[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.27,0:00:01.10,Default,,0000,0000,0000,,- [Instructor] Today in the gym,
Dialogue: 0,0:00:01.10,0:00:02.97,Default,,0000,0000,0000,,when my wife was doing dumbbell curls,
Dialogue: 0,0:00:02.97,0:00:05.40,Default,,0000,0000,0000,,I started wondering, see,\Nshe's putting a force
Dialogue: 0,0:00:05.40,0:00:07.20,Default,,0000,0000,0000,,on that dumbbell upwards, right?
Dialogue: 0,0:00:07.20,0:00:09.81,Default,,0000,0000,0000,,But does that force stay a constant
Dialogue: 0,0:00:09.81,0:00:12.45,Default,,0000,0000,0000,,as she moves the dumbbell up or not?
Dialogue: 0,0:00:12.45,0:00:15.09,Default,,0000,0000,0000,,Does it change? And if it does\Nchange, how does it change?
Dialogue: 0,0:00:15.09,0:00:17.76,Default,,0000,0000,0000,,Does it increase, does it\Ndecrease? What happens to it?
Dialogue: 0,0:00:17.76,0:00:18.59,Default,,0000,0000,0000,,Guess what.
Dialogue: 0,0:00:18.59,0:00:20.52,Default,,0000,0000,0000,,We can answer this question by the end
Dialogue: 0,0:00:20.52,0:00:23.43,Default,,0000,0000,0000,,of this video using Newton's second law.
Dialogue: 0,0:00:23.43,0:00:24.99,Default,,0000,0000,0000,,So let's start with a simpler example.
Dialogue: 0,0:00:24.99,0:00:27.00,Default,,0000,0000,0000,,We have a ice hockey ground over here,
Dialogue: 0,0:00:27.00,0:00:30.30,Default,,0000,0000,0000,,and there's a puck moving\Non top of it at some speed.
Dialogue: 0,0:00:30.30,0:00:33.54,Default,,0000,0000,0000,,If there are no frictional\Nforces acting on this,
Dialogue: 0,0:00:33.54,0:00:36.75,Default,,0000,0000,0000,,if you assume that, then the\Nforces acting on this puck
Dialogue: 0,0:00:36.75,0:00:40.71,Default,,0000,0000,0000,,would be balanced because in\Nthe horizontal, you can see
Dialogue: 0,0:00:40.71,0:00:43.14,Default,,0000,0000,0000,,that there are no forces\Nbecause we're ignoring friction.
Dialogue: 0,0:00:43.14,0:00:46.38,Default,,0000,0000,0000,,And in the vertical,\Nthe gravitational force,
Dialogue: 0,0:00:46.38,0:00:48.21,Default,,0000,0000,0000,,which is pulling down on\Nit is completely balanced
Dialogue: 0,0:00:48.21,0:00:50.79,Default,,0000,0000,0000,,by the force that the\Nground is pushing up on it,
Dialogue: 0,0:00:50.79,0:00:52.38,Default,,0000,0000,0000,,the normal force, they balance it out.
Dialogue: 0,0:00:52.38,0:00:54.84,Default,,0000,0000,0000,,And so since there are no\Nunbalanced forces acting
Dialogue: 0,0:00:54.84,0:00:56.58,Default,,0000,0000,0000,,on this puck from Newton's first law,
Dialogue: 0,0:00:56.58,0:00:59.40,Default,,0000,0000,0000,,we know that this thing will\Ncontinue its state of rest,
Dialogue: 0,0:00:59.40,0:01:02.04,Default,,0000,0000,0000,,or in this particular case,\Nthe state of uniform motion.
Dialogue: 0,0:01:02.04,0:01:04.80,Default,,0000,0000,0000,,So it'll continue to move\Nwith that same velocity.
Dialogue: 0,0:01:04.80,0:01:06.87,Default,,0000,0000,0000,,But now comes the question,
Dialogue: 0,0:01:06.87,0:01:09.33,Default,,0000,0000,0000,,what if there was an\Nunbalanced force acting on it?
Dialogue: 0,0:01:09.33,0:01:11.19,Default,,0000,0000,0000,,What happens because of that?
Dialogue: 0,0:01:11.19,0:01:12.60,Default,,0000,0000,0000,,Well, let's find out.
Dialogue: 0,0:01:12.60,0:01:15.00,Default,,0000,0000,0000,,For that, let's just whack\Nit with a hockey stick.
Dialogue: 0,0:01:15.00,0:01:16.80,Default,,0000,0000,0000,,No.\N(instructor laughing)
Dialogue: 0,0:01:16.80,0:01:19.86,Default,,0000,0000,0000,,So if I whack it to the\Nright, let's say in this case,
Dialogue: 0,0:01:19.86,0:01:23.49,Default,,0000,0000,0000,,I will now put an unbalanced\Nforce to the right.
