WEBVTT physics/high-school-physics/selhorst-jones 00:00:00.000 --> 00:00:05.000 Hi, welcome back to educator.com, today we are going to be talking about kinetic energy. 00:00:05.000 --> 00:00:10.000 Just to begin with, you have definitely been exposed to the idea that there is a lots of different types of energy out there. 00:00:10.000 --> 00:00:12.000 You probably have been hearing about this from a very young age. 00:00:12.000 --> 00:00:16.000 You have talked about this, almost as early as kindergarten. 00:00:16.000 --> 00:00:22.000 There is lots of different kinds of motion, energy, and ways for things to be energetic out there. 00:00:22.000 --> 00:00:32.000 Some examples: there is kinetic motion energy, gravitational potential, if you lift something up and drop it, or lift it even higher, it is going to wind up having more energy. 00:00:32.000 --> 00:00:46.000 Chemical energy, say hydrocarbons like gasoline inside of your car, light energy: the sun casting all that light energy at us, heat energy: if you are near fire, there is heat energy in the air. 00:00:46.000 --> 00:00:58.000 Nuclear energy: radioactive Uranium is going to be able to put out energy in a form, electricity: moving electrons, and even more things than what I am talking about right now. 00:00:58.000 --> 00:01:00.000 There is lots of different ways to have energy out there. 00:01:00.000 --> 00:01:13.000 It is really useful to us as physicists to be able to understand how energy works, and get a really strong grasp on it, if you are going to be able to solve a lot of things, and model stuff, to be able to have control over environment. 00:01:13.000 --> 00:01:16.000 Another idea you have probably been exposed to is this one: 00:01:16.000 --> 00:01:21.000 Energy cannot be created or destroyed, only changed from one type to another. 00:01:21.000 --> 00:01:23.000 The conservation of energy, you have almost certainly heard this. 00:01:23.000 --> 00:01:29.000 It is not possible to get rid of energy in a system, it is only possible to change it from one form to another. 00:01:29.000 --> 00:01:37.000 If you have a car driving along and hits the breaks, it is not that its kinetic energy goes into nothing, the kinetic energy goes into the squeal of the breaks. 00:01:37.000 --> 00:01:41.000 The friction of the brake pads, the friction of the tyres on the road. 00:01:41.000 --> 00:01:46.000 There is various different kinds of heat, the heat energy is what is going to come out of it instead. 00:01:46.000 --> 00:01:56.000 Or you could have that chemical energy burning, it is not that the chemical energy goes away, it gets turned into heat, it gets turned into sound, it gets turned into kinetic energy, if say it is a rocket or a space ship. 00:01:56.000 --> 00:01:58.000 All these sorts of things. 00:01:58.000 --> 00:02:02.000 We can think of energy through this metaphor of a bucket full of sand. 00:02:02.000 --> 00:02:06.000 We have got a bunch of different buckets, and each one has an amount of sand. 00:02:06.000 --> 00:02:10.000 Say we are looking at a rocket on a launch pad. 00:02:10.000 --> 00:02:13.000 It has some amount of chemical energy, it has got lot of chemical energy. 00:02:13.000 --> 00:02:29.000 All of the fuels in these rockets, it has got no kinetic energy, it is just right at the bottom, no kinetic energy, and it has got some amount of gravitational energy, it is currently sitting on a pad, and it is also, maybe we are at some high place, wherever we are launching it from. 00:02:29.000 --> 00:02:30.000 So, it is slightly higher up. 00:02:30.000 --> 00:02:33.000 What happens when we ignite those rockets? 00:02:33.000 --> 00:02:40.000 When we ignite the rockets, we use that chemical energy up, so we wind up burning all that chemical energy. 00:02:40.000 --> 00:02:51.000 And that energy instead cause the rocket to move upwards, we causes an increase in kinetic energy, it is now moving up. 00:02:51.000 --> 00:02:59.000 Now we have got an increase in kinetic energy, and the gravitational energy will go up as it moves up. 00:02:59.000 --> 00:03:07.000 But the total amount of sand, the amount that the chemical goes down by, is the total amount that the kinetic and gravitational go up together. 00:03:07.000 --> 00:03:11.000 If we look at the total amount of sand in all the buckets, all use the same. 00:03:11.000 --> 00:03:14.000 It is just we change the location from one bucket to the other. 00:03:14.000 --> 00:03:23.000 We either want our sand or energy in our chemical bucket or we want it in our kinetic bucket, or we want it in our gravitational bucket, but these are only different ways of swapping it around. 