So let's go ahead and get started. Looks like we have a light Friday crowd. Everyone getting ready for Memorial Day. Today, a little different. Hopefully you got the email.

I kind of decided this last minute, obviously. I decided yesterday. We're going to basically have an exciting day today of solving two problems. Yay! Good. And there are first what I would call real examples of the type of problems that certainly would be on the final.

And even with this next midterm coming up, the free response energy problem will almost definitely involve some kinematics or force or momentum. It'll be a multi-step problem.

So given the complexity of these, you may have noticed I posted, and I sent an email, fill in the blank five box solutions that you could print out and have with you today to go through and fill in as the lecture goes. Because as I go through these, I'm going to spend a lot of time, even more than usual, pointing out how to think about a problem like this.

How to approach. How to recognize what you're doing. And a little less time actually solving the details. So you'll really want to go look at the solution when it gets posted after class. But if you have that five box thing in front of you, as I discuss the different things

that come up, you can say, oh, that'll go here and it kind of can help you formulate things. So hopefully that'll help you out. Not much else I want to say at the beginning, so let's just get into it. What I will do, I'm going to give away the answer to two of the four clicker questions right now.

So we're going to do the first problem, which will involve using our force laws plus energy. And we're going to do a second problem that involves energy and kinematics. And I'm telling you this right at the beginning, not just to give away an

answer to a clicker question, but also so you can be prepared to watch for that. Let's do it, see how I recognize that that's what's going on. So let's go to our first problem.

This is a bouncing on a trampoline. And notice, this is one thing you have to watch for now that we've learned about springs. There are lots of things in the world that act like a spring. You saw in the homework, a number of you had emailed, made comments that homework 7.9 problem was a bit of a challenge.

The climber falling with the rope and then the rope stretching. The rope was a spring, it stretched, it had a spring constant. That one you'll see is actually much easier to visualize and understand with energy conservation. We did that in the work chapter, you should have used work done by gravity, work done by the spring.

If you go back and do it, pretty much it's this trampoline problem in a sense that I'm about to do. And do energy conservation, you'll see it's much easier. But we have to be prepared. And what does it mean to act like a spring? That means it obeys the force law, f equals minus kx.

And it has a potential energy if it's in the system of one half kx squared. And so the key thing is always the spring constant k. If I've got a spring, I've got to worry about k.

Now, this is actually telling you two pieces of information. If I just stand still on the trampoline, it's displaced a certain amount from its equilibrium. Once you know something is a spring, it will have an equilibrium position.

In the case of the trampoline, in case you were not familiar with them, that's when the trampoline is flat. So I can draw, this would be, I'm going to call it y, y equals zero, the equilibrium position. And then when I stand on it, the trampoline looks like this, and it's displaced y equals 0.2 meters down.

So that's the first thing I learned in this problem. Now, when I jump on the trampoline, the second thing that I'm doing, I'm told I go down 1.5 meters. And now I finally get to the question, how high above the flat trampoline do I get when I'm bouncing like that?

So do we all get the situation? This is the actual question. So what I'm being asked to find is basically the height above the flat trampoline.

Any questions on the problem? Good. Let's go to our first clicker question.

Is this one using kinematics, forces, energy, both A and C, or both B and C? So 54 percent of you paid attention to the first slide. We are going to be using forces and energy.

So let's spend a moment kind of taking this apart. The two motions that we're given, we're given a situation, right, that involves a single instant. We are still, and our motion's not changing.

So the first thing we know, if we're standing still, and now sometimes the velocity can be zero and changing, but if we're just standing still on the trampoline and nothing's happening, we're just hanging out, what do we know about our motion? What big physics thing do you immediately know?

Yeah, you know about gravity, you know something else, something about an object whose velocity is not changing. Shout it all together now? Acceleration is zero. You need to know that. These are the sort of things, and this is the heart of what causes the most problems in physics.

Recognizing these things is way more valuable than asking yourself when you start a problem, what equation should I use? If you go right to what equations you should use, you'll almost never get that right if you don't know what the motion is first. Being able to recognize when you have acceleration equals to zero,

when you have a given constant acceleration, or when it's moving in a circle so you have centripetal acceleration. That immediately tells you important stuff about the situation. We immediately know a equals zero, and we know we're at a single instant. That means there's no initial, final, right?

A key thing about energy, work energy, was always about the difference between the initial and final. Once you see these pieces, the only concept from the course that's really available to you is F equals ma.

