1 00:00:00,480 --> 00:00:02,671 - [Voiceover] This is Walter the penguin. 2 00:00:02,671 --> 00:00:05,113 And Walter the penguin gets bored in Antarctica, 3 00:00:05,113 --> 00:00:08,222 so he likes to run, jump, and then slide 4 00:00:08,222 --> 00:00:10,207 across the ice to a stop. 5 00:00:10,207 --> 00:00:13,503 But Walter is a clever and curious penguin, 6 00:00:13,503 --> 00:00:15,392 so while he's sliding, he's thinking 7 00:00:15,392 --> 00:00:18,159 about energy conservation and he's confused, 8 00:00:18,159 --> 00:00:20,744 'cause he knows that he starts off over here, 9 00:00:20,744 --> 00:00:22,695 with some amount of kinetic energy. 10 00:00:22,695 --> 00:00:24,319 But he knows that he ends over here 11 00:00:24,319 --> 00:00:27,455 with no kinetic energy since he slides to a stop. 12 00:00:27,455 --> 00:00:30,111 So he wonders, how can energy be conserved, 13 00:00:30,111 --> 00:00:32,895 when he seems to be losing kinetic energy? 14 00:00:32,895 --> 00:00:34,184 Now if you would have asked this question 15 00:00:34,184 --> 00:00:36,416 when we dealt with forces, you would have said, 16 00:00:36,416 --> 00:00:39,294 "Oh, well obviously, this penguin is coming to a stop 17 00:00:39,294 --> 00:00:41,751 "'cause there must be some amount of friction 18 00:00:41,751 --> 00:00:43,535 "between the penguin and the ice. 19 00:00:43,535 --> 00:00:45,510 "Maybe the ice is very slippery, 20 00:00:45,510 --> 00:00:47,791 "but it can't be frictionless, or this penguin 21 00:00:47,791 --> 00:00:49,750 "would probably keep sliding forever. 22 00:00:49,750 --> 00:00:51,248 "There might be some air resistance 23 00:00:51,248 --> 00:00:52,695 "causing the penguin to slow down 24 00:00:52,695 --> 00:00:54,304 "but it's probably mostly friction 25 00:00:54,304 --> 00:00:55,927 "between the penguin and the ice." 26 00:00:55,927 --> 00:00:58,103 But we're talking about conservation of energy here, 27 00:00:58,103 --> 00:01:00,985 so how can we put this idea in terms of energy? 28 00:01:00,985 --> 00:01:02,471 Well, the way we do it, is we just say 29 00:01:02,471 --> 00:01:04,423 that this force of friction is doing 30 00:01:04,423 --> 00:01:06,848 negative work on the penguin. 31 00:01:06,848 --> 00:01:08,080 And we know the work is negative 32 00:01:08,080 --> 00:01:10,175 because the force of friction is directed 33 00:01:10,175 --> 00:01:13,417 in the opposite direction to the penguin's motion. 34 00:01:13,417 --> 00:01:15,012 The other way we could see this is that, 35 00:01:15,012 --> 00:01:18,432 we could just use the formula for work done by any force. 36 00:01:18,432 --> 00:01:20,872 That formula's f d cosine theta. 37 00:01:20,872 --> 00:01:23,120 If we want to find the work done by the force of friction, 38 00:01:23,120 --> 00:01:25,927 we would plug in the force of friction for our force, 39 00:01:25,927 --> 00:01:28,561 the magnitude of it, times the distance 40 00:01:28,561 --> 00:01:31,439 the penguin slid to the right, and then this theta 41 00:01:31,439 --> 00:01:33,720 in cosine theta is always the angle 42 00:01:33,720 --> 00:01:36,767 between the force and the direction of motion. 43 00:01:36,767 --> 00:01:38,801 So this penguin's sliding to the right, 44 00:01:38,801 --> 00:01:40,280 the forces directed to the left, 45 00:01:40,280 --> 00:01:42,392 you might think that's zero, but that's not zero. 46 00:01:42,392 --> 00:01:45,765 Think about it, the angle between leftward 47 00:01:45,765 --> 00:01:48,308 and rightward is not zero, 48 00:01:48,308 --> 00:01:50,541 that's actually 180 degrees. 