Hello, miss. Hello Jackson. How are you? Great. Good. So we're working through our materials topic. And yesterday we revised hook law where the more you've add a force to a spring, the more it extends. So we have eficus kx and the energy stored is equal to a half fx or a half kx squared. So we have another equation for a half K X squared kbeing the stiffness of the spray. X squared. So hook law, if we had to define hook law, extension is directly proportional to the force applilied until the limit of proportionality is reached. So if you think of a graph for hooks law, the stress of the strain or the force over the extension ops, the force against the extension is a straight line relationship. If our spring is obeying hooklaw. And that's nutons. That's straight line relationship and the gradient will be. You're a spring constant. So different materials have different properties, glass, steel and rubber. They behave differently if you exert a force per unit area on them. Force stress is force over area, and the strain is the change in length. Over the length, the original length of the material. So they have very different properties. So steel wire will nearly obey hook 's law up to this point. And then there's a yield point. And then there's a breaking point here. So brittle materials like glass, we'll give this line. Weaker materials like rubber have a lower breaking stress. So if you stretch rubber and keep stretching, it will eventually break. Ductile materials, they can be drawn into a wire, so steal a metal is a ductile material. Rubber has the greatest elastic strain energy for a given stress, as it has the greater the greatest extension. So if we were to measure the stress of restrain for some steel wire, we would set up this equipment, wehave a marker here, so you can see by how much the wire gets longer, and you choose a microrometer to measure the diameter. So for Young modulus, the stress over the strain is the force over the area, over the change in length, over the original length, which is a force times extension over area times original length. And Young modulus is usually given the symbol e. So I thought, wedo some straightforward questions on hooks law first, and then on Young modulus today. So are you ready for this? Have you got a calculator handy? Good. So force is spring constant times extension. The elastic strain energy stored in a stretch spring is a half the force times the extension, which is a half kx squared. If we put kx into this equation, substitute that for that, we get a half kx squared kinetic energy and gravitational potential energy. Okay. A spring is stretched by a force of 15 nusions, and its extension is point two m. Calculate the spring constant of the spring. That's an easy one to start with. Ten. Yes what are the correct units for the spring constant? Jackson Newton Newton meters meters, Newton meters so it's nutions per meter. Nutions per meter okay that is nutions per meter so that maybe the answer is not not ten, maybe 50 divided by 0.2 that is newtons per meters. Yeah this the this is. 215 nutons per meter Yeah. Okay, number four, a force of 18 nusions is applied to a spring with a spring constant of 160 newtons per meter. What is the extension of the spring? Not upon five. Meters, Yeah, I think. No point 5m units are important, Jackson. A spring is stretched beyond its elastic limit. The graph shows the force extension characteristic for the spring. Calculate the spring constant of the spring. Estimate the energy stored by the spring before it passes its elastic limit. So the first thing we have to work out is the spring constant. Where does it go beyond its elastic limit? At what point would you say this spring is no longer obeying a proportional relationship? Relationship. So let's read the question again, Jackson. A spring is stretched beyond its elastic limit. The graph shows the force extension of the spring. At what point on the graph is the line no longer giving a direct proportional relationship? Would you say mark on the graph? That is the positive proportion. Up to which point. Take. This one. Yeah, Yeah. So how do we find the spring constant? So find k and then find energy stored in the stretch spring using either that equation or this equation. Maybe we can use the supreme constant equal to force divided by extension. Up to this point, Yeah. So. What is the spring constant for this spring using the graph? Spring constant the spring constant is 60, divided by nopoint two. Sorry 0.8 75. Good. 75. Newtons per meter. Good, then. You can use e equals a half fx to find the energy stored, or als a half kx squared. Estimate the energy stored by the spring before it passes its elastic limit. Which bit of this graph gives us the energy stored? Who? Energy storres look, and let's take potential energy equal to a half times. Force times x delta x square ared. Square maybe times not on a square square. Thomnineteen point two juice juice. Yeah point five times 75 times orange fish. What did you get? Yeah, I found it's a problem. 2424. I think that is 24. So we got 75 trying Yeah so they're they're reading the graph, a half kx, a half fx. So they're reading it to slightly more than here. That's fine. Good. A spring with the spring constant of k equals that is stretched by 15 cm. Calculate the plalastic potential energy stored in the spring. Astic potential energy, elastic potential energy. So firstly, we should use. No, no, sir. 15 cm 0.15 square times 0.5 times 200. 2.25. And units. What is energy? Yes. Spring constant so 0.5 times 200 times point 15. 2.25 jles Yeah. Number seven, a spring stores five Jews of elastic potential energy. When it is stretched, the spring constant k is 215 newtons per meter. What is the extension of the spring? Extension. 5225 times two divided by 250. Point 2m. 0.2m good. Let's try question eleven. As spring extends by 0.1m, when a force of this is applied, calculate the spring constant of the spring. 