We're here in Leiden in the Netherlands. Everything that you see around me was once completely submerged in water. The only reason why the Netherlands is still dry today is because a series of dikes have been constructed around the country. Now even with these dikes, we still need to constantly move water from low to high or as we learn to say by the end of this unit, we need to do work on the water to move at some place of low potential energy to places of high potential energy. Today, the dikes that surround the Netherlands help to keep its 17 million inhabitants safe but without human intervention, all the areas that you see in blue would be submerge. Since so much of the country is actually below sea level. Of course this setup isn't perfect, water constantly must be transported up and away from land. One tool, the Archimedes' screw uses power from the Dutch windmills to transport this water and keep the Netherlands dry. We learned about forces in the last unit but thinking in terms of forces may not be enough to understand this complex problem.
So here we have the Netherlands. Over here we have the land. Here's the sea and you can see that the sea levels a bit higher than the level of the land. We have this dike preventing the sea from overtaking the land and of course we have this beautiful little wind mill powering this Archimedes' screw. Now seems like this is going to be a very difficult task. We have a lot water to move. To get you starting to think about just how difficult this task could be, I want you to think about this question. What two factors contribute most to the difficulty of this task? So this windmill has a lot of work to do, what two factors contribute most? Maybe the total weight of water or the total amount of water that we have to lift. Maybe the cleanliness of the water is important. Maybe it's the distance, which I've labeled here as d, that we have to lift the water or maybe the temperature of the water is the most important thing or one of the most two most important thing. I want you to choose the best two answers here and then we can talk about that.
Well, it wasn't too precise in what I meant by difficulty. You might have chosen different answers than me, but I see that the cleanliness of the water wasn't very important and nor is the temperature. In fact, it's the total weight or total amount of water that we have to lift and the distance we have to lift it that most affected how difficult this task could be. We see this total weight and force we're going to need to supply and this distance we need to supply are both seem to be somehow important concepts. Qualitatively, we know that these are the important concepts, but how can we quantify difficulty?
We know when we are quantifying difficulty, it must somehow incorporate these two concepts of total amount or weight of water and the distance we have to lift it and so why don't we start by thinking about the last unit. Well, in the last unit, we learned about force and namely force is the cause of acceleration and we use this equation to describe acceleration. Well, let's try to analyze this situation using forces. Well, I have all this water here that I need to give rid of it that I need to pump up. This water have some mass, which means that I can draw it as a force diagram and to make things simple, I've represented the waters as a simple block of mass m. The block has a weight and the ground is supplying some normal force to keep the water up. We're using forces to analyze this situation and well it seems like total weight of water is taken care of. We've addressed that concept but we haven't talked about the distance we have to lift this water which also seems very important in analyzing or quantifying this difficulty of this task. How much work we're going to have to do. I've been using this word work kind of colloquially but it has a very precise meaning in physics and that's one we're going to learn today. We're also going to learn about energy and what is called the conservation of energy so even though these terms work and energy are relatively new once. We can go all the way back to the time of Archimedes to study their origins.
Over 2,000 years ago, Archimedes was obsessed with this simple question-- How can I lift a 100 kg mass to a height of 1 m? Okay so this might not have been exactly how we phrase the question, but this captures the essence of what Archimedes was thinking when he invented many of these simple machines. Naively, we might think the only way to answer these question of how to get this, let's say 100 kg rock unto this 1 m table. It is well, to pick it up and put it on the table. Archimedes, however, realized there's better way to solve this problem. One way is to use an incline plane and we analyzed the incline plane a little bit in the previous problem set. When we use an incline plane, we reduced the force we have to push with the one in expense. We have to push over longer distance before we have to lift 1 m but now we have to push a distance that's greater than 1 m. Another option is to use a system of pulleys. Pulleys actually have the same trade off that incline planes do. We may not have to pull quite as hard but we'll have to pull over a greater distance. Once again we see this force and distance trade off popping up. Another simple machine that can help us out is the lever. If we use the lever, we can exert a smaller force than we would have to if we we're just picking it up but once again we'll have to exert it over a greater distance just to lift this rock 1 m. This is very interesting, let's dig dipper into some of this simple machines and see if can quantify what's going on.
