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Lecture Comments (7)

1 answer

Last reply by: Professor Selhorst-Jones
Fri Jun 20, 2014 9:34 AM

Post by Thivikka Sachithananthan on June 16, 2014

for example 4, since the block was moving, why is there no kinetic energy on our equation?

1 answer

Last reply by: Professor Selhorst-Jones
Fri Jun 20, 2014 9:25 AM

Post by Thivikka Sachithananthan on June 16, 2014

For example 3, how come gravity is not negative?

2 answers

Last reply by: Abdelrahman Megahed
Fri Jan 11, 2013 7:44 PM

Post by Abdelrahman Megahed on November 30, 2012

FOR EXAMPLE 3:

How come when you solve for k using conservation of energy you get 490N/m.
Ebefore+W=Eafter
mg(Xequilibr)+0.5k(Xequilibr)^2= mg(Xfinal)+ 0.5(Xfinal)^2


0=MG(0.2)+0.5K(0.2)^2

K=490

Energy: Elastic Potential

  • If you deform an elastic object (such as a spring or rubber band), it will resist the deformation, attempting to return to its original shape.
  • The amount of deformation (x) is measured compared to its original shape. An object with no deformation has x=0. A 0.1m spring stretched to 0.12m has the same deformation as a 1.0m spring stretched to 1.02m: x=0.02m.
  • Different objects will resist deformation at different rates. We show this with the spring constant: k (units in [N/m]).
  • An elastic object resists deformation with a force of

    F
     

    spring 
    = −k

    x
     
    .
    The negative denotes that the force always points opposite to the deformation (x).
  • Deforming an elastic object is a way of storing energy. The amount of potential energy in a spring is
    Espring = 1

    2
    kx2.
    [When working with energy, displacement is no longer measured as a vector. Now it is simply x, the length of the displacement. This is because energy is stored whether we stretch or compress. The direction of deformation doesn't matter: just the magnitude of deformation.]
  • By the conservation of energy, we can look at the entire energy of the system at the start and end:
    Esys,  start + W = Esys,  end.
    [Remember, positive work puts energy into the system, while negative work takes it out.]

Energy: Elastic Potential

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Introduction to Elastic Potential 0:12
    • Elastic Object
    • Spring Example
  • Hooke's Law 3:27
    • Hooke's Law
    • Example of Hooke's Law
  • Elastic Potential Energy Formula 8:27
    • Elastic Potential Energy Formula
  • Conservation of Energy 10:17
    • Conservation of Energy
  • You Ain't Seen Nothin' Yet 12:12
    • You Ain't Seen Nothin' Yet
  • Example 1: Spring-Launcher 13:10
  • Example 2: Compressed Spring 18:34
  • Example 3: A Block Dangling From a Massless Spring 24:33
  • Example 4: Finding the Spring Constant 36:13

Transcription: Energy: Elastic Potential

Hi, welcome back to educator.com, today we are going to be talking about elastic spring energy.0000

Spring energy or elastic potential energy, and we are also going to use this as a combination to talk about everything we have learned in energy so far.0005

Let us dive right in.0011

Just to start off, when we are talking about springs or any elastic object, elastic object is something that tend to return to its original shape when deformed.0013

Lots of objects have this behavior: springs, rubber bands, metal (when you deform metal to a slight amount, it wants to spring back into its original shape).0021

Of course there is extremes, if you deform something past a certain point, its plasticity fails it, and it is no longer able to return to what it was before.0031

But assuming it do not go past that really extreme case, you will be fine, and the rules that we are going to talk about will work out.0039

We are going to be able to talk about anything elastic, but in this case, we will just use a spring for all our examples, but this holds true for anything that has this really strong elastic property.0045

If you have any problems about bungee cords or rubber bands, where you know that it is going to involve some sort of elastic constant, you will be able to use the stuff that we are going to be talking about today.0055

For right now, we are going to refer to everything as springs, but it does not mean it is only limited to using for springs.0065

Let us go back to looking at springs very carefully now.0072

If we fix one end of a spring in place, and the spring is just sitting there, in will be in a relaxed position and no force will be exerted, you have a spring sitting on a desk, it does not do anything, it just sits there.0075

