 Hi, I'm Zor. Welcome to user education. Today I would like to solve a few problems related to kinetic energy. Now this lecture is part of the whole course called Physics for Teens, presented on Unisor.com. I do suggest you to watch this lecture from the Unisor.com. It's a free website which contains the whole course in logical sequence of lectures plus every lecture is supplemented with textual description, basically like a textbook and there are exams for those who would like to challenge themselves. Also the same website contains prerequisite course for this one which is called mass proteins because as you know physics cannot actually survive without mass. So, well, especially calculus and vector algebra, that's what you probably will need. The most from the mass course. So, back to energy, kinetic energy and the few problems which I have. Okay, now let's start with very simple one. So, you have a car of mass m which accelerates during the time t from speed 0 to speed and some kind of a maximum speed and acceleration is given. So, mass is given, time is given and acceleration is given. So, my question is what's the work performed by the engine of this car and what's the kinetic energy at the very end of this acceleration. All right, so let's just do it one thing at a time. First of all, speaking about work. Now, work is equal to force times distance. Well, in case we're talking about straight motion along the straight line with a constant force. Now, force is constant because the mass is assumed constant. We ignore actually some losses of mass for the gas which we are burning, right? So, generally speaking, the mass is the same and acceleration we have already said it's constant. So, the force is constant. So, if the constant force acts on the distance s, you've got this formula for the work, right? That's the definition basically. Now, do we know f? Yes, it's m times a, this is the second law of Newton. Now, speaking about the distance, well, distance if you are traveling from speed 0 to certain maximum speed with a constant acceleration a, then it's equal to a t square over 2. This is simple kinematics and if you forgot about this, just go to kinematics part of this course where everything is explained. So, we know this, right? Okay, so let's calculate what is the work. Work is equal to m a square t square divided by 2, correct? Multiply this by this. Now, speaking about kinetic energy at the very end, well, kinetic energy at the very end is equal to m v max square over 2. Again, that's basically what's derived in the previous lecture which defined what is kinetic energy. Now, what is the v max? What is the maximum speed if you have accelerated with a constant acceleration during the time t? Well, obviously, v max is equal to a times t, which again is basic kinematics. So, m square a square t square divided by 2. So, we have exactly the same expression as expected. Energy is very much related to work. So, the work which we are spending accelerating the car is basically transformed into the kinetic energy at the very end of this process. So, that's the answer for the work and for kinetic energy. Next problem. Okay, now we are, instead of accelerating, we are decelerating the car. So, let's assume that the car moves with a speed v. And then the driver sees some kind of an obstacle in front of him. So, he basically breaks down. So, the speed goes down and he stops completely after distance s. Now, we are assuming again that the mass of the car is constant, although I do not specify it in this particular case. What I do need is, I need the coefficient of friction. Now, why am I talking about coefficient of friction? Well, think about it. If you press the brakes, then the wheels of the car basically stop rotating. And the car moves forward by inertia and what is actually slows the car down? Well, the force of friction of the wheels against the ground. So, friction is very much involved here. That's the only force actually which acts on the car, which basically slows down from the v to zero. And that's why the coefficient of friction is very important. So, what I'm right now asking is, what is this coefficient of friction? If all we know is initial speed and the distance during which we have slowed down to zero. Well, first of all, let's just think about it. Friction is the only force. Now, is it a constant force? Well, we know that the friction is equal to weight, which is m times g, where g is acceleration of the free fall, and m is the mass, times coefficient of kinetic friction. Now, we are assuming that this is the constant, although unknown. Well, g is again acceleration of the free fall, which is constant. On Earth, it's 9.8 meters per second square. Mu is, again, that's a coefficient of friction, that's a constant. So, f is constant. f is constant, m is constant. That means that our acceleration, well, deceleration in this case, negative acceleration, is constant as well. Right? So, negative acceleration is equal to force divided by mass, which is g times mu. So, if we know this acceleration, we can definitely find out what's the coefficient of the kinetic friction. All right? So, all we need to do is find out acceleration if we know the distance and initial speed. Okay, this is simple algebra related to kinetics. So, first of all, we know that s is equal to At square over 2, where t is the time which we spent slowing down from v to 0. On the other hand, since A is a constant, v is equal to At. Right? Well, again, A, in theory, is negative because f is negative. So, obviously, if I'm talking about absolute values, let's just talk about absolute values only, everything is positive. So, these are two equations with two variables from which we can find A and t, the time which we have spent to slowing down and the acceleration, well, deceleration if you wish. All right? So, that's easy. First of all, we can determine time from here, which is v over A. Substitute it into this. We get s is equal to A times t square, which is v square divided by A square and 