go so well, now would be a great time to go back and review the more detailed lesson on springs and if it went great, well terrific.
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Go ahead and keep moving on; you are probably ready for the AP level questions.
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Thanks so much for your time everyone and make it a great day.