WEBVTT physics/ap-physics-c-mechanics/fullerton
00:00:00.000 --> 00:00:03.200
Hello, everyone, and welcome back to www.educator.com.
00:00:03.200 --> 00:00:11.100
I'm Dan Fullerton, and in this lesson we are going to talk about conservation of energy, one of the most useful concepts in the entire course.
00:00:11.100 --> 00:00:17.000
Our objectives include stating and applying the relation between work and mechanical energy.
00:00:17.000 --> 00:00:21.900
Analyzing situations in which an object's mechanical energy is changed by external forces.
00:00:21.900 --> 00:00:26.100
Applying conservation of energy and analyzing the motion of objects.
00:00:26.100 --> 00:00:31.800
Solving problems that call for applications of both conservation of energy and Newton’s laws of motion.
00:00:31.800 --> 00:00:36.200
Let us start by talking about conservation of mechanical energy.
00:00:36.200 --> 00:00:40.700
That is considered a single conservative force doing work on a closed system.
00:00:40.700 --> 00:00:48.700
In that case, we know that the work done by that force is that change in its kinetic energy by the work energy theorem.
00:00:48.700 --> 00:00:55.900
We also know that the work done by the conservative force is the opposite of the change in its potential energy.
00:00:55.900 --> 00:01:01.000
If those two are equal, we could then set the change in kinetic energy must be equal to
00:01:01.000 --> 00:01:08.200
the opposite of the change + the change in potential energy.
00:01:08.200 --> 00:01:17.500
Or rearranging it slightly, change in kinetic energy + change in potential energy must equal 0 for some close system.
00:01:17.500 --> 00:01:24.500
If we expand this out a little bit or really saying is the initial energy must be equal to the final energy.
00:01:24.500 --> 00:01:39.100
Or kinetic initial + potential initial must equal kinetic final + the final potential energy and conservation of mechanical energy.
00:01:39.100 --> 00:01:42.100
What if you have non conservative forces?
00:01:42.100 --> 00:01:48.000
Non conservative forces change the total mechanical energy of a system but not the total energy of the system.
00:01:48.000 --> 00:01:54.300
The work done by the non conservative force is typically converted to some sort of internal or thermal energy.
00:01:54.300 --> 00:02:05.000
If our total energy is kinetic + potential + the work done by non conservative forces
00:02:05.000 --> 00:02:14.100
and the total mechanical energy we are going to define is just the kinetic + the potential.
00:02:14.100 --> 00:02:16.400
Let us look at some examples here.
00:02:16.400 --> 00:02:20.400
We have an object of mass M, it falls from some height h.
00:02:20.400 --> 00:02:26.500
Find its speed prior to impact using conservation of mechanical energy and resistance.
00:02:26.500 --> 00:02:29.300
You could do this with the kinematics you already know.
00:02:29.300 --> 00:02:33.700
I recommend after we do this problem here with energy, take a minute and try it with kinematics.
00:02:33.700 --> 00:02:37.000
Make sure you get the same answer.
00:02:37.000 --> 00:02:46.900
We have our object there at B, it is going to fall some distance to a new position.
00:02:46.900 --> 00:02:52.700
It has some kinetic + potential energy initial.
00:02:52.700 --> 00:02:59.000
There it has some final amounts of kinetic energy + potential energy.
00:02:59.000 --> 00:03:06.000
It changes its height by an amount H.
00:03:06.000 --> 00:03:13.300
Writing our formula for conservation mechanical energy because we are only dealing with conservative forces here or neglecting air resistance.
00:03:13.300 --> 00:03:22.700
The initial kinetic + the initial potential energy must equal the final kinetic + the final potential energy.
00:03:22.700 --> 00:03:32.100
If it starts at rest of the initial kinetic to 0 and the initial potential energy is MGH assuming more uniform gravitational field.
00:03:32.100 --> 00:03:37.600
Its final kinetic energy is ½ M × its final velocity².
00:03:37.600 --> 00:03:45.200
Its final potential energy is 0 because it sets final point what we are calling 0.
00:03:45.200 --> 00:03:57.500
We can simplify this and divide M both sides to say that gh=V² /2 or velocity² =2 gh.
00:03:57.500 --> 00:04:05.800
If we want just the speed that is going to be √2gh.
00:04:05.800 --> 00:04:18.100
Alright, moving on, a jet fighter with a mass of 20,000 kg goes through the sky in an altitude of 10,000m with the velocity of 250 m/s.
00:04:18.100 --> 00:04:34.000
We can find its total energy as the sum of its gravitational potential energy and its kinetic will be MGH + ½ Mv²
00:04:34.000 --> 00:04:53.700
or 20,000kg × 10 m/s² × 10,000m + ½ × its mass × squared of its velocity 250².
00:04:53.700 --> 00:05:03.600
Complies that its total energy is 2.63 × 10⁹ joules.
00:05:03.600 --> 00:05:14.400
That jet dives to an altitude of 2000m and drops in height 8000m, find the new velocity of the jet.
00:05:14.400 --> 00:05:36.400
Total energy is gravitational potential + kinetic which is MGH + ½ MV² which implies if we solve for V MV² must equal to × our total energy – MGH.
00:05:36.400 --> 00:05:45.300
Or V will be equal to 2 × our total energy - MGH all divided by mass.
00:05:45.300 --> 00:05:48.400
We have to take the square root of that.
00:05:48.400 --> 00:06:06.300
When I substitute in my values, that is going to be √2 × our total energy 2.63 × 10⁹ joules - our mass 20,000kg G 10 m/s².
00:06:06.300 --> 00:06:18.300
Our height now 2000m all divided by the mass 20,000kg.
00:06:18.300 --> 00:06:24.800
I come up with the speed of about 472 m/s.
00:06:24.800 --> 00:06:33.800
The jet trading in the altitude for speed keeping the same total mechanical energy.
00:06:33.800 --> 00:06:36.800
Let us take a look at our pendulum example.