If you want to learn anything, now you may not need to learn anything about this situation. I might have just told you about this for fun. But if you're going to use this part of the problem, the only concept that can be connected to it, that we've learned, is Newton's second law.

Because it's at a single instant in time, while you're standing there, and it's involving zero acceleration. Kinematics is about moving from one place to the other, energy is about a before and after, and we'll see momentum conservation is about a before and after. Work has a delta x. Impulse, which we're going to learn next week, has a delta t.

So this is a big thing to recognize. So right away, not even dealing with the question or the answer, you should be able to recognize, okay, I've got a part of my problem, if I need to use this part, I'm going to be using F equals ma in some way. And then follow that away in your brain.

Now, the question is about how high you're going. Right? And it's got an initial state, where you're stretched down, and a final state, where you're up some height, h final.

Now, moving from some position and flying through the air to some height, what are the possible ways you might find heights when you're thrown up into the air? There's actually two. We've learned three concepts so far, they were on the multiple choice.

Kinematics, forces, and conservation of energy, or work energy, the more general one. Of our three concepts, work equals change and energy, kinematics and forces, which two would allow you to solve for, given some initial condition, what final height you get to, possibly. Energy can certainly do that in principle, because it's about a before and an after.

And what else can do it? Kinematics, the first week or second week, how many problems did you do, throwing something up in the air and asking how high it went, right? Now, why in this problem is kinematics going to be very hard to use? What have we kind of added here that makes it interesting?

This whole business with the spring, right? A kinematics problem is always about, if you know v naught, then you can find h, right? That's what kinematics is telling you about. Here, we're starting somewhere, we're going to generate a v naught when we leave the trampoline,

but we have all this initial stuff. So that's where we really immediately go, okay, let's better start with energy. We might find, as you start to set up the problem, that you actually need to use three ideas. This might be a problem where you need to use energy to get the beginning part and kinematics later.

Now, it's not. You can do the whole thing with energy. But there are times where you might end up using all three of these. I want to pause there. Any questions about how I kind of attacked this and broke it into these pieces? Because that, when you're faced with one of these problems, if you don't take two to three minutes or five minutes at the beginning to break down

what are the motions I'm dealing with, what are the most likely physics I'm dealing with, it's really, really hard to make progress on these. If you just try and solve them from the beginning matching equations. Yes, and kinematics? Okay, what I did was, let's suppose you said, oh, I'm going to use kinematics to solve this, right?

Well, first of all, you need two things. I need constant acceleration, and my equations are all about delta x equals something and v final equals something, and they all have a v naught in them, right? If I look over here, yeah, my v naught here is zero.

I kind of have a v naught, but that's not the right v naught. Kinematics is all about being launched into the air and only having gravity. And for this whole motion here, I have a spring. And the big thing about the spring, right, is f equals minus kx, not constant acceleration.

The acceleration changes because the spring force changes. If the force changes, a changes. So once I throw in that spring, a spring is a good hint that you're either going to use forces or energy, right? It kind of eliminates kinematics for that part of the problem. Now, we'll see in our next problem, we have a spring that launches something,

and then the second half we use kinematics. So you could have ended up having to use it here, but because we're going in a straight line, we know the motion, it's all straight, and we know all we care about is our final height, we're pretty confident height comes into mgh, we should be able to get it. So that's why we're going to start with energy. If it doesn't give us enough, yes, we'll have to look for something else.

But we want to start there. Any other questions? Good question. So let's start it having decided that we're going to use energy. So I'm going to redraw things.

So for the motion, I've got my spring coming down, and I'm going to do two things, and this is the one time you're allowed to do this. I'm going to call this h equals zero for the gravity part of the problem,

but I'm going to call this x equals zero for the spring. And you'll see why, and I'm allowed to do that. And I'm going to go to some height h, that's where I end up,

which, if you read the problem, the answer I actually want is the h relative to the flat point of the spring. So I could do lots of different things here. I could put h equals zero right here if I wanted, and then I have a negative height. But I always find it easiest to put it right at the bottom of where I want to be. Now, this is the motion.

I've already decided I'm going to use energy, so I must pick a system. So let's go back to our PowerPoint slides, and, oh wait, I didn't have one for system, okay?

Let's pick our system and then ask that question, right? I'm going to pick the trampoline, the person, and the earth. Pretty much everything.

Nice thing for this, I know the net force is zero, so now I know there's no work, so I'm actually using conservation of energy. I don't have to use work equals change in energy. They're basically the same thing, but it's nice to be specific. If you look at all my solutions, you really want to look at those strategies and you'll see that.

