Welcome to a new problem. This time it involves forces in motion. So think about you know, the Y axis and the ex oxes. So we have an object that eyes being acted upon by a bunch of forces. But there is one specific force which is, um, acting the ex direction. And that force happens to be Alfa X. Why on DH then I had. So it's a specific force that's acting in the ex direction. Even though these object happens to have a bunch of forces are acting really And it depends on the A position off the object exposition off the object, okay, depends on the exposition of the object. The other thing that happens is that our Alfa has given us two point five zero mutants, but meters squared and so want to know what's the work done by the force if it's moving through different scenarios. So the first scenario is a where you know, if you think about the x and y axis like that, we do have the object starting up the origin and then it moves its not at the origin, but it's at the point where X equals to zero and why equals two three? So that's maybe somewhere right there. And then it's moving parlor to the X axis. So it's moving that way, and it stops at a point where X equals to two. So this is kind of like the displacement of the object from the point zero to the point, too along the line, Why equals to remember what done if if the forces variable we do have variable force means is that the work done by this force is going to be the interval off half times the changing position, Infinitesimal change in position. So part, eh, in part, gave you apply that form low going to get had the work done along this y axes being equivalent to the integral ofthe force in the ex direction times the ex This's the forces given us the Alfa X. Why, uh hi hut! And then times the X of course, the certain Constance which we can pull out. That's the awful end, the wide the reason why Why's a constant? You can see why equals two three never changes. It stays the same. So we treat that as a constant. And so we left with X t X I had is just the unit Vector doesn't make any difference in this case. And so our are for happens to be if we do the integral delta y the integral of X is X squared all over too. You do have limits off a integration. You can think of them as going from zero to two. For example. Then we can do that. Khun sake. Zero to two. How far is two point five zero Newton? The meter squared and then our x zero point not zero point. Sorry, I need to change that real quick. So our ex happens to be not our expert. Why happens to be three three point point zero zero meters, as you can see and then, uh, ex is too. So two point zero zero that's going to be squared all of that divided by two. That's the first part of the problem. Second part, we have to subtract zero. And the reason why we subtract there is because you see this lower limit. It's a special more limit if you plug it in is going to give us there. So eventually if you punching those numbers, you get fifteen point zero zero. Jules, The second case of the problem is such that the same again we have. We have ah y axis, which is vertical. Growing up was like Parton and we also have an X axis. And we're assuming that initially the object starts at the point two meters and then it's, um, right here where y zero and access to meters and it moves in the white direction so it's going upwards. That's gonna be a problem because it goes all the way up, up until where? Why equals two. Three. But we have a challenge. Remember, the force we had was inthe e X direction. So anything in the white direction it's not going to give us any type of what But we could still do some computations and validate our thinking. We know that it is the variable force. It's an integral you know, f vex Thie X and this example So we'LL deal with the ex component fast will say our DX. Unfortunately, since the object stays along the two meters are the ex with zero and so regardless of what we do with the integral, was still going to get zero jewels. The reason why our delta X csis here because we stay at two meters going up with in the wind direction we have, you know, f of x times g y Or we could say we have, um, half the y coastline ninety because S o the wide direction is perpendicular to the ex direction. And so we used cause I ninety is a dark product. And that's another problem because the co sign of ninety happens to be zero even though we have danger call. So this is going to be zero. So the total work done in the X and the why we'll just be zero George. So if you're moving along this point from this point up until that point, you know, changing your ex and the force is only directed in the ex direction That way So we have a problem. Get zero Jules In the final part of the problem, we have a different scenario so again will still sketch the X and y axis. It helps us visually understand how the object is moving. So the object starts at the origin. You can see that and then it moves on the line. Why equals to one twenty five acts. So this is our line right here. Why it was to one point five x. It moves up until a specific point where it's X coordinate happens to be two meters and it's why co ordinate happens to be three meters. Okay, and that's the white cord in it. So we still want to find out. You know, what's the walk down? The force that's doing all of this is our for X y. I had, um and we're told that why it was to one point five. So the force becomes Alfa X one point five x high heart makes her life easier. This is our for one point five times x squared, because if you multiply this too So we've changed the former off that force and it still has to obey the law into goal formula for work, which is f the eggs have happens to be how far one point five x squared dx we finding the integral that on the, you know, limits of integration again. You're going from zero to two if you think of it that way. So we're taking out the one point five how far outside the integral sign and then we have into gold off that from zero to one point five. How far, if you could recall from the previous page is two point five zero. What's no ofthis? Two point five zero Newtons Tomato squared. So we go back and plug that in two point five zero mutants per meter squared. And then we do the integral we get executed off a three limits of integration from zero to two. Simplify this problem. We have one point five Alfa, which is two point five zeron eaten Kameda squared times x Next member was two meters right here we have two meters cubed over three reason rights to its. Like that to me the sign simp signal if you want. Um, that's going to subject zero. And the reason for that is if you plug in this zero inside off the X, it eliminates the whole problem. So we only have zero there. We don't have to show him the computations. And so finally, if you plug in those numbers, you jet ten jewels and you could check you. You'LL walk again, you end up getting potentials. So come back to the problem. The problem had three sections. We had a specific force, which was a variable force. And therefore, you know what? Um, that is computed. Using a variable force requires that you use the integral. The limits of integration depends on each and every problem. For example, for this one, it's at zero to two. That's why we have to redo the integral we end up with explored over to the Alfa, wise of Constant. Because why is not changing? And then when you do the integral you get fifteen. Jules, next parts is something we end up with zero total. And the reason for that is along the X there's no motion so d. X with zero. So we don't have any energy along the why the wind direction is perpendicular to the direction of the force course in nineteen zero. And so, um, the last part of the problem we have a different scenario where the object is moving from the origin up until the intersection ofthe X equals two and and why it was too, uh, three. And so we do a replacement. Why? It is one point five x we do they enter? Go again. Because what we're using a variable force. And finally we solve and get ten. Jules, hope you enjoy the problem. Feel free to send me any questions or comments and have a wonderful day occupy.