00:06:36.800 --> 00:06:42.700
The pendulum comprises the light string of length L swings mass M back and forth.
00:06:42.700 --> 00:06:46.800
We have look at this previously but let us highlight a couple things here.
00:06:46.800 --> 00:06:49.500
There is the length of our pendulum L.
00:06:49.500 --> 00:06:58.600
If we went here, we already calculated this in our last lesson and we said that this must be L cos θ.
00:06:58.600 --> 00:07:08.700
This amount h= L - L cos θ or L 1 – cos θ.
00:07:08.700 --> 00:07:11.700
We have done that derivation before.
00:07:11.700 --> 00:07:18.700
As this mass swings back and forth in the pendulum, it is transferring kinetic and potential energy.
00:07:18.700 --> 00:07:23.200
Its highest point it has the maximum potential energy.
00:07:23.200 --> 00:07:29.500
Its lowest point, its potential energies at a minimum and it is all kinetic energy.
00:07:29.500 --> 00:07:33.600
And swings to the other side transfers that kinetic energy back to potential energy.
00:07:33.600 --> 00:07:38.700
For a split second that stops but it is at its highest point.
00:07:38.700 --> 00:07:54.300
We can write an equation here for the maximum potential energy is MGH which is going to be MG × our height L1 - cos θ.
00:07:54.300 --> 00:07:57.000
Which means assuming you are in a system we are not dealing
00:07:57.000 --> 00:08:02.100
with any non conservative forces when we get to this point it is all kinetic energy.
00:08:02.100 --> 00:08:14.800
Our maximum kinetic energy which is ½ Mv² has to be equal to MG L 1 - cos θ.
00:08:14.800 --> 00:08:21.300
We can solve for the velocity of our pendulum at that lowest point when we got the maximum kinetic energy.
00:08:21.300 --> 00:08:23.400
Let us take a second to do that.
00:08:23.400 --> 00:08:39.500
If ½ mv² equals MGL × 1 - cos θ and that implies that we can divide the M to both sides ½ V² must be equal to GL × 1 – cos θ.
00:08:39.500 --> 00:08:48.200
Which implies then the V² =2g L × 1 – cos θ or V is just going to be equal to
00:08:48.200 --> 00:08:51.400
and let us make sure we know that that is our maximum velocity.
00:08:51.400 --> 00:08:59.600
It is going to be equal to 2 gL 1 - cos θ square root.
00:08:59.600 --> 00:09:07.200
If they wanted to make a graph of this as we look at the displacement in the X vs. Energy in the Y.
00:09:07.200 --> 00:09:12.300
Our total energy must remain constant by law of conservation of energy.
00:09:12.300 --> 00:09:14.600
There is our total.
00:09:14.600 --> 00:09:20.100
We have our highest potential energies and where the maximum displacements.
00:09:20.100 --> 00:09:26.600
We will draw our little points in here and at our lowest point the middle point no displacement
00:09:26.600 --> 00:09:31.100
we have maximum kinetic energy and minimum 0 potential energy.
00:09:31.100 --> 00:09:34.200
We will put our kinetic energy the maximum here.
00:09:34.200 --> 00:09:38.900
The furthest displacements that is not moving so it is kinetic energy to 0 out here.
00:09:38.900 --> 00:09:44.200
Our kinetic energy graph might look something like this.
00:09:44.200 --> 00:09:46.900
As best as I can eyeball that in their.
00:09:46.900 --> 00:09:56.700
There would be our kinetic energy and our potential energy might look something like that.
00:09:56.700 --> 00:10:05.500
At any point, wherever we happened to be the total sum of the potential + the kinetic has to equal our total.
00:10:05.500 --> 00:10:11.500
We have that maintain that total because we are not dealing with any non conservative forces.
00:10:11.500 --> 00:10:13.900
Take a look at another example.
00:10:13.900 --> 00:10:18.000
The diagram below shows a cart possessing 16 joules of kinetic energy
00:10:18.000 --> 00:10:22.100
traveling on a frictionless horizontal surface to a horizontal spring.
00:10:22.100 --> 00:10:30.000
If the cart comes to rest after compressing the spring a distance of 1m what is the spring constant of the spring ?
00:10:30.000 --> 00:10:38.100
The initial kinetic energy must be equal to the final spring elastic potential energy
00:10:38.100 --> 00:10:44.100
because when the kinetic energy become 0 we put all of energy and compressing the spring.
00:10:44.100 --> 00:10:52.300
16 joules must equal to ½ KX² assuming it follows Hooke’s law.
00:10:52.300 --> 00:11:09.100
K must be 32 joules / X² which is going to be 32 joules / 1m² or 32 N/m.
00:11:09.100 --> 00:11:12.200
There is a spring constant.
00:11:12.200 --> 00:11:14.100
Lets do another example.
00:11:14.100 --> 00:11:20.900
A popup toy has a mass of 0.02kg and the spring constant of 150 N/m.
00:11:20.900 --> 00:11:25.700
A force is applied to the toy to compress the spring 0.05m.
00:11:25.700 --> 00:11:31.600
Calculate the potential energy stored in the compressed spring.
00:11:31.600 --> 00:11:34.800
Alright potential energy in the compressed spring.
00:11:34.800 --> 00:11:51.000
Make it follow Hooke’s law is ½ KX² which would be ½ × our spring constant 150 N/ m × our spring compression 0.05²
00:11:51.000 --> 00:11:59.700
which is going to be 0.1875 joules.
00:11:59.700 --> 00:12:02.200
Let us take this one step further.
00:12:02.200 --> 00:12:07.600
The toy was activated all the compressed springs potential energy is converted a gravitational potential energy.
00:12:07.600 --> 00:12:10.300
As it pops up it goes up into the air.
00:12:10.300 --> 00:12:15.400
Find the maximum height to which the toy is propelled.
00:12:15.400 --> 00:12:21.900
Our initial potential energy must equal our final potential energy.
00:12:21.900 --> 00:12:26.600
No kinetic in there because it at rest that both its initial and its final positions.