49 00:01:50,541 --> 00:01:54,300 So this angle would be 180 right here, or pi radians. 50 00:01:54,300 --> 00:01:57,316 And cosine of 180 is going to give you a negative one, 51 00:01:57,316 --> 00:02:00,702 so the work done by the force of friction on this penguin 52 00:02:00,702 --> 00:02:03,193 is going to be negative f k d. 53 00:02:03,193 --> 00:02:04,892 Negative the force of friction, 54 00:02:04,892 --> 00:02:07,533 times the distance the penguin slid to the right. 55 00:02:07,533 --> 00:02:09,729 But this still doesn't answer Walter's question. 56 00:02:09,729 --> 00:02:12,012 Where did the kinetic energy go? 57 00:02:12,012 --> 00:02:14,701 Friction may have done negative work on this penguin, 58 00:02:14,701 --> 00:02:16,556 but where did that energy end up? 59 00:02:16,556 --> 00:02:17,934 And you probably have a good idea, 60 00:02:17,934 --> 00:02:20,069 'cause when two surfaces rub together, 61 00:02:20,069 --> 00:02:22,036 some of that energy of motion is going to get 62 00:02:22,036 --> 00:02:26,131 transformed into thermal energy in those two surfaces. 63 00:02:26,131 --> 00:02:28,253 In other words, this sheet of ice 64 00:02:28,253 --> 00:02:30,285 is going to have a little more thermal energy, 65 00:02:30,285 --> 00:02:31,932 it's going to heat up just a little bit. 66 00:02:31,932 --> 00:02:34,437 And Walter's feathery coat is going to heat up 67 00:02:34,437 --> 00:02:35,812 just a little bit, and they're going to have 68 00:02:35,812 --> 00:02:37,372 more thermal energy to end with, 69 00:02:37,372 --> 00:02:38,620 than what they started with. 70 00:02:38,620 --> 00:02:40,836 Just like when you rub your hands together vigorously 71 00:02:40,836 --> 00:02:42,972 on a cold day to get warm, you're turning 72 00:02:42,972 --> 00:02:45,491 some of that kinetic energy into thermal energy 73 00:02:45,491 --> 00:02:46,875 that warms up your hands. 74 00:02:46,875 --> 00:02:49,043 And you might be like, "Alright, that's all well and good, 75 00:02:49,043 --> 00:02:50,730 "but how do we put this all together? 76 00:02:50,730 --> 00:02:52,873 "I mean I've got this idea of work over here, 77 00:02:52,873 --> 00:02:55,259 "I've got this idea of kinetic energy over here. 78 00:02:55,259 --> 00:02:57,897 "What unifying framework lets me think about this 79 00:02:57,897 --> 00:02:59,362 "all in one package?" 80 00:02:59,362 --> 00:03:01,443 And that would be the statement of conservation of energy, 81 00:03:01,443 --> 00:03:04,724 so we can say that the initial energy of any system 82 00:03:04,724 --> 00:03:08,416 plus any external work that's done on that system 83 00:03:08,416 --> 00:03:11,730 has to equal the final energy of that system. 84 00:03:11,730 --> 00:03:13,963 In other words we could say that Walter the penguin 85 00:03:13,963 --> 00:03:16,132 started with kinetic energy, so Walter starts 86 00:03:16,132 --> 00:03:19,185 with 1/2 m v squared, and we don't have to worry 87 00:03:19,185 --> 00:03:21,154 about gravitational potential energy 88 00:03:21,154 --> 00:03:22,721 'cause Walter's not changing his height, 89 00:03:22,721 --> 00:03:24,946 he's just sliding straight along the ice 90 00:03:24,946 --> 00:03:26,279 at a horizontal level. 91 00:03:26,279 --> 00:03:28,352 And we can add the external work that was done, 92 00:03:28,352 --> 00:03:29,497 which we've just figured that out. 93 00:03:29,497 --> 00:03:31,518 We know the external work would be the work 94 00:03:31,518 --> 00:03:34,230 done by friction, so we'd have a minus, 95 00:03:34,230 --> 00:03:36,861 'cause it was negative work, f k d, 96 00:03:36,861 --> 00:03:38,951 and it's negative again because this force 97 00:03:38,951 --> 00:03:41,182 is taking energy out of the system. 