Dialogue: 0,0:01:23.49,0:01:26.13,Default,,0000,0000,0000,,What will happen? Well,\Nwe can probably guess it.
Dialogue: 0,0:01:26.13,0:01:28.11,Default,,0000,0000,0000,,That puck's velocity will now be higher.
Dialogue: 0,0:01:28.11,0:01:29.67,Default,,0000,0000,0000,,It'll just get blasted off over there.
Dialogue: 0,0:01:29.67,0:01:31.59,Default,,0000,0000,0000,,So its velocity will increase.
Dialogue: 0,0:01:31.59,0:01:34.92,Default,,0000,0000,0000,,In other words, it will accelerate.
Dialogue: 0,0:01:34.92,0:01:38.67,Default,,0000,0000,0000,,Ooh, this means when there's\Nan unbalanced force acting
Dialogue: 0,0:01:38.67,0:01:40.40,Default,,0000,0000,0000,,on a an object, in other words,
Dialogue: 0,0:01:40.40,0:01:45.40,Default,,0000,0000,0000,,if there is a non-zero net\Nforce acting on an object,
Dialogue: 0,0:01:45.48,0:01:47.88,Default,,0000,0000,0000,,which is the same thing as\Nsaying an unbalanced force,
Dialogue: 0,0:01:47.88,0:01:49.71,Default,,0000,0000,0000,,but whenever this net\Nforce acts on an object,
Dialogue: 0,0:01:49.71,0:01:50.82,Default,,0000,0000,0000,,what does it do?
Dialogue: 0,0:01:50.82,0:01:52.68,Default,,0000,0000,0000,,It accelerates our puck.
Dialogue: 0,0:01:52.68,0:01:55.98,Default,,0000,0000,0000,,The puck undergoes, or the\Nobject undergoes an acceleration.
Dialogue: 0,0:01:56.91,0:02:00.03,Default,,0000,0000,0000,,This is the essence of\NNewton's second law.
Dialogue: 0,0:02:00.03,0:02:02.04,Default,,0000,0000,0000,,Now all we gotta do is\Nanalyze the situation
Dialogue: 0,0:02:02.04,0:02:03.00,Default,,0000,0000,0000,,even more carefully
Dialogue: 0,0:02:03.00,0:02:05.76,Default,,0000,0000,0000,,and see if we can concretize\Nthis relationship.
Dialogue: 0,0:02:05.76,0:02:06.93,Default,,0000,0000,0000,,So let's do that.
Dialogue: 0,0:02:06.93,0:02:08.46,Default,,0000,0000,0000,,The first question we could have is yeah,
Dialogue: 0,0:02:08.46,0:02:10.59,Default,,0000,0000,0000,,so a net force causes an acceleration,
Dialogue: 0,0:02:10.59,0:02:13.14,Default,,0000,0000,0000,,but how long does that acceleration last?
Dialogue: 0,0:02:13.14,0:02:14.55,Default,,0000,0000,0000,,Well, let's see.
Dialogue: 0,0:02:14.55,0:02:17.76,Default,,0000,0000,0000,,When the stick hits the puck,
Dialogue: 0,0:02:17.76,0:02:19.44,Default,,0000,0000,0000,,that's when it starts accelerating,
Dialogue: 0,0:02:19.44,0:02:22.77,Default,,0000,0000,0000,,which means as long as\Nthe stick is in contact
Dialogue: 0,0:02:22.77,0:02:25.29,Default,,0000,0000,0000,,with the puck, as long as\Nit's in contact with it,
Dialogue: 0,0:02:25.29,0:02:28.50,Default,,0000,0000,0000,,like right now here, it's during that time
Dialogue: 0,0:02:28.50,0:02:30.33,Default,,0000,0000,0000,,there will be acceleration.
Dialogue: 0,0:02:30.33,0:02:33.00,Default,,0000,0000,0000,,But what happens once it loses contact?
Dialogue: 0,0:02:33.00,0:02:35.94,Default,,0000,0000,0000,,Once it loses contact, again,\Nnet force goes to zero.
Dialogue: 0,0:02:35.94,0:02:38.64,Default,,0000,0000,0000,,And now coming back to Newton's first law,
Dialogue: 0,0:02:38.64,0:02:43.11,Default,,0000,0000,0000,,it'll continue moving with\Nthat same increased velocity.
Dialogue: 0,0:02:43.11,0:02:45.90,Default,,0000,0000,0000,,This means the acceleration only happened
Dialogue: 0,0:02:45.90,0:02:48.45,Default,,0000,0000,0000,,during this time when the hockey stick
Dialogue: 0,0:02:48.45,0:02:49.68,Default,,0000,0000,0000,,was in contact with it.
Dialogue: 0,0:02:49.68,0:02:52.68,Default,,0000,0000,0000,,In other words, the\Nacceleration lasts as long
Dialogue: 0,0:02:52.68,0:02:55.44,Default,,0000,0000,0000,,as the net force lasts.
