Hello. In this video, we're going to work on how we can obtain the work and energy relationship from the Newton's second law. From the given Newton's second Law, F equals ma, we are going to take an integral on both sides over the displacement. That's how we can get the work and energy relationship. To begin with, let's just define how we can define the work. The work, a small work is going to be defined as a force multiply by the inner product of the displacement dr. So the work done over the state one to two is going to be integral of the dU over one to two, and dU integral Fdr over one to two. Note that the F by Newton's law is going to be ma. So we're going to replace the F as an ma here. Then a by definition is what? The dv over dt. Then here you can combine this one together, dr over dt, which is the definition of the V. So the original Fdr form is going to be changed by the mvdv. So if you take the integral of the V over dv from state one to two, that's where we get the kinetic energy, 1.5mv square from state two to one. So that's how we get work done. It's going to be a kinetic energy change, which is work energy relationship. So note that work done is the kinetic energy change. Now, suppose that the forces, there are many different forces, like F1 is applied to the body and F2 and normal force and the gravity and the spring force. We have a special name for the work done by the gravity and the spring force, which is potential energy. So other than the potential energy, we name it as just work done by external forces as U prime from state one to two. So this equation will give you the familiar form such as you have a kinetic energy at state one and potential energy at state one. Addition of external work done is going to be same as the potential energy at state two. This is another form of work energy relationship. But note that by definition, work done is the kinetic energy change is the original form of the work-energy relationship. This one is sort of a modified form by plugging in work done by the gravity and the spring, switch it as potential energy here. When there's no external work done by the external forces, then dynamical energy, which is a combination of potential and the kinetic energy, is conserved. That's conservation of mechanical energy, what you have learned in high school Physics or the University Physics. So there are steps like the work energy problems are. First, you should define the coordinate, as same as what we have done for the first part of Chapter three. draw the free body diagram. Free the body and all the contacts, because that's how the force exerted. Except a field force like a gravity force, and forces on the left and acceleration on the right. That's a pictorial relationship, and just transfer that as a mathematical form, which is the equations of motion. Now to solve the work and energy, you just take the integral on both sides of the equations of motion, and that's how you can get work-energy relationship. Let's work on the example. There is a block A of mass small M here, and then there is another block of B, which has a mass capital M. Okay? Those two masses are connected by the cable through the pulley. Now the system is released from rest. So initially there is no velocity here. The block B is also attached to the spring, which is on stretch in the beginning. The static and kinetic friction coefficients of the surface and the block A is given as mu-s and mu-s. Now, find the distance S traveled by the block A before it comes to rest. So initially, it's been under rest and then when I release it, it seems like the B is heavier than the other. So maybe the whole structure will move down. Okay? Then it will come to rest at the end ultimately, and then how are the displacements traveled by block A or the block B as well? Okay. So first step is define the coordinate. Since we have two masses, you should define two coordinate, one per each. So this one, I'm going to set the Cartesian coordinate parallel to the slope. Then for the B, it's been constrained for the vertical motion. So I'm going to set the vertical positive downward as Z. Okay? However, so the block A and the block B's, are connected by the same string here, right? So the displacement by the A is going to be same as displacement by the B. So instead of introducing new variable, I'm going to set the same variable as X, because those are same displacement. Okay? Once you obtain the free body diagram and equations of motion, the last step is you have to take the integral on both sides to obtain the work and energy relationship. This is one way. Also, since we have derived the relationship between the state one, T1 and V1, it's going to be same as T2 and V2 with the summation of external work done. This is sort of a formula that you can use, right? However, the second one, just plugging all the initial kinetic energy and potential energy to obtain the final energy is kind of simple one. You just plug in the value to the formula which we have worked on for the high school Physics and now the university Physics. I would strongly recommend you to take the displacement integral to obtain the work and energy relationship here, as any student, and then use this formula, the easy one, to check your answer is correct. Let's work on take the integral procedures. So first we should obtain the free body diagram. So free the body in how many contacts? The surface contact and the string. So I have two forces with the contact and one force with the string and the gravity will generate the acceleration X and Y. For this one, I have a two contact, up and bottom. So I have a tension in the spring force and the gravity will generate the acceleration downward. All those direction is coherent to the direction you set it as a positive when you set the coordinate direction. Now, from those pictorial relationship, I can obtain the equations of motion for the small mass, T minus friction and gravity components will generate the x direction or acceleration in the y as well. I can also obtain the equations of motion for the larger block, which has a minus T minus kx, and the gravity will generate the acceleration. In this problem, whenever the whole system is released, those block has been goes up and this one will compress the spring. So it's been all known. However, if suppose that this mass is a lot heavier than the other, what would happen? Whole system will move to the left hand side. So you never know if the whole structure is stucked, or move, or what. So to check if the system moves to the right hand side or left hand side, what should you do? If the structure moves together, that means this accelerations are the same. So you can put the a_x as a same acceleration here, and then see if a is a positive, then it means a whole structures are left to the right