1 00:00:03,670 --> 00:00:06,970 Let's now explore an example of a force in which the work done 2 00:00:06,970 --> 00:00:09,710 it is not path independent. 3 00:00:09,710 --> 00:00:12,080 And the classic example is the friction force. 4 00:00:12,080 --> 00:00:14,650 So let's consider the following setup. 5 00:00:14,650 --> 00:00:18,130 Suppose we have a horizontal surface with friction 6 00:00:18,130 --> 00:00:20,080 and we have an object. 7 00:00:20,080 --> 00:00:21,470 And we're moving this object. 8 00:00:21,470 --> 00:00:23,170 So let's choose an origin. 9 00:00:23,170 --> 00:00:26,470 We'll call this plus x, our i hat direction, 10 00:00:26,470 --> 00:00:28,510 it's all going to be one-dimensional motion. 11 00:00:28,510 --> 00:00:31,540 And we're going to move this object from an initial 12 00:00:31,540 --> 00:00:32,960 to a final state. 13 00:00:32,960 --> 00:00:35,110 And were going to move it directly 14 00:00:35,110 --> 00:00:38,020 in a straight line from the initial to the final state. 15 00:00:38,020 --> 00:00:40,660 And this will be our path 1. 16 00:00:40,660 --> 00:00:42,130 And in our second case, what we'd 17 00:00:42,130 --> 00:00:44,430 like to contrast with that, is that we'd 18 00:00:44,430 --> 00:00:50,200 like to move the object out to a point xa 19 00:00:50,200 --> 00:00:54,520 and then back to the final point. 20 00:00:54,520 --> 00:00:57,340 So this is our path 2. 21 00:00:57,340 --> 00:00:59,650 Has two legs. 22 00:00:59,650 --> 00:01:03,070 And we'd like to compare the work done on these two paths. 23 00:01:03,070 --> 00:01:07,930 So for path 1 we'll begin by calculating-- our force here 24 00:01:07,930 --> 00:01:10,930 is the kinetic friction force. 25 00:01:10,930 --> 00:01:14,900 And the kinetic friction force remember is, in this case, 26 00:01:14,900 --> 00:01:17,140 it's going to oppose the motion. 27 00:01:17,140 --> 00:01:21,039 So we have force kinetic for path 1, 28 00:01:21,039 --> 00:01:28,910 and that is minus mu k mg in the i-hat direction. 29 00:01:28,910 --> 00:01:31,870 And so when we do the integral for the work 30 00:01:31,870 --> 00:01:37,660 from x initial to x final, this is path 1, 31 00:01:37,660 --> 00:01:44,080 then we have minus mu k mg i-hat dotted into-- Now 32 00:01:44,080 --> 00:01:46,330 what is the ds for this path? 33 00:01:46,330 --> 00:01:50,550 It's simply dx i-hat, so dx i-hat. 34 00:01:50,550 --> 00:01:55,810 Notice we're not putting any sign into dx. 35 00:01:55,810 --> 00:01:58,780 The sign will show up in terms of our end 36 00:01:58,780 --> 00:02:00,260 points of our integral. 37 00:02:00,260 --> 00:02:01,660 So we do the dot product here, we 38 00:02:01,660 --> 00:02:05,320 have i-hat dot i-hat, that's 1. 39 00:02:05,320 --> 00:02:08,550 And so this interval, we can pull out all the constants, 40 00:02:08,550 --> 00:02:11,140 mu k mg. 41 00:02:11,140 --> 00:02:15,280 We're just integrating dx from x initial to x final. 42 00:02:15,280 --> 00:02:22,480 And so we get mu k mg times x final minus x initial. 43 00:02:22,480 --> 00:02:25,910 Now for path 2 we have two separate integrals. 44 00:02:25,910 --> 00:02:30,730 So for path 2 we'll just show the first part 45 00:02:30,730 --> 00:02:34,270 where we're going from x initial to xa. 46 00:02:34,270 --> 00:02:39,820 Then the friction force is opposing the motion. 47 00:02:39,820 --> 00:02:43,810 And we always just write dx in terms of the coordinate system, 48 00:02:43,810 --> 00:02:48,910 dx i-hat because you'll see that the signs show up in the end 49 00:02:48,910 --> 00:02:50,140 points of the integral. 50 00:02:50,140 --> 00:02:52,660 And then when we're coming back-- 51 00:02:52,660 --> 00:02:56,890 I'll put that in a different color and I'll put it below it. 52 00:02:56,890 --> 00:03:00,400 So when we come back, notice the friction force 53 00:03:00,400 --> 00:03:03,960 is going to change direction. 54 00:03:03,960 --> 00:03:06,160 ds will still be written that way 55 00:03:06,160 --> 00:03:09,820 but pay close attention to the end points of the integral. 56 00:03:09,820 --> 00:03:11,960 So now what we have is two integrals. 57 00:03:11,960 --> 00:03:16,990 So W is the integral from x initial to xa. 58 00:03:16,990 --> 00:03:20,050 And now I'm going to take the dot products here directly. 59 00:03:20,050 --> 00:03:22,390 It's the same friction force, we still 60 00:03:22,390 --> 00:03:29,170 have this same integral, which is minus mu k mg dx. 61 00:03:29,170 --> 00:03:31,750 Now here's where it's a little bit tricky. 62 00:03:31,750 --> 00:03:40,480 Notice on this path fk is plus mu k mg i-hat. 63 00:03:40,480 --> 00:03:45,730 And so when we dotted into dx we have a plus sign, 64 00:03:45,730 --> 00:03:48,430 we'll just continue that integration here, 65 00:03:48,430 --> 00:03:56,841 of mu k mg dx from xa to x final. 66 00:03:56,841 --> 00:03:59,050 Both of these integrals are straightforward integrals 67 00:03:59,050 --> 00:03:59,550 to do. 68 00:03:59,550 --> 00:04:07,030 This is minus mu k mg xa minus x initial. 69 00:04:07,030 --> 00:04:15,310 And over here, we have a plus mu k mg x final minus x initial. 70 00:04:15,310 --> 00:04:20,120 Notice x final minus xa, rather, is negative. 71 00:04:20,120 --> 00:04:23,650 And so both of these integrals are negative, as we expect. 72 00:04:23,650 --> 00:04:27,280 And so what we see here is that we have two pieces, 73 00:04:27,280 --> 00:04:32,230 so minus 2 mu k mg xa. 74 00:04:32,230 --> 00:04:36,476 And then we have that other piece, mu k mg 75 00:04:36,476 --> 00:04:40,750 x final minus x initial. 76 00:04:40,750 --> 00:04:52,970 Hang on, this is actually a plus sign. 77 00:04:52,970 --> 00:04:55,740 So our answer is very different because the 78 00:04:55,740 --> 00:04:59,860 displaced the amount that we've traveled is different. 79 00:04:59,860 --> 00:05:03,520 So what we see here is an example of a force which 80 00:05:03,520 --> 00:05:07,500 is the work done is not path independent 81 00:05:07,500 --> 00:05:10,330 but depends on the path taken from the initial 82 00:05:10,330 --> 00:05:12,055 to the final states.