250Yeah units for spring constant. What's the correct unnewance perpermeters good nuance permeter you will lose Marks x Jackson if you don't use units, unfortunately, so nuons per meter. Number twelve a spring stores 20 jels of elastic potential energy. If the spring constant is 180 nutons per meter, calculate the force required to supply the spring with this amount of energy. Also require to supply spray. So. To require 20 shows, 20 tends to divided by 180. So the answer is not one, four, seven newtons. 0.47m. Good. So you've got the extension f equkx. So we're asked what force will give that 100. So if we know the spring constant and we know the extension, we can work out the force. Calculate the force required. So you got to that point and then f equals kx will give us the force required. 13, a spring with a spring constant of 300 nutons per meter is compressed by 40 cm cubes. So we're compressing the spring this time, just like lying on a bed, a two kilogram object is placed on top of the spring and launched vertically. When the spring is released, calculate the elastic potential energy stored in the spring. E equals. So a half times 0.4 square ared times a half times 324. So I think it's 20 42 choose Yeah point five times 300 times 0.4 squared. Good, using the energy calculated in a, determine the maximum height the object reaches, assume all the elastic potential energy is converted to gravitational potential energy. What's our equation for gravitational potential energy? Mgh, good. Mj H M is two kilograms, J is 9.8 kilograms, muters per kilograms. So and we got 24 jewels, didn't we? 24 juice. Yeah. So and an H maximum height here we want to know about the H 24 divided by no point two also sorry, two divided by 9.81. 1.22. Units. Meters natagain you will 1.2. Yeah you will lose Marks if you forget your units after you've done all the hard work. Okay, we're doing well. A spring with a spring constant of 500 nutons per meters compressed by 30 cm. A 300 gram object is placed on top of the spring, launched horizontally when the spring is released. Calculate the elastic potential energy stored in the spring. What to do that? 22.5 energy measure didn't use the energy calculator, determine the maximum velocity of the object when all the elastic potential energy is converted to kinetic energy. So 22.5Jules. Equal to a half. Mv square. And we have to convert. So 22.5 times two, divided by node point three square, no cube. Square roots. 12.2. 12.24Yeah kinetic energy. What is the velocity? So what's the velocity measured in. Speed what speed measured in neas per second? Okay, good. Okay. So then remember, we have Young modulus, which helps us to describe properties of a material depend whatever its length or cross section tional area is. So Young modulus e. Equals sigma over epsilon. Equals four serarea. Over change in length. Over original length. So again, if you're stretching a piece of glass, if you're stretching a steel wire, you can work out a lot using the Young modulus, the strength, it's like the rigithe rigidity, the stiffness of your strength, the strength of it, its resistance to be able to be deformed. So Young moduluis stress over strain. And stress is force over area. So measured in Pascal newtons per meter squared. And extension doesn't have units. So. A wiir experiences a stress of 210 megapascals and a strain of 8.0x ten to the minus four. Remember, there's no units here because it's a ratio. Calculate the Young modulus of the wire. Strength stress ymodels models equal to suggest divided by string. 210 divided by eight times ten to the minus four. 2.63 times ten to the. Ten cuos of five. Now check check your working. Jackson. You've missed something here. One of the numbers you've put in one times ten, two os of five. Divided by eight times cencuthe length four. And equal to 2.63 times ten to the eight. Megar is ten to the six, isn't it? Yeah, I get 2.6 by ten to the eleven. So two, that's not right. Okay rise, let's try number two. I get for this one I get 2.63 by ten to the. Eleven. So mega is ten to the power of six and divided by that. Okay, let's try number two, a 2.8m wiof cross sexual area. This experiences a force of ten. Uit extends 1 cm. Calculate the stress experienced by the wire. So stress. Is foresoarea now experience that? Extends by one centimeters ter, so that is not one. Not points. No, two points, eight. One divide about. One stress is one. So force ten divided by area. Ten divided by 2.5x minus seven. So stress is for servarea. So that's ten nutions divided by cross section area 2.5 by ten to the minus seven. Okay, that is stress. So I get four by ten to the seven pascals. The strain is change in length 1 cm over original length 2.8. So the strain. North Point, north one. Over 2.8. So we get. 320 57 times ten two is the minus three. Thank you. And calculate the Young s okay. 1.12 times ten to the ten. And the younmodulus has the same units as pascals Yeah so. 1.3. PaaS scps okay, rice. You do this one now, Jackson, number three, calculate the stress first calculate the strain, and then calculate the Young modulus, just like before. Remember, stress is force over area. Strain is changing length over length. And the Young modulus is stress over strain, the first stly first one, a wire diameter 1.2 mm, 1.2 times ten to the minus three square times, pi times a half. And then this number, we use two divided by this number and we can get 8.84 times ten to the five. Which one? First of. So two divided by so 1.2. So the next one, the strain experienced by the wire, initially it is 1.2 mm, and now it's 0.5. So 0.5 divided by 1.2, that is 0.62, 0.42, 0.42. 0.42. Yeah straand, what is your Young modulus then? 8.84 times ten to the five divided by 0.42. 