Here we have the simple pulley--it's called a simple pulley because the rope goes from the rock wounds over the pulley and right into the person's hand. There's only one, let's call it connection point to the rock. If we want to draw a force diagram for this rock as we are lifting, we have the weight going downwards and that will be balanced by the tension, label T in the rope. Remember in a rope, the tension is constant throughout. So, what do you think what causes the tension in the rope? And this is a sort of subtle question. There's a lot of things you could say or contributing tension in this rope. But I want to know of these three options, what is the best answer? Is the weight of this person, is it how hard this person is pulling, or is it the normal force on the rock? Choose the best answer.
Well, it can't be the normal force on the rock, the rock is in the air. The ground isn't pushing up on it and there is normal force. I say the best answer was the pulling force. This person is pulling down on the rope. Depending on how hard they pull, they can control the tension in this rope. So we know that the tension is equal to this pulling force. And it's always going to be true no matter how many pulls to that, and if this tension force is balancing the weight, which it is, well then, these forces are also equal to the weight (mg). And (mg) is about 1000 N because the (m) is 100 kg and (g) is 10 m/s², actually 9.8 but we're using 10. This doesn't seem like a great improvement. We're still using a force equal to the weight to lift this object. Why not just pick it up? Well, that's why this is called the simple pulley.
More interesting than the simple pulley is the compound pulley. Now this picture is better than my drawing over here and you can kind of see what's going on. By using this clever mechanism, we actually let the string pull on the object twice. This string is pulling upwards as is this string. If we want to do a force diagram on the subject, we still have (mg) going down but now the tension goes up twice. One representing each of these arrows . What would the tension have to be now in order to balance this 1000 N weight? Enter your answer in Newtons here.
We have two tension forces, so each has to be half as big. A force of 500 N twice will balance the 1000 N force. Now that's pretty amazing, of course, there is a caveat. If I'm going to lift this 1 m high, it turns out I'm actually going to have to pull down over a distance of 2 m. Check out the link below this video to understand why this is the case.
We said that when we want to lift a 100 kg or 1000 N rock by 1 m, We have at least a couple of ways to do it. With a single simple pulley, we just have to pull with the force of 1000 N over a distance of 1 m. For a compound pulley with two attachment points, we call that a double pulley, we have to pull with a force of 500 N over a distance of 2 m. Can you complete this table for the case of a triple and a quadruple pulley?
Let's look at a triple pulley. We'll first see have a triple pulley work. We would wrap around this top pulley. Go down around the bottom, wrap once more around the top and then connect somewhere in the middle here. Now we can see that the tension is pulling upwards one, two, three times. The force diagram for that is well, the weight points down and we have three arrows upwards each representing that tension. Since the weight is a 1000 N, these each balance 1/3 of that and so we get a force of 333 N. Of course the consequence is that your going have to pull three times further than you did in the single case. You can do similar thinking to solve for the quadruple pulley. Now you may already be seeing a very amazing pattern emerging from this data. Before we explicitly point it out there, why don' we calculate one more method for lifting this rock by 1 m, let's look at an incline plane.
Here we have an incline plane with a person pushing this, let's say rock, We saw in the last unit that to analyze a block in an incline plane instead of writing mg like normally do going straight down, we can break it into components, which we call perpendicular and parallel components. For the case of a nice smooth frictionless plane, this parallel force is balanced by the pushing force of the guy who I just erased, sorry guy and the perpendicular force is balanced by the normal force of the surface of this plane pushing up against the block. We show in the problem set that the perpendicular force can be expressed as the weight times the cosine of this angle, which we have called α and the parallel force here can be expressed as this mgsin α. Can you tell me for the case of a 100 kg block and angle of 30°, what would this parallel force of gravity, the force tending to pull the block down the incline plane. What would that be equal to?
Well, this isn't too bad. We just plug in our known numbers 100 kg10 m/ssin 30°. Plug these all in. We find a parallel force of 500 N.
Now take a look at this force diagram, this force is going down the plane must be balance assuming it's not accelerating by the force going up the plane. Put another way, the push force and that's the force that the guy pushing on the block was exerting must also be equal to 500N. Just like in the previous examples of the pulleys, we don't just want to know the force we have to exert to move this object up to a height of 1 m. We want to know the distance over which we must exert that force. Can you use your knowledge of trigonometry and geometry to tell me over what distance must we observe this 500 N force to move it from the bottom down here all the way to the top.