00:12:26.600 --> 00:12:38.600
Our initial potential energy the spring potential energy was 0.1875 joules that must be equal to its final gravitational potential energy mgh.
00:12:38.600 --> 00:13:05.700
Or the height is 0.1875 joules / MG which is 0.1875 joules / mass is 0.0kg × G 10 m / s² and I come up with the height of about 0.9375m.
00:13:05.700 --> 00:13:07.600
Taking a look at another one.
00:13:07.600 --> 00:13:14.100
A car initially travels 30 m/s and that slows uniformly as it skids to stop after the brakes are applied.
00:13:14.100 --> 00:13:24.300
Sketch a graph showing the relationship between kinetic energy of the car as it is being brought to a stop in the work done by friction and stopping the car.
00:13:24.300 --> 00:13:35.300
Let us make a couple axis here.
00:13:35.300 --> 00:13:45.000
We have kinetic energy on the Y and we have the work done by the force of friction on the X.
00:13:45.000 --> 00:13:50.900
Our initial velocity was 30 m/s, our final velocity is 0.
00:13:50.900 --> 00:13:57.500
Since it is doing this uniformly, it is pretty easy to see our starting point for kinetic energy is going to have to be at the maximum.
00:13:57.500 --> 00:14:02.700
It is going to have some final point 0.
00:14:02.700 --> 00:14:11.600
Because we are looking at the work done by friction on the x, all the work done by friction is what slowing it down.
00:14:11.600 --> 00:14:21.900
Therefore, our graph is just going to look like that.
00:14:21.900 --> 00:14:24.700
Let us do a block on the ramp problem.
00:14:24.700 --> 00:14:31.700
The 2 kg block sliding down a ramp from a height of 3m above the ground reaches the ground with the kinetic energy of 50 joules.
00:14:31.700 --> 00:14:37.200
Find the total work done by friction on the block as it slides down the ramp.
00:14:37.200 --> 00:14:43.800
Let us make ourselves a diagram here.
00:14:43.800 --> 00:14:53.800
Give ourselves a nice little ramp to play with.
00:14:53.800 --> 00:15:07.300
On the ramp, we have a 2kg block and the height of our ramp 3m.
00:15:07.300 --> 00:15:17.300
If you want a total work done by friction at the top here the potential energy due to gravity, there is no kinetic it is at rest.
00:15:17.300 --> 00:15:26.100
At the bottom, it has kinetic energy at the bottom + we have whatever work was done by friction.
00:15:26.100 --> 00:15:36.300
This implies that the work done by friction must be the gravitational potential energy at the top - the kinetic energy at the bottom
00:15:36.300 --> 00:15:47.500
which is going to MG h at the top - the kinetic energy at the bottom which is given as 50 joules.
00:15:47.500 --> 00:15:57.900
The work done by friction is its mass 2 kg × acceleration due to gravity 10 m/s² ×
00:15:57.900 --> 00:16:08.100
the height 3m -50 joules or 20 × 360 -50 is just going to 10 joules.
00:16:08.100 --> 00:16:15.000
10 joules must be the total work done by friction on the block as it slides down the ramp.
00:16:15.000 --> 00:16:17.100
Let us do a multipart problem here.
00:16:17.100 --> 00:16:23.500
Andy, the adventurous adventurer while running from evil bad guys in the Amazonian rain forest
00:16:23.500 --> 00:16:32.800
saving the world from impending doom of course, trips, falls, and slides down the frictionless mudslide of height 20m is depicted here.
00:16:32.800 --> 00:16:38.700
Once he reaches the bottom of a mudslide however he flies horizontally off a 50m cliff.
00:16:38.700 --> 00:16:43.100
How far from the base of the cliff Andy land?
00:16:43.100 --> 00:16:44.500
A multipart problem.
00:16:44.500 --> 00:16:46.000
Let us see.
00:16:46.000 --> 00:16:50.600
Here at the top he has all gravitational potential energy.
00:16:50.600 --> 00:16:56.500
If we then convert then do is kinetic energy here we can find the velocity with which he goes the side of a cliff
00:16:56.500 --> 00:17:03.900
and then this becomes a projectile motion problem for the object launch horizontally off the cliff with the height of 15m.
00:17:03.900 --> 00:17:09.600
We can break it up into two parts, the mudslide piece and the cliff piece.
00:17:09.600 --> 00:17:13.800
Let us do this where we have ourselves a little bit more room.
00:17:13.800 --> 00:17:28.300
As we look at gravitational potential energy at the top is MGH must be equal to ½ Mv² when we get down to this point ½ mv².
00:17:28.300 --> 00:17:47.500
Therefore, the velocity when we get down here is going to be √2gh which is going to be √2 × 10 m/s² × that height difference 20 or 20 m/s.
00:17:47.500 --> 00:17:53.500
He is going to go off horizontally with 20 m/s how far does he land there?
00:17:53.500 --> 00:17:56.500
That is a kinematics problem.
00:17:56.500 --> 00:17:57.900
Let us take a look.
00:17:57.900 --> 00:18:04.600
Horizontally, when we look at our kinematics problem we have some horizontal velocity 20 m/s
00:18:04.600 --> 00:18:08.200
that is not going to change for neglecting air resistance.
00:18:08.200 --> 00:18:13.500
We do not know how long he is in the air so we cannot figure out yet how far he travels.
00:18:13.500 --> 00:18:16.000
We will call that D.
00:18:16.000 --> 00:18:21.100
We have to analyze the vertical motion in order to find out how long is in the air.
00:18:21.100 --> 00:18:24.400
We will call down the positive y direction.
00:18:24.400 --> 00:18:27.600
The initial vertically 0.
00:18:27.600 --> 00:18:32.000
The final vertically we do not know and frankly do not care a lot.
00:18:32.000 --> 00:18:39.400
The change in Y is going to be 15m from our problem, as you look at just this window as he goes over the cliff.
00:18:39.400 --> 00:18:49.400
Our acceleration is 10 m/s² positive because we call down the positive y direction and we are looking for T.