98 00:03:41,182 --> 00:03:43,118 We could set that equal to the final energy, 99 00:03:43,118 --> 00:03:45,381 but Walter ends with no kinetic energy, so we'd say 100 00:03:45,381 --> 00:03:48,190 that there's no final kinetic energy to end with, 101 00:03:48,190 --> 00:03:50,366 there's no energy in our system to end with, 102 00:03:50,366 --> 00:03:52,054 in which case we find out, 103 00:03:52,054 --> 00:03:54,278 something that might be obvious to you is that 104 00:03:54,278 --> 00:03:56,951 1/2 m v squared, the initial kinetic energy 105 00:03:56,951 --> 00:03:59,726 that Walter started with, has to equal, 106 00:03:59,726 --> 00:04:01,943 if we add to both sides f k d, 107 00:04:01,943 --> 00:04:04,762 the magnitude of the work done by friction. 108 00:04:04,762 --> 00:04:05,917 But some people might object, they might say, 109 00:04:05,917 --> 00:04:08,692 "Wait a minute, we just said there was thermal energy 110 00:04:08,692 --> 00:04:11,300 "to end with, how come we didn't include that 111 00:04:11,300 --> 00:04:12,948 "in our final energy?" 112 00:04:12,948 --> 00:04:16,362 And the reason is, that in this calculation here, 113 00:04:16,362 --> 00:04:19,812 we assumed that Walter, and only Walter, 114 00:04:19,812 --> 00:04:21,517 was part of our energy system. 115 00:04:21,517 --> 00:04:24,920 In other words, Walter, and only his motional energy, 116 00:04:24,920 --> 00:04:26,870 his kinetic energy, was the only energy 117 00:04:26,870 --> 00:04:29,350 we were keeping track of, that's why we said that, 118 00:04:29,350 --> 00:04:32,029 initially, there was just Walter's kinetic energy. 119 00:04:32,029 --> 00:04:35,429 And this sheet of ice was external to our system, 120 00:04:35,429 --> 00:04:37,918 not part of our system, that's why it exerted 121 00:04:37,918 --> 00:04:41,304 a negative external work, removed the energy 122 00:04:41,304 --> 00:04:44,775 from the system, and Walter ended up with no kinetic energy. 123 00:04:44,775 --> 00:04:47,078 But there's an alternate way to go about this calculation. 124 00:04:47,078 --> 00:04:49,712 You could say, "Alright, instead of just considering 125 00:04:49,712 --> 00:04:52,581 "Walter and Walter alone to be part of our system, 126 00:04:52,581 --> 00:04:55,735 "let's go ahead and include the ice as part of our system." 127 00:04:55,735 --> 00:04:58,270 And all those places where the thermal energy can go, 128 00:04:58,270 --> 00:05:01,542 like Walter's feathery coat, if we include all the places 129 00:05:01,542 --> 00:05:04,590 energy can go, then there won't be any external work. 130 00:05:04,590 --> 00:05:07,085 So an alternate way to solve these problems, 131 00:05:07,085 --> 00:05:09,590 is to use this same formula, but now, 132 00:05:09,590 --> 00:05:12,302 Walter and the ice are both part of our system. 133 00:05:12,302 --> 00:05:14,597 Our system would still start with the kinetic energy 134 00:05:14,597 --> 00:05:16,918 that Walter had at the beginning, that doesn't change. 135 00:05:16,918 --> 00:05:19,277 But now there would be no external work, 136 00:05:19,277 --> 00:05:21,263 not because force of friction isn't acting, 137 00:05:21,263 --> 00:05:22,837 there's still a force of friction, 138 00:05:22,837 --> 00:05:26,533 but that's an internal force between objects in our system. 139 00:05:26,533 --> 00:05:29,109 'Cause Walter and the ice are part of our system. 