1 00:00:00,000 --> 00:00:01,857 - [Instructor] There's a hamburger sitting right in 2 00:00:01,857 --> 00:00:03,883 front of you, but you don't want it. 3 00:00:03,883 --> 00:00:05,668 Maybe it's because you're a vegetarian, 4 00:00:05,668 --> 00:00:07,412 maybe you're already full, maybe you found someone that 5 00:00:07,412 --> 00:00:09,596 needs it more than you do, so you're gonna push this 6 00:00:09,596 --> 00:00:13,372 hamburger to the right, with a force of four newtons 7 00:00:13,372 --> 00:00:15,377 and you're gonna do this and you're gonna push this 8 00:00:15,377 --> 00:00:19,168 a distance of five meters to the right. 9 00:00:19,168 --> 00:00:20,494 So a question we get asked would be 10 00:00:20,494 --> 00:00:23,130 how much work did we do in passing 11 00:00:23,130 --> 00:00:26,260 this hamburger to the person to our right? 12 00:00:26,260 --> 00:00:29,077 Now since this is a pretty simple problem we can just plug 13 00:00:29,077 --> 00:00:32,049 straight into the work formula, but I'm not gonna do that. 14 00:00:32,049 --> 00:00:34,477 I'm gonna show you an alternate way to think about this, 15 00:00:34,477 --> 00:00:35,907 because what we learn from this alternate 16 00:00:35,907 --> 00:00:37,554 approach is gonna help us in more 17 00:00:37,554 --> 00:00:40,475 challenging, complicated work examples. 18 00:00:40,475 --> 00:00:41,519 So I ask you to imagine this. 19 00:00:41,519 --> 00:00:43,043 Instead of just plugging straight in, 20 00:00:43,043 --> 00:00:44,811 consider the fact that we exerted a force of 21 00:00:44,811 --> 00:00:47,720 four newtons for the entire five meters 22 00:00:47,720 --> 00:00:50,165 that we pushed this hamburger to the right, 23 00:00:50,165 --> 00:00:53,101 so if we were to plot what the force was 24 00:00:53,101 --> 00:00:56,271 on our hamburger as a function of its position, 25 00:00:56,271 --> 00:00:57,430 it would look something like this. 26 00:00:57,430 --> 00:01:00,088 So if it started at zero it moved five meters to the right 27 00:01:00,088 --> 00:01:02,370 and we exerted a constant force of four newtons, 28 00:01:02,370 --> 00:01:04,480 That's why this is a horizontal line. 29 00:01:04,480 --> 00:01:07,232 It's horizontal, that means it was a constant amount 30 00:01:07,232 --> 00:01:09,518 of force, and the reason I'm doing this is because 31 00:01:09,518 --> 00:01:12,619 there's gonna be a geometrical significance to this work. 32 00:01:12,619 --> 00:01:13,452 Check it out. 33 00:01:13,452 --> 00:01:14,652 We can just use the work formula. 34 00:01:14,652 --> 00:01:18,548 We know work is the force in the direction of motion 35 00:01:18,548 --> 00:01:21,774 times the displacement or you could say the entire force 36 00:01:21,774 --> 00:01:24,778 times displacement times cosine theta. 37 00:01:24,778 --> 00:01:27,153 If you don't like this cosine theta, this just makes sure 38 00:01:27,153 --> 00:01:30,072 that you're singling out the force in the direction 39 00:01:30,072 --> 00:01:32,986 of motion and only that component of force that's 40 00:01:32,986 --> 00:01:35,946 directed in the direction of motion but for our hamburger 41 00:01:35,946 --> 00:01:38,182 example, the force already was in the direction 42 00:01:38,182 --> 00:01:40,306 of motion so we don't really need this. 43 00:01:40,306 --> 00:01:43,411 In other words the angle here would be zero. 44 00:01:43,411 --> 00:01:46,035 Whenever you take cosine of zero, you just get one 45 00:01:46,035 --> 00:01:48,321 and one times anything is just that thing so whenever 46 00:01:48,321 --> 00:01:50,236 your force is already in the direction of the 47 00:01:50,236 --> 00:01:52,408 motion of the object you don't really need that 48 00:01:52,408 --> 00:01:54,897 cosine theta and we don't need it here. 49 00:01:54,897 --> 00:01:57,916 So I'm gonna say that the force here was four newtons 50 00:01:57,916 --> 00:02:00,439 so we can calculate four newtons was the force for 51 00:02:00,439 --> 00:02:03,576 the entire displacement of five meters 52 00:02:03,576 --> 00:02:05,736 and we get the work that we did on the hamburger 53 00:02:05,736 --> 00:02:08,283 was positive 20 joules of work. 54 00:02:08,283 --> 00:02:09,998 Now if you're clever you might be like 55 00:02:09,999 --> 00:02:13,531 wait, four newtons times five meters, look it, 56 00:02:13,531 --> 00:02:16,145 that's just the area of this rectangle. 