 In physics, work is a force applied over a distance. So intuitively, work measures the amount of effort it takes to move something a distance, right? Work is a scalar quantity. It's not a vector quantity like velocity or force. The direction doesn't matter. It doesn't matter whether you're pushing your car to the left or to the right or up or down. Moving a heavy object like a car takes the same amount of work independent of the direction, right? If you were in a parking lot and you pushed it around Southern Utah University's campus back to that same spot again, in terms of displacement, the net displacement would be zero. It's like no distance was measured compared to stopping and starting there. But definitely work was exerted in that process. Drying in a circle takes work. It's without a direction there. So work is the amount of effort it takes to move something a distance. And we can actually compute work very quickly using vectors. So if a constant force F is applied to an object and moves the object in a straight line, some distance D where this is a direction, right? It's a length and a direction, then the work is gonna equal the dot product between the force vector and the distance vector. So dot products give you the work between force and distance, which we saw from the law of cosines that F dot D is the same thing as take the magnitude of F, the magnitude of D. So the magnitude of F would be how much force is being applied. D here would be the distance has to be traveled and then take cosine of theta where theta is the angle between F and D. So we measure the angle between these forces because they don't have to be in the same direction the distance in the force. So let me give you an example of that. So a shipping clerk pushes a heavy box across the floor. He applies a force of 64 pounds in a downward direction that makes an angle of 35 degrees with the horizontal. If the package has moved 25 feet, how much work is being done by the clerk? Let me draw you a picture to help us illustrate what's going on here. So we have the ground, we have a box. This box needs to move 25 feet. This is our distance vector. We have distance equals 25 I, like so. What about the force vector? Well, the force vector because the man is probably taller than the box, that's typically what happens here. So he's exerting the force in a downward direction, right? But with respect to the horizontal, this angle is gonna be 35 degrees, like so. And he's applying a force of 64, 64 pounds. So you'll notice that if the man's pushing at an angle of 35 with the horizontal, since he's pushing across the floor, which is horizontal, that's the angle between them, right? And so we then see by the formula above, the work is equal to the magnitude of the force, which is 64 times the distance, 25, times cosine of the angle between them, which is 35 degrees, for which we then can put that in a calculator, of course, 25 times 64 is 1600, but you're gonna need a calculator eventually for the cosine of 35 degrees. So putting that in a calculator, you end up with a work of 1,311 foot-pounds. Foot-pounds is a measurement of work. One foot-pound is the effort it takes to move one pound, one foot. So we get about 1,300 foot-pounds to push that box 25 feet. So we can very easily compute the work if it's given in a trigonometric form using the formula F times D times cosine of theta. But we can also do the same thing if it's an algebraic form, and actually I think in the end, algebraic form is gonna prove to be simpler. It always is when it comes to vectors. Suppose that a force is given as 35i minus 12j pounds, and we're pushing an object up a ramp, all right? The resulting distance vector, the displacement vector is gonna be 15i plus 4j. This is gonna be measuring feet. What's the work between them? Now, this one seems a little bit weirder to set up because we're describing both the force and the displacement vector in terms of these unit vectors here. This is the algebraic form of the vectors. It makes sense to be like, oh, I applied 64 pounds in a downward direction of such and such angles. The trigonometric form of vectors is more natural to describe, but when it comes to calculations, it's much easier to use these right here. But we have some interpretation, right? So this means it's 35 pounds to the left or positive 12 pounds downward, right? This one right here means it's 15 feet to the right, four feet up. So you see a picture like this. So don't worry about the conversion. I skipped over the conversion here, but you can convert from the trigonometric form of vectors to algebraic forms. If you wanna find the work, it's just the dot product of F and D right here. So we're gonna get 35i minus 12j dot. We get 15i plus 4j, like so. So multiply together the horizontal, we get 35 times 15. And then multiply together the horizontal, we get negative 12 times four, like so. Multiply these things out. 35 and 15 gives you 525. Four and 12 gives you 48. So we get 525 minus 48, and we end up with 480 foot pounds of work. Like so. And so whether it's easy to interpret the algebraic form of the vectors, sure, the geometric interpretation of these vectors is a little bit more challenging. But in terms of computing the dot product of work, it's so much easier. So this is a very important application of the dot product in physics. Computing work can be done by taking the dot product of the vectors. And you can do that either in the trigonometric form or in the algebraic form. Algebraic form is definitely gonna be easier, but it does require translation between trigonometric and algebraic that we skipped on this exercise right here. And so this brings us to the end of lecture 31, which also brings us to the end of our chapter nine about vectors in our lecture series here. So thanks for watching everyone. If you learned anything about vectors or trigonometry in these videos, please give those videos a like. If you wanna see more videos like this in the future, of course, subscribe to the channel. And as always, if you have any questions, please post them in the comments below and I'll be glad to answer them to help improve your mathematical learning.