WEBVTT
00:00:01.380 --> 00:00:04.970
In this video, we’re going to talk about the conservation of energy.
00:00:05.230 --> 00:00:08.920
We’ll learn what this principle says, what its limits are.
00:00:08.970 --> 00:00:12.710
And we’ll get some practical experience working with energy conservation.
00:00:13.230 --> 00:00:17.810
To start out, imagine that you and a friend are busy shooting a music video.
00:00:18.210 --> 00:00:29.800
Standing on a single-story height by a driveway where a car is parked, you see it would make a great shot if you jumped off this height onto the roof of the car and then from there onto the driveway.
00:00:30.240 --> 00:00:38.230
The only possible issue you see, as this is a low-budget video production, is that you might damage the car and incur expenses.
00:00:38.690 --> 00:00:45.080
To better understand whether getting this shot is a good idea, we’ll want to know more about the conservation of energy.
00:00:45.760 --> 00:00:54.910
The conservation of energy means that, within an isolated system, the total energy at any one time equals the total energy at any other time.
00:00:55.400 --> 00:01:01.040
Written as an equation, we can say that the initial system energy is equal to the final system energy.
00:01:01.670 --> 00:01:07.410
This principle is simple to remember, as long as we keep in mind the limits under which it applies.
00:01:07.840 --> 00:01:15.400
The main thing to keep in mind is that energy conservation requires an isolated system, one where no energy enters or leaves.
00:01:15.880 --> 00:01:19.500
This idea of a system can best be explained through an example.
00:01:19.990 --> 00:01:24.430
Let’s say you hold a ball in your hand and drop it from some height above ground.
00:01:25.040 --> 00:01:30.930
As we normally think about this scenario, in our system, we might include the ball, your hand, and the earth.
00:01:31.550 --> 00:01:41.150
And in that case, we could show that the gravitational potential energy of the ball when it’s in your hand is equal to the kinetic energy of the ball just as it hits the ground.
00:01:41.570 --> 00:01:45.670
In other words, we could show that the energy of the ball is conserved.
00:01:46.230 --> 00:01:48.890
But what if our system was not isolated?
00:01:49.280 --> 00:01:56.610
Imagine that there was a vertically downward moving wind that, as soon as the ball was released, helped push it towards the ground.
00:01:57.090 --> 00:02:02.540
In that case, the ball would speed up more than its initial potential energy will lead us to think.
00:02:03.100 --> 00:02:07.920
When it hit the ground, its kinetic energy would exceed its initial potential energy.
00:02:08.090 --> 00:02:10.310
It would seem as though energy is not conserved.
00:02:10.920 --> 00:02:15.160
But that’s only because we violated a condition of conservation of energy.
00:02:15.420 --> 00:02:19.780
Our system, in the case of the wind coming in, is not an isolated system.
00:02:19.810 --> 00:02:21.060
We’ve added energy.
00:02:21.700 --> 00:02:27.610
When we look at the expression for energy conservation, we realize this covers all kinds of energy.
00:02:27.990 --> 00:02:35.210
Broadly speaking, energy comes in two main types: kinetic or energy of motion and potential energy.
00:02:36.030 --> 00:02:41.100
We can expand our energy conservation expression to include these two main types.
00:02:41.360 --> 00:02:49.560
So we can say that the initial kinetic and potential energy of a system together is equal to the final kinetic and potential energy of that system.
00:02:50.080 --> 00:02:55.560
And here initial and final are any two times in the evolution of the system.
00:02:56.150 --> 00:03:00.990
Let’s get some practice applying this conservation principle through a few examples.
00:03:01.970 --> 00:03:04.060
A wagon sits at the top of a hill.
00:03:04.280 --> 00:03:12.350
The wagon is given a push that increases its speed negligibly but is just sufficient to set the wagon in motion down a straight slope.
00:03:12.770 --> 00:03:21.890
The wagon rolls 53.9 meters down the slope, which is inclined 16.5 degrees below the horizontal, and reaches the bottom of the hill.