00:03:23.000 --> 00:03:29.000 Or may be what will happen is, the rocket would run into a satellite, now we are having two system interact. 00:03:29.000 --> 00:03:38.000 Now, swapping from the kinetic energy of the rocket, it is getting that kinetic energy swapped into the kinetic energy of the satellite, because the satellite gets knocked out of the orbit, and it hurdles away. 00:03:38.000 --> 00:03:55.000 These sorts of things, it is not that we destroy energy or we change it inherently, it just gets changed from form to form, or from one object to another object, it gets transferred from systems, or gets transferred from types with in a system, or it gets transferred from types between systems. 00:03:55.000 --> 00:04:01.000 But it does not get lost, does not get destroyed, it is always there, it is moved around in different ways. 00:04:01.000 --> 00:04:12.000 Now, we will talk a little bit more about what specifically, why we can guarantee this, why we believe in it, why we think that this is one of the fundamental laws of the universe, but that is for thermodynamics, and that is while down the road. 00:04:12.000 --> 00:04:20.000 But for now we can trust in the fact that the conservation of energy works, and that is going to be incredibly key to be able to solve all sorts of different things. 00:04:20.000 --> 00:04:24.000 It gives us a lot of power over the understanding of how the world around us works. 00:04:24.000 --> 00:04:26.000 What is energy? 00:04:26.000 --> 00:04:30.000 We have talked about it having these different forms, we have not really talked about what it is. 00:04:30.000 --> 00:04:42.000 It is hard to come with a truly rigorous definition that we can actually make use of, and be able to both play with it, and do cool things and be able to talk about it in a deep rigorous way, all the same time. 00:04:42.000 --> 00:04:47.000 That is really hard, so we are going to sidestep that, we will be able to think of energy as the capacity to do work. 00:04:47.000 --> 00:04:51.000 Being able to put work in, being able to make a change in the world. 00:04:51.000 --> 00:04:59.000 Energy is your ability to change the system around you to build a change in environment, or change yourself with in the environment, it is to do work. 00:04:59.000 --> 00:05:00.000 Work is change in energy. 00:05:00.000 --> 00:05:03.000 So, what is work? 00:05:03.000 --> 00:05:09.000 If work is what energy can do, work is a lot easier to define in fact. 00:05:09.000 --> 00:05:11.000 Work is just the transfer of energy from one type to another. 00:05:11.000 --> 00:05:17.000 Doing work is just the act of transferring energy, whereas the quantity if work is the amount of energy transferred. 00:05:17.000 --> 00:05:41.000 If we had two buckets, and we wind up moving a whole bunch of stuff from one bucket to the other bucket, the amount that we move over, the quantity, is the quantity of the work, and doing work, this is the act of doing work, it is the moving it over. 00:05:41.000 --> 00:05:44.000 Work is just moving energy from bucket to bucket. 00:05:44.000 --> 00:05:46.000 That makes sense. 00:05:46.000 --> 00:05:49.000 Wait!, that was a circular definition. 00:05:49.000 --> 00:05:58.000 I defined energy with work, and work with energy, but in fact I did not really define either, that is not allowed, you cannot just make circular definitions and expect anyone to trust you. 00:05:58.000 --> 00:06:04.000 You should not do that, you want to be able to have something where you can stand in, and at least make a definite claim. 00:06:04.000 --> 00:06:13.000 You got me, that is a really good point, we really should be having something that is a meaningful statement that we can really push against and test and try out. 00:06:13.000 --> 00:06:16.000 We do not want circular definitions, because they are not scientific statements. 00:06:16.000 --> 00:06:20.000 But, in my defence, we have actually defined work. 00:06:20.000 --> 00:06:29.000 Work is the transfer of energy, work is the motion of energy form one of our theoretical bucket to another theoretical bucket, not theoretical but metaphorical. 00:06:29.000 --> 00:06:35.000 You move energy from one type to another, form one system to another, and that is work. 00:06:35.000 --> 00:06:38.000 But what we did not really define is, we did not really define energy. 00:06:38.000 --> 00:06:40.000 We did not do that rigorously. 00:06:40.000 --> 00:06:54.000 To be honest, to finding energy precisely, in a really rigorous meaningful way, like push against and really scientifically talk about, and it to really mean something, is kind of outside the scope of this course. 