Well let's see. This is a right angle. This height is 1 m and this side over here is opposite the 30° angle. We want to know about this distance, which is the hypotenuse like. Why not that the sin 30 must be equal to the opposite side over the hypotenuse. We call this unknown side while this is called x and look what we find. This distance d is equal to 2 m. We're on the verge of something amazing here.
We have talked about three methods of lifting a rock of 100 kg by 1 m. If we just want to lift the rock, well our force has to balance the weight, which means you're going to need a 1000 N force, and we're going to have to exert that force of a 1m, exactly the height of the table. If we use the pulley, we had a few options. One was a simple pulley with one connection point. We'd have to exert a force of 1000 N over the full distance of 1 m. It was a double pulley, we have to exert half that force, but over twice the distance and so on. If we use an incline plane, well if the angle is 30°, we have to exert a force of 500 N over a distance of 2 m. There's a deep and fundamental pattern emerging here. For all of these seemingly entirely different ways of moving this block up by a height of 1 m something is constant, the force times the distance is always equal to 1000. This quantity force times distance is so important that we're going to give it its own name--we're going to call it work.
Let's use this idea of work and for now I'm going to call work F Δx. It just means some force times some change in position. This x could also be y. I'm just going to use x for now and let's think of a very simple way of moving water from here up. Now, one thing we can do is well, why not install a ladder. We can take our bucket down to where this water is sort of leaking in, fill it up, carry it up. You need to tell me, how much work is required to lift a 5 kg bucket of water 7 m? Enter your answer here and I'll tell you the unit. This is big J which stands for joules and that's just the unit for work--well, get more into that later. Be careful. The answer is not just 5 7. Okay. Enter your answer there.
Well let's think. If we have 5 kg bucket of water, I know that bucket has a weight, mg. I'm going to exert some force and this force actually must balance mg. In the case of a 5 kg bucket, mg would be 5 10, 50 N as with the force I had to supply to hold the bucket. Well work is just force times this distance of 7 m, that's equal to 350 J. And what's truly amazing about work is it doesn't matter if we try to use a pulley or a compound pulley or an incline plane, we have to do this much work to lift the water.
Okay, so we've taken our bucket all the way up to the top here that took 350 J of work and now we need to carry it to 4 m so that we can dump it into the sea. Now I want to know, how much work do we do in carrying this 5 kg bucket 4 m over to the sea. Now, I have to issue a little disclaimer here. This is much trickier problem than it looks like at first. You'll probably get it wrong and that's okay. Part of the purpose here is to get you a little bit frustrated because what I'm going to teach you in the solution video is so important that I really want you to remember.
The answer is that you do a zero work in moving this bucket 4 m to the side. Now of course, there's no way you should have or could have no met. Don't worry if you got that wrong.
Why was there no work done in the previous example? What happened? Well, our definition of work, this is the product of a force and Δx, a distance. Well here, while I'm holding my 5 kg bucket, I'm definitely exerting a force of 50 N and I exerted over a distance of 4 m. Why is the answer not 50 4? Well, the reason is I was dishonest with you. I need to give you a little more instruction on how to use this equation. Noticed that when I write force, I have to write 50 N that's the size but I write an up arrow to represent the direction. When I write Δx, 4 m is the distance, but that distance was to the right. Both of these quantities have both a size 50, 4, and the direction upward to the right, associate with them. When I make this multiplication, I have to very careful on how I do it. Specifically, I have to take what I'm going to call the parallel component of force with the distance. Now, this idea of parallel component is a little hard to explain and a little hard to understand at first. Why don't we go ahead and do an example and I think you'll see what I mean.
What do I mean with parallel force? Let's think of Peter pulling his sister, Annie, in a cart. He has his rope here, and it's connected to the cart, and the rope actually makes an angle of 30°. Well, let me give you some other information. The mass of the cart and Annie together is 25 kg. The force that Peter is pulling with is 10 N. And, I want to know if Peter pulls Annie for 15 m, how much work does he do. Well, you might be tempted to say, well work, it's just force times distance. He is pulling with a force of 10 N over a distance of 15 m. Shouldn't that be 150 J? The answer is no. We shouldn't be using this entire force of 10 N, just the parallel component. Let me show you what I mean. I'm going to have to erase Peter and Annie now because this drawing is just getting too cluttered. Instead, Annie is going to become a box. Peter is pulling with a force, we'll call Fp, and as always, there is going to be some normal force. Now, I know the direction of motion of this car is going to be along the street, this way. That means I want to break this force into components, and this is something we know how to do. Here, I've broken it into what we can call a parallel component because this line is parallel to the direction of motion, and there's a perpendicular component, and I remember that this angle is 30°. Now, can you tell me if Peter pulls Annie for 15 m, how much work does he do? Enter your answer here.