140 00:05:29,109 --> 00:05:31,349 So there's no external work done now. 141 00:05:31,349 --> 00:05:33,446 That might be a new or confusing idea to some people, 142 00:05:33,446 --> 00:05:36,725 so let me just say, if there's forces between objects 143 00:05:36,725 --> 00:05:40,292 within your system, then those forces cannot exert 144 00:05:40,292 --> 00:05:42,249 external work and they cannot change 145 00:05:42,249 --> 00:05:44,373 the total energy of your system. 146 00:05:44,373 --> 00:05:47,511 Only forces exerted on objects within your system 147 00:05:47,511 --> 00:05:50,151 from outside of your system, can change 148 00:05:50,151 --> 00:05:52,317 the total energy of your system. 149 00:05:52,317 --> 00:05:54,813 So when this ice was not part of our system, 150 00:05:54,813 --> 00:05:58,540 it was exerting an outside external force on Walter, 151 00:05:58,540 --> 00:06:00,365 and the energy of our system changed. 152 00:06:00,365 --> 00:06:03,166 We started with kinetic energy, we ended with no energy. 153 00:06:03,166 --> 00:06:05,828 But now that the ice and Walter are part of our system, 154 00:06:05,828 --> 00:06:08,118 this force of friction is no longer external. 155 00:06:08,118 --> 00:06:11,740 It's internal, exerted between objects within our system, 156 00:06:11,740 --> 00:06:14,324 and so it does not exert any external work. 157 00:06:14,324 --> 00:06:16,386 It's just going to transform energies 158 00:06:16,386 --> 00:06:19,204 between different objects within our system. 159 00:06:19,204 --> 00:06:20,881 So that's why we write this zero here, 160 00:06:20,881 --> 00:06:23,588 there'd be no external work done if we choose the ice 161 00:06:23,588 --> 00:06:25,547 and Walter as part of our system. 162 00:06:25,547 --> 00:06:27,626 And this would have to equal the final energy, 163 00:06:27,626 --> 00:06:29,163 and we know where this energy ends up. 164 00:06:29,163 --> 00:06:31,518 It started with kinetic energy, and it ends 165 00:06:31,518 --> 00:06:34,221 as this extra thermal energy in the snow, 166 00:06:34,221 --> 00:06:35,902 and Walter's feathery coat. 167 00:06:35,902 --> 00:06:37,717 So I could write that as e thermal. 168 00:06:37,717 --> 00:06:39,903 But I know how much thermal energy was generated, 169 00:06:39,903 --> 00:06:43,087 this just has to equal the amount of work done by friction. 170 00:06:43,087 --> 00:06:45,806 So even though this work done is not external, 171 00:06:45,806 --> 00:06:49,029 it still transfers energy between objects 172 00:06:49,029 --> 00:06:51,478 within our system, so when we write that the work 173 00:06:51,478 --> 00:06:54,383 was negative f k d down here, we mean that 174 00:06:54,383 --> 00:06:58,110 the force of friction took f k d from something 175 00:06:58,110 --> 00:06:59,958 and turned it into something else, 176 00:06:59,958 --> 00:07:01,182 and that's all we need up here. 177 00:07:01,182 --> 00:07:03,286 We need an expression for thermal energy. 178 00:07:03,286 --> 00:07:06,341 But if friction took f k d and turned it 179 00:07:06,341 --> 00:07:08,775 into something else, the thing it turned it into 180 00:07:08,775 --> 00:07:11,831 was the thermal energy so that value of f k d, 181 00:07:11,831 --> 00:07:15,007 that magnitude of the work done, was how much energy 182 00:07:15,007 --> 00:07:17,084 ended up as thermal energy. 183 00:07:17,084 --> 00:07:18,877 People might find that confusing, they might be like, 184 00:07:18,877 --> 00:07:20,660 "Wait a minute, why do we have this 185 00:07:20,660 --> 00:07:22,685 "with a positive here and not a negative?" 