00:18:49.400 --> 00:18:58.700
Finding a kinematic equation I probably look at δ Y=V initial T + ½ ay T².
00:18:58.700 --> 00:19:02.900
We have this nice little helper that V initial 0.
00:19:02.900 --> 00:19:18.300
Δ y is ½ at² or T is going to be 2 δ Y / a square root which is 2 × 15 m / 10 m s² square root.
00:19:18.300 --> 00:19:23.400
√30/10² is about 1.73s.
00:19:23.400 --> 00:19:29.400
The time we can now filling up here horizontally 1.73 s.
00:19:29.400 --> 00:19:36.300
To figure out how far it goes the δ X or I think we call that D
00:19:36.300 --> 00:19:53.500
is just going to be horizontal velocity × time or 20 m/s × 1.73s should give us about 34.6m.
00:19:53.500 --> 00:19:59.200
Distance here 34.6m.
00:19:59.200 --> 00:20:05.500
A multipart problem where we had to pull in conservation of energy.
00:20:05.500 --> 00:20:09.100
A rollercoaster problem.
00:20:09.100 --> 00:20:15.000
A rollercoaster car begins at height H above the ground and completes a loop along its path.
00:20:15.000 --> 00:20:23.800
In order for the car to remain on the track at the loop, what is the minimum value for h in terms of the radius of the loop capital r.
00:20:23.800 --> 00:20:32.300
I'm going to recognize here first that this is a kinetic initial + potential initial energy.
00:20:32.300 --> 00:20:33.800
Kinetics going to be 0.
00:20:33.800 --> 00:20:36.900
The potential energy there is going to be all gravitational.
00:20:36.900 --> 00:20:45.300
When we get to the highest point here, we have kinetic final + potential final because that is what we are worried about.
00:20:45.300 --> 00:20:49.600
This distance is r and that distances is r.
00:20:49.600 --> 00:20:55.300
2r is our diameter.
00:20:55.300 --> 00:21:06.000
As I look at this, our initial gravitational potential energy MGH must be equal to our kinetic and gravitational here.
00:21:06.000 --> 00:21:19.300
That is ½ M V final² + mgh here where our height now is going to be 2r the radius + a radius the diameter 2r above the ground.
00:21:19.300 --> 00:21:30.100
MG × 2r this implies then as we multiply through here let us get rid of our M we can divide all those out
00:21:30.100 --> 00:21:43.800
and I can also multiply that by 2 to say that the left hand side 2 gh is going to be equal to V final² + 4gr.
00:21:43.800 --> 00:21:57.100
Or solving for V final² that is just going to be 2g × (h- 2r).
00:21:57.100 --> 00:21:59.700
There is V final².
00:21:59.700 --> 00:22:06.000
Let us come look at the condition, we need to have with the cart at this highest point for it not to fall off the loop.
00:22:06.000 --> 00:22:11.700
Its centripetal acceleration has to be greater than the acceleration due to gravity otherwise it is going to fly off.
00:22:11.700 --> 00:22:23.500
We can set that condition by saying the centripetal acceleration V² / r has to be greater than or equal to G our acceleration due to gravity.
00:22:23.500 --> 00:22:35.400
Rearranging this, V² then must be greater than equal to gr but V² we just said was 2g × (h-2r).
00:22:35.400 --> 00:22:45.900
We could write then that 2g × (h-2r) has to be greater than or equal to gr.
00:22:45.900 --> 00:23:00.500
Which implies then h -2r must be greater than or equal to r/2 factoring out the g.
00:23:00.500 --> 00:23:07.700
Therefore, getting h all by itself h has to be greater than or equal to 5r / 2.
00:23:07.700 --> 00:23:14.900
As long as h is more than 2 ½ × r you are in great shape.
00:23:14.900 --> 00:23:17.400
This is assuming you are frictionless so I would not build it right there.
00:23:17.400 --> 00:23:20.800
You do not want to put some leeway in there if you are a roller-coaster designer.
00:23:20.800 --> 00:23:28.800
A little bit of a safety net in there because there is going to be some friction and some other issues that come into account.
00:23:28.800 --> 00:23:32.000
H is greater than or equal to 5 r/2.
00:23:32.000 --> 00:23:34.900
Let us do the problem with the bungee jumper.
00:23:34.900 --> 00:23:39.800
Alicia a 60 kg bungee jumper steps off the 40m high bridge.
00:23:39.800 --> 00:23:44.500
The bungee cord behaves like a spring with K =40 N / m.
00:23:44.500 --> 00:23:47.000
Assume there is no slack in the cord.
00:23:47.000 --> 00:23:57.200
We want to find the speed of Alicia at the height of 15 m above the ground when she is 30m above the ground and how close she gets to the ground.
00:23:57.200 --> 00:24:05.500
I'm going to make a quick diagram here of where we are going to know our speeds.
00:24:05.500 --> 00:24:17.000
If we are down to the ground here, we want to know the speed at a height of 15m above the ground let us call that A that is 15m above the ground.
00:24:17.000 --> 00:24:23.300
We want to know the speed of the height of 30m above the ground we will call that B.
00:24:23.300 --> 00:24:27.300
This is another 15m there.
00:24:27.300 --> 00:24:33.100
He is stepping off the 40m high bridge so there is our starting point.
00:24:33.100 --> 00:24:39.400
There is another 10m there and we want to know how close she gets to the ground down here as well.
00:24:39.400 --> 00:24:46.600
Let us start on part A, find the speed of the jumper the height of 15m above the ground or here at A.
00:24:46.600 --> 00:24:53.200
Give ourselves a little bit of room, we can take a look and say that the gravitational potential energy
00:24:53.200 --> 00:25:05.500
when she is at the top must be equal to a gravitational potential energy at point A + spring potential energy at point A + kinetic energy point A.
00:25:05.500 --> 00:25:09.300
Which implies after speed we want kinetic energy all by itself.