But, if we compress the spring, if we push the spring in, it will resist it, it has a force against it, and the amount that we compress it, is going to cause it to have more force.0086

If we compress a little bit, it is going to give us a little bit of force back.0097

But if we compress with a lot, it is going to give us a lot of force in response.0101

The more you compress it, or the more you stretch it, the more force you will get.0105

Exact same thing happens if we are stretching it.0109

Similarly, if we pull the string, we get a resistance of another force going in the opposite direction.0111

So, if we pull a little bit, we get a little bit in response.0115

Small pull results in a small force, but a large pull results in a large force.0119

The more you deform the spring, the more force you are going to get back, the harder it is going to try to get back to its original position.0126

The force will always point in the direction opposite to displacement.0134

If you displace the spring, if you compress the spring, it will resist compression by going in the opposite direction.0139

If you pull it, if you displace it apart, it is going to resist by effectively pushing in.0144

If you push, it pulls, of you pull, it pushes.0153

Of course it becomes kind of meaningless, because you think of pull as something that goes away, but what we really mean is that the force is going to be the opposite to that force that you put on it.0156

Not equal forces, but the opposite direction, magnitudes will depend on how far it has been compressed or pulled.0165

Our displacement is based on how far we do it, but each spring is going to have a different amount of force.0171

It stands to reason that some springs, and some objects are going to be more forceful than another.0180

Consider we had this spring in a mechanical pencil, versus the spring in a car garage door, or the springs for keeping up the suspension in a truck, these are going to be totally different things that we are looking at.0184

We are going to have to deal with in some way, with different kinds of springs having different forces connected.0195

We know we have to use the displacement quantity somehow, and we are also going to have to use something intrinsic to the spring that we are working with.0201

If we put this all together, we will be able to get this idea empirically verified.0207

We are not going to do this here, but it has been empirically verified, which is how we know it is true.0213

The fact that scientists have gone out and tested it, it was a good idea that seemed likely, and people went out and actually tested it, and that is how we know that Hooke's Law works.0219

We call it Hooke's Law after Robert Hooke who was a pioneering English scientist who studied springs in the late 1600's and he described it first, so we named the law after him.0228

From this we got the fact that, the force of the spring will be equal to, -kx.0238

The negative is there, because 'x' is how much we compressed it from equilibrium, so 'x' is the displacement that the spring undergoes from its relaxed state.0243

'x' is how much, if we start off with some base amount, then 'x' is the amount that we wind up compressing it to.0252

So, this amount here, would be 'x'.0264

But there is also 'k', which is something that is inherent to the spring.0266

If we are dealing with a big, hard, powerful spring, then it is going to have a spring constant that is much larger, it is N/m .0270

How much the compression results in force, newtons.0280

Big 'k' means that we are going to have a really strong spring, small 'k' means flimsy spring.0282

The negative there is to make sure that we flip the direction of our vector, because 'x' is the displacement vector, so to put a negative tells us, go in the opposite way.0289

-kx, something that includes the intrinsic nature of the spring, and also says go in the opposite direction, of how far you have pulled.0298

We know how far is the displacement, but we also know the direction.0305

-kx, is the force exerted by a spring on an object.0309

Say we had a vertically aligned spring, and we already tested it and found out that k = 200 N/m, so our spring is like this.0316

On top of it, we place a mass of 1.2 kg.0327

What is going to happen next is, this is going to go to equilibrium, and the spring is going to wind up getting compressed some amount.0331

Same mass, but now the spring is compressed.0342

The spring is resisting it, it pushes down, and then it stops.0348

What do we have in equilibrium?0352

We know in equilibrium, the forces involved have to be canceling one another out.0353

Force of gravity is still there, so we definitely know that there has to be a spring force involved.0357

What is pulling down on the box?0361

The box is being pulled down by mg, so 1.2×g.0363

What is resisting that pull, is the force of the spring, which is, = -kx.0369

The negative in this case is negative to mg, but we already got that, because we drew this in.0378

Since we drew in the vector, we can think of this as a positive vector, it is now up to us to deal with the vectors.0384

In this case, we know that, mg - kx = 0, but we can also see that those two vectors have to equal length, they have to have equal magnitudes, and they point in the opposite direction.0389