2. All right? So, this is out. So, v square divided by 2A from which A is equal to v square divided by 2S. Fine. We found A and that's why mu is equal to A divided by g, which is v square divided by 2GS. That's the answer. That's our coefficient of friction. Now, notice very important thing, by the way, that it does not depend on the mass. I told that the mass is constant, but I did not specify the mass. Mass is unknown and doesn't really matter what the mass is. If you have initial speed and you have the distance which you spent slowing down from that initial speed to zero, that's sufficient to find out the coefficient of friction. Obviously, if you have a more massive car, it will take basically a longer distance to slow it down, right? So, that's why. All right. Next. Next is the following. You have, on a thread, this is a vertical, on a thread you have some kind of an object, point object, obviously, of mass m and the thread is of the length l. And it performs circular movements. So, you're holding the thread at the top and it just rotates around the vertical. Now, what is known is the length of the thread, mass, and this angle. No other forces right now are actually acting on this particular object, right? So, what we have to do is we have to find out what is the kinetic energy of this particular object. Well, kinetic energy of this particular object is m v square over 2, where v is its linear speed. So, basically what we have to do is we have to find the linear speed using whatever parameters I have. Now, you probably have noticed, if you did yourself this type of an experiment, the faster it rotates, the more horizontal the line on which this object is hanging becomes. So, it's almost like, whenever you're doing it very, very fast, it will be almost 90 degrees. But if it's a slow, then it will be much closer to the vertical, right? So, the angle is very important. And again, the angle is very much dependent on the speed. The faster the speed, the bigger the angle, closer just to 90 degrees. All right, so how can we approach this particular thing? All right, first of all, we know that the linear speed is actually equal to radius times angular speed, right? We know it from, again, from kinematics of rotational motions. All right, we also know that whenever we have a rotation of the object, there is always the centripetal force and centripetal acceleration. What is centripetal force? Well, that's the force actually which forces this object. Instead of going straight, it goes back to the center to maintain the same position of the center. So, it's a tension of this thread. In this case, it's not the tension of this thread, it's a component, horizontal component. So, there is always a centripetal force and centripetal acceleration, which is equal to r omega square. Again, this is the formula from kinematics of rotational movement. You can go to a corresponding lecture in this course where I explain why it is. Well, or this is the same thing, it's v square, which is r square omega square divided by r, right? Now, what we also know is the acceleration is equal to force of centripetal force divided by mass, because this is just Newton's law. And all we have to do right now, therefore, is to find the centripetal force. From the force, we will find acceleration. From acceleration, we find v square and v square is participating in expression for energy. Okay, well, r, also r, r is simple, r is this one. So, r is equal to l sin phi, that's simple. Now, the f is slightly more difficult. Here is how we will do it. Now, this is the force of the gravity, which is m times g, right? Where g is acceleration of the free fall, m is a mass. Now, this is the tension of the thread. And this tension should actually be represented as the sum of two forces. One of them is vertical acting against the gravity. And another is centripetal, so it's this way. And this is the centripetal force. So, this force, which is the tension, is supposed to be represented as the sum of two forces, two vectors. This one, which is equal in magnitude but opposite in direction to the gravity. And this one, which actually keeps it on the orbit, keeps it on the circular trajectory. This is the force, which is actually the centripetal force here. How can we determine it? Well, very simply. Now, this is parallelogram. So, this is also phi. Now, this is mg. So, this force f is equal to mg times tangent phi. All right, fine. So, we've got that. And that's actually all we need right now. Everything else is just trivial manipulation with the formulas. So, what we do is, from this, we determine v square, which is equal to f times r divided by m, which is equal to f mg tangent phi times r. r is l sine phi and divided by m. So, that's my v square. And since I found v square, I can find kinetic energy, which is equal to m v square, which is g tangent phi l sine phi divided by 2. That's the answer. That's my kinetic energy. And again, obviously, it depends on the phi, on the angle. The higher the greater the angle, both sine and tangent are increasing as the angle increasing. If phi is equal to zero, it's zero. So, basically, it means that the object is hanging vertically. It doesn't really perform any movement because the angle is zero, right? And the greater the angle, the more we have here. Now, what's important in this particular case is to understand that, you see, if my angle goes closer to 90 degrees, you see tangent of 90 degrees, tangent is sine over cosine, right? The sine is equal to one, but the cosine is equal to zero. So, basically, my energy is going to infinity in this particular case because the cosine will be in the denominator because the speed must actually go to infinity, which means that we can never achieve 90 degrees exactly. We can achieve 89 or 89.9 or whatever, but not 90 degrees exactly. And obviously, it all depends on the strength of the thread because the faster you are rotating, the stronger the force on my picture, this force which goes along the thread, it's supposed to be greater