00:25:09.300 --> 00:25:23.000
Kinetic energy at point A is going to be the gravitational potential energy at the top – the gravitational potential energy at A - the spring potential energy at A.
00:25:23.000 --> 00:25:49.800
Which implies then, we will replace kinetic energy at A day with ½ M × v at a² so ½ mva² = Mg δ y -1/2 K × δ Y² Mg -1/2 K δ Y².
00:25:49.800 --> 00:25:56.500
Which implies then the VA² equals we will multiply by 2/ M.
00:25:56.500 --> 00:26:15.500
The right hand side we get 2g δ Y - K / M δ Y² which implies that va² = 2 × 9.8 m/s²
00:26:15.500 --> 00:26:21.400
because we do not want here running into the ground in your calculations.
00:26:21.400 --> 00:26:39.600
× our δ Y is 25m to get from the pop to point A - K / M 40/60 × δ Y 25² which is going to be equal to 73.3m²/s².
00:26:39.600 --> 00:26:48.500
Or the velocity at A must be 8.6 m/s.
00:26:48.500 --> 00:26:53.600
We found the first part the speed of Alicia at the height of 15m above the ground.
00:26:53.600 --> 00:27:00.900
Let us take a look and see what we can find in the next part of problem when she is 30 m above the ground.
00:27:00.900 --> 00:27:05.400
She has fallen 10m at this point.
00:27:05.400 --> 00:27:14.300
Gravitational potential energy at the top equals gravitational potential energy at B + the spring potential energy at B +
00:27:14.300 --> 00:27:22.300
the kinetic energy at B which implies then that the kinetic energy B equals the gravitational potential energy
00:27:22.300 --> 00:27:32.700
at B - the gravitational potential energy at B - the spring potential energy at B
00:27:32.700 --> 00:27:46.900
which implies again that ½ M V at B² = MG × change in position δ y -1/2 K δ Y².
00:27:46.900 --> 00:28:05.000
Or VB² = 2g δ y - K / M δ y² which implies then that vb²-= 2 × 9.8.
00:28:05.000 --> 00:28:18.300
Our δ y at 10m - K / M still 40/60 × δ y 10m² or 129m² /s².
00:28:18.300 --> 00:28:28.900
Taking the square root of that velocity at B must be 11.4 m/s.
00:28:28.900 --> 00:28:33.300
Finally, how close does she get to the ground?
00:28:33.300 --> 00:28:39.000
This occurs when the velocity and kinetic energy are equal to 0.
00:28:39.000 --> 00:28:51.900
Potential energy at the top equals a gravitational potential energy at the bottom + the spring potential energy at the bottom.
00:28:51.900 --> 00:29:03.100
Which implies that our change in gravitational potential energy must equal the stored potential energy and the spring.
00:29:03.100 --> 00:29:11.700
Or MG δ Y = ½ K δ Y².
00:29:11.700 --> 00:29:17.600
Which implies then, let us get rid of that two at about ½ by multiplying by two.
00:29:17.600 --> 00:29:33.700
We can also divide by that by δ Y 2Mg = K δ Y or δ Y = 2 Mg / K.
00:29:33.700 --> 00:29:48.100
Which is 2 × Alicia is 60kg g 9.8 m/s² /spring constant 14 N/m which is 29.4m.
00:29:48.100 --> 00:30:00.700
Be careful here, that means if δ Y =29.4m the jumper must be 40m - that from the ground that are lowest points.
00:30:00.700 --> 00:30:18.600
40 -29.4 is going to be equal to 10.6m from the ground.
00:30:18.600 --> 00:30:21.000
I think to the point where we can do a couple simple problem.
00:30:21.000 --> 00:30:27.700
Let us do a couple of old AP free response problems to finish up this unit here.
00:30:27.700 --> 00:30:33.500
Let us start by taking a look at the 2002 Mechanics free response number 3 problem.
00:30:33.500 --> 00:30:39.400
I will give you a minute, you can go download the problem at the address above or google search it.
00:30:39.400 --> 00:30:43.800
As we look at that one, I want you to take a minute, look here and see if you cannot solve it at your own
00:30:43.800 --> 00:30:49.100
as you pause the video, give it a minute or 2 then come back here and check your answers
00:30:49.100 --> 00:30:54.700
or if you get stuck, use this to help you get past that sticking point.
00:30:54.700 --> 00:30:59.000
As we look at 2002 free response 3.
00:30:59.000 --> 00:31:09.700
Question A, says it wants us to sketch the graph of the potential energy vs. X where it has given us the formula for potential energy.
00:31:09.700 --> 00:31:16.300
We will create a graph for these sorts of things by now.
00:31:16.300 --> 00:31:20.900
Of course, you will graph it very carefully plugging the points.
00:31:20.900 --> 00:31:24.500
I am going to try and give a rough idea of where we are at.
00:31:24.500 --> 00:31:33.600
I can plot points of we know it must be 2 at 0 point and when we have a value of X of 2 where the value of 1.
00:31:33.600 --> 00:31:35.100
We are somewhere here.
00:31:35.100 --> 00:31:47.800
Our graph is going to look something like that.
00:31:47.800 --> 00:31:50.300
You got that general shape.
00:31:50.300 --> 00:31:57.500
Now B says, determine the force associated with the potential energy function given above.
00:31:57.500 --> 00:32:00.400
We are given potential energy we want force.
00:32:00.400 --> 00:32:11.500
Force is - du DL which in this case is going to be - D / Dx of our equation 4/2 + X.
00:32:11.500 --> 00:32:14.300
We can pull the 4 out that is a constant.
00:32:14.300 --> 00:32:32.500
That force equals -4 derivative with respect to X of 1/2 + x which I can write is -4 derivative with respect to x of X + 2⁻¹
00:32:32.500 --> 00:32:47.900
or the force is going to be equal 24 × x + 2⁻² which is 4/ x + 2².
00:32:47.900 --> 00:32:51.300
There is part B.
00:32:51.300 --> 00:32:58.100
Taking a look here at C, suppose the object is released from the origin, find the speed at x = 2m.