Dialogue: 0,0:02:55.44,0:02:58.26,Default,,0000,0000,0000,,Okay, next, let's think\Nabout what would happen
Dialogue: 0,0:02:58.26,0:03:01.11,Default,,0000,0000,0000,,if the net force was higher?
Dialogue: 0,0:03:01.11,0:03:03.48,Default,,0000,0000,0000,,For that let's imagine\Nwe whacked it harder.
Dialogue: 0,0:03:03.48,0:03:04.77,Default,,0000,0000,0000,,What's gonna happen now?
Dialogue: 0,0:03:04.77,0:03:07.77,Default,,0000,0000,0000,,Or you can imagine it'll get\Nblasted off even more faster,
Dialogue: 0,0:03:07.77,0:03:09.03,Default,,0000,0000,0000,,even faster, right?
Dialogue: 0,0:03:09.03,0:03:10.77,Default,,0000,0000,0000,,Which means it'll have a higher velocity
Dialogue: 0,0:03:10.77,0:03:12.33,Default,,0000,0000,0000,,when it loses contact.
Dialogue: 0,0:03:12.33,0:03:15.75,Default,,0000,0000,0000,,Ooh, that means there'll\Nbe bigger acceleration.
Dialogue: 0,0:03:15.75,0:03:17.73,Default,,0000,0000,0000,,So if the net force is larger,
Dialogue: 0,0:03:17.73,0:03:19.95,Default,,0000,0000,0000,,it means you'll have\Na larger acceleration.
Dialogue: 0,0:03:19.95,0:03:22.89,Default,,0000,0000,0000,,If the net force is smaller,\Nyou get a smaller acceleration.
Dialogue: 0,0:03:22.89,0:03:26.22,Default,,0000,0000,0000,,In other words, we see\Na direct relationship
Dialogue: 0,0:03:26.22,0:03:30.39,Default,,0000,0000,0000,,between acceleration and the net force.
Dialogue: 0,0:03:30.39,0:03:32.61,Default,,0000,0000,0000,,All right, what else can we deduce?
Dialogue: 0,0:03:32.61,0:03:35.10,Default,,0000,0000,0000,,Hey, let's think about the direction.
Dialogue: 0,0:03:35.10,0:03:37.62,Default,,0000,0000,0000,,What is the direction of the acceleration?
Dialogue: 0,0:03:37.62,0:03:40.62,Default,,0000,0000,0000,,Well, in this case, the\Nnet force is to the right,
Dialogue: 0,0:03:40.62,0:03:42.54,Default,,0000,0000,0000,,and our puck's velocity
Dialogue: 0,0:03:42.54,0:03:45.00,Default,,0000,0000,0000,,is also increasing towards the right.
Dialogue: 0,0:03:45.00,0:03:47.64,Default,,0000,0000,0000,,So in this case, the\Nacceleration is to the right.
Dialogue: 0,0:03:47.64,0:03:49.71,Default,,0000,0000,0000,,So in this case, if the\Nnet force is to the right,
Dialogue: 0,0:03:49.71,0:03:51.17,Default,,0000,0000,0000,,the acceleration is to the right.
Dialogue: 0,0:03:51.17,0:03:54.33,Default,,0000,0000,0000,,What would happen if the\Nnet force was to the left?
Dialogue: 0,0:03:54.33,0:03:56.79,Default,,0000,0000,0000,,So let's imagine we whack that puck now
Dialogue: 0,0:03:56.79,0:03:59.37,Default,,0000,0000,0000,,to the left, what would happen?
Dialogue: 0,0:03:59.37,0:04:00.75,Default,,0000,0000,0000,,Well, we can again imagine the puck
Dialogue: 0,0:04:00.75,0:04:02.82,Default,,0000,0000,0000,,would now get blasted off to the left.
Dialogue: 0,0:04:02.82,0:04:04.77,Default,,0000,0000,0000,,But let's look at it carefully.
Dialogue: 0,0:04:04.77,0:04:07.95,Default,,0000,0000,0000,,Since the puck is already moving\Nto the right, if we push it
Dialogue: 0,0:04:07.95,0:04:11.28,Default,,0000,0000,0000,,to the left, now we're gonna slow it down.
Dialogue: 0,0:04:11.28,0:04:13.68,Default,,0000,0000,0000,,The puck will come to a stop first.
Dialogue: 0,0:04:13.68,0:04:16.98,Default,,0000,0000,0000,,It'll happen very quickly\Nthat we won't even see it.
Dialogue: 0,0:04:16.98,0:04:19.74,Default,,0000,0000,0000,,But it has to happen before\Ngoing to the left, right?
