 In this video, I'm going to do an example of how to solve a problem using conservation of energy. We have a block of mass 1 attached to a spring, then we have the block connected with the cable, our frictionless pulley, to a block of mass m2 which is hanging here. And the question is, if I'm letting go from the initial position, at what position delta x will the block eventually come to a stop? I'm going to be starting with the shared general set up for conservation of energy, which is that energy final is energy initial plus the work done. Now, what are my forms of final energy here? I could have, we're doing mechanics, I could have kinetic energy final. I could have potential energy final, and for the initial energy the same, I could have kinetic energy initial plus potential energy initial plus any work done by any force. Now, in the problem set up, we were assuming that the block was initially addressed and will come to rest again, so we have a kinetic energy final of zero. And we have a kinetic energy initial of zero. What is the potential energy at the end? So let's just assume for a second that I put my zero potential energy level here. This is zero, meaning when I'm below the zero at the end, I'm at minus mgh potential energy. So I'm going to have minus mg delta x potential energy in total. The block on the top does not change in height, so I'm going to not consider its potential energies at all, as they will be just the same on both sides. But however, I can consider with the block at the top is that there will be spring potential energy as the spring will be extended by the distance delta x. I have plus one half k delta x square as my final energy. And that must be equal to my initial kinetic energy, which we also consider to be zero as we're not moving initially. And then my potential energy initially, if we take height zero as being the upper green line, would be zero for the right block. And again, would be zero for my block number one here because we're not changing height. And the initial spring potential energy, if we're starting at the un-extended or unstretched spring, will be zero as well. Now comes the question of who is doing work. If we do a little free body diagram of the left block, we get that the normal force is up, gravity is down, and I will have the friction going to the left, there is any, and I will have the spring force also going to the left, holding me back, while the whole system is traveling to the right. So I can see that between my normal force and the distance of travel, there is a 90 degree angle, so we cannot do any work. The same is true for gravity and the delta S cannot do any work. The spring force here has an angle of 180 and definitely will do work. However, remember the important thing is for a conservative force. If we consider the potential energy attributed to a conservative force, then we do not consider the work done by that conservative force in the same equation, otherwise we account for it twice. So here we already took in the potential spring energy, therefore we will not consider the work done by the spring. So the only work done on M1 is the work done by the friction, as it is in opposite direction. That means we have an angle of 180 degrees, cosine 180 is minus one. So therefore I will have minus the friction times the displacement, just the delta X. What forces are doing work on M2? If I do that free body diagram, I will see the following. So I have M2, I have the force of gravity going down, the force of tension going up, which actually reminds me, I completely forgot that there is a tension here going to the right. The two tensions, if my pulley's friction has always been equal. So in the case of my second mass, my force of gravity does work, but now the same thing as we have for the force of the spring, if we consider the potential energy of that conservative force, then we should not consider the work of that force. So in this case we considered the potential energy of gravity, therefore we do not consider the work done by gravity. Now what about the work done by the tension? Here the work done by the tension will be tension times the delta X, and it will be negative, because the tension is up while we are actually moving downwards to S. And in this case here we have a tension to the right, traveling at delta, so plus tension delta S. So basically those two works cancel each other out, so we don't have to include it in the equation. We could, but it will just cancel each other out. So let me rewrite this a bit more clean. So we have minus mg delta X plus one half k delta X square is minus the force of friction times delta X. Now I want to solve for delta X, and according to math there are two solutions. One of the solutions being the one where delta X is zero, and we know that's the initial position, so that's definitely for physics not the solution. So what I'm going to do here, I'm going to divide by delta X, as I know the delta X equals zero cannot be my final answer, that was my initial answer or my initial condition. So with this I will get minus m, I actually should be specific, this is m2 times g plus one half k delta X equals minus the force of friction. So if I solve this for delta X, delta X will be minus the force of friction plus mass 2 times gravity divided by one half of spring force and that's my final answer. Now let's see if this makes any sense. So the bigger the friction is, the shorter the distance is that the block will slide until it comes to a stop, which kind of makes sense. The bigger my mass 2 or my gravitational constant is, the further the system will go down, this also makes sense, and then if my spring is stiffer, we will create a bigger force, the less I'm going to be sliding to the right two. Now if every problem like this where actually the friction is zero, then this one simply falls out and you have mass 2 times gravity over one half of the spring constant. And if you need numbers for the friction, in this case it will be sliding, so you could always plug in that the friction is, in this case, the normal force times the mu k and the normal force, you get it from this free bi-diagram by doing the sum of all forces in y direction, must be zero.