00:32:58.100 --> 00:33:00.900
That sounds like a conservation of energy problem.
00:33:00.900 --> 00:33:08.400
We know its change in potential energy is its potential energy at 0 - its potential energy at 0.2 and
00:33:08.400 --> 00:33:15.800
all that has to be equal be turned into the change in energy be turned into its kinetic energy ½ mv².
00:33:15.800 --> 00:33:19.800
Which implies that U of 0 is 2, U of 2 is 1.
00:33:19.800 --> 00:33:28.000
2 -1 is going to be 1 joule = ½ f MV².
00:33:28.000 --> 00:33:35.900
We can solve that for V that means 2/ M is equal to V².
00:33:35.900 --> 00:33:43.800
V is going to be equal to V² is 2/m which is 2/0.5 or 4.
00:33:43.800 --> 00:33:51.100
V is √4 or 2m/s.
00:33:51.100 --> 00:33:53.000
That did not seem so bad.
00:33:53.000 --> 00:33:59.000
Let us move on and take a look at part D.
00:33:59.000 --> 00:34:02.300
In a lab you are given a glider of a mass ½ kg on the track,
00:34:02.300 --> 00:34:08.400
the gliders acted on by the force determined in B and your goals to determine the validity of your theoretical calculation.
00:34:08.400 --> 00:34:13.100
We can get to pick some equipment to help us do this.
00:34:13.100 --> 00:34:16.700
What we are going to do is pick the equipment and outline a procedure we will use.
00:34:16.700 --> 00:34:21.900
They are probably a bunch of different ways you can do this but what I'm thinking about
00:34:21.900 --> 00:34:28.000
is I would probably take one of those photo D timers at probably two of them.
00:34:28.000 --> 00:34:33.300
And set the photo gate a small distance apart right here that x =2m.
00:34:33.300 --> 00:34:38.300
If you measure the distance between the gates with the meter stick and you have obtained that time for the gliders travel
00:34:38.300 --> 00:34:46.600
between the gates you could then use V = D/T to determine its velocity and confirm your previous findings.
00:34:46.600 --> 00:34:52.800
You would probably need a photo gate timer.
00:34:52.800 --> 00:35:00.300
We need a meter stick, I think it will be is set for equipment if you follow that procedure.
00:35:00.300 --> 00:35:06.200
The trick is to make sure that you measure the distance between gates when you really close to that the 2m mark.
00:35:06.200 --> 00:35:09.700
Determine that the time and then you can get the velocity.
00:35:09.700 --> 00:35:14.200
Make sure you are in the ballpark to confirm your findings.
00:35:14.200 --> 00:35:15.600
That is D and E.
00:35:15.600 --> 00:35:22.500
In E, make sure you explain that in a complete sentence fairly clearly.
00:35:22.500 --> 00:35:27.200
Let us go on to our next free response question from 2007.
00:35:27.200 --> 00:35:33.100
Let us look at free response 3, a spring and glider question with the photo gate again.
00:35:33.100 --> 00:35:36.100
It looks familiar.
00:35:36.100 --> 00:35:40.500
The apparatus they show you here is used to study conservation mechanical energy.
00:35:40.500 --> 00:35:46.900
It looks like they have extended the spring and have the speed of a glider in the extension squared the speed and all of that.
00:35:46.900 --> 00:35:54.400
They say assuming no energy is lost, write the equation for conservation and mechanical energy that would apply here.
00:35:54.400 --> 00:36:07.500
We only have kinetic energy is going to be turned into spring potential energy which implies that ½ MV² is going to be equal to ½ KX².
00:36:07.500 --> 00:36:12.600
But it tells us in the problem that K is 40 N /m.
00:36:12.600 --> 00:36:24.300
We can write that ½ Mv² is equal to ½ × 40 X² or 20 x².
00:36:24.300 --> 00:36:29.300
There is an equation of conservation of energy that would apply to this problem.
00:36:29.300 --> 00:36:32.500
For part B, it looks like we are getting into some graphing again.
00:36:32.500 --> 00:36:38.200
Plot V² vs. X² label the axis, units, and scale.
00:36:38.200 --> 00:36:44.700
Plotting points should be easy points on the AP exam.
00:36:44.700 --> 00:36:54.300
We will give ourselves some axis and you have got to label the scale and things like that.
00:36:54.300 --> 00:37:07.000
As by I did this and have X² and m² on the x, I have v² and m² for s² on the Y.
00:37:07.000 --> 00:37:15.600
We have 0.005, 0.01, 0.015, and so one for the x.
00:37:15.600 --> 00:37:25.300
In the y, I did 1, 2, 3, 4, 5, 6 something like that.
00:37:25.300 --> 00:37:29.700
I am not going to the plot the points myself point by point but when we do this you plot the points and
00:37:29.700 --> 00:37:37.100
draw a best fit line should get something that looks fairly linear and like it goes through 00.
00:37:37.100 --> 00:37:39.200
My line looks something like that.
00:37:39.200 --> 00:37:40.600
We are not connecting points.
00:37:40.600 --> 00:37:43.600
We are drawing the best fit line.
00:37:43.600 --> 00:37:48.100
We plotted points labeled the axis including units and scale.
00:37:48.100 --> 00:37:57.400
And moving on the part C, draw the best fit line and use the best fit line to obtain the mass and the glider.
00:37:57.400 --> 00:38:03.800
For C, to determine the mass and the glider I am first going to recognize here that the slope is going to K/M.
00:38:03.800 --> 00:38:24.300
As I look at that, because as we do our plots here let see we have our equation ½ MV² = ½ 40x².
00:38:24.300 --> 00:38:36.200
For plotting V² vs x² we have ½ MV² = 20x².
00:38:36.200 --> 00:38:45.700
Multiply that by 2 mv² = 40x² or if V² = 40/M or k/Mx².
00:38:45.700 --> 00:38:52.000
It fits the form Y = MX where X is x² our y is V².