We can also think of this as just, mg = kx, because we know that to be in equilibrium, they have to be the same magnitudes, and they are pointing in opposite directions.0409

So, mg = kx, we know m,g,k; plug in numbers, we get, 1.2×9.8 = 200x, x = 0.059 m.0417

But in this case, we have to pay attention to, which way did we go?0446

We went this way, so x is considered positive, because for some reason we decided to make gravity positive up here, we made it positive, so unlike usual, we got x being positive, it is really just saying, it went down by a 0.059 m.0449

You could also think of this being a negative value, if we were going in the normal direction as positive-negative, but once again, like usual, it is up to us to pay attention to how the coordinates are.0466

If we pay attention to what we mean by the diagrams thought, our problems we will be able to work out, the important thing is that you know what you are doing.0475

In this case, if we look carefully at the diagrams, we see that it is compressed, so it has gone down by 'x' amount.0480

So, we made mg positive for this case, and we made kx the negative one, but if we look at it as arrows and vectors, we know that those two arrows are pointing the opposite directions, and they have equal magnitudes, we can start off with mg = kx, and we just have to know that our 'x' involves this arrow pointing down.0489

Now we have got Hooke's law.0508

Hooke's law, we can definitely understand at this point.0510

Force of the spring = -kx, where 'x' is how much it has been compressed, and 'k' is the spring constant.0512

Now, we want to be able to turn this into energy, if you compress the spring, and then hold it in place, you have done work, you put energy into the system.0520

If you compress the spring and hold it in place, and then you let go down the road, it is going to pop out, and you are going to be able to have energy stored, you will be able to cause a ball to be shot off of a launcher, you would be able to have a piece of lead pushed forward in your mechanical pencil.0530

Anything that you can wind up storing energy in a string for, all sorts of co-uses you can do for this, bungee jumping for example.0544

We are going to need a formula to be able to talk about elastic potential energy.0551

Sadly, we do not have mechanical tools to do that yet.0555

We need calculus.0557

But we can still use this formula.0558

The formula is that the energy in our spring = (1/2)kx2, where 'x' is the amount of displacement off of equilibrium.0561

Remember, it has to be measured from what the spring would be if it were at rest.0575

Also, it needs to be noted that just like with energy of gravity, we have to compare two different spring values.0579

We start off with a compressed spring, and then we compress it even more, what we wind up doing is, we put in energy, but we have not put in energy form rest to our ending compression, we put in energy from our beginning compression to our ending compression.0585

So, we will have to pay attention to starting compression versus ending compression.0600

If our ending compression is rest, we are doing well, we know we just have to do (1/2)kx2, but if you compare two compressed values, then you are going to have to actually compare them by coming up with two different values using the formula.0604

If we are going to be able to use anything in energy, we have to rely on conservation of energy, the fact that we can trust energy to always stay the same in the universe.0617

So, Esystem (start) + work = Esystem (end), we have said this repeatedly , and it still remains true.0625

This case, it means that the work, positive work implies that the energy in the system has gotten larger, work that is positive means energy going into the system.0633

Work that is negative means energy taken out.0643

So friction would always be negative, but if we were to have a spring and then push down, we do work into it, which means that the system have more energy stored in it.0646

Remember, this equation continues to hold true, as long as you are learning about energy.0656

Any form of energy that you learn about, it is going to wind up still having this true.0660

The important thing is to pay attention to all of the energy changes that are involved.0663

All the energy changes, all the work that is involved, and then you are going to be able to use this formula.0668

If you come up with some of it, but forget to put in other parts, say if you put in your start in height, but you forget to put in your ending height, it is not going to work.0672

If we know we are on a flat surface, we can forget about both if them because it is the same amount of energy stored, on both the left side and right side.0681

But, if it changes, you are going to have to make sure you account for it in your equation.0688

That is the only way conservation of energy works, we have to pay attention to what changes.0692

We have assumed that all the particles have stayed there, all the particles have stayed constant, nothing changed forms, nothing has been burned, no chemical have been used to create energy or take away energy, so we can trust that chemical energy on both sides is the same in this case, so we never had to pay attention to it.0697