So now, which type of energies do we need? Having picked that system, talk to the person next to you and determine which type of energies you need. All but C. Well, because, you know, there's at least two of them, and so you might as well go with that, right?

Notice there's no friction in this problem, right? And this one here, if I was to pick that, that has to do with our energy dissipated, with our force of friction times distance, but there's no friction here, so I don't have to worry about that one.

I do have both the spring and gravity, because I have the earth in the system. This happens because I put the earth there. If I did not, I would have work done by gravity, because I'd have a force on the system. Likewise, I put the trampoline, which in this case is our spring, so it has its energy.

Having said that, let's make our table of energies. And what I want you to do is, because I think I gave it in my little pre-handout thingy,

what is my initial kinetic energy in this case? This is the easy one, everyone. Initial kinetic energy? Zero. Final kinetic energy? Zero. Why?

What is the number one reason, going back to the beginning of the quarter, that our speed is zero in the final situation? We're at the top of our jump, and that is the key.

If I'm going to answer how high I jump, that means I'm at the top, and so we all learned and recall that the speed and the direction we're moving, and we're only moving vertically here, is zero. So, notice nowhere did it say that it was zero, right?

Okay. Yay. There we go.

Is everything else still running?

Relax for a moment, talk to your neighbor. Oh, and we're back.

Yeah, the real problem there was I thought in my head today, I haven't had any technical difficulties yet, and I forgot to knock on wood. And sure enough, PowerPoint hangs. Now, we're back, live.

Potential energy from the spring. What's our initial potential energy of our spring? One half kx squared. And notice, what is x? It's our 1.5 meters that we're down.

And notice, it's x squared. So, for potential energy, it doesn't matter whether I call this negative 1.5 or positive 1.5, because I'm going to square it. What's my final potential energy of the spring? Zero. Zero. Why? The trampoline, we're off the spring, it's back to flat.

Now, notice in this case what I've done. What's my potential energy of gravity initially? Zero. Zero. And what's my final? No, no, gravity. mgh. mgh. And it's this h here, as defined in my picture.

So, you'll be very careful to let me know how your variables are defined. So, now I set my initial energy equal to my final energy, because we recall my work was zero. And I'm very happy, I have an expression for the height now.

And if you look at what was given in the problem, I know x, right? x equals 1.5 meters. I always know g, that's my 10 meters per second squared. And my mass was 80 kilograms. So, my only issue is I don't know k.

At this point, because it's now been another 10 minutes into the exam, I may panic. But, I may also remember what I did at the beginning. What did I say to myself when I first looked at the words in this problem? Is there another part to the problem? Yes. There was my fourth part. Notice what I said on Monday and Wednesday's lecture.

Energy only tends to give us one equation. In most interesting problems, there's at least two unknowns. And there's two unknowns, because we're often connected to some other part of the problem. So, in this case, the other part of the problem that we're connected to, right?

Was that initial a equals zero. And now I'm going to switch to y, y equals 0.2 meters. And I'm going to call this y equals zero.

And I'm going to make y positive in the up direction. So, notice the pieces that came in, right? I decided I'm going to use f equals ma. So, I need to identify a. I must have a coordinate system. And now what's the next thing I will do?

If I'm doing a force problem, what will I do? Draw my free body diagram. What's the free body diagram for me when I'm standing on the trampoline and it's down at the bottom?

On what down? So, I have my weight down. And what force is up on me? We can call it a number of different things here, right? In the old beginning, we might have called it the normal force, because I'm standing on the trampoline. I might have called it the force from the trampoline. But my trampoline is acting like a spring.

So, I'm going to call it my spring force, just to remind me that I have a formula for it, minus ky. Notice, here's one of the dangers when an object acts like a spring. If I draw, and don't do this, if I also draw a normal force, that's wrong.

There's just this one force between me and the trampoline. And that's how you remember it, right? Newton's third law tells you that for every two objects, there's one force between them. So, when I'm standing on the trampoline, there is one force from the trampoline on me.

That force is given, in this case, I would normally call it the normal force, but it's given by the fact that I'm treating it as a spring, so I now know a formula for it, and it's minus ky. Now, I know it has to be up, because the spring went down. Notice, minus ky works perfectly, because I'm actually at y equals negative 0.2 meters.