57 00:02:16,145 --> 00:02:18,817 So notice that this line is a straight line in this 58 00:02:18,817 --> 00:02:21,555 force graph just forms a rectangle in here 59 00:02:21,555 --> 00:02:24,745 and all we did, we took four newtons, that was just 60 00:02:24,745 --> 00:02:28,237 the height of this rectangle and we multiplied by 61 00:02:28,237 --> 00:02:31,851 five meters and that was just the width of this rectangle 62 00:02:31,851 --> 00:02:33,402 and if you multiply a height times width, 63 00:02:33,402 --> 00:02:36,071 you know what you get, you get the area of a rectangle, 64 00:02:36,071 --> 00:02:39,710 so what we found was when the force is constant 65 00:02:39,710 --> 00:02:42,566 one way to find the work done is by using the work formula 66 00:02:42,566 --> 00:02:44,623 but another way to find the work done is just to find 67 00:02:44,623 --> 00:02:47,206 the area enclosed by the graph. 68 00:02:48,061 --> 00:02:51,119 So from this line that determines the force 69 00:02:51,119 --> 00:02:54,778 down to this x-axis, if you find the area, 70 00:02:54,778 --> 00:02:56,173 that's gonna equal the work done. 71 00:02:56,173 --> 00:02:58,563 Now you might be like, alright, well, 72 00:02:58,563 --> 00:03:01,147 fat lot of good that does us, we can 73 00:03:01,147 --> 00:03:02,888 already find it with this formula. 74 00:03:02,888 --> 00:03:05,356 Why would I ever want to know that the work is 75 00:03:05,356 --> 00:03:08,043 equal to the area under a force graph? 76 00:03:08,043 --> 00:03:09,123 I'll show you why. 77 00:03:09,123 --> 00:03:10,895 Because in this case, yes, it was easy, 78 00:03:10,895 --> 00:03:13,277 we had a formula, we could just plug the force in here, 79 00:03:13,277 --> 00:03:15,388 we could just plug the displacement in there, 80 00:03:15,388 --> 00:03:17,643 we get our value, but think about this. 81 00:03:17,643 --> 00:03:19,598 What if our force was not constant? 82 00:03:19,598 --> 00:03:21,841 What if we had a varying force? 83 00:03:21,841 --> 00:03:24,141 In that case, what force would we plug in here? 84 00:03:24,141 --> 00:03:25,191 It'd be changing. 85 00:03:25,191 --> 00:03:26,854 One way to handle that scenario is 86 00:03:26,854 --> 00:03:28,291 using calculus, but if you don't know 87 00:03:28,291 --> 00:03:30,098 calculus that doesn't do you any good. 88 00:03:30,098 --> 00:03:32,218 Fortunately, there's another way to determine 89 00:03:32,218 --> 00:03:34,828 the work done by a varying force 90 00:03:34,828 --> 00:03:37,501 and that's to take this idea seriously. 91 00:03:37,501 --> 00:03:38,540 That's why I showed you this. 92 00:03:38,540 --> 00:03:41,679 It turns out the area underneath any force versus 93 00:03:41,679 --> 00:03:44,509 position graph is gonna equal the work, 94 00:03:44,509 --> 00:03:47,518 not just ones where the force is constant, 95 00:03:47,518 --> 00:03:49,219 even where the force is varying, 96 00:03:49,219 --> 00:03:52,019 if you can find the area underneath that graph, 97 00:03:52,019 --> 00:03:53,334 that's a quick and easy way to 98 00:03:53,334 --> 00:03:55,759 get the work done by that force. 99 00:03:55,759 --> 00:03:56,772 So what I mean is this. 100 00:03:56,772 --> 00:03:59,532 Instead of just pushing with a constant four newtons 101 00:03:59,532 --> 00:04:02,045 for the entire duration of this trip, 102 00:04:02,045 --> 00:04:04,599 let's say you started pushing with four newtons 103 00:04:04,599 --> 00:04:07,197 but you were getting tired and your force was diminishing 104 00:04:07,197 --> 00:04:09,664 and you were pushing with a weaker and weaker force 105 00:04:09,664 --> 00:04:11,797 until your force became zero newtons. 