AP Physics 1 & 2 Springs

Hello everyone and welcome back to Educator.com.0000

I am Dan Fullerton and in this mini-lesson, we are going to go through page 1 of the APlusPhysics worksheet on springs.0002

You can find the link to the worksheet down below the video.0009

With that, let us get started.0012

Question 1 -- In an experiment, a student applied various forces to a spring and measured the springs corresponding elongation.0014

The table below shows his data.0021

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).0024

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.0032

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.0048

There is my general look at the data.0068

Number 2 says draw our best fit line.0073

To draw our best fit line, you are looking to have roughly the same amount of points above and below the line.0075

I would recommend using as straight edge, so it should look something kind of like that.0079

Number 3 -- Using your graph, calculate the spring constant of the spring showing all work.0087

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.0094

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...0110

...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.0123

Moving on to Number 4 -- A 10 N force is required to hold a stretched spring 0.2 m from its rest position.0144

What is the potential energy stored in the stretched spring?0155

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.0159

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) × x².0174

I can divide (x) and the square, therefore, potential energy will be 1/2 F × (x)...0188

...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 1.0200

Number 5 -- A 5 N force causes a spring to stretch 0.2 m.0218

What is the potential energy stored in the spring?0223

Well, we have a force of 5 N. We have a stretch of 0.2 m.0226

Let us try this one a different way.0233

Let us find the spring constant first, which is F/x, or 5 N/0.2 m, which should be 25 N/m.0235

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...0248

...25 × 0.5 × 0.2² will give us a value of 0.5 J -- Answer Number 2.0264

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.0279

How much elastic potential energy is stored in the car's spring during this process?0287

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.0292

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.0315

Number 7 -- The graph represents the relationship between the force applied to each of two springs (A) and (B) and their elongations.0328

What physical quantity is represented by the slope of each line?0336

The slope of the force versus elongation graph, remember, is the spring constant.0339

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)?0349

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 × x².0365

We can divide out that (x) to say that potential energy is 1/2 Fx.0379

Now we have 1 kg mass suspended from each of these.0389

If each mass is at rest, how does the potential energy stored in spring (A) compare to the potential energy stored in spring (B)?0393

As I look at this, we need to figure out really what (x) is going to be.0401

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/k².0407

So I could also write this as -- divide out the (k) and that will be F²/2 k.0424

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.0434

(A) has less potential energy. That was a kind of an around about way to get to the answer on that one.0455

Hopefully that gets you a great start on springs.0462

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.0465

Go ahead and keep moving on; you are probably ready for the AP level questions.0473

Thanks so much for your time everyone and make it a great day.0477