00:03:22.370 --> 00:03:27.860
If friction is negligible, what speed is the wagon moving when it reaches the bottom of the hill?
00:03:28.490 --> 00:03:33.110
We can call the speed of the wagon when it reaches the bottom of the hill 𝑣 sub 𝑓.
00:03:33.300 --> 00:03:37.200
And we’ll start on our solution by drawing a diagram of the situation.
00:03:37.910 --> 00:03:43.980
A wagon sits perched at the top of a long flat hill whose length we’ve called 𝐿.
00:03:44.020 --> 00:03:46.070
It’s 53.9 meters.
00:03:46.510 --> 00:03:53.320
The hill is inclined at an angle below the horizontal — we’ve called 𝜃 — given as 16.5 degrees.
00:03:53.750 --> 00:03:59.710
We’re told the wagon is given a very slight push to set it in motion and then rolls down the hill.
00:03:59.710 --> 00:04:04.830
And when it reaches the bottom of the hill, we want to solve for its speed, 𝑣 sub 𝑓.
00:04:05.460 --> 00:04:09.470
In this scenario, our system consists of the wagon and the hill.
00:04:09.710 --> 00:04:15.790
And since energy is neither added to nor taken away from this system, we can say that it’s conserved.
00:04:16.370 --> 00:04:26.090
Applying the general statement of energy conservation to this scenario, we’ll use our freedom to decide when the initial moment and when the final moments are.
00:04:26.600 --> 00:04:34.440
Choosing the initial moment to be just when the cart is pushed at the top of the hill, and the final moment when it just reaches the bottom of the hill.
00:04:35.010 --> 00:04:44.250
We can expand this conservation expression to say that the kinetic energy plus potential energy initially is equal to the kinetic energy plus potential energy finally.
00:04:44.940 --> 00:04:47.870
At the outset, the wagon is not in motion.
00:04:47.900 --> 00:04:50.370
So its initial kinetic energy is zero.
00:04:50.860 --> 00:05:01.640
And if we choose the height at the bottom of the hill to be our reference level of zero, we can say that the final potential energy of the cart is also zero, since it’s at zero height.
00:05:02.070 --> 00:05:06.560
So the wagon’s initial potential energy is equal to its final kinetic energy.
00:05:06.970 --> 00:05:10.540
The wagon’s potential energy is entirely gravitational.
00:05:10.740 --> 00:05:15.870
And we recall that gravitational potential energy is equal to 𝑚 times 𝑔 times ℎ.
00:05:16.250 --> 00:05:23.460
Applying that relationship, we write 𝑚 times 𝑔 times the initial height of the wagon equals KE sub 𝑓.
00:05:24.000 --> 00:05:29.310
We also recall that an object’s kinetic energy equals half its mass times its speed squared.
00:05:29.890 --> 00:05:33.260
We can write this in for our KE sub 𝑓 expression.
00:05:33.680 --> 00:05:37.830
Looking at this equality, we see that the mass of the wagon appears on both sides.
00:05:37.860 --> 00:05:38.860
So it cancels out.
00:05:39.380 --> 00:05:42.060
We want to solve for the final speed of the wagon.
00:05:42.090 --> 00:05:49.890
So we rearrange this equation and see the wagon’s final speed is equal to the square root of two times 𝑔 times ℎ sub 𝑖.
00:05:50.230 --> 00:05:55.780
The acceleration due to gravity, 𝑔, we take to be exactly 9.8 meters per second squared.
00:05:56.300 --> 00:06:04.580
Looking at our diagram, we see that the height of the hill, ℎ sub 𝑖, relates to the hill length, 𝐿, and the angle of inclination, 𝜃.
00:06:05.120 --> 00:06:09.580
In particular, we can write that ℎ sub 𝑖 is equal to 𝐿 times the sin of 𝜃.
00:06:09.920 --> 00:06:15.280
So we substitute 𝐿 sin 𝜃 in for ℎ sub 𝑖 in our expression for 𝑣 sub 𝑓.