If you got this right, good work--the way we've solved this is by saying, well, we want to know the parallel component. The cosine of 30 tells me the ratio adjacent to hypotenuse, and that gives me a parallel force of 8.7 N. Now, I just multiply this by 15, and I get a value of about 130 J. Let's do one more problem so that you can practice this on your own.
For this problem, let's imagine Justin here is mowing the lawn. His lawnmower has a mass of 7 kg, and when he mows, he pushes down with a force of 20 N and angle here of 60°. If Justin has to push the lawnmower a total distance of 150 m to completely mow this lawn, how much work does he do? Enter your answer here.
Well, as always, I'm going to try this a little more simply. Here's my lawnmower, the weight, the normal force, and the push force that comes in like this. But that's not how I would like to draw my forces. I'd like to draw them coming from the object, not going into it. I'm going to draw like this and it has a value of 20 N. I'm going to break that into a parallel component and a perpendicular one, and then I'm going to remember that this angle here is 60°. Well, I know that this is my equation for work and Δx, I know is 150 m. Now, I just need to know F parallel. Well, looking at this right triangle here, I can see F parallel at the adjacent side and carrying out the math, I find F parallel is 10 N, multiplying this by 150 m, I get total work of 1500 J. You got this? Very good. Let's go back to our man, moving water along the dike to see exactly what was going on. So in this case, the man was lifting upwards with some force, but the direction of motion is this way. Actually, let me just extend this arrow, so that we can see this is a right angle. When our vectors are perpendicular in this way, no component of force, whatever will be parallel to Δx. In fact, whenever the force is perpendicular to the displacement of the Δx, no work is done. One thing we haven't thought about, is what happens when the force is in the opposite direction of the displacement?
When the parallel component of force is in the opposite direction as the displacement, work is negative. Here' s a guy climbing up some stairs. He's sad because he's blue, and here's a lady carrying a bucket down some stairs. For this question, I just want you to tell me who is doing the positive work and who is doing the negative work? The person going downstairs, check the appropriate box and the person going upstairs, check the appropriate box.
This guy is moving his bucket upwards and he is exerting a force that is also somehow upwards. I can take some component of this and it would be parallel to this. Same direction means positive. For the woman on the other hand, well, she still lifts up to hold the bucket upright but she is taking it downstairs. Opposite direction means negative work. And this sort of makes sense, it's in line with our intuition. It's hard to carry heavy things upstairs. To do so, you need to do work on that. Taking things downstairs is much easier. The work done there is negative. This thing wants to go downstairs. I know we're talking a lot about work but that's only because it's such a vital concept. Stick with me a few more minutes and you're going to encounter something amazing.
Earlier in the unit, we talked about how much work this man does when he carries the bucket of full of water up this ladder. And we calculated that it was 350 J. Now, it's tempting to say we're done here and we've studied all the work going on but there's another fore-set play--let's look at the diagram. It's not just the man pulling on the bucket, he's pulling against gravity. In this situation of a 5 kg bucket being lifted at 7 m, can you tell me how much work is done by gravity?
Well, our force is 15 N and that force was downwards. Our displacement are Δx with 7 m upwards. We can multiply these numbers but we have to remember these are in opposite directions. The work done by gravity is -350 J. The total work is 0. Now, anytime you see a zero in physics, you just start to wonder what's going on? Is it a coincidence? Probably not. Let's think of few examples where work is done and see if we can come up with any where the total work is not 0.
I want you to consider these five situations. Lifting a brick by carrying it up some steps at constant speed, walking this bucket up a ladder at constant speed, pushing a crate up an inclined plane at constant speed, or using a pulley to lift a rock at constant speed, or Galileo standing at the edge of the Leaning Tower of Pisa dropping an object. I want you to tell me which of these situations would there be nonzero total work done on the object in question? Check the box next to the appropriate object.
