1 00:00:00,240 --> 00:00:06,000 So let us look at this example, what I've drawn here is a roller coaster, so the blue line is the 2 00:00:06,000 --> 00:00:14,490 path that the roller coaster will follow from a high point at 20 meters at point A to a lower point 3 00:00:14,490 --> 00:00:15,450 at point B. 4 00:00:16,570 --> 00:00:22,860 Initially at point, the roller coaster is stationary, doesn't move, so it has zero kinetic energy, 5 00:00:23,230 --> 00:00:28,540 but because it is at a height of 20 meters, it has a potential energy that we can calculate because 6 00:00:28,540 --> 00:00:32,950 we know the mass of this roller coaster cord is 300 kilograms. 7 00:00:33,760 --> 00:00:40,570 Now, the question is, what is the velocity of this roller coaster when it reaches point B, we can 8 00:00:40,570 --> 00:00:44,650 ignore friction and we can ignore the air resistance. 9 00:00:46,270 --> 00:00:55,800 So let us investigate this court in terms of its energy at point and at point B, so at point A. the 10 00:00:55,810 --> 00:00:59,280 question stated that the court is stationary. 11 00:00:59,440 --> 00:01:03,850 So we know that the kinetic energy, the first position will be zero. 12 00:01:04,880 --> 00:01:11,990 But it is at a height of 20 meters relative to point B, so we can calculate a potential energy which 13 00:01:11,990 --> 00:01:18,770 is equal to engage, which is equal to 300 multiplied by nine point eight, one multiplied by twenty, 14 00:01:19,340 --> 00:01:23,730 which gives us thirty eight thousand eight hundred and sixty joules. 15 00:01:24,410 --> 00:01:26,030 That's the same unit for energy. 16 00:01:27,610 --> 00:01:28,720 At point B. 17 00:01:30,050 --> 00:01:31,250 It has now reached. 18 00:01:32,490 --> 00:01:39,690 A zero distance from the reference line, so the potential energy will be zero, but it will have a 19 00:01:39,690 --> 00:01:40,910 certain velocity. 20 00:01:41,160 --> 00:01:48,840 So there will be a kinetic energy and that kinetic energy is calculated by one half in the square. 21 00:01:48,870 --> 00:01:51,800 Now, we don't know what is V that is what we want to calculate. 22 00:01:52,650 --> 00:01:59,390 So let us write down the principle of energy conservation, kinetic energy one plus potential energy. 23 00:01:59,400 --> 00:02:02,610 One is kinetic energy two plus potential energy to. 24 00:02:04,100 --> 00:02:07,140 So zero, this is a point a plus. 25 00:02:07,280 --> 00:02:14,720 Fifty eight thousand eight hundred and sixty is equal to one half in the squared plus zero. 26 00:02:14,750 --> 00:02:18,200 That would be at point B because we know the mass. 27 00:02:18,200 --> 00:02:25,940 We can substitute that for him and we can solve for the velocity at point B to be nineteen point eight 28 00:02:25,940 --> 00:02:27,300 meters per second. 29 00:02:28,310 --> 00:02:35,960 This is a fairly simple example, but it illustrates the principle and the principle is that the total 30 00:02:35,960 --> 00:02:41,810 amount of energy in the system stays the same, but it is exchanged between different forms. 31 00:02:41,810 --> 00:02:45,040 It is exchanged between kinetic energy and potential energy. 32 00:02:45,650 --> 00:02:50,690 And so at point A. we knew what was the total amount of energy because it was only potential energy. 33 00:02:51,350 --> 00:02:54,040 And at point B, we knew the potential energy is zero. 34 00:02:54,770 --> 00:03:01,400 So that energy at point B was the kinetic energy and therefore we could solve for the velocity of.