00:06:54.000 --> 00:07:06.000 It is actually kind of tough to talk about what it really means to be able to have a really deep understanding of that, and there is a whole bunch more to learn about energy before we can really have a very strong perfect meaning, to be able to truly say that. 00:07:06.000 --> 00:07:18.000 But that is okay, we know enough intuitively, we are able to talk about the idea of energy, you know about having rocket fuel, you know about something moving very fast, you know about something being in a high location or stretching out a string. 00:07:18.000 --> 00:07:25.000 Any of the different ways of storing energy, or seeing a system that has a lot of energy, you got a really good intuition, and that is enough. 00:07:25.000 --> 00:07:33.000 You will be able to rigorously define and being able to talk about things in a really specific way, for a later advance Physics course, that starts to matter. 00:07:33.000 --> 00:07:37.000 For now, we are going to be still be able to do plenty of stuff with what we have got, 00:07:37.000 --> 00:07:39.000 We have got plenty to work with. 00:07:39.000 --> 00:07:46.000 Being able to define it rigorously, it is something to do down the road, when you study a whole bunch more Physics. 00:07:46.000 --> 00:07:54.000 With all that behind us, let us talk about how to derive a formula, useful mathematical formulae, we can talk about things quantitatively. 00:07:54.000 --> 00:07:57.000 How can we get the amount of energy in something's motion. 00:07:57.000 --> 00:07:58.000 Kinetic energy. 00:07:58.000 --> 00:08:03.000 From the theoretical point of view, we have seen that putting work into a system is the same as putting energy in. 00:08:03.000 --> 00:08:05.000 But we have not quantified it yet. 00:08:05.000 --> 00:08:08.000 Let us fix it. 00:08:08.000 --> 00:08:13.000 Consider a simplified picture of work, where the force F is parallel to the displacement. 00:08:13.000 --> 00:08:33.000 That means that θ = 0, so from here on we are just going to pretend that cos θ = 1, we are just going to get rid of it, so for this case, work = force × distance. (normally, work = force × distance × cos θ) 00:08:33.000 --> 00:08:40.000 We have got a block of mass 'm', and it is on some magical frictionless surface and it starts at rest. 00:08:40.000 --> 00:08:44.000 This is how things are to begin, let us talk about from here. 00:08:44.000 --> 00:08:46.000 Consider the following three formulae from all of our previous work. 00:08:46.000 --> 00:09:08.000 Definition of how we got the force, force = ma, also from kinematics, vf^2 = vi^2 + 2ad (d is distance), and what we just defined in our last section, work = fdcos θ = fd, because we are looking at a simplified picture. 00:09:08.000 --> 00:09:21.000 From this, we get, we can talk about the acceleration, because we want to be able to slope things in to our velocity, so we can talk about what velocity means, we need to be able to get a relationship between 'fd' and velocity. 00:09:21.000 --> 00:09:33.000 What we are going to be doing is to mathematically massaging this, until we are able to get an expression that shows us something that can connect work and this new formula would become kinetic energy. 00:09:33.000 --> 00:09:41.000 We are going to say that is work, because work is the amount of transferred energy, if we put work into something that starts off with no kinetic energy, that will tell us what the formula for kinetic energy is. 00:09:41.000 --> 00:09:50.000 If we want to get acceleration, we get that, a = F/m . 00:09:50.000 --> 00:09:56.000 Also, vi = 0, we said that this thing started at rest, so we can knock that out. 00:09:56.000 --> 00:10:07.000 For ease, vf = v. 00:10:07.000 --> 00:10:29.000 From there, v^2 = 2ad, we have 'a' up here, plug that in, v^2 = 2F/m × d, we are really close, we got an 'F' and a 'd' already. 00:10:29.000 --> 00:10:42.000 We move that m and 2 over, multiply both sides by m, we get, mv^2, and divide both sides by 2, we get, (1/2)mv^2 = Fd = work. 00:10:42.000 --> 00:10:57.000 So the work, if we start off with no kinetic energy, if we out work into the system, that is all going to become kinetic energy, because we have got nothing resisting it, no friction, it is just freely moving, then all of that just turns into (1/2)mv^2. 00:10:57.000 --> 00:11:02.000 The amount of kinetic energy in the system, is (1/2)mv^2. 00:11:02.000 --> 00:11:21.000 Now, from our derivation, we get this formula, the amount of energy in a moving object, Ekinetic = (1/2)mv^2, notice this is using speed, we cannot square velocity, squaring a vector does not mean anything, but we can square a magnitude. 