The answer is this one, and now, I'm not going to go through all of these other four cases. I'll leave that to you to show that in each of these other four cases, the total work done when you really include everything--gravity, pushing etc., really must be nonzero because in each of these, I said, happened at constant speed. Here, when the object falls, well, that's clearly not constant speed. We know it accelerates. Now, we're getting somewhere. This case is so special out of these five because the motion is changing in a very specific way. When we keep track of the total work which I'll abbreviate with a little T down here, we find that it causes the object's kinetic energy to change which I will label like this. This is so important that I'm going to write it. Work causes changes in kinetic energy. Now, we don't know what kinetic energy is quite yet. Stay tuned one second but this statement is known as the work energy theory and it's vital to physics. Now, let's talk about what this whole energy nonsense is.
So what is kinetic energy? Well, it's ½ mv². I hope that right now you're seething. I hope you're writhing in your chair and saying, Andy, that's not what kinetic energy is. That's an equation that describes kinetic energy. What is kinetic energy? Oh, what is kinetic energy? Well, it's the energy associated with motion. Well, I hope now you were saying something like this, because using energy to define kinetic energy isn't helping anything. Well, this question has a short answer. The short answer is, I don't know what energy is, but luckily, I know what energy does. If you've had enough of this philosophical mumbo jumbo, why don't we go ahead and actually solve some problems. We had this idea, this work energy theorem. It says that the total work on an object is equal to it's change in kinetic energy. and since this is the equation for kinetic energy, well the change is just this. The F means final velocity and Vi means initial velocity. So why don't we solve a problem that we've solved previously using motion considerations, but this time, let's solve it using the work energy theorem. So, I want to know what is the ball's final velocity, if it's dropped from rest? You can enter your answer here, and I know I haven't actually shown you how to use this equation yet but you might be able to figure it out just by looking here, and if you can, I think you still have the skills to answer this question.
Now, you could have solve this problem using the equation of motion that we learned about in unit 2 or you could have use our new super cool technique of the work energy theory, which is how I'm going to solve this one. For work energy theory, first we need the total work and that is given by this equation. Well, let's look at the forces on the ball. Well, this is an easy one, there is only gravity pointing downwards. Okay, and Δx, well, we're just going to use this distance 57 meters. W = mg 57 and that is also equal to the change in kinetic energy. And notice that an amazing but necessary thing happens, the mass cancels out. I can also multiply both sides by 2. Oh and don't forget that the initial velocity was zero since it was released from rest, and now, we have this equation. I just take the square root of both sides, and I get the final velocity of 34 m/s. So, now, we have another tool to solve this problem. We'll find there's actually places where this tool will solve problems that our previous tools couldn't. I think I teased you a little too much when I said, I don't know what energy is, but I know what energy does. Let's dove into that a little more deeply.
This is what energy does or I should say doesn't do--it never ever changes. The total amount of energy in the universe stays the same. Energy is what we call in physics--a conserved quantity. I told you before that I don't know what energy is and I don't need to. I know that there is a certain number that if I calculate it once--let nature do its thing and then calculate it again, I will have the exact same number. This is an amazing fact and one that gives us incredible predictive power. If I could calculate the total energy in the universe today--something that I probably couldn't do-- how much energy would there be tomorrow--more, less or the same.
Well, that's easy. I just told you. The total amount never changes. They'd be the same. That's a trickier version of this quiz.
Same question, but now I want to know in 10 billion years--more, less or the same.
Were you fooled? It's not that tricky. The answer is still the same. Energy is an amazing quantity, but before we can start making calculations using it, we're going to need to learn about something called potential energy.
Let's see--we've learned that work and energy have an intimate relationship namely that total work causes a change in kinetic energy and we see the energy is conserved. The total amount never changes. Now, what's really fascinating is that even though the total amount never changes, energy can change forms and that's when potential energy comes in. We can think of this as the potential to do work. Now, I can think of what sort of situations involved potential energy. It's often the case that things that are very frightening have high potential energy. Which of these situations, tell me what you think has more potential energy? A rock teetering in a high cliff versus a low, a 12-volt car battery versus a small AA battery with 1.5 volts, a lit stick of dynamite versus a lit twig, and a drawn bone arrow versus a relax one.