hand side, and minus means everything goes to the left hand side. So this is by combining the equations of motion. You can check the direction of the block A moves left or right. Now, coming back to the original problem, what we are supposed to know is, how long does the block A travels before it stopped? So let's define the state 1 as the initial one, and status 2 as the final one over the travel distance S, and take the integral of the equations of motion. I have three equations of motion. I can take the integral for anyone, but I'd like to take the integral of the first one and the third one because I want to eliminate the unknowns, the tension t through the cable, and if I sum them up, what I could have is integral of the friction, gravity, and the spring force and another gravity for the large mass, which will give the integral for the Ma which is a kinetic energy change. So all the work done by the friction force, and gravity, and the spring force are same as kinetic energy change, which is work energy relationship, and what is that was at the final state, everything has come to rest. So the final V is going to be zero. So what I have is only the work done by the friction, gravity, and spring force all matched out. So by solving this equations about the S, I can figure that out how long this block has been traveled. Let's solve the same problem with the easier method, like just using the formula. The initial T_1 and V_1, and additional work done is going to be same as T_2 and V_2. State one is like this, and the state two, this has been traveled up, this has been traveled down. What's going to be T_1 and V_1? Let's assume all the reference for the potential energy is set to be zero at one, and work done is due to the friction. The friction is the only external force that do the work, and I know the N here from the equations of motion. Now, what's going to be the energy at the second state? Second state, its comes to rest, so there's no kinetic energy, and then the potential energy has been moved up for the block A this much, and block B, it's been down over this much. Also, the spring has been compressed, so there is a spring potential energy. So if you equate this, you can obtain the relationship about the S equation, second-order equation, and you can find out the travel distance S. So it's pretty simple, you just plugged in all to the formula and easily get the answer. Is this answer equivalent to the way that you obtained it by integrating the equations of motion? Yes, those are equivalent. So it's your choice using the simple formula or just integrate the equations of motion, but I would strongly recommend you to try both and compare your answer. Let's solve another problem. There is a particle of mass m, here is released. It's been moved through the guide, the light blue guide. This is a straight region and this is the curved region, and it moves along the curve and the spring is initially natural length is 0.8R. Since the radius is capital R, what it means is, initially, this has been stretched, the spring has been initially stretched. So when it's released from rest from point A, what's going to be the velocity at point B, point C, and point D. That's the problem that you're supposed to solve. Even though this is a simple like a single mass, I'm going to set the coordinate two different way for this linear region as a Cartesian, and this circular region using the polar. So I'm going to set the two different coordinates here, and I've obtained a free-body diagram, equations of motion. I have a two options. Either I can integrate the equations of motion to obtain the work and energy relationship, or I can simply plug in all the values to this formula. Which one is easier? Definitely this one. However, as a ME student, you are supposed to be able to do both and compare your answer. In this problem, I'm going to set the easier step first and we are going to handle about integral approach in the ME question session. Let's go with the easier way. So I'm going to just use the formula here. So I should define what's going to be a status A, a status 1 and status 2. If I'm focusing on the movement from A to B, A is going to be status 1 and B is going to be status 2. If I'm interested in from A to C, C is going to be then the status 2. So initial kinetic energy is zero because it's at rest, and initial potential energy, because if I set this point as zero gravity potential energy line, reference line, there is a potential energy and there is a spring energy exists because it's been initially stretched. Since it's a frictionless guide, there's no work done throughout the whole process. Like A, B, C, to D, there is no external forces, so the energy at the A is going to be same as energy at the B, which is the velocity at B, and the potential energy here is only the spring potential energy because I said the gravitational energy reference at B, or at C, you have another kinetic energy here it will be V_c here, and the potential energy is by the spring potential and the gravitational potential. So by equating energy, total energy is going to be same as A, and B, and C, you will be able to find the velocity at each point. Is this equivalent to the result that you would obtain by integrating the equations of motion? Yes, that's what I'm going to show you later. So this is the easy way to obtain the velocity at each state using the formula. Going back to the problem, you are actually figured that out the velocity changes over each point as well as the normal force. Great, so normal force changes from point A, to point B, and point C, and D? Why? Why does normal force change? Remember, normal force is actually the reaction force usually in response to the gravity, or spring forces, or the motion in the structure, collision occurs so there's the reaction force from the structure, that's a reaction force. So N changes implies that there is a gravity or gravitational impact, or the spring effect, or the motion effect, will vary over the point B, C, and D. So by examining the equations of motion, we will examine how normal forces vary over the position. If you only just use the simple formula, there is no physical interpretation for those force and motion relationship, so that's why I strongly recommend you to work on equations of motion and integral of this to obtain the work and energy instead of just simply using this formula. Especially as a mechanical engineering student, you should be able to understand the physics underlying the problem. That's for this video. We simply go through how we can obtain the work and energy relationship by integrating the equations of motion, and have worked on several examples. In the next video, we are going to work on another example for the work and energy using integral approaches. Thank you for listening.