We only had to attention to what you know is going to change.0714

So, it is important to pay attention to what does change.0717

Otherwise you will not be able to trust using this formula.0720

But, this is always going to be true for everything you use, so it is a great formula, it is great to pay attention to it.0723

Just think, what we started with, plus the changes is what we end with; makes sense!0728

Finally, I just want to mention that, we have definitely learned a lot of useful formulae, and we have got a really good chance of understanding how energy interacts with physical world, but you should not think that it is all of it.0732

Like I was talking about chemical energy, there is a whole bunch of kinds of energy that we have not talked about, lot of them are not going to have real simple formulae that we can plug into equations, so there is a lot more complicated stuff going on here.0743

This is far from the end of what you can learn about energy.0755

There is lots to learn, and I encourage you to go and learn more, because there is all sorts of interesting stuff out there, we are setting the stage for future exploration, later Physics courses, if you decide to continue to learn more about energy.0758

But, at this point, we have got a real good start, we have got a real good understanding of how the world works around us in terms of energy, and there is a huge amount of stuff that we can do with the stuff we have learned so far.0768

We have a real chance to tackle the hardest problems involving energy, andn we are going to see some really interesting applications in the set of examples.0777

These are going to be the most advanced energy examples we have seen yet, and will be a chance to really stretch our muscles.0784

First example, first big strong example: We have got a mass less spring launcher, (the reason why we are talking about all these springs being mass less, is because the spring had some mass, and it is going to take some its energy to move around.)0792

So we have to pretend, we have to be able to assume that we can treat the spring as mass less, and in most cases, the mass of the spring is very low compared to the mass of the object dealt with.0809

It is not that unreasonable.0819

But, if it was a really big heavy spring, and the object was not that heavy, it is something that we have to take into account.0820

The mass of the spring is going to have some effect on what is happening.0825

That is why all our problems are mass less, to be able to just jump into being able to using energy conservation.0828

But if you knew that the spring was not mass less, it is an even harder kind of example that you have to pay attention to, you really have to think about it0833

Back to our example, a mass less spring launcher, with spring constant, k = 900 N/m, located above the ground.0841

If we put a ball of mass 0.8 kg into the launcher, compress the spring by 0.2 m, and then the ball also starts at 8 m after compression.0859

Disregarding air resistance, what is the ball's speed on impact with the ground?0860

We are going to put it in a launcher, we are going to compress it by this amount, and we know that once it is compressed, it is going to have a height of 8 m above the ground, and then we are going to let it lose, and it is going to shot off of the launcher, and it is going to take an arc, then come down to the ground, and hit the ground.0865

First question: What about, will the orientation of the launcher affect the impact speed?0882

What would happen if we had the orientation like this?0887

If we had the orientation like this, the ball might shoot up, get to a different height, and then fall back down.0891

But, remember, what is the energies that we are going to be dealing with.0899

Energy we are going to be dealing with, is the height involved in the launcher, the amount of spring energy involved in the launcher, does it have any velocity in the beginning?0902

No, it does not have any velocity in the beginning.0910

Do we have to worry about work being lost?0912

Do we have to worry about friction taking place?0914

No, we do not have to worry about that, because we have been told that we can disregard air resistance, so we know that what begins in the beginning, is going to be the height, and the energy in the spring.0915

The angle has nothing to do with it.0924

The speed, because we are going to solve for speed, using (1/2)mv2 is going to wind up being the same in any case.0926

We should pay attention to the fact that, we are going to get totally different velocities at the end.0934

If we had an even flatter one, it would have, we would have totally different velocities depending on the orientation.0939

But, the amount of speed in the object, is going to be the same, it is just that the angle will change.0947

We can trust in the fact that the speed, the impact speed will be the same no matter how we do it.0954

If it were the impact velocity, then we would have to pay attention to the orientation, then we would have to pay attention to the arc being made, we would have to start caring about these sorts of things.0959

But, in our case, we know that we are only being asked for the impact speed, so it is enough to solve for it using energy real easy.0967

What do we know here?0974

We know that, Energy(start) + Work involved = Energy(end).0977

Right off the bat, we know that, work = 0, because we know that there is no friction, and the ball is not going to do anything, and no force is being applied to the ball, once it is already in flight.0983

