WEBVTT physics/high-school-physics/selhorst-jones
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Hi, welcome back to educator.com, today we are going to be talking about kinetic energy.
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Just to begin with, you have definitely been exposed to the idea that there is a lots of different types of energy out there.
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You probably have been hearing about this from a very young age.
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You have talked about this, almost as early as kindergarten.
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There is lots of different kinds of motion, energy, and ways for things to be energetic out there.
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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.
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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.
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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.
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There is lots of different ways to have energy out there.
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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.
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Another idea you have probably been exposed to is this one:
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Energy cannot be created or destroyed, only changed from one type to another.
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The conservation of energy, you have almost certainly heard this.
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It is not possible to get rid of energy in a system, it is only possible to change it from one form to another.
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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.
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The friction of the brake pads, the friction of the tyres on the road.
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There is various different kinds of heat, the heat energy is what is going to come out of it instead.
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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.
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All these sorts of things.
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We can think of energy through this metaphor of a bucket full of sand.
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We have got a bunch of different buckets, and each one has an amount of sand.
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Say we are looking at a rocket on a launch pad.
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It has some amount of chemical energy, it has got lot of chemical energy.
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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.
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So, it is slightly higher up.
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What happens when we ignite those rockets?
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When we ignite the rockets, we use that chemical energy up, so we wind up burning all that chemical energy.
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And that energy instead cause the rocket to move upwards, we causes an increase in kinetic energy, it is now moving up.
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Now we have got an increase in kinetic energy, and the gravitational energy will go up as it moves up.
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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.
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If we look at the total amount of sand in all the buckets, all use the same.
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It is just we change the location from one bucket to the other.
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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.
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Or may be what will happen is, the rocket would run into a satellite, now we are having two system interact.
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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.
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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.
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But it does not get lost, does not get destroyed, it is always there, it is moved around in different ways.
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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.
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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.
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It gives us a lot of power over the understanding of how the world around us works.
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What is energy?
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We have talked about it having these different forms, we have not really talked about what it is.
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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.
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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.
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Being able to put work in, being able to make a change in the world.
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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.
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Work is change in energy.
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So, what is work?
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If work is what energy can do, work is a lot easier to define in fact.
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Work is just the transfer of energy from one type to another.
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Doing work is just the act of transferring energy, whereas the quantity if work is the amount of energy transferred.
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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.
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Work is just moving energy from bucket to bucket.
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That makes sense.
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Wait!, that was a circular definition.
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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.
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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.
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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.
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We do not want circular definitions, because they are not scientific statements.
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But, in my defence, we have actually defined work.
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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.
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You move energy from one type to another, form one system to another, and that is work.
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But what we did not really define is, we did not really define energy.
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We did not do that rigorously.
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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.
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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.
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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.
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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.
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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.
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For now, we are going to be still be able to do plenty of stuff with what we have got,
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We have got plenty to work with.
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Being able to define it rigorously, it is something to do down the road, when you study a whole bunch more Physics.
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With all that behind us, let us talk about how to derive a formula, useful mathematical formulae, we can talk about things quantitatively.
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How can we get the amount of energy in something's motion.
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Kinetic energy.
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From the theoretical point of view, we have seen that putting work into a system is the same as putting energy in.
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But we have not quantified it yet.
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Let us fix it.
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Consider a simplified picture of work, where the force F is parallel to the displacement.
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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)
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We have got a block of mass 'm', and it is on some magical frictionless surface and it starts at rest.
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This is how things are to begin, let us talk about from here.
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Consider the following three formulae from all of our previous work.
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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.
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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.
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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.
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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.
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If we want to get acceleration, we get that, a = F/m .
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Also, vi = 0, we said that this thing started at rest, so we can knock that out.
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For ease, vf = v.
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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.
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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.
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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.
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The amount of kinetic energy in the system, is (1/2)mv^2.
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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.
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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.
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Ekinetic = (1/2)mv^2 .
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Now we have got a formula, now we can really get our hands dirty.
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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.
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So, at the moment, we are just hoping that work and energy are going to wind up having the same units.
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Let us see if that is the case.
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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.
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Force is newtons, and we got metre, we need to figure out what newtons are!
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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!
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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.
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Work is joules, energy is joules, joules is energy.
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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.
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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.
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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.
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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.
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All the types of energy at once.
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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.
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Energy that is transferred out is negative work.
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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.
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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.
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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.
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One last thing to talk about: How does friction work!
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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.
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Friction reduces the energy in the system.
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By how much?
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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.
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If it goes this distance, then the work is just going to be equal to, (-) × (magnitude of friction) × (distance), that is it.
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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.
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Friction always points backwards, and always gives a negative value, and always saps the energy out of the system we are looking at.
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Ready for our examples.
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First off, real easy one: Block of mass 10 kg is at rest on a frictionless surface.
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It has a horizontal force of 20 N, applied to it for a distance of 25 m.
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What is its velocity afterwards?
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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.
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How much work is put into the system?
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The work = force × distance × cos θ = fd (since parallel) = 20 N × 25 m = 500 J, going into the system.
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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).
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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.
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From here, we just solve for what 'v' is, we know what 'm' is.
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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 .
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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.
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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.
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Example 2: Two identical cars of mass 1500 kg are driving directly towards one another at 15 m/s, and -15 m/s.
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They crash into one another in a horrible screaming trash of metal, they come to a complete rest after impact.
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How much energy is released during the crash?
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At the end, we have got 0.
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So, whatever energy beginning is, that is going to be our answer.
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How much energy is in the system in the beginning?
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How much is it in each car?
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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.
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(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.
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That is a pretty fair amount.
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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.
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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.
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So that is how much energy gets released during the crash.
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That is a fair bit of energy.
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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.
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If that is the case, then we are going to have to change what the velocity is.
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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.
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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.
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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.
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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.
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Those are reasonable freeway speeds, and those are like normal city speeds.
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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.
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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.
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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.
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Then acted upon by a force of 35 N at an angle of 20 degrees above the horizontal for a distance of 15 m.
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What is its speed afterwards?
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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.
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The work = fdcos θ = 35 × 15 × cos(20) = 493.3 J.
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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.
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What is the energy in the beginning?
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Remember, it did not start at rest this time, this time, it had an initial velocity of 4 m/s.
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So, we have to include that.
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(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!