 Welcome to Mechanics 3. In this video, we develop the concepts of work and energy. So far in this series, we've examined in some detail two mechanical systems, a freely falling mass and a mass oscillating on a spring. For both systems, we've found expressions giving the masses position and velocity as functions of time. Consider the freefall experiment. In order to set it up, we have to lift the mass into place. Suppose the mass is 1 liter of water as shown here. We know that it takes effort, let's call it work, to lift the water up some height. According to the concepts we've developed so far, we say this is because gravity pulls down on the water with a force m times g, where m is the mass and g is the gravitational acceleration, about 9.8 meters per second squared. To overcome this, I have to pull up with an equal force, and then exert this force through some distance h. A reasonable mathematical definition of the work I do is the product of the force times the distance. If I'm lifting twice the mass through the distance, I'm doing twice the work. If I'm lifting the mass through twice the distance, I'm also doing twice the work. So, the work required to lift the water is mgh. We define one joule, the unit of work, as one Newton of force acting through one meter of displacement. Its symbol is capital J, and we write one joule equals one Newton meter. A Newton is one kilogram meter per second squared. So in base units, the joule is one kilogram meters squared per second squared. Since velocity is meters per second, we see that the units of work are the same as the units of mass times velocity squared. Let's use the work concept to analyze the free fall experiment. To set up the experiment, we raise a mass m to a height h above the ground. This requires work mgh. Then we let the mass fall. In video one, we saw that after a time t, the distance fallen is x equals one half gt squared. And the velocity is gt. When the mass has fallen a distance x, its height above the ground is h minus x. Let's call the work required to raise the object to this height its quote potential energy, denoted by the letter u. We write u equals mg times h minus x. Substitute one half gt squared for x. Then distribute the mg factor. We get u equals mgh minus one half m quantity gt squared. The first term is the work we did to set up the experiment. In the second term, substitute v for gt. Rearranging, we have w equals one half mv squared plus u. All three terms have units of joules, the unit of work, and of what we'll call energy, as in the potential energy u. This expression leads us to define the kinetic energy of the mass denoted by the letter k as one half mv squared. This is the energy associated with motion. Then our expression reads, the work we put into the system, w, equals the object's kinetic energy, k, plus its potential energy, u. Which sum to its total energy, e. Because w is just a number, not a function of time as k and u are, the total energy e is a constant. This expresses the so-called conservation of energy. As time goes on, the form of energy may change, say from potential energy to kinetic energy. But the total energy remains constant, and is equal to the work we did to set up the initial conditions of the system. Suppose we raise a one kilogram mass to a height of two meters. The required work is just a bit less than 20 joules. And this is the initial potential energy of the mass. If we let the mass fall, the potential energy starts decreasing from this initial value, while the kinetic energy starts increasing from zero. Their sum, the total energy of the mass, remains constant, equal to the initial work done on the system. As time goes on, the potential energy decreases at an ever-increasing rate, while the kinetic energy increases at an ever-increasing rate. Eventually, kinetic energy exceeds potential energy. And when the mass reaches the ground, all the energy of the system takes the form of kinetic energy. Yet the entire time the total energy of the system remains constant. Except that, when the mass hits the ground, this elegant relation falls apart. The kinetic energy of the mass, maybe after a bounce or two, seems to vanish. And we end up back where we started, with a mass at rest on the ground, with zero potential energy, and zero kinetic energy. What happened to the energy that was put into the system through our work, and conserved during the free fall? We'll come back to this question in a future video. Now let's look at the other dynamical system we've studied, a mass on a spring. Let's consider the work required to initialize this system by stretching or compressing the spring. Suppose the spring is stretched, and we are applying a force to offset the spring force. Then we stretch the spring an additional distance delta x. The work done, call it delta w, is approximately the force f times the distance delta x. Approximately, because the force changes as the spring stretches. If delta x is very small, this change is very small, and the approximation is very good. The force is, ideally, given by Hooke's law, f equals a spring constant k times x. The small amount of work, f times delta x, is the area of the green rectangle. If we were to stretch the spring from x equals zero to x equals x zero, the total work done would be, approximately, the sum of the areas of all the green rectangles. This is approximately the area of the orange triangle. And the approximation gets better the smaller delta x is. Letting delta x become infinitesimally small, and using the formula for a triangle's area, we find the total work w is one half the base x zero times the height k x zero, or one half k x zero squared. This idea of approximating the area of a curve by the sum of the areas of small rectangles is the basis for integral calculus. We write that w equals the integral from zero to x zero of kx dx equals one half k x zero squared. Now let's analyze the mass on a spring system. We take x equals zero to be the equilibrium position of the mass, where it can remain motionless. We assume its initial displacement is x zero. The spring's force on the mass is minus kx. As time goes on, the mass moves. As we saw in the previous video, its position is x equals x zero cosine omega t. And its velocity is minus x zero omega sine omega t, where omega squared equals the spring constant k over the mass m. The work we put into the system to move the mass to its initial position is w equals one half k x zero squared. Let's define the potential energy, u, of the mass at an arbitrary position x, to be the work required to move the mass to that position from the equilibrium position. u equals one half k x squared. And we assume our previous expression for kinetic energy, k equals one half mv squared. Let's see if the total energy, e, equals k plus u, is conserved. Plugging in the expression for v and x, this becomes one half m x zero squared omega squared sine squared omega t, plus one half k x zero squared cosine squared omega t. Plugging in omega squared equals k over m, and factoring we end up with e equals one half k x zero squared times the quantity sine squared plus cosine squared of omega t. The sine squared plus cosine squared of any number is one. So indeed, the total energy is constant and equal to one half k x zero squared, which equals the work we did to put the system in its initial state. So, as the mass oscillates, there are times when potential energy decreases and kinetic energy increases, and times when kinetic energy decreases and potential energy increases. But at all times there's some, the total energy, remains constant. In its initial position at the top of the oscillation, all energy is potential. As the mass travels downward, the spring pushes downward, so the mass accelerates downward, and the kinetic energy increases. It reaches a maximum at x equals zero, where the spring force disappears and the potential energy is zero. As the mass falls further, the spring pulls upward. This decelerates the mass, decreasing kinetic energy. At the same time, the stretching spring increases potential energy. At the bottom of the oscillation, the mass comes to rest. There is no kinetic energy, and once again all energy is potential. Then the upward half of the oscillation mirrors the downward half.