00:11:21.000 --> 00:11:44.000 So we use the magnitude, its energy is a scalar value, it is not a vector, direction does not affect the energy, you could be going a 100 miles/h to the north, to the east, to the south, up, down, does not matter, 100 km/h, 100 km/s, they are all going to wind up being directionless, it is all about the speed that they are going at. 00:11:44.000 --> 00:11:52.000 Ekinetic = (1/2)mv^2 . 00:11:52.000 --> 00:11:55.000 Now we have got a formula, now we can really get our hands dirty. 00:11:55.000 --> 00:12:08.000 Units: We talked about work being the quantity of energy moved, so if work is the quantity of energy moved, if that is the amount that we move, we are going to need energy to be the same as work, in terms of units. 00:12:08.000 --> 00:12:11.000 So, at the moment, we are just hoping that work and energy are going to wind up having the same units. 00:12:11.000 --> 00:12:13.000 Let us see if that is the case. 00:12:13.000 --> 00:12:41.000 Ekinetic = (1/2)mv^2, so what is m?, (1/2) is just a scalar, so we get rid of it, we got m is kg, velocity is m/s, but squared, m^2/s^2, now Fdcos θ , cos θ is a scalar, just a number, changing the amount, but it does not change the units. 00:12:41.000 --> 00:12:45.000 Force is newtons, and we got metre, we need to figure out what newtons are! 00:12:45.000 --> 00:13:08.000 F = ma, 'a' is m/s/s, mass is kg, that means, m/s/s × kg, this is the same as, m^2/s^2kg, everything checks out, we are happy! 00:13:08.000 --> 00:13:20.000 So the units make sense, so the world is safe, so we can just shorten things easily, we will call it a joule, a joule is both the measurement for energy and for the amount of work moved. 00:13:20.000 --> 00:13:26.000 Work is joules, energy is joules, joules is energy. 00:13:26.000 --> 00:13:35.000 Conservation of energy: To be able to really do anything with kinetic energy, we are going to need to talk about the fact that it stays the same, unless it is affected by its environment. 00:13:35.000 --> 00:13:43.000 We talked about the fact that energy cannot be made or destroyed, it is only transferred from different systems to different systems, or from different types to different types. 00:13:43.000 --> 00:13:54.000 Energy in a system must stay constant unless some of its energy is transferred, work, that is what work is, to another system, or another type in another system. 00:13:54.000 --> 00:14:10.000 But, if we look at all the types within a system, the energy of the system, so this is not just kinetic, but all the energies within the system at start, plus the work, the amount of that change, and that is equal to the energy of the system at the end. 00:14:10.000 --> 00:14:12.000 All the types of energy at once. 00:14:12.000 --> 00:14:28.000 Notice that energy transferred into the system, if you put work into the system, to increase its energy, that is represented as a positive value, positive work, environment does work on the system, force is going with motion, chemical energy being contributed, things like that. 00:14:28.000 --> 00:14:30.000 Energy that is transferred out is negative work. 00:14:30.000 --> 00:14:40.000 If work is taken out of it, if system does work on its environment, or the environment takes energy out of the system, say friction, then that is going to be negative work. 00:14:40.000 --> 00:14:48.000 Work is positive if it is going into the system, negative if it is coming out of it, it is going to be up to us to pay some attention to what is going on here. 00:14:48.000 --> 00:15:03.000 Energy of the system at the start + the work = energy of the system at the end, that is the conservation of energy, that guarantee is going to give us so much power in solving an entire new set of problems, allows to really get a good understanding of how the world works. 00:15:03.000 --> 00:15:06.000 One last thing to talk about: How does friction work! 00:15:06.000 --> 00:15:22.000 It is really easy in fact, form an intuitive stand point, friction is the environment taking energy out of it, or we can think of as the energy putting energy in the environment, in either case, the object's motion is going to be slow, it is going to lose its kinetic energy, it is not really going to gain anything, we are just going to get heat dispersal. 00:15:22.000 --> 00:15:25.000 Friction reduces the energy in the system. 