So, we can just knock out work.0992

What is the energy at the start?0994

Energy at the start is, (potential energy of gravity) + (potential energy of spring) = (its kinetic energy when it is about to impact), mgh + (1/2)kx2 = (1/2)mv2.0996

We know m, g, h, k, x; so the only thing left to find out is what our speed is.1021

We get, (0.8 ×9.8×8) + (1/2)×900×(0.2)2 = (1/2)×0.8×v2.1031

80.72 J is L.H.S, multiply by 2, 2×80.72 J / 0.8, v = sqrt(2×80.72 J / 0.8) = 14.2 m/s.1058

The important thing is, we knew what the energies involved at the beginning were, we knew what the potential energy was, we knew what the spring energy was.1089

We know no work was done to it, we know that at the end we are only going to have to be looking at the speed energy, the kinetic energy, because we are going to have its impact speed, so we know it is at a height of zero when it is impacting, and there is no spring around it, so we do not have to worry about compression, we know everything else remains the same.1095

So, that is it.1112

Example 2: Block of mass, m = 0.5, sitting on a horizontal table, which has a friction constant of μk = 0.55.1115

It begins at rest, goes against a mass less spring, (since mass less, we do not have to worry about how much energy is put into moving the spring around, how much force/work is put into moving the spring around, so we make it easy for ourselves by making it a mass less spring.)1125

We compress 'x' m, with spring constant, k = 470 N/m.1140

After we release the block, it slides 7 m before coming to a stop.1145

Right now, this is the moment, we are holding in place and let go.1149

Once we let go, it slides forward 7 m, at which point it comes to a dead stop, no speed, no velocity.1152

It goes along, and cause it to slow down.1168

Once it is moving, what force is going to be able to slow it down?1173

Friction, so friction is the force moving against it.1176

Remember, work = fd, our force here is friction, we know, d = 7 m that it slides.1179

That 'x' is included in that 7 m, because once we let go, it slides 7 m.1188

That 'x' is just the beginning of the 7 m, the 'x' between equilibrium, and where we compressed the spring, it is just the first part that it moves through of the 7 m that it slides.1196

At this point, we have enough to solve for 'x'.1206

We know, E(start) + work = E(end), what is the energy at end?1209

It is still at the same height, so mgh = 0 throughout.1219

Is it moving anymore?1224

It is not moving.1225

It is not moving.1226

Are there any springs involved anymore once it is out over here?1227

No, there is nothing out here, it is just sitting over here, so we know, E(end) = 0.1230

What is the energy at the start?1236

E(start) is just the energy that we put into the spring: (1/2)kx2 + work involved.1238

Is work positive or negative out here?1244

We know that it is going to have to be negative, because we are going to have to take this positive number, x2 is always going to be positive.1247

We are going to have to lower this down to zero.1254

How does it happen?1256

We know that friction is going to be going this way, so we have got, friction×d, distance is positive, friction is negative in terms of vectors.1257

Or we can think of it as, distance is negative and friction is positive, but that is weird, because normally we are going to think, going to the right is positive, so in this case we are going to have works as a negative number.1272

So, this is going to be, -(the size of the friction force)×d.1283

Or, we could think of this as, +work as friction vector, dotted with the distance vector, or + (work of the friction)×d×cos θ (angle between them, θ = 180 degrees).1292

In any case, the important thing to know that our work is going to have to be a negative number, and it going to be, friction×d for the magnitude, so at this point, we are ready to do it.1306

We got how much the friction force is, normal force = mg (only thing applied to it is gravity, table is canceling out gravity), so, friction(kinetic) = μk×FN.1315

So the work involved = (1/2)kx2 - μkFN×d = 0.1338

We know what 'k' is, we do not what 'x' is, we know μk, we know FN, we know 'm', 'g' and 'd'.1354

So, at this point, we know everything but one of them, so we solve for it.1374

x2 = 2×μkFNd / k, x = sqrt (2 × 0.55 × 0.5 × 9.8 × 7 / 470) = 0.283 m.1378

It is important to pay attention to the fact that, because 'x' was squared, this actually was +/-, it is up to us to know which direction does it go.1426