106 00:04:11,797 --> 00:04:14,502 This would be zero newtons of force when it hit this axis. 107 00:04:14,502 --> 00:04:17,142 Let's say the hamburger still made it five meters. 108 00:04:17,142 --> 00:04:17,975 You might be confused. 109 00:04:17,975 --> 00:04:19,720 How could it make it five meters 110 00:04:19,720 --> 00:04:21,880 if we're pushing with less force? 111 00:04:21,880 --> 00:04:24,377 Well it may have taken more time to get there, 112 00:04:24,377 --> 00:04:26,507 but let's say it still made it five meters. 113 00:04:26,507 --> 00:04:29,555 How do we find the work done by our force now? 114 00:04:29,555 --> 00:04:31,207 Well I'm still gonna make the claim that 115 00:04:31,207 --> 00:04:35,105 the work done is gonna be equal to the area underneath 116 00:04:35,105 --> 00:04:37,239 this force versus position graph, 117 00:04:37,239 --> 00:04:39,743 but you might be skeptical, you might be like wait a minute, 118 00:04:39,743 --> 00:04:41,632 all we really showed in that previous 119 00:04:41,632 --> 00:04:45,481 example was that the work done for a constant force 120 00:04:45,481 --> 00:04:47,494 was equal to the area underneath. 121 00:04:47,494 --> 00:04:50,804 How do I know for this varying force that the area 122 00:04:50,804 --> 00:04:53,819 underneath this graph is still gonna give me the work done? 123 00:04:53,819 --> 00:04:56,551 Well physicists and mathmeticians are clever. 124 00:04:56,551 --> 00:04:57,710 What they did is they said this, 125 00:04:57,710 --> 00:04:59,955 alright, they were like, the only thing we know 126 00:04:59,955 --> 00:05:03,308 is that for a constant force the area underneath 127 00:05:03,308 --> 00:05:05,724 is equal to the work, so let's just say this, 128 00:05:05,724 --> 00:05:09,086 instead of just considering this case where I diminish 129 00:05:09,086 --> 00:05:12,553 my force continuously, let's say I pushed with a whole 130 00:05:12,553 --> 00:05:16,489 bunch of constant forces but for a small displacement, 131 00:05:16,489 --> 00:05:19,263 so I started with four newtons but I pushed for like only 132 00:05:19,263 --> 00:05:21,859 10 centimeters with four newtons, and then I dropped that 133 00:05:21,859 --> 00:05:25,126 down to maybe 3.9 newtons and I pushed for another 134 00:05:25,126 --> 00:05:27,714 10 centimeters, and I keep doing this over and over, 135 00:05:27,714 --> 00:05:30,090 reducing my force but keeping it constant, 136 00:05:30,090 --> 00:05:32,614 and then dropping it again and here's why that's useful. 137 00:05:32,614 --> 00:05:35,631 Because the work done for this rectangular case 138 00:05:35,631 --> 00:05:38,374 is just gonna equal the area of all of these rectangles 139 00:05:38,374 --> 00:05:41,495 added up, right, because they're each constant forces 140 00:05:41,495 --> 00:05:43,486 so I know the work done is just equal to the area 141 00:05:43,486 --> 00:05:45,583 underneath, so if I add up the area of all these 142 00:05:45,583 --> 00:05:49,003 rectangles I get the work done during that rectangular 143 00:05:49,003 --> 00:05:52,018 process and here's the really clever idea. 144 00:05:52,018 --> 00:05:56,145 If we made these rectangles infinitesimally small 145 00:05:56,145 --> 00:05:59,432 they would add up to the total area just underneath 146 00:05:59,432 --> 00:06:01,977 this triangle, it'd be the exact same area. 147 00:06:01,977 --> 00:06:04,171 Now the area under this triangle would be exactly 148 00:06:04,171 --> 00:06:07,659 the same as the area under all of these rectangles 149 00:06:07,659 --> 00:06:10,432 and it'd be the exact same process we had before, 150 00:06:10,432 --> 00:06:12,848 because think about it, you'd be pushing with four newtons 151 00:06:12,848 --> 00:06:16,791 for like a millimeter but if it's infinitesimal I mean it's 152 00:06:16,791 --> 00:06:19,345 even smaller than that, but for the sake of just 153 00:06:19,345 --> 00:06:20,754 conceptually thinking about it, 154 00:06:20,754 --> 00:06:23,229 let's say you push with four newtons for a millimeter 155 00:06:23,229 --> 00:06:25,094 and then for the next millimeter you pushed 156 00:06:25,094 --> 00:06:27,653 with 3.99 newtons and then for the next 157 00:06:27,653 --> 00:06:31,050 millimeter you pushed with 3.98 newtons. 