00:38:52.000 --> 00:38:55.600
Our slope must be K/M as 40/M.
00:38:55.600 --> 00:39:05.400
We find our slope and when I find the slope I came up with about slope is K/M so m is k/slope.
00:39:05.400 --> 00:39:10.900
My slope was right around 201/s².
00:39:10.900 --> 00:39:29.400
Our mass which is going to K/slope becomes 40 N/m / 201/s² or 0.2kg.
00:39:29.400 --> 00:39:32.100
That covers C1 and 2.
00:39:32.100 --> 00:39:38.200
It looks like we move on the part D.
00:39:38.200 --> 00:39:44.100
For part D, the truck is not tilted at an angle θ when the spring is on stretch the center of the glider is at height h
00:39:44.100 --> 00:39:47.500
above the photo gate and the experiment is repeated.
00:39:47.500 --> 00:39:54.000
Assume no energy is lost write the new equation for conservation of mechanical energy that would apply.
00:39:54.000 --> 00:40:05.400
Potential energy in the spring + gravitational potential energy equals kinetic energy or ½ KX² + gravitational potential energy
00:40:05.400 --> 00:40:19.300
is going to be MGH where height is going to be h + X sin θ factor must equal ½ mv².
00:40:19.300 --> 00:40:25.900
Part B2, with a graph of V² vs. X² for this new experiment be a straight line.
00:40:25.900 --> 00:40:36.900
It looks now like V² is a function of both X and x².
00:40:36.900 --> 00:40:54.900
You no longer have that linear relationship to say no it is not going to be a straight line because V² is a function of both X and X².
00:40:54.900 --> 00:41:02.100
This from the ½ KX² factored this from our potential energy factor.
00:41:02.100 --> 00:41:05.300
To that finishes up to 2007 question.
00:41:05.300 --> 00:41:11.400
Let us take a look at the 2010 exam free response 1.
00:41:11.400 --> 00:41:20.800
A coffee filter question which is a very popular lab and physics courses for looking at retarding and drag forces.
00:41:20.800 --> 00:41:25.800
Here we are dropping some coffee filters we are measuring their terminal velocities.
00:41:25.800 --> 00:41:32.400
We have got a graph here in A of the mass of the filters vs. their terminal speed.
00:41:32.400 --> 00:41:35.000
Let us say what they want us to do.
00:41:35.000 --> 00:41:40.600
In A derive an expression relating the terminal speed to the mass.
00:41:40.600 --> 00:41:46.600
Going back to our lecture our lesson on retarding and drag forces.
00:41:46.600 --> 00:41:54.400
Draw a coffee filter free body diagram we have a weight down and it looks like the retarding forces proportional CV².
00:41:54.400 --> 00:42:03.600
At terminal velocity it is no longer accelerating, it is a constant speed that means the net force in the y direction must be 0.
00:42:03.600 --> 00:42:06.500
That means CV² must equal MG.
00:42:06.500 --> 00:42:15.500
CV² = MG and that V² that is V terminal².
00:42:15.500 --> 00:42:24.700
V term is just the square root of MG/your constant C.
00:42:24.700 --> 00:42:26.900
What do we do next?
00:42:26.900 --> 00:42:33.700
Assuming that function relationship, use the grid to plot a linear graph as a function of M to verify the relationship.
00:42:33.700 --> 00:42:38.300
Use empty boxes in the table to record calculated values you are graphing.
00:42:38.300 --> 00:42:42.900
Label the vertical axis as appropriate and place numbers on the axis.
00:42:42.900 --> 00:42:48.100
The first thing I would do is we know that is going to be proportional to V².
00:42:48.100 --> 00:42:52.100
I will add a row for V² vs M if we want to get a linear graph.
00:42:52.100 --> 00:42:58.700
To the table under terminal speed I would add a row for vt² and go calculate that.
00:42:58.700 --> 00:43:06.700
Your values should run from something near about 0.26 to about 1.12 as you fill that in.
00:43:06.700 --> 00:43:09.300
We are going to make our graph.
00:43:09.300 --> 00:43:29.300
Let us go to the next page to give ourselves lots of room here.
00:43:29.300 --> 00:43:41.700
We have VT² for Y, 10 m²/s² we will have a mass on our x in kg.
00:43:41.700 --> 00:43:48.100
I will leave it to you to choose appropriate scales I think 1 × 10⁻³, 2 × 10⁻³, and so on.
00:43:48.100 --> 00:43:54.100
In the Y 0.2, .4, .6 something like that.
00:43:54.100 --> 00:44:00.100
When you plot your points and then draw your best fit line you should get something that looks fairly linear.
00:44:00.100 --> 00:44:05.900
Do not connect your points draw the best fit line.
00:44:05.900 --> 00:44:08.300
It looks like we have done that there.
00:44:08.300 --> 00:44:12.300
We shown that that is linear if we have done everything correctly.
00:44:12.300 --> 00:44:15.500
For part B2, let us see what we got.
00:44:15.500 --> 00:44:19.300
Use your graph to calculate C.
00:44:19.300 --> 00:44:31.100
As I look at our graph, let us say we have VT² vs. M.
00:44:31.100 --> 00:44:35.100
It looks like our slope is going to be VT²/M.
00:44:35.100 --> 00:44:45.100
Our slope is VT²/M that is going to be g/C.
00:44:45.100 --> 00:44:51.100
If I find my slope and take a couple points on the line not θ points, find your slope.
00:44:51.100 --> 00:45:02.100
I found my slope I came up with the slope of about 215.9 m² /kg s².
00:45:02.100 --> 00:45:06.700
Hopefully, you are something remotely close to that must equal g/C.
00:45:06.700 --> 00:45:22.700
If you want C, C is just going to be g over that slope value and I came up with something around 0.0454 kg/m for my slope there in B2.
00:45:22.700 --> 00:45:25.900
That is how I would go about that piece.
00:45:25.900 --> 00:45:34.700
It looks like we are moving on to part C.
