1 00:00:00,000 --> 00:00:01,570 2 00:00:01,570 --> 00:00:03,600 One way to find the amount of work done 3 00:00:03,600 --> 00:00:06,920 is by using the formula Fd cosine theta. 4 00:00:06,920 --> 00:00:09,030 But this number for the amount of work done 5 00:00:09,030 --> 00:00:12,530 represents the amount of energy transferred to an object. 6 00:00:12,530 --> 00:00:14,550 For instance, if you solve for the work done 7 00:00:14,550 --> 00:00:16,980 and you get positive 200 joules, it 8 00:00:16,980 --> 00:00:20,960 means that the force gave something 200 joules of energy. 9 00:00:20,960 --> 00:00:22,530 So if you have a way of determining 10 00:00:22,530 --> 00:00:25,870 the amount of energy that something gains or loses, 11 00:00:25,870 --> 00:00:28,870 then you have an alternate way of finding the work done, 12 00:00:28,870 --> 00:00:32,330 since the work done on an object is the amount of energy 13 00:00:32,330 --> 00:00:33,850 it gains or loses. 14 00:00:33,850 --> 00:00:36,820 For instance, imagine a 50-kilogram skateboarder 15 00:00:36,820 --> 00:00:38,220 that starts at rest. 16 00:00:38,220 --> 00:00:40,700 If a force starts the skateboarder moving 17 00:00:40,700 --> 00:00:43,150 at 10 meters per second, that force 18 00:00:43,150 --> 00:00:46,280 did work on the skateboarder since it gave the skateboarder 19 00:00:46,280 --> 00:00:47,190 energy. 20 00:00:47,190 --> 00:00:50,240 The amount of kinetic energy gained by the skateboarder 21 00:00:50,240 --> 00:00:52,220 is 2,500 joules. 22 00:00:52,220 --> 00:00:55,820 That means that the work done by the force on the skateboarder 23 00:00:55,820 --> 00:00:58,670 was positive 2,500 joules. 24 00:00:58,670 --> 00:01:01,610 It's positive because the force on the skateboarder 25 00:01:01,610 --> 00:01:04,680 gave the skateboarder 2,500 joules. 26 00:01:04,680 --> 00:01:07,280 If a force gives energy to an object, 27 00:01:07,280 --> 00:01:09,980 then the force is doing positive work on that object. 28 00:01:09,980 --> 00:01:12,940 And if a force takes away energy from an object, 29 00:01:12,940 --> 00:01:15,580 the force is doing negative work on that object. 30 00:01:15,580 --> 00:01:17,510 Now imagine that the skateboarder, who's 31 00:01:17,510 --> 00:01:19,580 moving with 10 meters per second, 32 00:01:19,580 --> 00:01:22,930 gets stopped because he crashes into a stack of bricks. 33 00:01:22,930 --> 00:01:25,060 The stack of bricks does negative work 34 00:01:25,060 --> 00:01:27,750 on the skateboarder because it takes away energy 35 00:01:27,750 --> 00:01:28,910 from the skateboarder. 36 00:01:28,910 --> 00:01:31,180 To find the work done by the stack of bricks, 37 00:01:31,180 --> 00:01:34,004 we just need to figure out how much energy it took away 38 00:01:34,004 --> 00:01:34,920 from the skateboarder. 39 00:01:34,920 --> 00:01:38,500 Since the skateboarder started with 2,500 joules 40 00:01:38,500 --> 00:01:42,310 of kinetic energy and ends with zero joules of kinetic energy, 41 00:01:42,310 --> 00:01:45,650 it means that the work done by the bricks on the skateboarder 42 00:01:45,650 --> 00:01:48,050 was negative 2,500 joules. 43 00:01:48,050 --> 00:01:50,810 It's negative because the bricks took away energy 44 00:01:50,810 --> 00:01:52,050 from the skateboarder. 45 00:01:52,050 --> 00:01:54,580 Let's say we instead lift the bricks, which 46 00:01:54,580 --> 00:01:58,750 are 500 kilograms, upwards a distance of four meters. 47 00:01:58,750 --> 00:02:00,950 To find the work that we've done on the bricks, 48 00:02:00,950 --> 00:02:03,350 we could use Fd cosine theta. 49 00:02:03,350 --> 00:02:04,280 But we don't have to. 50 00:02:04,280 --> 00:02:06,460 We could just figure out the amount of energy 51 00:02:06,460 --> 00:02:08,240 that we've given to the bricks. 52 00:02:08,240 --> 00:02:10,000 The bricks gain energy here. 53 00:02:10,000 --> 00:02:12,650 And they're gaining gravitational potential energy, 54 00:02:12,650 --> 00:02:15,860 which is given by the formula mgh. 55 00:02:15,860 --> 00:02:17,680 If we solve, we get that the bricks 56 00:02:17,680 --> 00:02:22,800 gained 19,600 joules of gravitational potential energy. 57 00:02:22,800 --> 00:02:25,300 That means that the work we did on the bricks 58 00:02:25,300 --> 00:02:28,810 was positive 19,600 joules. 59 00:02:28,810 --> 00:02:32,550 It's positive because our force gave the bricks energy. 60 00:02:32,550 --> 00:02:35,970 This idea doesn't just work with gravitational potential energy 61 00:02:35,970 --> 00:02:37,260 and kinetic energy. 62 00:02:37,260 --> 00:02:39,500 It works for every kind of energy. 63 00:02:39,500 --> 00:02:42,900 You can always find the work done by a force on an object 64 00:02:42,900 --> 00:02:45,910 if you could determine the energy that that force gives 65 00:02:45,910 --> 00:02:47,755 or takes away from that object. 66 00:02:47,755 --> 00:00:00,000 [MUSIC PLAYING]