00:15:25.000 --> 00:15:27.000 By how much? 00:15:27.000 --> 00:15:36.000 It is as simple as using our formula for work, and our knowledge about friction, remember, if we have got something sliding along, friction always works this way. 00:15:36.000 --> 00:15:47.000 If it goes this distance, then the work is just going to be equal to, (-) × (magnitude of friction) × (distance), that is it. 00:15:47.000 --> 00:15:53.000 The work in the system will be negative because its losing it, it is just the friction force times the distance that the object travels. 00:15:53.000 --> 00:16:00.000 Friction always points backwards, and always gives a negative value, and always saps the energy out of the system we are looking at. 00:16:00.000 --> 00:16:02.000 Ready for our examples. 00:16:02.000 --> 00:16:13.000 First off, real easy one: Block of mass 10 kg is at rest on a frictionless surface. 00:16:13.000 --> 00:16:20.000 It has a horizontal force of 20 N, applied to it for a distance of 25 m. 00:16:20.000 --> 00:16:22.000 What is its velocity afterwards? 00:16:22.000 --> 00:16:38.000 Notice, if we were doing it the old way of kinematics, and F= ma, it will be a little bit more difficult, we will have to figure out what the acceleration is, and we will have to use that complicated, vf^2 = vi^2 + 2ad, whereas in this case, we just figure out the work, we plug into our formula, BOOM!, we have got it. 00:16:38.000 --> 00:16:40.000 How much work is put into the system? 00:16:40.000 --> 00:17:01.000 The work = force × distance × cos θ = fd (since parallel) = 20 N × 25 m = 500 J, going into the system. 00:17:01.000 --> 00:17:16.000 The system starts at rest, we have got the block starting at rest on a frictionless surface, so no energy is lost, so, (energy of the system at the beginning) + (work) = (energy of the system at the end). 00:17:16.000 --> 00:17:31.000 At the beginning, this is zero, there is no energy, it is sitting there still, 0 + 500 J = (1/2)mv^2, since we know it is all going to be velocity in the end. 00:17:31.000 --> 00:17:38.000 From here, we just solve for what 'v' is, we know what 'm' is. 00:17:38.000 --> 00:18:06.000 2 × 500 = 1000, 1000/m = v^2, sqrt(1000/m) = v, m is 10 kg, so, v = sqrt(1000/10) = sqrt(100) = 10 m/s . 00:18:06.000 --> 00:18:19.000 Do not even need a calculator for this one, because it is so easy, 10 m/s, and we are able to do it by just figuring out the work, and figuring out what the kinetic energy is, what the energy connected to that speed is. 00:18:19.000 --> 00:18:29.000 We know it starts off with no energy, and then we know that all the energy goes into its kinetic energy, so it is as simple as figuring out how much energy is put into the system, and we are done. 00:18:29.000 --> 00:18:38.000 Example 2: Two identical cars of mass 1500 kg are driving directly towards one another at 15 m/s, and -15 m/s. 00:18:38.000 --> 00:18:43.000 They crash into one another in a horrible screaming trash of metal, they come to a complete rest after impact. 00:18:43.000 --> 00:18:47.000 How much energy is released during the crash? 00:18:47.000 --> 00:18:52.000 At the end, we have got 0. 00:18:52.000 --> 00:18:58.000 So, whatever energy beginning is, that is going to be our answer. 00:18:58.000 --> 00:19:00.000 How much energy is in the system in the beginning? 00:19:00.000 --> 00:19:02.000 How much is it in each car? 00:19:02.000 --> 00:19:17.000 For one car, the Ekinetic = (1/2)mv^2, and we are going to assume that all the energy is in its kinetic energy, because it is just its motion here, so that makes sense. 00:19:17.000 --> 00:19:40.000 (1/2)mv^2, we know it is 1500 kg, we know what the mass is, we plug these things in, (1/2) × 1500 × 15^2, and the energy of one of the cars is going to be equal to 168750 J. 00:19:40.000 --> 00:19:42.000 That is a pretty fair amount. 00:19:42.000 --> 00:19:56.000 But we got to remember that, also the car is going to have that released, because it is in each one of the case, we do not just have that +15, we have that -15 velocity. 00:19:56.000 --> 00:20:20.000 Since it is speed that we are looking at, we know that they are each going to have the same kinetic energy, so for the crash, it is going to be , the energy of the car × 2, so 2 ×, Ekinetic(of one car), which is = 337500 J. 00:20:20.000 --> 00:20:27.000 So that is how much energy gets released during the crash. 00:20:27.000 --> 00:20:28.000 That is a fair bit of energy. 00:20:28.000 --> 00:20:44.000 Now, what would happen if we double their velocities, instead of driving at 15 m/s, and -15 m/s and hitting one another, they were at 30 m/s each when they crash into one another in completely opposite directions. 