158 00:06:31,050 --> 00:06:33,906 That's basically just you diminishing your force 159 00:06:33,906 --> 00:06:36,427 continuously and going through this exact 160 00:06:36,427 --> 00:06:38,461 process that we described earlier. 161 00:06:38,461 --> 00:06:41,516 But now we know the area under all these rectangles 162 00:06:41,516 --> 00:06:43,479 is gonna just equal the work done, 163 00:06:43,479 --> 00:06:45,868 but that's the same area and process 164 00:06:45,868 --> 00:06:48,401 that's just the area underneath this triangle. 165 00:06:48,401 --> 00:06:49,501 Now, you might not buy that. 166 00:06:49,501 --> 00:06:51,415 You might be like, wait a minute, look it, 167 00:06:51,415 --> 00:06:53,456 these edges are throwing everything off, 168 00:06:53,456 --> 00:06:54,867 look at this little edge, this shoots out 169 00:06:54,867 --> 00:06:57,851 a little too far, that's a little overestimating the area, 170 00:06:57,851 --> 00:06:59,332 and then you gotta little hole in here, 171 00:06:59,332 --> 00:07:00,817 that's underestimating the area, 172 00:07:00,817 --> 00:07:02,267 you got these all over the place. 173 00:07:02,267 --> 00:07:04,422 Is that really gonna be equal to the area? 174 00:07:04,422 --> 00:07:07,359 And it will be if you let these widths of the rectangles 175 00:07:07,359 --> 00:07:11,301 become infinitesimally small, the error you're incurring 176 00:07:11,301 --> 00:07:14,034 from all these little overshoots and undershoots 177 00:07:14,034 --> 00:07:16,556 are also gonna get infinitesimally small, 178 00:07:16,556 --> 00:07:18,856 and if the error is getting infinitesimally small 179 00:07:18,856 --> 00:07:21,112 that means that the area under all these 180 00:07:21,112 --> 00:07:24,992 infinitesimal rectangles is gonna exactly equal 181 00:07:24,992 --> 00:07:28,685 the area under this triangle which is great news for us, 182 00:07:28,685 --> 00:07:30,379 because that means that the area 183 00:07:30,379 --> 00:07:33,286 under any force versus position graph, 184 00:07:33,286 --> 00:07:34,965 it didn't have to be a triangle, 185 00:07:34,965 --> 00:07:37,918 we could use infinitesimally small rectangles to 186 00:07:37,918 --> 00:07:40,134 represent the area under any graph. 187 00:07:40,134 --> 00:07:43,393 That means the area under any force versus position 188 00:07:43,393 --> 00:07:47,794 graph is gonna equal the work done by that force 189 00:07:47,794 --> 00:07:50,857 and now we have a powerful tool to determine the work 190 00:07:50,857 --> 00:07:54,265 done for cases where the force is not constant. 191 00:07:54,265 --> 00:07:56,513 This formula here where you have work equals 192 00:07:56,513 --> 00:07:59,093 Fd cosine theta, this only gives you true 193 00:07:59,093 --> 00:08:01,435 answers if you use a constant force. 194 00:08:01,435 --> 00:08:04,076 If the force is constant you can use this equation, 195 00:08:04,076 --> 00:08:06,950 but if the force is varying you couldn't just use this, 196 00:08:06,950 --> 00:08:08,499 but now we derived this. 197 00:08:08,499 --> 00:08:09,339 This is great. 198 00:08:09,339 --> 00:08:11,779 This showed me that if I can find the area under 199 00:08:11,779 --> 00:08:14,531 my force graph, my force versus position graph, 200 00:08:14,531 --> 00:08:16,146 I can find the work done. 201 00:08:16,146 --> 00:08:17,920 So, in other words, for this case right here, 202 00:08:17,920 --> 00:08:20,060 if I push with four newtons on my hamburger 203 00:08:20,060 --> 00:08:22,228 and then I reduce that force to zero 204 00:08:22,228 --> 00:08:25,064 and it went five meters, how much work was done? 205 00:08:25,064 --> 00:08:26,090 Well now I know how to find it. 206 00:08:26,090 --> 00:08:28,417 I just need to get the area under this force versus 207 00:08:28,417 --> 00:08:30,575 position graph and it's a triangle 208 00:08:30,575 --> 00:08:32,443 and I know how to find the area of a triangle. 