00:06:15.650 --> 00:06:21.790
Looking at this expression, we’re given the length 𝐿, the angle 𝜃, and 𝑔 is a known constant.
00:06:21.820 --> 00:06:24.280
So we’re ready to plug in and solve for 𝑣 sub 𝑓.
00:06:24.790 --> 00:06:32.490
Entering these values on our calculator, we find that, to three significant figures, 𝑣 sub 𝑓 is 17.3 meters per second.
00:06:32.990 --> 00:06:36.320
That’s the speed of the wagon when it reaches the bottom of the hill.
00:06:36.940 --> 00:06:42.710
Now let’s look at an example where, in our initial condition, we have both potential and kinetic energy.
00:06:43.330 --> 00:06:47.990
A hiker being chased by a bear runs to the edge of a cliff and jumps off.
00:06:48.350 --> 00:06:53.420
The hiker is running horizontally at 4.5 meters per second when he reaches the cliff edge.
00:06:53.860 --> 00:06:57.540
The hiker’s speed when he hits the ground is 16 meters per second.
00:06:57.970 --> 00:07:00.880
What vertical downward distance did the hiker fall?
00:07:01.650 --> 00:07:08.680
If the hiker stepped forward off the cliff edge with negligible horizontal velocity, what speed would they hit the ground at?
00:07:09.510 --> 00:07:13.010
So we want to solve for the distance the hiker falls.
00:07:13.160 --> 00:07:20.190
And we also wanna solve for the speed the hiker would have if the hiker left the cliff edge with negligible horizontal velocity.
00:07:20.680 --> 00:07:22.630
We can call this distance ℎ.
00:07:22.630 --> 00:07:26.080
And we can label the speed they hit the ground at 𝑣 sub 𝑔.
00:07:26.450 --> 00:07:29.230
We’ll start by drawing a diagram of the situation.
00:07:29.550 --> 00:07:39.080
In the first scenario, our hiker is running away from the bear at a speed of 4.5 meters per second in the horizontal direction, before leaping off the edge of the cliff.
00:07:40.040 --> 00:07:46.330
We know that, under these conditions, the hiker’s overall speed when he hits the ground is 16 meters per second.
00:07:46.680 --> 00:07:50.470
And we want to use that information to solve for the height of the cliff, ℎ.
00:07:50.930 --> 00:08:02.150
If we approach this from an energy perspective, we can say that the hiker’s energy, kinetic plus potential, just as he leaves the edge of the cliff is equal to his energy just as he lands.
00:08:02.410 --> 00:08:04.970
Those are our initial and final moments.
00:08:05.370 --> 00:08:08.950
In this system of the hiker and the cliff, energy is conserved.
00:08:09.120 --> 00:08:17.090
So we can write that 𝐸 sub 𝑖 is equal to 𝐸 sub 𝑓 and expand that expression to include kinetic and potential energies.
00:08:17.670 --> 00:08:26.230
If we set our zero value for height at the base of the cliff, that means that when the hiker lands, he has no potential energy due to gravity.
00:08:26.490 --> 00:08:28.530
So that term goes to zero.
00:08:29.120 --> 00:08:39.330
It turns out that’s the only term we can eliminate because, initially, as the hiker leaves the cliff’s edge, he has both gravitational potential energy and kinetic energy due to his motion.
00:08:39.810 --> 00:08:52.000
Recalling the equation forms of gravitational potential energy, 𝑚 times 𝑔 times ℎ, and kinetic energy, one-half 𝑚𝑣 squared, we can expand our energy balance equation.
00:08:52.000 --> 00:08:58.730
So it reads one-half the hiker’s mass times 𝑣 sub 𝑥 squared, the horizontal speed as the hiker leaves the cliff.
00:08:59.080 --> 00:09:06.190
Plus the hiker’s mass times 𝑔 times the height of the cliff is equal to one-half the hiker’s mass times 𝑣 squared.