So, when I plug that in, I'm going to have k times 0.2 meters, which will be a positive number, which is up. So, you have to be careful with your minus signs that you don't end up with the wrong number of them, and it doesn't work out.

In this case, it will be obvious, because you would end up with something that makes no sense, because you just have to have two forces that add to 0, so they have to cancel. So, if I'm writing down Newton's Law, I would have minus ky minus mg equals 0. So, k equals mg over negative y, and when I plug that in, I get 4,000 newtons per meter.

And notice, I do get a positive number because of this effect, plugging in y equals minus 0.2 meters. If I get a negative number for k, I know I did something wrong with my minus signs, because k is always positive.

Any questions? Yeah. Well, it is. See, I put my minus sign right here. Notice, when I write gravity, right here, I'm just writing its magnitude. And this, that's why I wanted to do this particular problem.

I could have, it depends how you, how I want to say this. The arrow lets me know, in this case, the direction no matter what. With springs, it's helpful to be very careful. Remember that the formula has the minus sign in it.

And actually, I should be, I should leave my arrow here, because I'm really not doing the magnitude for the force of spring at this moment in case I have the direction wrong. So, I'm being very careful, I'm leaving that minus sign here, I'm keeping careful track of what y is, and when I'm all done, I say yes, I got it positive, it is up, I did it right.

With gravity, the magnitude is always just mg and it's always down. And so, I don't put the minus sign in until I write it in the f equals ma, because that's just the way I've done it all quarter, that's the way I always do it, because I go back and look which way was positive, and I have my arrow showing me it's down.

I know this could be very confusing and bad, but it is just, it's a fairly common convention in the books and among myself, and it's consistent with all the rest of my lectures. If I turned and made down positive, right, then I still would have done these two things, and now I would have plugged in over here a positive 0.2,

and I would have minus k times positive 0.2 plus mg equals 0. So, this is for down positive, and notice this still gives me that, which still gives me, I mean, I'm sorry, it doesn't give me that formula.

This still gives me that answer, because from here I still have k equals mg over 0.2, which gives me that. It isn't the best, because I do mix two different things, a magnitude with a non-magnitude, yes,

because that is where I picked h to be 0 at the beginning. And that, remember, we said this the other day, all that really matters is the change in potential energy,

so where potential energy equals 0 is arbitrary. You're free to pick, and notice I'm free to do the spring. This is the one case where I don't need the same coordinate system for both potential energies, because they're changes, they're not absolute values, and they're done separately.

So I can do my potential energies for my spring with a different coordinate system than I do my potential energies for gravity. And that's what I did here, notice, right? For the spring, 0 is here. For gravity, 0 is here.

And that also is very common. If you read a lot of the solutions when I post them for the homework, you'll see in the book that I'm doing that. Now sometimes they pick 0 the same everywhere, and then you get negatives for potential energy for gravity. You've got to have negative heights and positive heights. I always like to make h equal 0 at the lowest point I'm going to deal with,

whether it's the initial or final, whichever is lower, so that I don't ever have to deal with anything negative. But that's just me. Yes? Oh, I did.

I just took a step to do it, to make sure I was doing it right. Yeah, the more you do these problems, you'll realize, oh, there's two forces. So ky has to equal mg. And that's a magnitude. And my only choice is for k to equal mg over the magnitude of y.

And I'm done. But if you haven't done a lot of these or don't see that, if you do it step by step as we've done every force problem, you'll get there. And that's true of most of what we do now. As you've done more of these, particularly with the Newton Law's ones, you know, are you going to write fw and then write mg later?

No, just go right to mg, if you're comfortable with it. Now in your strategy, you're going to tell me you used the definition, or I'll wonder where mg came from. But that's the beauty of having a strategy. The beauty of using words, and that's another point I want to make. The strategy, again, is not there for some arbitrary reason.

The strategy is there so I know you can explain to me what you did. And I don't require paragraphs, I just need short sentences. And that's why your strategy should match your calculation. If you don't have time to do a calculation and you write a correct strategy, I still know you know how to do the problem. Right, and so if you tell me I'm going to use these definitions,

and then those expressions just show up, I'm good, right? I didn't need you to do each step. If you tell me you're using Newton's second law and the acceleration is equal to zero, and you just write down ky equals mg, I know where that came from. And I'm happy. So that's the advantage of it, and that's why they go together. If you're not sure, and you do all those steps, that's fine too.

Other questions? Yes. Right, so right here, y is the vector. k is just a positive constant always.