If we want to make this into a vector, it is going to be, -0.283, because remember, we know that our 'x' was going to the left.1435

So we know that we had a displacement in the direction that we decided to call negative.1443

However, in this case, what we know is that 'x', the length of x', the amount that it was shoved over, is 0.283 m.1448

We can think of it as the magnitude and the direction that you moved in, or you can just think of it as a vector, with a positive or negative in front of it, depending on it.1454

And if it were a two or three dimensional vector, it will be getting a little bit more complicated, but in our case it is one dimension, so it is enough to just have just +/- in front.1461

Important part though is , 0.283 m, and to know what direction it went in.1468

Next example: We start off with a block of mass 5 kg, and it is dangling from a mass less spring.1474

At rest, the block stretches the spring form its relaxed position by 0.2 m.1481

We start off knowing, we have been pulled 0.2 m.1486

If we wanted to, we can figure out what 'k' is.1491

That is very good, because that does not show up anywhere, and we will need it later.1493

At this point, we know that if we pull it down by 0.2 m, the pull of the spring is able to equally resist the pull of gravity, so that they are in equilibrium, they are perfect equal opposite.1496

At this point, at 0.2 m down, we got that gravity is canceled out by the spring force, that is what it means if the block stretches from its relaxed position by 0.2 m when it is at rest.1511

After it being at rest, the block is then grabbed and pulled a further 0.3 m, so the total here would be, it starts at 0.2, and we take it to 0.5 m.1523

We hold it at rest, so we know that it has no motion at that moment, and we let go, and it begins to move up.1533

You pull the spring down and you let go, and you are used to seeing that it is going to go up before falling back down, and then it will start to oscillate.1538

How high will it make it, what is the highest point it will make it to above its resting location, how high is it make it before it winds up being canceled out and pulled back down?1545

At the beginning, we do not know 'k', but that is what we can use this piece of information for.1556

We can solve for 'k', if we got this information about what it is like at rest, when it is at equilibrium.1562

Once we have got 'k', we can use energy on this, to figure out how high it is going to make it up.1569

Let us get to work.1574

First, we are going to solve for what 'k' is.1577

It goes down 0.2 m, so we know that kx - mg = 0 (up positive, down negative, kx fighting mg)1585

And this is at rest.1608

We do not 'k' yet, but we do know mg and x, we get, k = mg/x at rest, k = 5×9.8/0.2 = 245 N/m, which makes sense, because mg is in N, d is in m, so 245 N/m.1613

With that, we can go on to figure this out using energy.1647

If we pull it down, we have got some energy stored in it.1650

Any work be done on it over the course of its movement, any energy taken out of the system?1655

Is there any friction involved?, No, there is no friction involved.1660

Is there anything else, any other external forces acting on the system?1664

No, there is no other external forces, just the internal forces of the system, of energy in the form of gravity, which we can consider an internal force because it is dealt with your potential gravity, and the force of the spring, which is another internal force we are dealing with through the form of spring energy.1667

E(start) + work = E(end).1685

We do not lose any energy, or gain any energy from the environment around us, so we can knock out work.1693

So, E(start) = E(end).1700

What is E(start)?, we start off with some amount of potential energy, in this case, it is actually going to be a negative amount of potential energy, because the box is below what it becomes later.1703

Later on, it is going to go up, so that is going to have actually going to gain potential energy over the course of its movement up.1715

If you take a box and lift it up, it has to gain potential energy to get up there.1726

Either at the beginning we are going to have a negative potential energy, or at the end we are going to have a larger potential energy, a positive potential energy.1730

The important part is that the difference between the two is going to be, that it is going to have to gain the potential energy, the ending potential energy will be greater than the starting potential energy.1737

Also at the start, we are going to have the amount of energy we put into the spring.1748

What about the end, will there be energy in the spring?1752

Of course! If we are up here, we are not in equilibrium, now we have got a compressed spring, a tightly compressed spring, if we have gone past equilibrium.1755

It makes sense, if we pull it down and let go, we are going to get a little bit past equilibrium if we pull it far enough.1763