Dialogue: 0,0:04:19.74,0:04:21.99,Default,,0000,0000,0000,,Which means when you go from here to here,
Dialogue: 0,0:04:21.99,0:04:24.09,Default,,0000,0000,0000,,notice even though the puck\Nis moving to the right,
Dialogue: 0,0:04:24.09,0:04:26.58,Default,,0000,0000,0000,,it is slowing down, which\Nmeans the acceleration
Dialogue: 0,0:04:26.58,0:04:28.41,Default,,0000,0000,0000,,is to the left.
Dialogue: 0,0:04:28.41,0:04:29.85,Default,,0000,0000,0000,,So when the net force is to the left,
Dialogue: 0,0:04:29.85,0:04:32.13,Default,,0000,0000,0000,,we're seeing an\Nacceleration is to the left.
Dialogue: 0,0:04:32.13,0:04:34.62,Default,,0000,0000,0000,,After that, its velocity\Nmight increase to the left,
Dialogue: 0,0:04:34.62,0:04:37.92,Default,,0000,0000,0000,,which means again, the\Nacceleration is to the left.
Dialogue: 0,0:04:37.92,0:04:38.97,Default,,0000,0000,0000,,Ooh.
Dialogue: 0,0:04:38.97,0:04:41.07,Default,,0000,0000,0000,,So if the net force is to the left,
Dialogue: 0,0:04:41.07,0:04:42.39,Default,,0000,0000,0000,,the acceleration is to the left.
Dialogue: 0,0:04:42.39,0:04:43.77,Default,,0000,0000,0000,,If the net force is to the right,
Dialogue: 0,0:04:43.77,0:04:45.30,Default,,0000,0000,0000,,the acceleration is to the right.
Dialogue: 0,0:04:45.30,0:04:48.72,Default,,0000,0000,0000,,So the acceleration will\Nbe in the same direction
Dialogue: 0,0:04:48.72,0:04:51.09,Default,,0000,0000,0000,,as that of the net force.
Dialogue: 0,0:04:51.09,0:04:55.53,Default,,0000,0000,0000,,Okay, is there anything else\Nthat affects our acceleration?
Dialogue: 0,0:04:55.53,0:04:56.82,Default,,0000,0000,0000,,Well, let's see.
Dialogue: 0,0:04:56.82,0:04:58.14,Default,,0000,0000,0000,,If you come back over here,
Dialogue: 0,0:04:58.14,0:04:59.70,Default,,0000,0000,0000,,what if you use the same bat,
Dialogue: 0,0:04:59.70,0:05:01.44,Default,,0000,0000,0000,,whacked it with the same force,
Dialogue: 0,0:05:01.44,0:05:03.15,Default,,0000,0000,0000,,but instead of a puck,
Dialogue: 0,0:05:03.15,0:05:04.50,Default,,0000,0000,0000,,let's say there was a bowling ball
Dialogue: 0,0:05:04.50,0:05:05.70,Default,,0000,0000,0000,,moving with the same velocity.
Dialogue: 0,0:05:05.70,0:05:07.36,Default,,0000,0000,0000,,What would happen now?
Dialogue: 0,0:05:07.36,0:05:08.25,Default,,0000,0000,0000,,(instructor laughing)
Dialogue: 0,0:05:08.25,0:05:11.40,Default,,0000,0000,0000,,I'm pretty sure you can\Nfeel it in your bones now.
Dialogue: 0,0:05:11.40,0:05:14.88,Default,,0000,0000,0000,,It would be much harder\Nto stop that bowling ball
Dialogue: 0,0:05:14.88,0:05:17.19,Default,,0000,0000,0000,,and make it turn backwards, right?
Dialogue: 0,0:05:17.19,0:05:18.63,Default,,0000,0000,0000,,I mean, the same thing will happen.
Dialogue: 0,0:05:18.63,0:05:23.07,Default,,0000,0000,0000,,You will slow it down, but\Nit'll be much, much harder.
Dialogue: 0,0:05:23.07,0:05:26.31,Default,,0000,0000,0000,,It'll take a much longer\Ntime to slow it down,
Dialogue: 0,0:05:26.31,0:05:29.10,Default,,0000,0000,0000,,even though you're putting\Nthe same amount of force.
Dialogue: 0,0:05:29.10,0:05:30.93,Default,,0000,0000,0000,,So the net force has stayed the same.
Dialogue: 0,0:05:30.93,0:05:33.15,Default,,0000,0000,0000,,But what has happened to our acceleration?
Dialogue: 0,0:05:33.15,0:05:36.45,Default,,0000,0000,0000,,Since the velocity changed\Nover a much longer time,
Dialogue: 0,0:05:36.45,0:05:39.30,Default,,0000,0000,0000,,the acceleration became smaller.
Dialogue: 0,0:05:39.30,0:05:41.85,Default,,0000,0000,0000,,Hey, why did the\Nacceleration became smaller?
Dialogue: 0,0:05:41.85,0:05:43.50,Default,,0000,0000,0000,,What changed?