You can see, this is it. This is putting it all together. This right here will be the final in some version. And there will be two of these like this. Just to give you an idea, as you plan for the final,

ideally you should be able to do that problem in half an hour. I'm just giving you a gauge, right? Any other questions on that?

Excellent. What we're going to do now is stand up and stretch for two seconds, and then sit back down. Or sit in your seat and stretch, or talk to the person next to you. I don't have a demo to wake you up, so this is waking you up. Excellent. Now that we're all awake, we're going to build our rocket launcher.

So we're building a spring-loaded rocket launcher. And we're going to, we want one that when it launches things at 30 degrees, when it launches at 30 degrees, it travels 100 meters.

So, and we compress our spring 5 centimeters, we need to know what spring to buy to build it. That's our rocket launcher. And we know the mass of the rocket we're launching. So the obvious question. What type of physics is needed to solve this problem? Conservation of energy, kinematics, Newton's laws,

both A and B, or all of the above. If you strongly feel it's just A and C, you can also vote for all of the above and tell me later. Excellent. 64% of you remember that we do not need any of Newton's laws for this problem.

So this is one where we just need conservation of energy and kinematics. Little trickier. Let's step through to see where that happens. Now, actually the first one I'm going to do is right here. We are told to find k.

Now, very first thing you do when you're told to find the spring constant is realize there's only one of two ways you can do that. F equals minus kx, which is using Newton's law. Or potential energy of the spring is one half kx squared, which is energy.

Those are the only two places k shows up. We need to do one of these. Now, to use Newton's law, we need to know A and we need to know something about a single instant and to do a free body diagram. So, when our rocket, our little thing, is sitting on the spring

and it's been compressed five centimeters and it's at that angle of 30 degrees, what's the free body diagram for our rocket?

Draw a free body diagram for this rocket. I could ask that on a test.

Let's make it a little easier. Just so you know, your rocket launcher does have a little ramp for your rocket to be sitting on. Maybe it's a tube, maybe it's a ramp. The spring is at x equals zero when it hits the end of the ramp. So, it's like a platform launcher.

Okay. How many forces do we have in our free body diagram? A would be one, B would be two, C would be three, D would be four, E would be greater than four.

Go ahead and vote. Okay, ten more seconds because you should have already drawn your free body diagram.

Five, four, three, two, one. Pretty good. Half of you have three. So, we'll start with that because the next biggest is four and we'll see what we add.

So, the three that you have, tell me what they are. So, we've got the spring that's pushing up, weight coming down. What else do we got? Normal force. Okay, what else do we have?

I forgot to mention it's a frictionless rocket launcher. Oh yeah, so are we done now? Yeah, now why can't I just break into components, sum my forces and use f equals minus kx to find k? What am I missing?

If this is really going to work as a rocket launcher and there's really no other force in my problem, what should be happening right now? At this instant, it's accelerating. It's launching it. It can't be at rest. If it's at rest with this free body diagram, is it ever going to launch in the air?

No. I mean, I shouldn't say if it's at rest. If it is zero acceleration, it's never going to change its velocity. And if it's at rest with zero acceleration, it never launches. So, to hold it before launching, to compress it this amount, a real rocket launcher has another force that isn't even mentioned in the problem, holding it in place.

It's of some arbitrary value that when we let go, it launches. So that is why in this problem, despite the fact that I wasn't telling you explicitly, you should be able to recognize I can't use f equals ma because I do not know a.

I don't know what acceleration it has at the moment when it's compressed. It's a subtle but important point. And this is the sort of thing that does make these problems occasionally hard. Now, you might have done everything we just did and then realized as you started to write it down, you wrote minus kx minus mg sine theta equals ma.

Okay, I know x, I know g, I know m, I know theta. Uh oh, I don't know a. Well, go back to the problem. And you read the problem and you find absolutely no way to find a. You have two choices at this point in a physics problem.

You can say, oh wow, I'm screwed and keep trying to find a. Or you can say, oh, this must not be the way to do the problem. And luckily, there's only one other way. You really haven't wasted that much time. You really haven't. You've maybe spent five minutes drawing your free body diagram, writing that down, and now you move on.

That's one of the things that makes it both hard and easy. You may not know right away what the right way to do it is, but there's very few ways to do it. If you do it wrong, move on quickly to another one. And that was the main point I wanted to make with this problem. But now, having made that main point, let's move on.

So, in this problem, we have our launcher sitting here, compressed at 0.05 centimeters. And what we know, oops, sorry, meters, thank you.