Why would be much more frightened to be this guy than him, a 12-volt car battery certainly scares him more than a 1.5, this TNT seems like it's going to do a lot more damage in it's explosion and this was a burns, and this is certainly more intimidating than this. Now, it's not always the case. What's more frightening has greater potential energy. But in each of these situations, potential energy could somehow be converted into kinetic energy. If this rock would have fall, it would gain motions, speed as it fell to the ground. If this dynamite will explode, the shards will go flying off in all directions. If this bow were released, the arrow would fly through the air at high speed. This one is a little harder to see, but if I were to connect the wire between these two terminals, we actually get the motion of what's called electrons through the wire, likewise here but more motion here. This idea of energy transforming between potential energy and kinetic energy and even back again is one that is fundamental to physics, and I hope that by the end of this unit, you'll be seeing everyday occurrences in terms of their energy transformations. Now, all of these cases exhibited a different type of potential energy. This, we will call gravitational potential energy, burning and combustion and explosion is an example of chemical potential energy as is the battery, and a bone arrow is an example of elastic potential energy. In this unit, we're going to focus on making calculations with gravitational potential energy and the problems that we didn't talk about elastic. Keep watching to learn how to use this incredible physics tool.
Gravitational potential energy--well, the energy that's somehow stored in gravity. Well, what do I mean by that--well, we talked earlier about lifting a bucket of mass m up to a distance--let's call it h for height. Now, last time we actually assigned some values to m and to h and of course, we used g the acceleration to gravity to calculate this work that we had to do. This time we're going to leave our answer in terms of variables and so what I want to know is how much work do you do and that's you--not how much work does gravity do and not how much is the total work--how much do you do when you lift an object of mass m by a distance of h. I've given you four options. Try and choose the best one.
Well, the best answer is mgh--in fact, that's the only answer. Let's see why. Well, let's do a diagram of for example this bucket--there's mass m and weight mg. The force you exert must balance mg, so actually the magnitude of this force that you're pulling on the bucket with must be equal to mg. You'll know that this is the equation which describes work and this case, your force is mg and that's upwards which is also the direction of the Δx, but here Δx becomes h--so that's how we get that answer.
In fact this equation doesn't just describe the work required to move bucket of mass m of a height 8 h. This is the potential energy that a bucket has once it's at the site. For potential energy we use the letter u and since there's many type of potential energy, I'm going to use this subscript g to remind us we're dealing with gravitational potential energy. Now one thing I should point out is that this height in this equation seems to be really special. How did the bucket know it's potential energy should be related to its height above this water level? Why it could have been that high or why can the height have been negative? As when the bucket is below somewhere at this point. In fact all of this choices for our reference zero point potential of energy whether it's here or down here or even up here are totally valid. Where we choose our zero point is completely arbitrary, but once we choose our zero point we have to consistent while we're solving a problem and you'll see what I mean by all this when we keep solving more problems and now of course the question is, "Why is this idea useful?" "Why do I care about potential energy?" Well, why don't we solve the problem that you could not have solved 10 minutes ago.
This is a quick recap. Let's compare the u from 10 minutes ago to the u of now. The u who now knows about conservation of energy and gravitational potential energy. Well, even 10 minutes ago you could have solved the problem with let's say a ball dropping down in free fall. Ten minutes ago you could have solve that and now could have solve that. You know everything that you need to know to describe this motion. What about an incline plane? Well 10 minutes ago, I think you could have solve this. You know how to break forces down and solve for acceleration and so now I see you could solve that as well. When an incline plane is in a very realistic shape. How often a nature do you see a hill that's perfectly flat at one single angle, not very often. More often what you see in nature is weird strange shape like this one and if you have a block of mass m on a hill like this, could you calculate the speed of the block when it gets to the bottom. I think 10 minutes ago, you may not be able to do that but now you can and how are you going to do that? Well as long as you know the height of the hill, you know that as the block slides down here well, its potential energy would be described by this equation mgh. As h goes down, potential energy will go down, but energy is conserved. It can't be created. It can't be destroyed. If potential energy is going down, kinetic energy must be going up and in fact, at the very bottom of this hill once the block has gotten all the way down here h is zero, so all of the potential energy is gone and converted into kinetic energy. We can say that mathematically by saying that the potential energy the block had up here is equal to the kinetic it has down here or mgh=1/2mv² and what do you know the mass tells us. The mass of the block doesn't matter. We can solve this equation for v to get this wonderful answer. The velocity of this block at the bottom will be the √2gh. This answer is remarkable. It doesn't depend on the mass and Galileo knew motion shouldn't really depend on mass. It depends on this acceleration due to gravity as we expect it it's higher well the speed is higher. It also depends on the height of the hill. Slide it down a higher hill and the velocity is greater. This makes a lot of sense--now that your so much smarter than you were 10 minutes ago, time to put this skills to use and solve some problems.