Dialogue: 0,0:05:43.50,0:05:46.74,Default,,0000,0000,0000,,From the puck to the bowling\Nball, the mass changed,
Dialogue: 0,0:05:46.74,0:05:48.66,Default,,0000,0000,0000,,the mass increased.
Dialogue: 0,0:05:48.66,0:05:51.96,Default,,0000,0000,0000,,So this means mass also\Naffects the acceleration,
Dialogue: 0,0:05:51.96,0:05:53.40,Default,,0000,0000,0000,,but how does it affect it?
Dialogue: 0,0:05:53.40,0:05:56.19,Default,,0000,0000,0000,,Well, we saw that the\Nmass increased right now.
Dialogue: 0,0:05:56.19,0:05:58.23,Default,,0000,0000,0000,,What did that do to the acceleration?
Dialogue: 0,0:05:58.23,0:06:01.11,Default,,0000,0000,0000,,It decreased, and this\Nis kind of intuitive.
Dialogue: 0,0:06:01.11,0:06:04.20,Default,,0000,0000,0000,,The bigger the mass, the\Nharder it is to accelerate,
Dialogue: 0,0:06:04.20,0:06:06.66,Default,,0000,0000,0000,,meaning the smaller the acceleration,
Dialogue: 0,0:06:06.66,0:06:11.46,Default,,0000,0000,0000,,which means acceleration\Nhas an inverse relationship
Dialogue: 0,0:06:11.46,0:06:13.35,Default,,0000,0000,0000,,with the mass.
Dialogue: 0,0:06:13.35,0:06:16.56,Default,,0000,0000,0000,,So now, everything that we just\Nanalyzed about acceleration,
Dialogue: 0,0:06:16.56,0:06:18.63,Default,,0000,0000,0000,,its direction, its\Ndependency on the net force,
Dialogue: 0,0:06:18.63,0:06:19.71,Default,,0000,0000,0000,,how it depends on the mass,
Dialogue: 0,0:06:19.71,0:06:23.25,Default,,0000,0000,0000,,all of it can be put down in an equation.
Dialogue: 0,0:06:23.25,0:06:25.35,Default,,0000,0000,0000,,And that equation is pretty\Nmuch right in front of us.
Dialogue: 0,0:06:25.35,0:06:29.58,Default,,0000,0000,0000,,So the acceleration will\Nequal the net force divided
Dialogue: 0,0:06:29.58,0:06:31.77,Default,,0000,0000,0000,,by the mass.
Dialogue: 0,0:06:31.77,0:06:36.63,Default,,0000,0000,0000,,This is our Newton's second law.
Dialogue: 0,0:06:36.63,0:06:38.61,Default,,0000,0000,0000,,And look, the equation\Nis saying the same thing.
Dialogue: 0,0:06:38.61,0:06:40.02,Default,,0000,0000,0000,,Direct relationship between acceleration
Dialogue: 0,0:06:40.02,0:06:42.48,Default,,0000,0000,0000,,and the net force, inverse\Nrelationship between acceleration
Dialogue: 0,0:06:42.48,0:06:44.64,Default,,0000,0000,0000,,and the mass and the arrowheads are saying
Dialogue: 0,0:06:44.64,0:06:46.71,Default,,0000,0000,0000,,that acceleration and the\Nnet force will always be
Dialogue: 0,0:06:46.71,0:06:48.27,Default,,0000,0000,0000,,in the same direction.
Dialogue: 0,0:06:48.27,0:06:50.04,Default,,0000,0000,0000,,Isn't it amazing that we can pack all
Dialogue: 0,0:06:50.04,0:06:53.07,Default,,0000,0000,0000,,of that information in just\None beautiful equation?
Dialogue: 0,0:06:53.07,0:06:55.35,Default,,0000,0000,0000,,And of course, you may\Nhave seen this written
Dialogue: 0,0:06:55.35,0:06:58.62,Default,,0000,0000,0000,,as f equals ma in some\Nsources, it's the same thing.
Dialogue: 0,0:06:58.62,0:06:59.45,Default,,0000,0000,0000,,I like to write it this way
Dialogue: 0,0:06:59.45,0:07:03.39,Default,,0000,0000,0000,,because acceleration\Nis caused by the force.
Dialogue: 0,0:07:03.39,0:07:06.06,Default,,0000,0000,0000,,So once we decide the\Nforce is and the mass,
Dialogue: 0,0:07:06.06,0:07:08.37,Default,,0000,0000,0000,,then the acceleration gets fixed.
Dialogue: 0,0:07:08.37,0:07:12.39,Default,,0000,0000,0000,,But anyways, what will happen\Nif the net force is zero?
Dialogue: 0,0:07:12.39,0:07:15.06,Default,,0000,0000,0000,,What if we plug in over here zero?
Dialogue: 0,0:07:15.06,0:07:18.09,Default,,0000,0000,0000,,Well, then the acceleration\Nalso goes to zero.
Dialogue: 0,0:07:18.09,0:07:19.15,Default,,0000,0000,0000,,What does this mean?