We are going to have to deal with the amount of energy stored in the spring, and worst case, if we still put (1/2)kx2 on the ending amount, and it turns out that the x is zero, that we wound being at equilibrium, it will just get rid of itself, and we do not have to worry about the fact that it is there.1770

It is better to have a little more information then to turn out not needing it, than to forget to use it and lose the problem because of that.1784

One last thing: How are we going to decide height? How are we going to decide equilibrium?1791

We know that we already have equilibrium at zero.1796

We can have 'x' be, up will be positive, down will be negative.1798

But, what about height? What are we going to do about height?1802

Let is make height, for ease, also be zero there.1805

That means that we can set, x = h for this problem, it will make things a lot easier on us.1810

It is up to say our coordinate system, with the coordinate system defined by us, we have nowhere.1817

We have to impose a way of looking at the picture.1828

We have to impose an orderly system so that we can think about it, without that orderly system, we are lost in the wilderness.1831

We have to impose, 0 = height somewhere, we have to say what that base height is.1838

And why not put it at the equilibrium!1843

That means that the amount that we vary in our 'x' is going to wind up being exactly equal to the amount that we varied in our height, we can wind up tying those two things down, without having any way to transfer between the two, it is not like, height is going to be the maximum height that it achieves, so we have to figure that out, it is going to be really hard.1845

It will still be possible, but it will make the algebra so much more difficult.1861

Right now, it makes it really easy to say x = height, because we have established them as being, x = 0 here, and 0 = height there, so they are both the same amount, because they are both going to move by the same distance.1864

Now we are ready to get into it.1877

(1/2)kx2(start) + (how much energy is stored as P.E. in the beginning) = (1/2)kx2 (end) + mgh (end).1879

Substitute in numbers, (1/2)×245×(-0.5)2 + 5×9.8×(-0.5) = (1/2)×245h2(end) + 5×9.8×h.1917

x in the beginning is going to be negative since we are pulling it 'down', and greater on the right side since it is going up.1929

At this point, we can calculate everything on the left side.2025

Move everything to the right side, and solve, we get, 0 = 122.5h2 + 49h - 6.125.2033

That is enough to know what it is, but it is not enough to know directly, this is not like an easy algebra problem, where we just have height on its own, and we solve for both sides.2051

In this case, we have got a parabola, so we have to find out what the zeros of this parabola are.2063

We got a couple of choices: one , we can factor it, but it looks like a real beast to factor it by hand, since you have got 6.125, 49, 122.5 in there, this is not going to be just easy to come up with factors out of hand, so that option, pretty much gone.2068

We could use the quadratic formula, great to apply it where we have numbers that are very difficult to factor, because you plug in, it will come out, it might take a little bit longer, but it will definitely work, in this case, factoring is going to take way longer than the quadratic formula would, because it is going to be really hard to factor.2082

Final choice: If we had a graphing calculator, you could punch it into your graphing calculator, look at the parabola, and just find the zeros.2097

But, remember, in any of these cases, you are going to wind up getting two answers, so what are the two answers that we get for height?2104

We get, height = -0.5 m or 0.1 m, so which one of these is our answer?2110

If you look at -0.5, that is where we started, of course that is an answer.2123

If we go back to this very first line, it makes perfect sense that if we plug in 0.5 here, and -0.5 here, we have got the exact same expression on the left side and right side of our equation.2128

Of course they are going to be equal, that is going to be a solution because it is trivially a solution, it is obviously a solution, so -0.5 is no surprise, that is one of the places it will work out, it is one of the extreme cases.2142

What is the other extreme case?2155

The other extreme case is 0.1 m.2157

So, the 0.1 m is where it will get up to before it has to start falling back.2160

It is the other solution to our parabola, it is the other solution to our equation, so 0.1 m is the maximum height that it attains.2165

Final example: A block is placed at the top edge of a frictionless half pipe.2174

On the other side of the half pipe is an uncompressed, yet again, mass less spring, the block has a mass of 20 kg, and starts at a height of 3.7 m.2180

At rest, the bottom part of the spring is at a height of 2 m.2192

When released, the block will slide down the half pipe, compress the spring additionally by x = 0.5 m, so and then it will be forced back, because that is the maximum amount that it makes to.2197

If that is the maximum amount that it makes it, what do we know about just before it being forced back?2224