A great place to try out our skills with this new found tool of energy is on a roller coaster. If you ever look at a physics book, I promised you on the chapter on conservation of energy you will see some problems of roller coasters. The reason is because the energy transformations are very easy to see. Up here, we have Isaac, Newton, and Galileo let's say and their cart isn't moving yet, but they have a lot of potential energy because they're up high. As they go down that would be converted to kinetic energy. Start going up some of the kinetic energy is converted back to potential and so on as they go across the track. Using what you know about the conservation of energy, could you tell me, assuming that there is no friction on this track, we'll get to friction later in this unit, "What's the fastest speed that this roller coaster will reach?", and we'll call thet Vmax. Enter your answer here in m/s.
Well, let's see how do we solve this, first I want to identify where is that roller coaster will be going the fastest and the answer is when it's down here. This is the lowest point in the roller coaster and therefore the point with the lowest potential energy, which means the high is kinetic energy and let's see, down here, what kind of energy it does have. It has kinetic energy. I know it's going to be moving down here and it didn't have a little bit of potential energy if I'm using the ground as my reference point for zero, so let's see. Energy is conserved means energy appear equals energy down here. Roller coaster starts at rest, which means it starts with only potential energy. I think we're going to call this h₁ to distinguish it from this h over here. By the time it gets to the bottom, it still has a little bit of potential energy and it has kinetic. We know motion doesn't depend on mass and luckily the mass is cancelled out. I can do a little bit of massaging of this equation to get into this form for v² and to solve for velocity, I just take the square root and look at that. All that matters, the only way that height comes into this equation is through the change in height, h₁-h₂,. This is really pointing out the fact that we can choose our zero point anywhere and all that matters is the change in height. Well let's plug in our numbers, when I do this I get a very nice number, I get the √900 which is exactly 30 m/s.
Now I confess, I've been oversimplifying it because we haven't talked about friction. Now even though energy can never ever be created or destroyed, friction in a way removes energy from a system. Now what do I mean by that. Well before we ??? let's say a ball dropping from some height. Well, we can think of this as energy converting from gravitational potential to a kinetic energy as it moved in. All of the gravitational I want to connect, but in reality that's not what happens. All of the gravitational potential energy does not get convert to kinetic energy. How did friction do this? How does it remove energy if energy can't be created or destroyed? Well before when we look at, let's say a block sliding down a plane. We visualized this as gravitational potential energy getting converted to kinetic energy as the block moved down, but let's think about that. As this blocks slides down, it's rubbing against this plane. Go ahead and rub your hands together for a second and do it really hard. They get really hot. In fact, friction creates heat. This heat is another form of energy. If we start with a certain amount of gravitational potential energy, some of it gets converted to useful kinetic energy, but some gets lost as heat. The energy is still there, it's just that we can't really use it anymore. It's lost as heat to the atmosphere. Now depending on the material of this plane or other situational factors, the fraction of this energy that gets converted to heat can change, maybe a lot of this energy gets converted to heat maybe almost none of it. So let's consider these three cases. In the first case, we have an object falling and let's say it's falling to what's called vacuum so there's no air, it's just falling through empty space Another example of object sliding down this rough jagged surface. Another we have an object sliding down a relatively smooth surface. What I want to know is, which of this diagrams corresponds to which of this pictures. So in this diagram all of the potential energy is getting converted to kinetic energy and here most of it is getting converted to heat, but in here only a small portion gets converted to heat. Enter the appropriate number into the appropriate box.
Well, let's see. Number 3 seems pretty straightforward. When will we not lose any energy to friction? Well, if we drop something through a vacuum where there's no air, nothing to collide with--no energy is lost to friction. This very rarely happens. When is the most energy lost to friction. Well, on a rough surface, a lot of energy gets lost to friction. That leaves a smooth surface in which--yes, some energy does get lost to heat but not all the time.
Well, let's see. Where does this final energy come from? Let's look at this one for an example. Well, this total energy, the initial energy, minus whatever was left to heat. This answer.