00:09:06.430 --> 00:09:09.480
Where 𝑣 is the hiker’s overall speed when he lands.
00:09:09.930 --> 00:09:12.810
We see that the mass of the hiker appears in every term.
00:09:12.810 --> 00:09:14.110
So we can cancel it out.
00:09:14.590 --> 00:09:19.740
And since we want to solve for the height of the cliff, ℎ, we rearrange this equation to do so.
00:09:20.310 --> 00:09:30.730
And we find that ℎ is equal to 𝑣 squared minus 𝑣𝑥 squared all over two 𝑔, where we treat the acceleration due to gravity as exactly 9.8 meters per second squared.
00:09:31.220 --> 00:09:35.740
Since we know 𝑣 and 𝑣 sub 𝑥, we’re ready to plug in and solve for ℎ.
00:09:36.240 --> 00:09:43.330
When we do and enter these terms on our calculator, we find that, to two significant figures, ℎ is 12 meters.
00:09:43.710 --> 00:09:46.830
That’s the height of the cliff from which the hiker jumps.
00:09:47.390 --> 00:09:57.050
Next, we picture a slightly different scenario where, in this one, instead of the hiker running off the cliff’s edge at speed, the hiker jumps off from rest.
00:09:57.590 --> 00:10:01.570
We want to solve for the speed of the hiker when he hits the ground, 𝑣 sub 𝑔.
00:10:01.770 --> 00:10:05.200
And once again, our initial point will be when the hiker jumps.
00:10:05.200 --> 00:10:07.550
And the final point will be when the hiker lands.
00:10:07.950 --> 00:10:11.400
Just like before, in this system, energy is conserved.
00:10:11.730 --> 00:10:15.830
But unlike before, the initial kinetic energy of our hiker is zero.
00:10:16.300 --> 00:10:23.360
In this case, our final potential energy will again be zero since we let our height zero reference point be at the ground level.
00:10:23.850 --> 00:10:32.460
Our equation, therefore, reduces to PE sub 𝑖 equals KE sub 𝑓, where the potential energy is entirely due to gravitation.
00:10:32.790 --> 00:10:43.040
Referring once again to our gravitational potential energy and kinetic energy expressions, we write that 𝑚𝑔ℎ is equal to one-half 𝑚𝑣 sub 𝑔 squared.
00:10:43.510 --> 00:10:46.770
Once again, the mass of the hiker cancels out from both sides.
00:10:47.080 --> 00:10:55.600
And rearranging to solve for 𝑣 sub 𝑔, we find it’s equal to the square root of two times 𝑔 times ℎ, the height of the cliff.
00:10:56.070 --> 00:10:59.860
We know 𝑔 and we’ve solved for ℎ in the earlier part of the problem.
00:11:00.020 --> 00:11:02.470
So we’re ready to plug in and solve for 𝑣 sub 𝑔.
00:11:02.950 --> 00:11:09.160
When we enter these terms on our calculator, to two significant figures, it’s 15 meters per second.
00:11:09.610 --> 00:11:13.550
That’s the hiker’s speed when he reaches the ground after jumping from rest.
00:11:14.130 --> 00:11:17.430
Let’s summarize what we’ve learned about the conservation of energy.
00:11:18.610 --> 00:11:26.470
We’ve seen that energy conservation means that, within an isolated system, energy is neither created nor destroyed.
00:11:26.980 --> 00:11:33.450
Written as an equation, we’ve said that the initial energy in a closed system is equal to the final energy in that system.
00:11:34.020 --> 00:11:40.990
We’ve also seen that system energy can consist of both kinetic or moving energy as well as potential energy.
00:11:41.470 --> 00:11:51.700
This led us to expand our expression for energy conservation to say that initial kinetic plus potential energy is equal to final kinetic plus potential energy.
00:11:52.380 --> 00:11:58.600
And lastly, we’ve seen that an isolated system is one that neither gains nor loses energy.
00:11:59.280 --> 00:12:02.500
It’s within this sort of system that energy is conserved.