1 00:00:03,690 --> 00:00:06,660 We will now like to generalize the work energy theorem 2 00:00:06,660 --> 00:00:08,490 to more than one dimension. 3 00:00:08,490 --> 00:00:11,850 Remember that we defined our work as an integral 4 00:00:11,850 --> 00:00:17,130 from a to b of a force dot ds. 5 00:00:17,130 --> 00:00:20,370 And graphically let's look at that. 6 00:00:20,370 --> 00:00:24,450 If we start at A and we go to a point B, , 7 00:00:24,450 --> 00:00:29,310 and if at some point on this path we have a force F, 8 00:00:29,310 --> 00:00:32,640 and the displacement is always tangent to the force 9 00:00:32,640 --> 00:00:37,740 for an object M, we would like to now figure out how this 10 00:00:37,740 --> 00:00:41,100 integral is related to the change in kinetic energy. 11 00:00:41,100 --> 00:00:44,610 So first off, it's always important 12 00:00:44,610 --> 00:00:47,430 when analyzing a new result to understand where 13 00:00:47,430 --> 00:00:49,230 Newton's second law comes in. 14 00:00:49,230 --> 00:00:51,870 And for our point particle Newton's second law 15 00:00:51,870 --> 00:00:54,750 is that the force is equal to ma, 16 00:00:54,750 --> 00:01:00,930 which we can write as mass times the derivative of the velocity. 17 00:01:00,930 --> 00:01:05,250 And so our first step is to rewrite the work integral 18 00:01:05,250 --> 00:01:10,320 and now to incorporate Newton's second law into this integral. 19 00:01:10,320 --> 00:01:15,690 So this is no longer simply a concept of physics 20 00:01:15,690 --> 00:01:20,110 but it's combining the second law and our definition of work. 21 00:01:20,110 --> 00:01:22,530 Now what I'd like to do here is to note 22 00:01:22,530 --> 00:01:26,460 that the displacement of a particle 23 00:01:26,460 --> 00:01:30,160 is always equal to the velocity times dt. 24 00:01:30,160 --> 00:01:40,500 So if we put that result in here, AB of M dv dt dot v dt, 25 00:01:40,500 --> 00:01:45,539 you can see that we can cancel dt's, which is essentially 26 00:01:45,539 --> 00:01:48,450 taking the time out of this integral 27 00:01:48,450 --> 00:01:51,479 and writing the work entirely in terms 28 00:01:51,479 --> 00:01:56,910 of the small change in velocity, dv, scalar product 29 00:01:56,910 --> 00:01:58,500 with the velocity. 30 00:01:58,500 --> 00:02:02,220 And this is the quantity we would now like to calculate. 31 00:02:02,220 --> 00:02:06,120 So one way, when we ever want to calculate these dot products, 32 00:02:06,120 --> 00:02:08,050 we choose a coordinate system. 33 00:02:08,050 --> 00:02:10,470 So let's choose a Cartesian coordinate system, 34 00:02:10,470 --> 00:02:17,320 plus x plus y, with our unit vectors i-hat and j-hat. 35 00:02:17,320 --> 00:02:21,329 Now our task is to write down these various vectors. 36 00:02:21,329 --> 00:02:25,410 So for instance, dv can be written 37 00:02:25,410 --> 00:02:32,100 as dv x i-hat plus dv y j-hat. 38 00:02:32,100 --> 00:02:34,470 Now if we were to go into three dimensions 39 00:02:34,470 --> 00:02:36,780 we would have dv z k-hat but I'm just 40 00:02:36,780 --> 00:02:39,090 going to do this in two dimensions. 41 00:02:39,090 --> 00:02:46,250 Similarly, the velocity is vx i-hat plus vy j-hat. 42 00:02:46,250 --> 00:02:49,050 And now when we take the dot product, 43 00:02:49,050 --> 00:02:51,490 as we've learned in Cartesian coordinates, 44 00:02:51,490 --> 00:02:55,710 it's just the components dv x vx, 45 00:02:55,710 --> 00:02:57,750 and I'll just do a little parentheses there, 46 00:02:57,750 --> 00:03:02,580 plus dv y vy. 47 00:03:02,580 --> 00:03:10,320 And so now, our work integral, AB, 48 00:03:10,320 --> 00:03:24,450 is equal to the mass times dv x vx plus the mass times dv y vy. 49 00:03:24,450 --> 00:03:31,320 Now what we have here is to a sum in the [? integrant. ?] 50 00:03:31,320 --> 00:03:33,750 But by the rules of integration we 51 00:03:33,750 --> 00:03:38,250 can rewrite this integral as two separate integrals from A 52 00:03:38,250 --> 00:03:50,370 to B of M dv x vx and from A to B of M dv y vy. 53 00:03:50,370 --> 00:03:52,850 And now we can integrate each of these separately. 54 00:03:52,850 --> 00:03:55,530 And if you recall, in one dimension 55 00:03:55,530 --> 00:03:59,820 we used our integral formula, this is just vx squared. 56 00:03:59,820 --> 00:04:03,330 And when we integrate them what we have to look at 57 00:04:03,330 --> 00:04:06,300 is the end points of our integral. 58 00:04:06,300 --> 00:04:08,590 What we're integrating is a velocity, 59 00:04:08,590 --> 00:04:10,290 and we're going from some initial value 60 00:04:10,290 --> 00:04:11,910 to some final value. 61 00:04:11,910 --> 00:04:18,980 So our first integral simply becomes M vx final squared. 62 00:04:18,980 --> 00:04:27,570 I'll pull out the factor 2 minus vx initial squared, where A vx 63 00:04:27,570 --> 00:04:30,990 initial is the initial x component velocity of A 64 00:04:30,990 --> 00:04:33,840 in vx final is the final component of B. 65 00:04:33,840 --> 00:04:36,330 And I have a similar term here, which 66 00:04:36,330 --> 00:04:42,490 is vy final squared minus the vy initial squared. 67 00:04:42,490 --> 00:04:47,100 And when I combine our terms, vx final 68 00:04:47,100 --> 00:04:53,960 squared plus vy final squared minus one-half M 69 00:04:53,960 --> 00:05:00,280 vx initial squared plus vy initial squared. 70 00:05:00,280 --> 00:05:07,016 We've already shown that our kinetic energy, 1/2 M v 71 00:05:07,016 --> 00:05:13,660 dot v is 1/2 M vx squared plus vy squared. 72 00:05:13,660 --> 00:05:19,110 And so we see that this is just the final kinetic energy 73 00:05:19,110 --> 00:05:23,040 minus the initial kinetic energy. 74 00:05:23,040 --> 00:05:27,450 And so we can call that delta k. 75 00:05:27,450 --> 00:05:31,050 And what we have now is the work energy theorem. 76 00:05:31,050 --> 00:05:33,600 And I'll come back to my original statement 77 00:05:33,600 --> 00:05:40,810 that the work is equal to the change in kinetic energy. 78 00:05:40,810 --> 00:05:43,590 And this is a combination of our definition of work 79 00:05:43,590 --> 00:05:47,171 and Newton's second law.