These diagrams are kind of giving me a hint as to how we might incorporate energy lost to heat into our calculations involving conservation of energy. So actually what I want to know is which of these equations best relates the final energy, the initial energy, and the energy lost. So for example by final energy, I mean final useful energy--may be kinetic energy. Initial energy would be this initial potential energy, and energy lost-- because remember energy is never really destroyed--would be this energy lost to heat. Which of these equations describe the final energy?
Now, we're almost ready to tell the question that we started this unit with--how to keep the Netherlands dry. Let's do one process an example before we jump in. For this example--let's say you've dump 2 kg of water down a 12 m hill. It loses 40 J of energy to friction. How fast is that water going at the bottom of the hill? Add the answer here in meters per second.
The initial energy it has is entirely the potential energy, N which would be 2 kg, g which will of course be 10 or 9.8 but I'm using 10, and h which is 12 m. Now the initial energy isn't all going to be converted to kinetic energy. We're going to lose some of it--we're going to subtract some of it to friction. We're going to subtract 40 J in fact, and that quantity will be converted to kinetic energy. Now we can start plugging in--when I do the algebra out, I get an answer of 14 m/s.
One more concept that we're going to need is the idea of power. This is a word you've heard a lot like force or energy, but now, let's put a real definition to it. Power measures how quickly work can be done! And what I mean by that is, let's say if two people and there's a pile of rocks. Let's say, this first person is more powerful than the second person. Even though, if you give them bought a sufficient amount of time, they can move all these rocks up to, let's say, this basket here. If person 1 is more powerful, they'll do it faster. And so the equation for power is work divided by how long it takes to do that work, we call it Δt. Let's say this first person can lift one of these rocks, one of these 2 kg rock, a height of 15 m in 30 secs, what is their power? The units of a power are watts. Watt is just another term for joules per second. Enter your answer here.
Well, power is work over the change in time. How much work is done? Well, look at the rock. Well, the force we exert has to balance (mg), 2 kg10 is 20 N and the distance is 15 m. And since this is parallel to this displacement or this Δx, we can just multiply them. That means, if the work is equal to 20 times 15. We know the time is equal to 30 secs and the calculated power, we just do work divided by the change in time, 300/30, gives me a power of 10 watts. Not too bad. Now, let's solve our problem.
We want to keep the Netherlands dry, which means that when water comes into the area that's supposed to be land, we need to get it out. Now, there's a lot of factors that affect how quickly water can enter here. What I want to know is, if I can know how quickly water is entering this area over here, how many windmills will I need to pump it back out? And that is a matter of figuring out exactly how much one windmill can do? Each of these windmills has access to about 1000 W of wind power. Of that 1000 W, a lot of it is lost to friction. In fact, 400 W is lost as this rotates as the water sloshes along in here, There's many sources of friction here. We said the average dike is about 6 m high. I want to know how much water can each of these windmills move from down here all the say up to here in 1 day? This is a tricky problem but try your best and enter your answer here in kilograms. Remember, you can always go to the forums if you need help.
Well, here's how I solved it--this windmill has access to 1000 W of power, but it loses 400 W, so that means the usable amount of power is the difference, 600 W. Well, power is work over time and what's our time going to be if we're looking for how many kilograms of water we're going to move. Our time is going to be one day and when you do it at the unit conversion, you find that one day is equal to 86,400 seconds. We can solve this equation for the total work this windmill can do, multiplying both sides by 86,400 and we get a rather huge number of 51,840,000 joules. So this is the work that can be done in a single day and this work is just equal to the mass of water times (g) times (h) because we've already shown that when we lift an object, and it doesn't matter if that object is a stone or water by a height (h), this is the work that we have to do. Well, we know (g) and we know (h), if we solve this equation for (m), we find that our windmill can move a total of 864,000 kg of water per day. Now that we know this number, deciding on how many windmills to build, just depends on figuring out how quickly water is leaking in. Now, we've made some assumptions here and these are assumptions that I encourage you to talk about in the forums, but for now, I'd like to say, congratulations!
Wow! Congratulations! In this unit we've learned a powerful new predictive tool. The conservation of energy. We can learn about work and power. I can answer your problems like the problem of keeping the Netherlands dry in terms of their energy requirements. Next time, we're going to see how the conservation of energy will help us create incredible timekeeping devices that will let us answer the question-- how can we find our ships at sea.