If you get to the very apex of your movement, and then you have to fall back down, what do we know about that snap shot, that exact instant when you are at the very apex?2228

Your velocity is zero.2238

We know at the moment, when you are forced back, your velocity is zero.2240

If you take a quick example, that is not quiet the sane as this, we toss a ball into the air.2244

At some point, that ball is going to fall back to the earth.2250

If it is moving along, it is going to continue having that 'x' moving this way.2253

But, its y vector is going to get smaller, until eventually it is zero at the instant it begins to fall back down.2257

So, at the very top, we know it is zero.2268

So, at the apex, at the very switch over between moving in one direction, and coming back in one direction, you have to have a zero velocity.2273

So that is going to be a key point for us to understand in this problem.2281

We know that, m = 20 kg, height that it starts at is 3.7 m, the height that the bottom of the spring is at, is 2 m, and the x that it gets compressed, is 0.5 m, now we want to know what s the 'k' involved.2283

We know that, E(start) +work involved = E(end).2304

What is the work involved?2313

We are on a frictionless half pipe, it is a mass less spring, we can assume that there is no air resistance, because it is not moving that fast, or we did not mention it, we will make it easy for ourselves, we are going to remove all the cases that does work.2315

In a real engineering situation, we will have to start taking into account, what kind of friction is involved, how dense is the air, is it an area that has a lot of air pressure, it may be taking place in a vacuum, but it will have some amount of friction involved.2335

There are things that we have to care about in engineering, but we got a specific example problem where we know that it is frictionless, we know that there is no air resistance, we know that the spring is mass less, so we do not have to worry about work going into anything else than just our equation, so we can knock out work.2354

What is the energy that it starts with?2371

It starts off at a certain height.2372

Is it moving at height? Is energy stored in its motion?2374

No, it starts at rest the instant before it is let go, and starts to move.2378

At the beginning, we just have potential energy of gravity.2384

At the end, what are the things going to be involved?2388

We have compressed the spring by an amount, so we have got (1/2)kx2 + mgh(end) (there is no energy in velocity, since it is not moving, and total height winds up being 2.5 m up here.)2395

At this point, we are ready to do it, so we got, mgh(start) - mgh(end) = (1/2)kx2, we know what x, m, g, h(start), h(end) are, we just plug in.2422

So, [2×mgh(start) - mgh(end)]/x2 = k = 2×20×9.8×[h(start) - h(end)]/x2 = 2×20×9.8[3.7 - 2.5]/(0.5)2 (h(end) is, (where the bottom of the spring is) + (the 0.5 that it went up beyond that) because it compressed the spring 0.5 past equilibrium, so that to make it equal to equilibrium, 2 m, and a little bit further to 2.5 m off the bottom.)2443

Our final answer is, 1882 N/m = k.2544

Now, I left everything in the variable form very long time, because I think that the easiest way to do a problem is to get all of your thinking down with variables, because variables are easier to think about what does it mean in a general way.2562

Once you start throwing in numbers, you must wind up getting this mass of numbers that often, you cannot separate what the ideas are here.2575

But in this case, we are able to see that it is the difference between the starting potential energy, minus the ending potential energy, is going to be how much energy we have left over for our spring, and then 2x2 is what we got for, for how much the springs other components are, to get the 'k' that we trying to get to.2582

To get to 'k', we have to multiply by 2, and divide by x2, so I left it as that, and substituted in.2605

There is no reason you could not substitute things as you find, if it is easier for you, and you feel comfortable with, it will certainly work.2610

I recommend trying it this way every so often though, I certainly think it is easier.2615

To keep all your variables together at the end, and then do one long session of computations at the end, but sometimes that is going to wind up being a trouble for you.2619

If you wind up having difficulty with the order of operations, like if you did not do this operation first, you would be in hot water.2626

You should pay attention to what makes it easier for you, but I think this is the best way to do it.2633

In any case, we have gotten to our answer, 1882 N/m.2638

This is going to end the session on energy, this ends our set of lessons on energy.2643

I hope you learned a lot about energy, hope you got a much better understanding, there is a whole lot more that you can do with it.2647

But, that is for our future courses.2651