Dialogue: 0,0:07:19.15,0:07:22.26,Default,,0000,0000,0000,,Well, this means we have all\Nthe balance forces acting
Dialogue: 0,0:07:22.26,0:07:23.09,Default,,0000,0000,0000,,on an object.
Dialogue: 0,0:07:23.09,0:07:25.17,Default,,0000,0000,0000,,And if the acceleration is zero, it means
Dialogue: 0,0:07:25.17,0:07:27.27,Default,,0000,0000,0000,,that the velocity stays a constant.
Dialogue: 0,0:07:27.27,0:07:30.57,Default,,0000,0000,0000,,In other words, this\Nis Newton's first law,
Dialogue: 0,0:07:30.57,0:07:32.73,Default,,0000,0000,0000,,which says an object\Ncontinues to stay at rest
Dialogue: 0,0:07:32.73,0:07:34.50,Default,,0000,0000,0000,,or in uniform motion.
Dialogue: 0,0:07:34.50,0:07:36.33,Default,,0000,0000,0000,,That is zero acceleration, right?
Dialogue: 0,0:07:36.33,0:07:39.15,Default,,0000,0000,0000,,When there are no unbalanced\Nforces acting on it.
Dialogue: 0,0:07:39.15,0:07:42.63,Default,,0000,0000,0000,,So notice, Newton's first\Nlaw is just a special case
Dialogue: 0,0:07:42.63,0:07:44.91,Default,,0000,0000,0000,,of Newton's second law,
Dialogue: 0,0:07:44.91,0:07:48.90,Default,,0000,0000,0000,,which means this equation is\Nencompassing both the second
Dialogue: 0,0:07:48.90,0:07:50.55,Default,,0000,0000,0000,,and the first law as well.
Dialogue: 0,0:07:50.55,0:07:53.40,Default,,0000,0000,0000,,And finally, speaking\Nabout Newton's first law,
Dialogue: 0,0:07:53.40,0:07:56.10,Default,,0000,0000,0000,,what we also noticed over\Nhere is bigger the mass,
Dialogue: 0,0:07:56.10,0:07:57.45,Default,,0000,0000,0000,,smaller the acceleration.
Dialogue: 0,0:07:57.45,0:08:01.26,Default,,0000,0000,0000,,In other words, if the mass\Nis bigger, it is harder
Dialogue: 0,0:08:01.26,0:08:02.40,Default,,0000,0000,0000,,to change its velocity.
Dialogue: 0,0:08:02.40,0:08:05.61,Default,,0000,0000,0000,,It's much harder to do\Nthat, which means objects
Dialogue: 0,0:08:05.61,0:08:08.94,Default,,0000,0000,0000,,that have more mass have more inertia.
Dialogue: 0,0:08:08.94,0:08:10.84,Default,,0000,0000,0000,,That's something again we\Nlearned in Newton's first law.
Dialogue: 0,0:08:10.84,0:08:12.63,Default,,0000,0000,0000,,Inertia is the property due
Dialogue: 0,0:08:12.63,0:08:15.57,Default,,0000,0000,0000,,to which objects continue to stay at rest
Dialogue: 0,0:08:15.57,0:08:18.03,Default,,0000,0000,0000,,or continue to stay in\Nuniform motion, isn't it?
Dialogue: 0,0:08:18.03,0:08:19.01,Default,,0000,0000,0000,,It can fight acceleration,
Dialogue: 0,0:08:19.01,0:08:21.54,Default,,0000,0000,0000,,and we can now see what\Ninertia depends on.
Dialogue: 0,0:08:21.54,0:08:24.00,Default,,0000,0000,0000,,Inertia is the mass.
Dialogue: 0,0:08:24.00,0:08:27.27,Default,,0000,0000,0000,,More the mass of an\Nobject, more than inertia,
Dialogue: 0,0:08:27.27,0:08:29.97,Default,,0000,0000,0000,,harder it is to accelerate.
Dialogue: 0,0:08:29.97,0:08:32.02,Default,,0000,0000,0000,,Newton's second law could arguably be
Dialogue: 0,0:08:32.02,0:08:35.64,Default,,0000,0000,0000,,the most important equation\Nof all of classical physics.
Dialogue: 0,0:08:35.64,0:08:37.89,Default,,0000,0000,0000,,I say classical physics\Nbecause we now know
Dialogue: 0,0:08:37.89,0:08:40.98,Default,,0000,0000,0000,,that if objects are moving\Nvery close to speed of light,
Dialogue: 0,0:08:40.98,0:08:42.66,Default,,0000,0000,0000,,then this breaks down, it doesn't work.
Dialogue: 0,0:08:42.66,0:08:43.50,Default,,0000,0000,0000,,Now we'll have to resort
Dialogue: 0,0:08:43.50,0:08:45.42,Default,,0000,0000,0000,,to Einstein's theory of relativity.