and greater. The tension has its own limits, obviously. So, that's also a restriction. The faster you are moving, the stronger should be your thread. All right, so that's it for this particular problem. And we have one more. Three down, one to go. Now, this is about pendulum. So, you have a pendulum, so it goes this and this, not circular movement, but movements back and forth, back and forth within the plane, all right? So, we have a pendulum and we also have the length and the mass. All right. Now, here is a side note. Now, if you go to this particular course to the chapter to a lecture which is about pendulum, and they do suggest actually to review maybe this lecture again, the problem is that the real exact equation for pendulum movement and the movement is actually a function of the angle of time. Well, it cannot be expressed in normal formulas which we know because it goes to some kind of a differential equation which does not really solvable in regular algebraic formulas. However, if these oscillations of the pendulum are small, very close to the vertical, then approximation of this formula is the following. So, square root of G which is acceleration of the free fall divided by length of the thread multiplied by T, that's what goes under cosine. Phi zero is initial angle from which we start the motion. So, first we initially put it at the angle phi zero and let it go. And then under the force of gravity it goes into oscillation and this is the function of time. This function is derived again in the lecture dedicated to the pendulum. And again, I would like to make sure that you understand this is an approximation only for oscillations which are very close to vertical, very small oscillation of the pendulum. So, maybe it's a very long thread and then it goes to a very small distance from the neutral state. Well, that's unfortunate. I mean, we would like actually to express it in a formula which is like always true, not an approximation, but we will just deal with so-called small oscillations which can be expressed in this particular formula, albeit approximately. Okay, now what's necessary to do is to find out the kinetic energy at the lowest point. So, whenever the pendulum goes through this point, that's obviously the point where its linear speed is maximum. This is linear speed zero, then we let it go. It goes accelerates because the force of gravity pushing it this way and that's why there is an acceleration component here. Gravity this way, tension that k and the result of these two forces goes along the trajectory and it speeds up to this point. Now starting from this point, it slows down because the force will be this way against the movement. So, from this point, the beginning, force goes along the movement. After this, tension and gravity are exactly opposite and equal in magnitude. So, there is no additional push, but starting from movement to this direction, to this half of this oscillation, then the force would be against the movement and that's why it will be slowed down until the top position and on this end and then goes back again, accelerating, decelerating. So, we are interested in kinetic energy at the very bottom, where the speed is the maximum considering this is the formula, approximate as it is, but it's a formula. Now, if you go to this lecture about the oscillation of pendulum, you would also find the formula for a period. Well, this is actually simple, because the period of cosine is 2 pi and if you have a period of the function k times x, it's 2 pi divided by x, right? So, that's why 2 pi, this is square root of g over l, this is l over g. So, this is obviously the period, it comes from the properties of the cosine. So, we know the period. Since we know the period, we basically know the time from the beginning until it reaches this maximum position. Now, what is the period of cosine? It goes this way and then this way. So, if we are interested only in this piece, that's basically t divided by 4, which is pi over 2 square root of l divided by g. Now, using this, we can obviously find the angular speed. Now, what is angular speed? If we have a dependency of the angle of time, angular speed is the derivative of this, right? Now, derivative omega of t is equal to first derivative of time, of angle by time, which is... So, derivative of constant goes out, obviously. Of cosine is a minus sign and then I have to multiply by derivative of the inner function. So, it would be minus phi 0 square root of g over l sine of square root of g over l t. So, that's my function of angular speed. Since I have angular speed, I can find out linear speed. Linear speed of t is equal to r times radius. Well, radius is l. So, I can just put straight l times angular speed, right? So, linear speed is always radius times angular speed. So, we've got that, which is equal to minus phi 0. I multiply by l. This is square root of l in the denominator. So, it will be square root of g l times sine of square root of g over l t. And at this moment of time, whenever we are crossing this point, we have to substitute instead of the time t lowercase t. We have to substitute the value t over 4, which is this one. And that's why b maximum would be equal to minus phi 0 square root of g l times sine of square root of g l times phi over 2 times square root of l over g. So, this cancels this. Sine of p over 2 is 1. So, what we have is only this. So, what's my energy? Kinetic energy is equal to m v max square, which is phi 0 square times g times l divided by 2. That's it. That's my energy at the very, very bottom of this oscillation. Well, that's it for today. I do suggest you to maybe read the same lecture on the website. I mean, read textual description of this lecture, where I present the problems. And in some cases, full solution, in some cases, a very brief solution and an answer. But I do recommend you to try to solve all these problems just by yourself and check the answer against whatever is provided in the lecture. Well, that's it. Thanks very much and good luck.