00:20:44.000 --> 00:20:49.000 If that is the case, then we are going to have to change what the velocity is. 00:20:49.000 --> 00:21:22.000 We know that the crash at double speed (in red, since it is going to be a lot more dangerous), is going to be, 2 × (1/2)mv^2 = 1500 × 30^2 = 1350000 J. 00:21:22.000 --> 00:21:35.000 Notice that 1350000 is way more than double 337500 J, in fact it is quadruple, because it is going up with the squared of the velocity, the squared of the speed here. 00:21:35.000 --> 00:21:42.000 It is not just about going double the speed, that means you have quadruple the energy, and that is why freeway accidents are so dangerous. 00:21:42.000 --> 00:21:57.000 Because everyone is traveling at such a high speed, it is more dangerous, it is 4 times more dangerous in terms of the energy given out, to have an impact when you are driving at 40 m/s than at 20 m/s. 00:21:57.000 --> 00:22:02.000 Those are reasonable freeway speeds, and those are like normal city speeds. 00:22:02.000 --> 00:22:15.000 So, it is much more dangerous to get in a collision on the freeway, it is simply because the energy involved to stop those cars, at any reasonable rate is going to take a whole lot of more energy, so the energy is a lot more dangerous for the freeway impact. 00:22:15.000 --> 00:22:23.000 And that is why it is so important to be careful on driving on the freeway, it is because it is way more dangerous, potentially, if you get into an accident. 00:22:23.000 --> 00:22:40.000 Example 3: What if we got a block of mass 20 kg, once again sliding on a frictionless table, no air resistance, nothing like that, so it is just about the energy being put into it, with vi = 4 m/s. 00:22:40.000 --> 00:22:46.000 Then acted upon by a force of 35 N at an angle of 20 degrees above the horizontal for a distance of 15 m. 00:22:46.000 --> 00:22:47.000 What is its speed afterwards? 00:22:47.000 --> 00:22:58.000 We do not have to worry about the force of friction, so we do not have to worry about the normal force, so we do not have to break down that force of 35 N into its components, we just have to figure out how much work does it put into our system. 00:22:58.000 --> 00:23:27.000 The work = fdcos θ = 35 × 15 × cos(20) = 493.3 J. 00:23:27.000 --> 00:23:41.000 Conservation of energy formula, we know that the energy at the beginning in our system + work put in = the energy at the end in our system. 00:23:41.000 --> 00:23:42.000 What is the energy in the beginning? 00:23:42.000 --> 00:23:48.000 Remember, it did not start at rest this time, this time, it had an initial velocity of 4 m/s. 00:23:48.000 --> 00:23:49.000 So, we have to include that. 00:23:49.000 --> 00:24:25.000 (1/2)mvi^2 + 493.3 = (1/2)mvf^2 (work is positive because the environment acted on the object, not the object losing energy to the environment), (also, these are actually the magnitudes, since we are talking about speeds.) 00:24:25.000 --> 00:24:50.000 Plug things in, (1/2) × 20 × 4^2 + 493.3 = 653.3 J. 00:24:50.000 --> 00:25:17.000 So, that means our energy at the end, is equal to just the motion in the velocity, just the energy in the velocity, Ekinetic, is going to be, (1/2)mvf^2 = 653.3 J. 00:25:17.000 --> 00:25:59.000 We solve the algebra, we get, vf = sqrt(2 × 653.3 / 20) = 8.08 m/s. 00:25:59.000 --> 00:26:20.000 We know what the starting energy is, we know how much work goes into the system, we put those things together, that gives us the ending energy, and then we figure out what energies are going to be being used in the end, it is just the kinetic energy, that is the only energy that is going to be in our system at this point, so we know that the total ending energy is equal to the kinetic energy at the end. 00:26:20.000 --> 00:26:23.000 And we just solve for it. 00:26:23.000 --> 00:26:26.000 Final one: This one is going to take a bunch of ideas. 00:26:26.000 --> 00:26:31.000 Block of mass 3 kg, is resting on a surface, where it does have friction. 00:26:31.000 --> 00:26:38.000 Friction coefficients are, μs = 0.8, and μk = 0.4. 00:26:38.000 --> 00:26:47.000 It is acted upon by a force of 100 N, at an angle of 40 degrees below the horizontal, so it is pushing down on to the block. 00:26:47.000 --> 00:26:48.000 Does it move? 00:26:48.000 --> 00:26:54.000 The force acts on it for a distance of 50 m, what is its speed after 50m, how far will it slide afterwards? 00:26:54.000 --> 00:26:58.000 First things first, we need to figure out if it is able to move. 00:26:58.000 --> 00:26:59.000 How do we do that? 00:26:59.000 --> 00:27:09.000 To be able to figure out if it is able to move, we need to compare the force acting on it horizontally, to the maximum static frictional force, that it has. 00:27:09.000 --> 00:27:14.000 First, we are going to break this down, into its component pieces. 