209 00:08:32,443 --> 00:08:34,005 So the area under this triangle is 210 00:08:34,006 --> 00:08:37,657 1/2 base times height, well so I'll have 1/2, 211 00:08:37,657 --> 00:08:41,716 the base here is gonna be five because this is five meters, 212 00:08:41,716 --> 00:08:45,202 so I have five meters and the height of this triangle 213 00:08:45,202 --> 00:08:48,891 is four newtons and so I've got four newtons times five 214 00:08:48,891 --> 00:08:52,903 meters times a 1/2 is gonna give me positive ten joules 215 00:08:52,903 --> 00:08:56,968 of work done during this process where my force diminished 216 00:08:56,968 --> 00:08:59,793 continuously, and that's why this idea is so powerful, 217 00:08:59,793 --> 00:09:02,674 because if your force versus position graph happens 218 00:09:02,674 --> 00:09:05,296 to take a shape that you know how to find the area of, 219 00:09:05,296 --> 00:09:08,577 like a rectangle or a triangle or some combination 220 00:09:08,577 --> 00:09:11,093 of rectangles and triangles, then you can quickly 221 00:09:11,093 --> 00:09:13,527 get the work done without having to know anything 222 00:09:13,527 --> 00:09:16,279 about calculus because you just get the area underneath. 223 00:09:16,279 --> 00:09:18,326 Let me just clear up any confusion. 224 00:09:18,326 --> 00:09:20,834 You don't actually have to draw any rectangles. 225 00:09:20,834 --> 00:09:24,624 That was just to prove this relationship. 226 00:09:24,624 --> 00:09:27,152 Once you know this relationship you don't need to draw 227 00:09:27,152 --> 00:09:30,361 or think about those little infinitesimal rectangles. 228 00:09:30,361 --> 00:09:32,243 That was just how we were proving to ourselves 229 00:09:32,243 --> 00:09:34,841 that even for cases where the force was varying 230 00:09:34,841 --> 00:09:37,994 the area underneath, it's still gonna equal the work. 231 00:09:37,994 --> 00:09:39,828 So there's one more thing you gotta be careful about. 232 00:09:39,828 --> 00:09:42,291 When we say the area underneath the graph 233 00:09:42,291 --> 00:09:44,961 is equal to the work done, we mean the area 234 00:09:44,961 --> 00:09:49,622 from the line that represents the force to that x-axis, 235 00:09:49,622 --> 00:09:51,294 and you might be like, well of course 236 00:09:51,294 --> 00:09:53,452 that's what you mean, what else could you mean? 237 00:09:53,452 --> 00:09:55,381 Well sometimes it gets a little unclear when you 238 00:09:55,381 --> 00:09:56,947 have a case like this. 239 00:09:56,947 --> 00:09:58,668 So let's say your force started down here. 240 00:09:58,668 --> 00:10:00,632 So we'll say we first started pushing on our hamburger 241 00:10:00,632 --> 00:10:02,865 to the left and then we're like oh there's 242 00:10:02,865 --> 00:10:04,156 no one there, awkward, we'll start pushing 243 00:10:04,156 --> 00:10:05,984 to the right, so our force starts off 244 00:10:05,984 --> 00:10:08,175 negative and then it becomes positive. 245 00:10:08,175 --> 00:10:10,850 Now if we wanted to find the total work done during 246 00:10:10,850 --> 00:10:13,740 this entire process, how would we do it? 247 00:10:13,740 --> 00:10:16,357 We can still use this idea that the work done 248 00:10:16,357 --> 00:10:19,508 is gonna equal the area underneath the curve, 249 00:10:19,508 --> 00:10:21,406 but for this first portion what do I do? 250 00:10:21,406 --> 00:10:23,643 How do I find the area underneath this portion? 251 00:10:23,643 --> 00:10:25,477 Some people are like underneath, alright, I'm gonna 252 00:10:25,477 --> 00:10:26,824 go underneath, we'll keep going this way, 253 00:10:26,824 --> 00:10:29,837 how much area's down here, but that's just crazy. 