Dialogue: 0,0:08:45.42,0:08:46.26,Default,,0000,0000,0000,,On the other extreme,
Dialogue: 0,0:08:46.26,0:08:48.66,Default,,0000,0000,0000,,if we consider extremely tiny particles,
Dialogue: 0,0:08:48.66,0:08:51.24,Default,,0000,0000,0000,,like subatomic particles like electrons,
Dialogue: 0,0:08:51.24,0:08:53.52,Default,,0000,0000,0000,,protons, and neutrons,\Nwell, even over there,
Dialogue: 0,0:08:53.52,0:08:55.65,Default,,0000,0000,0000,,turns out Newton's laws don't work.
Dialogue: 0,0:08:55.65,0:08:57.48,Default,,0000,0000,0000,,So even over there, it breaks down.
Dialogue: 0,0:08:57.48,0:08:59.46,Default,,0000,0000,0000,,But as long as you don't\Ngo to such extremes,
Dialogue: 0,0:08:59.46,0:09:01.05,Default,,0000,0000,0000,,this equation will work for us.
Dialogue: 0,0:09:01.05,0:09:02.52,Default,,0000,0000,0000,,So now let's see if we can apply this
Dialogue: 0,0:09:02.52,0:09:04.11,Default,,0000,0000,0000,,to our original question.
Dialogue: 0,0:09:04.11,0:09:06.06,Default,,0000,0000,0000,,When she just started moving the dumbbell,
Dialogue: 0,0:09:06.06,0:09:08.25,Default,,0000,0000,0000,,dumbbell's velocity was increasing.
Dialogue: 0,0:09:08.25,0:09:09.93,Default,,0000,0000,0000,,After that, let's say\Nthere was a small phase
Dialogue: 0,0:09:09.93,0:09:11.85,Default,,0000,0000,0000,,during which the velocity was constant,
Dialogue: 0,0:09:11.85,0:09:14.49,Default,,0000,0000,0000,,and finally, when the\Ndumbbell is about to stop,
Dialogue: 0,0:09:14.49,0:09:16.23,Default,,0000,0000,0000,,its velocity is decreasing.
Dialogue: 0,0:09:16.23,0:09:18.63,Default,,0000,0000,0000,,So now the question is how\Ndo we figure out what happens
Dialogue: 0,0:09:18.63,0:09:20.76,Default,,0000,0000,0000,,to the force that she's\Nputting on the dumbbell?
Dialogue: 0,0:09:20.76,0:09:22.86,Default,,0000,0000,0000,,Well, let's apply Newton's second law.
Dialogue: 0,0:09:22.86,0:09:25.74,Default,,0000,0000,0000,,For that, let's first think\Nabout the acceleration.
Dialogue: 0,0:09:25.74,0:09:28.71,Default,,0000,0000,0000,,Well, over here, we are dealing\Nwith increasing velocity.
Dialogue: 0,0:09:28.71,0:09:31.35,Default,,0000,0000,0000,,Therefore, the acceleration is upwards
Dialogue: 0,0:09:31.35,0:09:33.51,Default,,0000,0000,0000,,in the same direction as it's moving.
Dialogue: 0,0:09:33.51,0:09:35.28,Default,,0000,0000,0000,,Then we have a constant velocity,
Dialogue: 0,0:09:35.28,0:09:38.16,Default,,0000,0000,0000,,which means the acceleration is zero.
Dialogue: 0,0:09:38.16,0:09:40.53,Default,,0000,0000,0000,,Finally, we have a decreasing velocity,
Dialogue: 0,0:09:40.53,0:09:43.35,Default,,0000,0000,0000,,since the dumbbell is still\Ngoing up, decreasing velocity,
Dialogue: 0,0:09:43.35,0:09:44.87,Default,,0000,0000,0000,,which means acceleration must be down
Dialogue: 0,0:09:44.87,0:09:47.19,Default,,0000,0000,0000,,in the opposite direction.
Dialogue: 0,0:09:47.19,0:09:49.50,Default,,0000,0000,0000,,Now, because we know the\Ndirection of the acceleration,
Dialogue: 0,0:09:49.50,0:09:51.21,Default,,0000,0000,0000,,we can figure out the\Ndirection of the net force.
Dialogue: 0,0:09:51.21,0:09:54.36,Default,,0000,0000,0000,,It has to be exactly\Nthe same, upwards here,
Dialogue: 0,0:09:54.36,0:09:56.37,Default,,0000,0000,0000,,zero here, downwards here.
Dialogue: 0,0:09:56.37,0:09:57.99,Default,,0000,0000,0000,,We are applying Newton's second law,
Dialogue: 0,0:09:57.99,0:10:00.21,Default,,0000,0000,0000,,the direction part over here.
Dialogue: 0,0:10:00.21,0:10:03.90,Default,,0000,0000,0000,,Now, finally, this is the\Ndirection of the net force.