186 00:07:22,685 --> 00:07:25,276 Well in this work formula, this negative is just saying 187 00:07:25,276 --> 00:07:28,198 that the force of friction is taking energy from something. 188 00:07:28,198 --> 00:07:29,635 If you take energy from something, 189 00:07:29,635 --> 00:07:31,156 you're doing negative work on it. 190 00:07:31,156 --> 00:07:33,965 If I gave energy to something, I'd be doing positive work. 191 00:07:33,965 --> 00:07:36,004 So this negative sign in the work done, 192 00:07:36,004 --> 00:07:39,116 just means that the force of friction took this much energy 193 00:07:39,116 --> 00:07:42,740 from something, and turned it into thermal energy. 194 00:07:42,740 --> 00:07:45,331 So when we want to write down how much thermal energy 195 00:07:45,331 --> 00:07:47,484 did we end with, well, we ended with 196 00:07:47,484 --> 00:07:48,939 the amount that we took. 197 00:07:48,939 --> 00:07:52,796 So we took f k d, the thermal energy ended with f k d. 198 00:07:52,796 --> 00:07:55,516 And I can still set this equal to the kinetic energy 199 00:07:55,516 --> 00:07:58,748 that Walter started with, and I get the same formula 200 00:07:58,748 --> 00:08:01,156 I ended up with over here, because I had to. 201 00:08:01,156 --> 00:08:02,700 Because we're describing the same universe 202 00:08:02,700 --> 00:08:06,605 and the same situation, so no matter what story you tell, 203 00:08:06,605 --> 00:08:08,891 you should get the same physics in the end. 204 00:08:08,891 --> 00:08:11,742 And we do, but some people prefer one to the other. 205 00:08:11,742 --> 00:08:13,667 Some people like thinking of friction 206 00:08:13,667 --> 00:08:16,446 as a negative external work, and not including 207 00:08:16,446 --> 00:08:19,708 the energy within the surfaces as part of their system. 208 00:08:19,708 --> 00:08:21,550 And some people like including those surfaces 209 00:08:21,550 --> 00:08:24,174 as part of their energy system, and just including 210 00:08:24,174 --> 00:08:26,886 that thermal energy on the e final side. 211 00:08:26,886 --> 00:08:30,125 Which is fine, you can do either, you just can't do both. 212 00:08:30,125 --> 00:08:32,332 Either the surface is part of your system, 213 00:08:32,332 --> 00:08:34,164 and you include it in your final energy, 214 00:08:34,164 --> 00:08:36,057 or the surface is not part of your system, 215 00:08:36,058 --> 00:08:37,827 and you include it as external work. 216 00:08:37,827 --> 00:08:39,908 But you can't say it does external work 217 00:08:39,908 --> 00:08:42,380 and it gains some final energy over here 218 00:08:42,380 --> 00:08:44,821 because it's got to be either part of your system, 219 00:08:44,821 --> 00:08:46,444 or not part of your system. 220 00:08:46,444 --> 00:08:48,685 So long story short, you can basically just think 221 00:08:48,685 --> 00:08:51,380 about the thermal energy generated by friction, 222 00:08:51,380 --> 00:08:55,470 as f k d, this is a formula that'll let you solve 223 00:08:55,470 --> 00:08:57,764 for the amount of thermal energy generated 224 00:08:57,764 --> 00:08:59,853 when two surfaces rub against each other. 225 00:08:59,853 --> 00:09:01,788 And we can take this idea a little further. 226 00:09:01,788 --> 00:09:04,398 The force of kinetic friction is going to be equal to 227 00:09:04,398 --> 00:09:06,373 the coefficient of kinetic friction 228 00:09:06,373 --> 00:09:09,396 times the normal force between the two surfaces. 