254 00:10:29,837 --> 00:10:31,975 That goes on infinitely. 255 00:10:31,975 --> 00:10:34,321 I did not do infinite work on this cheeseburger. 256 00:10:34,321 --> 00:10:35,154 No. 257 00:10:35,154 --> 00:10:37,660 The area I'm talking about is gonna be this area right here, 258 00:10:37,660 --> 00:10:40,093 so I gotta take all this area right here. 259 00:10:40,093 --> 00:10:41,431 That area is finite. 260 00:10:41,431 --> 00:10:43,146 It's from this line that represents 261 00:10:43,146 --> 00:10:45,146 the force to the x-axis. 262 00:10:46,159 --> 00:10:47,642 That's the area we're talking about, 263 00:10:47,642 --> 00:10:50,316 just like over here the area we'd be talking about 264 00:10:50,316 --> 00:10:53,762 is from this line that represents the force to the x-axis. 265 00:10:53,762 --> 00:10:55,272 It's all of this area. 266 00:10:55,272 --> 00:10:56,848 So how would I find these areas? 267 00:10:56,848 --> 00:10:57,742 They're both triangles. 268 00:10:57,742 --> 00:11:01,069 I can do this, so this one's gonna be 1/2 base times height, 269 00:11:01,069 --> 00:11:03,994 so it's gonna be 1/2 the base, this base right here is 270 00:11:03,994 --> 00:11:07,199 one meter, so it'd be one meter times the height. 271 00:11:07,199 --> 00:11:08,659 I've got negative height. 272 00:11:08,659 --> 00:11:09,492 Look it, it's down here. 273 00:11:09,492 --> 00:11:11,132 It's negative two newtons. 274 00:11:11,132 --> 00:11:13,009 That means I'm gonna have a negative area. 275 00:11:13,009 --> 00:11:13,842 Is that okay? 276 00:11:13,842 --> 00:11:15,665 Yep, that just means I did negative work 277 00:11:15,665 --> 00:11:17,582 on this cheeseburger during that time 278 00:11:17,582 --> 00:11:19,837 so I'm negative one joule. 279 00:11:19,837 --> 00:11:22,152 During the first meter of me exerting force 280 00:11:22,152 --> 00:11:24,839 on that hamburger, I did negative one joule of work, 281 00:11:24,839 --> 00:11:25,928 and then for this portion of the 282 00:11:25,928 --> 00:11:27,394 trip I'll find the area underneath. 283 00:11:27,394 --> 00:11:30,470 It's also a triangle, so 1/2 base times height. 284 00:11:30,470 --> 00:11:33,801 I'll have 1/2, the base this time is not three, 285 00:11:33,801 --> 00:11:35,874 the base of this triangle is only two, because it goes from 286 00:11:35,874 --> 00:11:39,861 one to three, so that's two meters times the height, 287 00:11:39,861 --> 00:11:42,204 well that is four newtons, so I'll multiply 288 00:11:42,204 --> 00:11:43,958 by four newtons and then I get for that 289 00:11:43,958 --> 00:11:47,253 portion of the trip, I did four joules of work. 290 00:11:47,253 --> 00:11:49,872 So, positive four joules of work for this portion, 291 00:11:49,872 --> 00:11:52,445 negative one joule of work for that portion. 292 00:11:52,445 --> 00:11:55,646 The total work done for the entire three meters 293 00:11:55,646 --> 00:11:59,272 would be four joules minus one joule 294 00:11:59,272 --> 00:12:01,840 and that would be positive three joules. 295 00:12:01,840 --> 00:12:04,550 So, recapping, if your force is constant you can 296 00:12:04,550 --> 00:12:06,764 just use this formula to find the work done. 297 00:12:06,764 --> 00:12:09,230 It's just Fd cosine theta, but you can also find 298 00:12:09,230 --> 00:12:12,615 the work done by determining the area underneath 299 00:12:12,615 --> 00:12:15,538 a force versus position graph and this is useful 300 00:12:15,538 --> 00:12:19,075 because this works even when the force is varying 301 00:12:19,075 --> 00:12:22,138 which would render this equation somewhat unusable 302 00:12:22,138 --> 00:12:24,334 but if the shape of the graph is one that you know 303 00:12:24,334 --> 00:12:26,303 how to find the area of, you can still find the work 304 00:12:26,303 --> 00:12:28,394 done by determining the area underneath 305 00:12:28,394 --> 00:00:00,000 the force versus position graph.