Okay, we know we have some spring kinetic energy that we're going to have to find k. And we're going to have to equate it to other energies. And at this point, we realize there may be some potential energy of gravity, there may be some kinetic energy. We'll come back more and do it carefully.

And we realize a big thing we'll be missing is the speed that it leaves with. But we have the second part of the problem. And the second part of the problem talked about the fact that it needs to be able to travel 100 meters. Right? And if I have a v naught at 30 degrees, and I know it's going to go 100 meters, what type of physics am I doing?

Kinematics. Now, in this problem, you'll see that we have to be a little careful and make it clear. 0.05 centimeters is very, very small compared to 100 meters. I should have been very careful and said, and it lands 100 meters at the same height at which it was launched.

Right? Or you might worry about that little bit of a height difference, but you'd see that wouldn't come in as a big number. But, you know, I might have said 100 meters and land on top of a wall 10 meters high. There could be other ways that I have to do it. Or clear a wall 5 meters high, 80 meters away. There'd be other ways that this problem could be phrased. But now I know I have an energy part here and a kinematics part here.

And with the energy, I do my usual thing that I always do with energy. I pick my system. And in this case, I'm going to pick the spring, the rocket or the mass, and the earth.

And for our final clicker question, what types of energy do we now consider? Look at this. It's very similar to the other one.

Five more seconds. And we're done. Excellent.

And it's all but C. There's no friction again. Now, notice something to think about as you're studying for the exam. I could have a rocket launcher that, as it launches, there's a 40-newton force of friction acting on the rocket as it leaves the tube.

Pretty easy for you to add that in after this point. You might want to consider looking at the various problems from today's lecture and where and how could friction show up. An interesting question. So, anyway, I've got my initial, I've got my final.

Let's try and fill that in. Kinetic energy. Initial is what? Zero. Final. No. What am I doing when I leave my rocket launcher? I'm going with speed V. So, my final is? One-half. Okay, you've got to be much better with your energy formulas here.

One-half mv squared. Thank you. Potential energy of the spring. Initial. I know it's Friday, just make my day. One-half? Thank you. You don't get to leave until you answer these questions. Final? Excellent.

Potential energy of gravity. Initial. Let's take h equals zero where we start. Zero. Zero. Thank you. Final? mgh. So, the only other little trick in this problem? You've got to make yourself a triangle. 30 degrees. 0.05 meters. So, h equals 0.05 times sine of 30.

Right? Because it's going up the ramp at an angle. I put that all together. I have my one-half kx squared equals my one-half mv squared plus mgh. So, I need v and I can find k.

Because k equals m quantity v squared plus 2gh over x squared. So, that's the one I need. I make my lovely kinematics triangle, which you all remember. Go back and review kinematics.

V naught one-half, v naught t one-half gt squared. And, I go 100 meters. And, I do this at an angle of 30 degrees. I look at this. Notice, here is a big hint that you needed kinematics. Because it does have an initial and a final. It has where we start and it has where we land.

You might try to use energy here. The problem is, unlike going straight up in the air, where it's all in one direction. Remember, projectile motion is a parabola. We're going along a curve. We have both our x and y direction. You have to worry about these kind of separately. And, there's really very, very few ways to do this.

Other than using your kinematics, knowing the time of flight, and working from there. That's a big hint it's kinematics. Not to mention that the only force on it is gravity at this point. And, so I look at this. Tangent 30 works well for me. Because it's one-half gt squared over 100 meters.

So, that will give me a time of 3.4 seconds. And, then once I have the time, lots of ways to get the velocity. But, 100 divided by cosine of 30 times time gives me the velocity or speed of 34 meters per second. Which I plug in here.

And, I get my k of 9.2 times 10 to the fourth newtons per meter. So, that's the calculation right there. Notice, the calculation is not actually that long. A little longer than just a simple energy problem.

Right, if I told you I build a rocket launcher that has a launch speed of 60 meters per second. What's k? Much easier to do. Before you all leave. Because we still have three minutes. I know it feels like the end. Any questions?

These two problems. Plus what we do next week. Will be the key to the... Notice we've now done... I did two problems where I used almost every concept except friction from the course. In just two problems. In 50 minutes. Your exam will be two hours.

It will have two problems like this. And, it will have roughly 10 short answer problems. So, this is what you want to prepare for in the next three weeks. Two and a half weeks. Have a good weekend. I do have office hours as always. In Rolland Hall. In Frederick Reines Hall.