229 00:09:09,396 --> 00:09:10,558 So we could rewrite this. 230 00:09:10,558 --> 00:09:13,782 This thermal energy term can be rewritten as 231 00:09:13,782 --> 00:09:16,486 mu k times f n times d. 232 00:09:16,486 --> 00:09:19,109 And you might say, "Well that's not all that remarkable, 233 00:09:19,109 --> 00:09:21,469 "it just looks even worse than it did before." 234 00:09:21,469 --> 00:09:24,607 But if you keep going, you realize that the normal force 235 00:09:24,607 --> 00:09:26,823 for something just sitting on a surface, 236 00:09:26,823 --> 00:09:28,799 is just going to be m times g. 237 00:09:28,799 --> 00:09:30,767 And we still multiply by the d, 238 00:09:30,767 --> 00:09:32,551 but now that we've replaced the normal force 239 00:09:32,551 --> 00:09:35,773 with m g, we notice that the mass cancels. 240 00:09:35,773 --> 00:09:37,223 And this should blow your mind! 241 00:09:37,223 --> 00:09:40,223 This means, no matter what the mass of this penguin is, 242 00:09:40,223 --> 00:09:43,030 if he starts with the same speed as some other penguin 243 00:09:43,030 --> 00:09:44,519 that's more or or less massive, 244 00:09:44,519 --> 00:09:46,776 he'll slide the exact same distance. 245 00:09:46,776 --> 00:09:48,366 Now some people will object, they'll be like, 246 00:09:48,366 --> 00:09:50,600 "Wait, a really massive penguin's going to have 247 00:09:50,600 --> 00:09:53,469 "a lot of inertia, it really wants to keep moving, 248 00:09:53,469 --> 00:09:55,000 "it should slide farther." 249 00:09:55,000 --> 00:09:56,790 But other people would say, "No, no. 250 00:09:56,790 --> 00:09:59,414 "The less massive penguin should slide farther, 251 00:09:59,414 --> 00:10:01,694 "because it has less frictional force." 252 00:10:01,694 --> 00:10:03,173 But that's why it doesn't matter, 253 00:10:03,173 --> 00:10:06,526 these two confounding effects exactly cancel. 254 00:10:06,526 --> 00:10:08,695 In other words, the more massive penguin 255 00:10:08,695 --> 00:10:11,670 does have more inertia, and has more friction. 256 00:10:11,670 --> 00:10:14,270 And the less massive penguin has lass inertia, 257 00:10:14,270 --> 00:10:16,949 but it has less friction, so the mass ends up canceling, 258 00:10:16,949 --> 00:10:19,502 and all penguins, no matter what their mass are, 259 00:10:19,502 --> 00:10:22,885 slide the same amount if they start with the same speed. 260 00:10:22,885 --> 00:10:24,668 And this also means that two cars, 261 00:10:24,668 --> 00:10:27,837 a really tiny Smart Car and a huge SUV, 262 00:10:27,837 --> 00:10:29,821 if they've got the same tires, 263 00:10:29,821 --> 00:10:31,797 they'll have the same coefficient of friction, 264 00:10:31,797 --> 00:10:33,301 and if they start with the same speed 265 00:10:33,301 --> 00:10:36,468 and slam on their brakes, they'll both skid to a stop 266 00:10:36,468 --> 00:10:38,149 in the same distance. 267 00:10:38,149 --> 00:10:39,436 Again, a lot of people would think 268 00:10:39,436 --> 00:10:42,668 that the really massive SUV has to slide farther, 269 00:10:42,668 --> 00:10:45,380 but that massive SUV that has more inertia 270 00:10:45,380 --> 00:10:48,726 also has more friction, so it stops in the same distance 271 00:10:48,726 --> 00:10:52,158 as the smaller Smart Car, that has less inertia, 272 00:10:52,158 --> 00:10:54,102 and less frictional force. 273 00:10:54,102 --> 00:10:55,733 So even though the coefficients of friction 274 00:10:55,733 --> 00:10:57,716 would be the same for both cars, the force 275 00:10:57,716 --> 00:11:00,398 of friction's going to be bigger for the larger car. 276 00:11:00,398 --> 00:11:02,724 Alright, so to wrap this up, let's do an example problem. 