For more information, please see full course syllabus of High School Physics

For more information, please see full course syllabus of High School Physics

### 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

The negative denotes that the force always points→F

spring= −k →x. __opposite__to the deformation (→x). - Deforming an elastic object is a way of storing energy. The amount of potential energy in a spring is

[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.]E _{spring}=1 2kx ^{2}. - By the conservation of energy, we can look at the entire energy of the system at the start and end:

[Remember, positive work puts energy into the system, while negative work takes it out.]E _{sys, start}+ W = E_{sys, end}.

### 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

### High School Physics Online Course

I. Motion | ||
---|---|---|

Math Review | 16:49 | |

One Dimensional Kinematics | 26:02 | |

Multi-Dimensional Kinematics | 29:59 | |

Frames of Reference | 18:36 | |

Uniform Circular Motion | 16:34 | |

II. Force | ||

Newton's 1st Law | 12:37 | |

Newton's 2nd Law: Introduction | 27:05 | |

Newton's 2nd Law: Multiple Dimensions | 27:47 | |

Newton's 2nd Law: Advanced Examples | 42:05 | |

Newton's Third Law | 16:47 | |

Friction | 50:11 | |

Force & Uniform Circular Motion | 26:45 | |

III. Energy | ||

Work | 28:34 | |

Energy: Kinetic | 39:07 | |

Energy: Gravitational Potential | 28:10 | |

Energy: Elastic Potential | 44:16 | |

Power & Simple Machines | 28:54 | |

IV. Momentum | ||

Center of Mass | 36:55 | |

Linear Momentum | 22:50 | |

Collisions & Linear Momentum | 40:55 | |

V. Gravity | ||

Gravity & Orbits | 34:53 | |

VI. Waves | ||

Intro to Waves | 35:35 | |

Waves, Cont. | 52:57 | |

Sound | 36:24 | |

Light | 19:38 | |

VII. Thermodynamics | ||

Fluids | 42:52 | |

Intro to Temperature & Heat | 34:06 | |

Change Due to Heat | 44:03 | |

Thermodynamics | 27:30 | |

VIII. Electricity | ||

Electric Force & Charge | 41:35 | |

Electric Fields & Potential | 34:44 | |

Electric Current | 29:12 | |

Electric Circuits | 52:02 | |

IX. Magnetism | ||

Magnetism | 25:47 |

### 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)kx ^{2}, 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)kx ^{2}, 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, E _{system} (start) + work = E_{system} (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 it*0833

*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)mv ^{2} 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)kx ^{2} = (1/2)mv^{2}.*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×v^{2}.*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 μ _{k} = 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)kx ^{2} + 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, x ^{2} 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) = μ _{k}×F_{N}.*1315

*So the work involved = (1/2)kx ^{2} - μ_{k}F_{N}×d = 0.*1338

*We know what 'k' is, we do not what 'x' is, we know μ _{k}, we know F_{N}, we know 'm', 'g' and 'd'.*1354

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

*x ^{2} = 2×μ_{k}F_{N}d / 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)kx ^{2} 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)kx ^{2}(start) + (how much energy is stored as P.E. in the beginning) = (1/2)kx^{2} (end) + mgh (end).*1879

*Substitute in numbers, (1/2)×245×(-0.5) ^{2} + 5×9.8×(-0.5) = (1/2)×245h^{2}(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 = 122.5h ^{2} + 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)kx ^{2} + 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)kx ^{2}, we know what x, m, g, h(start), h(end) are, we just plug in.*2422

*So, [2×mgh(start) - mgh(end)]/x ^{2} = k = 2×20×9.8×[h(start) - h(end)]/x^{2} = 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 2x ^{2} 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 x ^{2}, 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

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