00:27:14.000 --> 00:27:32.000 Actually, we probably would be better off by looking at it in the other point of view, because we know what the 40 degrees are, so, we look at it, over here, so, 100 N is the hypotenuse, 40 degrees here, so cos(40) × 100 will give us what the horizontal action is. 00:27:32.000 --> 00:27:45.000 That is going to be 76.6 N, and the vertical is going to 64.3 N, pointing down, pointing to the right. 00:27:45.000 --> 00:27:57.000 If we want to split this into a vector, we know that the force vector is going to be 76.6 N, and -64.3 N. 00:27:57.000 --> 00:28:05.000 Before we are able to figure out what the normal force acting on the block is, we need to figure out what the force of gravity is, because this block is being supported by the table. 00:28:05.000 --> 00:28:09.000 If the block is not falling through the table, then that means all the vertical forces on it are in equilibrium. 00:28:09.000 --> 00:28:19.000 The normal force, the force contributed by the table is going to have to beat out, not just the force of gravity now, but also the force of pushing down into block. 00:28:19.000 --> 00:28:21.000 What is the force of gravity? 00:28:21.000 --> 00:28:38.000 Force of gravity = mg = 3 × 9.8, plug that in later. 00:28:38.000 --> 00:28:45.000 If we want to know what the normal force is, we know that the normal force is going to have to be equal to cancelling out both of those. 00:28:45.000 --> 00:29:00.000 It is going to have to be positive, pointing in the up direction, it is going to be equal to 63.3, so it can cancel out the force pushing down on the block, and then also, plus mg, so it can cancel out gravity. 00:29:00.000 --> 00:29:04.000 We put those together, and that gives us 93.7 N. 00:29:04.000 --> 00:29:07.000 So, the normal force is 93.7 N. 00:29:07.000 --> 00:29:11.000 Now we are ready to calculate what our maximum static friction is. 00:29:11.000 --> 00:29:33.000 Maximum friction static = μs × FN = 0.8 × 93.7 = 75.0 N. 00:29:33.000 --> 00:29:36.000 That is the maximum friction static. 00:29:36.000 --> 00:29:45.000 Now, we need to compare that to the force of our pushing on the block, what is the force pushing on the block? 00:29:45.000 --> 00:29:58.000 Its horizontal force is 76.6 N, is greater. 00:29:58.000 --> 00:30:05.000 That means, YES!, it moves. 00:30:05.000 --> 00:30:13.000 That means , we can actually pay attention, we do have to care about the rest of the problem, because we are able to beat out static friction. 00:30:13.000 --> 00:30:16.000 Now, let us work on everything else. 00:30:16.000 --> 00:30:24.000 If friction static is able to be defeated, then we can now figure out how much work gets put into it by the force. 00:30:24.000 --> 00:31:03.000 The work of the force = magnitude of force × magnitude of displacement × cos θ = 100 × 50 × cos(40) = 3830 J. 00:31:03.000 --> 00:31:10.000 Now, remember, this entire time it is moving, it is also being hit by friction. 00:31:10.000 --> 00:31:13.000 We have got friction fighting it. 00:31:13.000 --> 00:31:17.000 The whole time it is moving that 50 m, it is being fought by friction. 00:31:17.000 --> 00:31:24.000 So, it moves along, it slides along, but as it is sliding, it is also continually losing energy to friction. 00:31:24.000 --> 00:31:34.000 The block is giving energy to its environment, it is gaining energy from the force that the environment is putting into it, but it is also winding up giving heat energy through friction. 00:31:34.000 --> 00:31:37.000 So, it is going to have that much work coming out if it. 00:31:37.000 --> 00:31:44.000 It moves along that 50 m, and it is going to continue moving, because the force is able to beat out the force of friction. 00:31:44.000 --> 00:31:48.000 But, once the force stops acting on it, it is going to slide. 00:31:48.000 --> 00:31:49.000 How far will it slide? 00:31:49.000 --> 00:31:54.000 It is going to take all of the friction to sap all of the kinetic energy out of it. 00:31:54.000 --> 00:32:01.000 Once it has lost all of its kinetic energy, it will be still, if we can figure out how far that distance is, we will know what the slide is. 00:32:01.000 --> 00:32:04.000 What is the speed at 50 m? 00:32:04.000 --> 00:32:11.000 We need to first figure out how much work has friction done over those 50 m. 00:32:11.000 --> 00:32:43.000 That is going to be, force × distance × cos θ = μk × FN × 50 m, (cos θ = 1, since perfectly parallel.) 