277 00:11:02,724 --> 00:11:03,965 Let's get rid of all this. 278 00:11:03,965 --> 00:11:06,311 Let's say Walter steps it up, he's going 279 00:11:06,311 --> 00:11:08,797 to the extreme games of sliding and he's going 280 00:11:08,797 --> 00:11:11,333 to slide down, starting at the top of this ramp, 281 00:11:11,333 --> 00:11:14,678 that's completely icy, this ramp has no friction. 282 00:11:14,678 --> 00:11:17,087 But Walter has no fear so he goes to the top, 283 00:11:17,087 --> 00:11:19,614 it's four meters tall, he starts at rest. 284 00:11:19,614 --> 00:11:22,535 Walter slides down, speeds up, but now 285 00:11:22,535 --> 00:11:24,671 he hits a patch that does have friction 286 00:11:24,671 --> 00:11:26,477 so Walter slides across this patch, 287 00:11:26,477 --> 00:11:29,548 over some distance, d, and then comes to a stop. 288 00:11:29,548 --> 00:11:31,999 And if we know Walter started four meters high, 289 00:11:31,999 --> 00:11:35,909 and the coefficient of friction along this path is 0.2, 290 00:11:35,909 --> 00:11:39,543 we could figure out what was the distance Walter slid 291 00:11:39,543 --> 00:11:42,135 on this horizontal surface before coming to a stop. 292 00:11:42,135 --> 00:11:42,984 So let's do this, we're going to use 293 00:11:42,984 --> 00:11:45,768 conservation of energy, same formula we did before, 294 00:11:45,768 --> 00:11:48,014 the initial energy of our system 295 00:11:48,014 --> 00:11:51,125 plus any external work done, that is to say, 296 00:11:51,125 --> 00:11:54,934 energy added to or subtracted from our system, 297 00:11:54,934 --> 00:11:58,318 has to equal the energy that we end up with in our system. 298 00:11:58,318 --> 00:12:00,126 Now what do we want to pick as our system? 299 00:12:00,126 --> 00:12:02,598 Personally, I like choosing everything that could 300 00:12:02,598 --> 00:12:04,543 get energy as part of my system. 301 00:12:04,543 --> 00:12:06,807 That way, I never really have to worry about any 302 00:12:06,807 --> 00:12:10,232 external work, because every force will just be internal. 303 00:12:10,232 --> 00:12:14,367 So let's choose Walter, the Earth, the ice, the snow, 304 00:12:14,367 --> 00:12:16,870 the incline, everything's going to be part of our system. 305 00:12:16,870 --> 00:12:18,728 And let's consider our initial point to be 306 00:12:18,728 --> 00:12:21,607 when Walter started at rest at the top of this incline. 307 00:12:21,607 --> 00:12:24,976 And our final point over here, where Walter slid to a stop. 308 00:12:24,976 --> 00:12:27,110 So what kind of energy does our system start with? 309 00:12:27,110 --> 00:12:29,294 Well Walter was up here, he was at rest, 310 00:12:29,294 --> 00:12:31,448 so he had no kinetic energy, but he did have 311 00:12:31,448 --> 00:12:35,448 gravitational potential energy, so that's m g h. 312 00:12:36,607 --> 00:12:38,360 And again, there was no external work done 313 00:12:38,360 --> 00:12:40,077 'cause even though there was friction down here, 314 00:12:40,077 --> 00:12:43,598 I'm going to include this surface and Walter's feathery coat 315 00:12:43,598 --> 00:12:46,134 as part of our system, so there was no external work done 316 00:12:46,134 --> 00:12:48,902 but, there will be thermal energy generated, 317 00:12:48,902 --> 00:12:50,838 and it'll end up in our system. 318 00:12:50,838 --> 00:12:53,110 So the thermal energy generated is going to be 319 00:12:53,110 --> 00:12:56,312 the force of friction, times the distance that Walter slid 320 00:12:56,312 --> 00:12:58,759 across the surface that had friction, 321 00:12:58,759 --> 00:13:01,260 so I'm not going to include this surface right here, 322 00:13:01,260 --> 00:13:02,461 'cause there was no friction there. 