Dialogue: 0,0:10:03.90,0:10:06.33,Default,,0000,0000,0000,,We want to know what happens to the force
Dialogue: 0,0:10:06.33,0:10:07.68,Default,,0000,0000,0000,,that we are putting on the dumbbell
Dialogue: 0,0:10:07.68,0:10:09.57,Default,,0000,0000,0000,,or she's putting on the dumbbell actually.
Dialogue: 0,0:10:09.57,0:10:10.53,Default,,0000,0000,0000,,How do we do that?
Dialogue: 0,0:10:10.53,0:10:13.44,Default,,0000,0000,0000,,Well, let's look at all the\Nforces acting on the dumbbell.
Dialogue: 0,0:10:13.44,0:10:16.11,Default,,0000,0000,0000,,Well, we know that there's\Ngravitational force acting
Dialogue: 0,0:10:16.11,0:10:17.64,Default,,0000,0000,0000,,on the dumbbell all the time.
Dialogue: 0,0:10:17.64,0:10:18.89,Default,,0000,0000,0000,,That force is a constant,
Dialogue: 0,0:10:19.74,0:10:22.35,Default,,0000,0000,0000,,and therefore, our force is\Nin the opposite direction
Dialogue: 0,0:10:22.35,0:10:23.73,Default,,0000,0000,0000,,of the gravitational force.
Dialogue: 0,0:10:23.73,0:10:26.76,Default,,0000,0000,0000,,Now, in this case, when she's\Njust lifting the dumbbell,
Dialogue: 0,0:10:26.76,0:10:30.21,Default,,0000,0000,0000,,if the net force needs to be upwards,
Dialogue: 0,0:10:30.21,0:10:32.79,Default,,0000,0000,0000,,that means her force must be larger
Dialogue: 0,0:10:32.79,0:10:33.81,Default,,0000,0000,0000,,than the gravitational force.
Dialogue: 0,0:10:33.81,0:10:36.00,Default,,0000,0000,0000,,Only then her force will win out,
Dialogue: 0,0:10:36.00,0:10:38.43,Default,,0000,0000,0000,,giving a net upward force, right?
Dialogue: 0,0:10:38.43,0:10:40.44,Default,,0000,0000,0000,,Okay, what about over here?
Dialogue: 0,0:10:40.44,0:10:42.60,Default,,0000,0000,0000,,We want the net force\Nto be zero over here.
Dialogue: 0,0:10:42.60,0:10:44.04,Default,,0000,0000,0000,,How can that happen?
Dialogue: 0,0:10:44.04,0:10:46.89,Default,,0000,0000,0000,,Ah, her force has to be exactly the same
Dialogue: 0,0:10:46.89,0:10:48.33,Default,,0000,0000,0000,,as gravitational force
Dialogue: 0,0:10:48.33,0:10:52.47,Default,,0000,0000,0000,,because only then the forces get balanced.
Dialogue: 0,0:10:52.47,0:10:54.06,Default,,0000,0000,0000,,Finally, what happens over here?
Dialogue: 0,0:10:54.06,0:10:55.65,Default,,0000,0000,0000,,Well, we want the net\Nforce to be downwards,
Dialogue: 0,0:10:55.65,0:10:57.57,Default,,0000,0000,0000,,which means we want gravity to win.
Dialogue: 0,0:10:57.57,0:11:00.81,Default,,0000,0000,0000,,That means her force must be smaller
Dialogue: 0,0:11:00.81,0:11:02.22,Default,,0000,0000,0000,,than the gravitational force.
Dialogue: 0,0:11:02.22,0:11:04.56,Default,,0000,0000,0000,,Look, even though we did\Nsimplify it a little bit,
Dialogue: 0,0:11:04.56,0:11:05.79,Default,,0000,0000,0000,,I mean, I'm not really sure
Dialogue: 0,0:11:05.79,0:11:07.88,Default,,0000,0000,0000,,that her dumbbell was moving\Nat a constant velocity,
Dialogue: 0,0:11:07.88,0:11:11.97,Default,,0000,0000,0000,,but once we simplified a little\Nbit, we were able to analyze
Dialogue: 0,0:11:11.97,0:11:13.47,Default,,0000,0000,0000,,what happened to the force
Dialogue: 0,0:11:13.47,0:11:14.73,Default,,0000,0000,0000,,that she was putting on the dumbbell.
Dialogue: 0,0:11:14.73,0:11:18.03,Default,,0000,0000,0000,,It went on decreasing as\Nthe dumbbell moved upwards.
Dialogue: 0,0:11:18.03,0:11:19.14,Default,,0000,0000,0000,,Isn't that incredible
Dialogue: 0,0:11:19.14,0:11:21.30,Default,,0000,0000,0000,,how we used Newton's second law do that?
Dialogue: 0,0:11:21.30,0:11:22.29,Default,,0000,0000,0000,,Amazing, isn't it?