00:32:43.000 --> 00:32:50.000 One last thing to figure out: which direction is friction working?, it is working negative, it is pulling away from this the whole time. 00:32:50.000 --> 00:32:53.000 Our work value for the friction is going to be negative. 00:32:53.000 --> 00:33:03.000 So, -0.4 × 93.7 × 50 = -1874 J. 00:33:03.000 --> 00:33:12.000 It gains 3830 J from the force acting on it, but it loses 1874 J through the force of friction acting on it. 00:33:12.000 --> 00:33:14.000 What is its starting kinetic energy? 00:33:14.000 --> 00:33:19.000 Its starting kinetic energy is nothing, it sits still at first. 00:33:19.000 --> 00:33:27.000 (Energy at the beginning = 0) + (the work) = (Energy at the end). 00:33:27.000 --> 00:33:33.000 The work and now our ending snap shot is at this 50 m line, we want to figure out what is the velocity there. 00:33:33.000 --> 00:33:57.000 The work that goes in, 3830, the work that comes out is 1874, and that is = the kinetic energy at that moment, because all of our energy, all of our work, is going into kinetic energy, (1/2)mv^2, where we are talking about the speed really. 00:33:57.000 --> 00:34:39.000 We figure out what that is, we get, 1956 = (1/2)mv^2, solve for v (speed, this whole time we know that v is single dimension, so that is not a problem), v = sqrt(2 × 1956 / 3), solving, we get that it is sliding at a very fast 36.1 m/s, at the end of it push. 00:34:39.000 --> 00:34:45.000 Now we know how fast it is sliding at the end of the push, how far will it slide after that? 00:34:45.000 --> 00:34:55.000 Now we need to figure out, the only thing working on it now, is we have got, E(beginning) + Work = E(end). 00:34:55.000 --> 00:35:08.000 So, for the first one, the one on the left, over here, we had a beginning snap shot, the beginning snap shot was still, and the ending snap shot was the very end of the push. 00:35:08.000 --> 00:35:19.000 For this one over here, we are talking about, beginning snap shot is the end of the push, when it already has all that energy stored in it, and its ending snap shot is just as it comes to rest, once it is still again. 00:35:19.000 --> 00:35:50.000 We know that, E(beginning) + Work = E (end), E(beginning) = 1956, we can also figure that out by doing (1/2)m(36.1)^2 because that is how much energy is, but that is going to wind up being 1956, because we already solved for that). 00:35:50.000 --> 00:36:00.000 So, 1956 + work = 0, (when still, it has no energy.) 00:36:00.000 --> 00:36:05.000 That means, our work = -1956. 00:36:05.000 --> 00:36:06.000 What is the work in this case? 00:36:06.000 --> 00:36:09.000 The work is just friction, remember? 00:36:09.000 --> 00:36:14.000 We know this is friction, because the only thing acting on it now is the work of friction. 00:36:14.000 --> 00:36:16.000 What is the distance that is slides? 00:36:16.000 --> 00:36:27.000 We know that friction is negative, so, μk × FN, but now the normal force is actually a different number. 00:36:27.000 --> 00:36:53.000 FN is equal to just gravity at this point, because now we no longer have it pushing on it, so, -μk × mg × d = -1956. 00:36:53.000 --> 00:37:05.000 We know what μk and FN are, we do not know what the distance is, that means we only have one variable to solve for, so we can make these positive on both sides. 00:37:05.000 --> 00:38:02.000 d = 1956 / (mg μk) = 1956 / (3 × 9.8 × 0.4) = 166.3 m, this time it slides a very long distance. 00:38:02.000 --> 00:38:11.000 It gets pushed for 50 m, and that whole time it has got a very large additional frictional force contributed by the force pushing down. 00:38:11.000 --> 00:38:19.000 But once we remove that, it is going to be able to slide longer, because it does not have to deal with the force pushing down on it the whole time now. 00:38:19.000 --> 00:38:30.000 The normal force change between the two different worlds we are looking at, in the second world, we did not have that force pushing down on it now, so we now have a different normal force, and that is a definite trap that you could fall into. 00:38:30.000 --> 00:38:33.000 If you forget to change that, at which point you get a different answer. 00:38:33.000 --> 00:38:39.000 Bu tin this case, we caught that, we know that it is going to be 'mg' is just the normal force at this point. 00:38:39.000 --> 00:38:57.000 μk × FN × d is the work done by friction, so now we need to figure out how much distance it goes, and once it completes that slide, it will manage to take out all of the kinetic energy, and it will come to a rest or come to a zero kinetic energy. 00:38:57.000 --> 00:39:03.000 So, we get 166.3 m, and there is our answer, that is how far it slides after you stop pushing it. 00:39:03.000 --> 00:39:07.000 Hope you enjoyed it!