323 00:13:02,461 --> 00:13:04,511 I'm only including this distance right here. 324 00:13:04,511 --> 00:13:06,287 And we know that the force of friction 325 00:13:06,287 --> 00:13:08,662 is going to be the coefficient of friction 326 00:13:08,662 --> 00:13:11,256 times the normal force, and then we still multiply 327 00:13:11,256 --> 00:13:13,095 by the distance that Walter slid. 328 00:13:13,095 --> 00:13:16,439 And because Walter was sliding over a horizontal surface, 329 00:13:16,439 --> 00:13:19,302 the normal force on Walter is just going to be equal 330 00:13:19,302 --> 00:13:21,671 to the force of gravity on Walter. 331 00:13:21,671 --> 00:13:24,527 So we can say m g h is going to equal 332 00:13:24,527 --> 00:13:26,382 the coefficient of kinetic friction and then 333 00:13:26,382 --> 00:13:29,102 f n, we're going to replace with m g, 334 00:13:29,102 --> 00:13:31,607 since the normal force is just equal to m g. 335 00:13:31,607 --> 00:13:33,518 And I still multiply by d. 336 00:13:33,518 --> 00:13:35,742 Again, something miraculous happens, 337 00:13:35,742 --> 00:13:38,095 the m's end up canceling, Walter could have been 338 00:13:38,095 --> 00:13:40,254 100 kilograms or 2 kilograms. 339 00:13:40,254 --> 00:13:41,551 Would have slid the same amount. 340 00:13:41,551 --> 00:13:43,279 Finally I can solve this for d. 341 00:13:43,279 --> 00:13:45,248 I'm going to say that d is going to equal 342 00:13:45,248 --> 00:13:47,127 g h, oh, actually. 343 00:13:47,127 --> 00:13:48,983 Turns out the g's cancel too! 344 00:13:48,983 --> 00:13:51,008 Could have done this on the moon, wouldn't have mattered. 345 00:13:51,008 --> 00:13:53,184 Even if the g, the gravitational acceleration, 346 00:13:53,184 --> 00:13:55,933 was different, Walter would still slide the same amount. 347 00:13:55,933 --> 00:13:58,599 So I'm just going to get that d equals h, 348 00:13:58,599 --> 00:14:00,384 and then I divide both sides by this 349 00:14:00,384 --> 00:14:03,245 coefficient of kinetic friction, and there's my result. 350 00:14:03,245 --> 00:14:04,559 It's really simple, in other words, 351 00:14:04,559 --> 00:14:07,087 h was four meters, and then I divide by 352 00:14:07,087 --> 00:14:09,742 the coefficient of friction, which was 0.2, 353 00:14:09,742 --> 00:14:12,440 and I get that Walter's going to slide 20 meters 354 00:14:12,440 --> 00:14:14,056 before coming to a stop. 355 00:14:14,056 --> 00:14:15,904 So recapping, when there's a force of friction 356 00:14:15,904 --> 00:14:18,990 acting on an object, you can use conservation of energy 357 00:14:18,990 --> 00:14:20,990 by treating that frictional surface 358 00:14:20,990 --> 00:14:23,207 as part of your system, in which case, 359 00:14:23,207 --> 00:14:26,655 you would include it as a thermal energy on the final side. 360 00:14:26,655 --> 00:14:29,765 Or, not including that surface as part of your system. 361 00:14:29,765 --> 00:14:32,216 In which case, you would include the same term 362 00:14:32,216 --> 00:14:34,575 with a negative sign as the external work done 363 00:14:34,575 --> 00:14:37,311 on your system, regardless of what you do, 364 00:14:37,311 --> 00:14:40,102 the thermal energy generated in such a situation 365 00:14:40,102 --> 00:14:41,567 is going to be the force of friction 366 00:14:41,567 --> 00:00:00,000 multiplied by the distance through which the object slides.