2.1 times ten to the six. Let's check this then. So 1.8 by ten the six. This isn't dried a stress. I wonder why not? Mason, pi R squared. Did you make it millimeters 1.2 by ten to the minus three, divided by tutic from the diameter to get the radius squared times pi. Maybe, Yeah, maybe. Is there some problem in calculation squared times pi? Okay, try this again. See if you get this answer with the stress. And then divide force by this number gets your stress. And then the strain again millimeters, so ten to the minus five. And then when we have that divided by that, we get this, which is larger than the number we originally get got. Let's try and take this next one slowly. 2.1m wiof diameter that experiences a stress of this and a strain of this. Calculate the force experienced by the wire. So we can use the stress is force if we know the area. Okay. So let's work it out over here. Question four is the one we're doing. Okay. So we know stress. Stress is force over area. Work out the area, the cross section area, Jackson. First area diameter firstly that is not point 38 mm and ten to the -3m, correct, divided by two and this number square times. Pi, yes. And then and then we Yeah, I see 5.5 point three times ten to the four divided by this number. 4.67 times ten to eleven. Pascal. Three I and six. Yes. So times 53 by ten to the six as it's mega pascals. So I guess. Six nusions. Six nusions Yeah. These aren't hard. You just have to work through them carefully calculate the new length of the wire at this stress. So calculate the new length of the wire at this stress. So we have to work out the change in. So if we know the Young modulus. Calculate the new length of the wire at this stress. Okay. Let's the new length of that of this one. We know the string. Oh yes, so we know 2.1 times seven times ten two is the minus four. Seven x minus four times. So the final answer is. 1.1 point 47 times ten two zero minus three. So calculate the new length of the wire at this stress. So we have to add the original length to the extension. So we get 2.1. Plus. 1.47. My tent, okay. Yeah. So that is 2.1m. So adding 1.47 mm to it, calculate the Young modulus of the wire. So the Young modulus is the stress over the strain. Stress and strain were given in the question. So stress 53 by ten to the six divided by seven x. Minus four. 7.57 by ten to the ten. Yeah 7.57 by ten to the ten pascals. Okay, let's try question five. Question five, try question five. Jackson, a piece of steel is tested, so we don't know the Young modulus. We know the cross section area. We know the force. We know the initial length and we know the extension, calculate the stress as experienced by the steel, so the stress is forover the area, calculate the extension. The strain. And then find the Young modulus. So off you go. First place first is 3000 divided by five times ten to the minus six. Six times ten to eight. Scgood. Strain is one by ten to the minus three and we use another. Length. So that is one times ten to the minus three. Divided by divided by. 0.3 ten 3.33 times ten to the minus three. And then calculate the Young modulus. Remember, it's the stress divided by that number 1.8 times ten to the eleven. Good. Number five, a financial question. Good. So we look at a graph question. So the graph is a one of stress against strain, strain by ten to the minus three, stress in mega pascals. So mega is ten to the six and strain is by ten to the minus three of the number. Okay, so you need to pay attention to these hair hours of ten. Often an unknown material is tested to find its Young modulus. The stress and strain are measured and the graph below is produced. Use the graph to calculate the Young modulus of the unknown material. Can you see that? Is that easier to see? Yes, moless stress and strength. Times ten to the minus three. Is is that minus three? Okay, minus five times ten to the minus three. And. 4.5 times ten to the six. Divided by. Five times ten two or minus three equal to nine times ten to the eight. I get nine times ten to the ten. I took 450 by ten to the mega is by ten to the six. So that is eight Yeah ten two. Big. So here's a force extension graph. This is in millimeters. This is in nusions. An unknown material is tested to find its Young modulus. The force extension are measured and the graph below generated. Use the graph to calculate the Young modulus of the unknown material. The diameter of the wire is 0.4 mm, and the length is this. Slowly should have found that appoint. A diameter is the length is 2.8. This point. So that's about 6.9. By ten to the minus three. 19. Was that a 609? Well, I'm reading there. Maybe that is 6.8. Okay 6.8. Okay. Six point in that one that is 3030 divided by. Six. Times. 6.8 times ten to the minus eight minus three. 4411.8. And that is Young smoless no. So remember constant. So the Young modulus is force times change in length over area times original length. So. It's force times. Length, which are given over extension change in length times area. Let me just double check that I'm getting mixed up now. Yeah, course times change in length. Over. This is wrong. I was right. Force times change in length, no change in length. And that' S L dot ught. So if we've worked out. If we know force is 30, change in length is 6.8 by ten to the minus three. So. Over 6.8 by ten to the minus three. Times area. Now the question, we know the diameter, so we have to work out the cross section area. So 0.2 by ten to the minus three. Squared times pi. And then original length is 2.8 times 2.8. If we work out all of that, we'll get the Young modulus. So 30 times 2.8 equals divided by. 6.8 by ten to the minus three times 0.2 by ten to the minus three. Oh, squared my calculator didn't like something I did. Yeah, I see. That is 1.8 times ten to the minus three. Yeah, I see. So we've worked if we work at the cross section area, so we have to fill in all of this f times length over cross section area times extension, which we read off the graph and we get this. Okay. Okay. Good work today, Jackson. This is, this is.