00:45:34.700 --> 00:45:39.700
Sketch and approximate graph of speed vs. Time from the time the filters are released up to
00:45:39.700 --> 00:45:54.200
the time T equals capital T when the fall in some distance Y.
00:45:54.200 --> 00:45:58.200
We have got V here and we have got time here.
00:45:58.200 --> 00:46:05.800
We know they are going to start at 0 and they are going to approach some asymptote.
00:46:05.800 --> 00:46:13.000
Our terminal velocity VT and that happens right about we get right about when we get to time T.
00:46:13.000 --> 00:46:20.000
That would be a sketch of a graph and for C2 it says suppose you had a graph like that had a numerical scale and each axis.
00:46:20.000 --> 00:46:24.300
How can you use the graph to approximate the distance Y?
00:46:24.300 --> 00:46:31.900
If I wanted to find that distance Y that travels in time T that is going to be the area in the graph under the graph.
00:46:31.900 --> 00:46:43.300
That is going to be area under graph from T equal 0 to T equals capital T.
00:46:43.300 --> 00:46:50.200
Finding all of this area in here we will not even sketch that in a little bit.
00:46:50.200 --> 00:46:54.700
To get a little bit better feel for what we are talking about there.
00:46:54.700 --> 00:47:07.300
From T =0 to T = T and that is the distance traveled Y.
00:47:07.300 --> 00:47:09.500
You can do that by integrating under that.
00:47:09.500 --> 00:47:15.200
You can take a look at if you had a nice grid paper you could count up a bunch a little grids with the scale the figure that out.
00:47:15.200 --> 00:47:21.500
But really it is taking the area as the final important answer.
00:47:21.500 --> 00:47:28.500
In D, determine an expression for the approximate amount of mechanical energy dissipated δ E
00:47:28.500 --> 00:47:34.300
and air resistance during the time the stack falls at distance Y or Y is greater than the capital Y.
00:47:34.300 --> 00:47:39.300
Express your answer and those terms of those variables have already given us.
00:47:39.300 --> 00:47:49.700
We can use conservation of energy to do that initial potential + initial kinetic energy must equal final potential + final kinetic energy.
00:47:49.700 --> 00:48:00.000
That δ in energy must be the final - the initial whenever that missing pieces which is the amount of energy dissipated due to air resistance.
00:48:00.000 --> 00:48:10.000
That is just going to be our MGY at this Y point – ½ MV terminal².
00:48:10.000 --> 00:48:17.400
There would be our expression for that energy.
00:48:17.400 --> 00:48:20.900
I think that finishes up 2010 question.
00:48:20.900 --> 00:48:31.300
Let us do one more here, we have got the 2013 exam mechanics question 1.
00:48:31.300 --> 00:48:34.700
Amazingly, we have another glider on the track with the spring.
00:48:34.700 --> 00:48:39.700
Maybe you are seeing a trend here.
00:48:39.700 --> 00:48:45.500
Take a minute and download the question look it over and give it a shot and comeback here.
00:48:45.500 --> 00:48:51.200
We have got our data what is the first thing that is going to ask us to do.
00:48:51.200 --> 00:48:57.800
On the axis below, plot the data points and draw a smooth curve that best fits the data.
00:48:57.800 --> 00:49:08.900
Once again, more graphing, it is very important stuff.
00:49:08.900 --> 00:49:11.200
There is my graph.
00:49:11.200 --> 00:49:16.900
We have velocity in m/s vs. Time in seconds.
00:49:16.900 --> 00:49:26.700
My plot when I go to plot the points looks kind of like that.
00:49:26.700 --> 00:49:32.300
Something close and label your scales on the axis and all that important stuff.
00:49:32.300 --> 00:49:36.600
For part B, that says a student wishes to use the data to plot position.
00:49:36.600 --> 00:49:40.000
Describe a method the student can use to do this.
00:49:40.000 --> 00:49:43.100
It is very similar from our last question.
00:49:43.100 --> 00:49:59.100
I plot the area under the VT curve as a function of time.
00:49:59.100 --> 00:50:04.300
Lots of ways you can do that, you can break this up into a bunch of little boxes and figure out how many boxes
00:50:04.300 --> 00:50:07.700
you have covered at each point in time and plot those data points.
00:50:07.700 --> 00:50:10.600
That might be the easiest way to graphically do it.
00:50:10.600 --> 00:50:13.800
You could fit a curve to it and actually integrate.
00:50:13.800 --> 00:50:22.600
A couple different options but the general idea is plotting the area under the curve as a function of time to get your position time graph.
00:50:22.600 --> 00:50:28.500
On the axis below now, sketch of position as a function of time for the glider labeling in the intercepts.
00:50:28.500 --> 00:50:32.400
Assuming we did this we are going to draw with that should look like.
00:50:32.400 --> 00:50:45.900
Let us go here and give ourselves another graph.
00:50:45.900 --> 00:50:53.000
We are plotting position as a function of time, position in meters vs. time.
00:50:53.000 --> 00:50:57.600
When I do this, my graph is going to look something like this.
00:50:57.600 --> 00:51:11.100
It is going to start there and as we get to about 0.79s or so, when we get to this value of about 0.25m.
00:51:11.100 --> 00:51:13.500
0.79s it becomes fairly linear.
00:51:13.500 --> 00:51:16.400
It is concave upward and then we go in a linear fashion.
00:51:16.400 --> 00:51:21.800
We no longer have any force on it.
00:51:21.800 --> 00:51:27.700
You can plot the data points yourselves that come up with something like that.
00:51:27.700 --> 00:51:35.100
Then we move on to part C, find the time at which the glider makes contact with the bumper at the far right.
00:51:35.100 --> 00:51:38.500
Let us give ourselves a little room here for this one.
00:51:38.500 --> 00:51:47.000
For part C, we can use our kinematics where X= X initial + VT.
00:51:47.000 --> 00:51:56.300
We know V is our slope or 0.5 m/s.
00:51:56.300 --> 00:52:08.200
That is going to be 2 when we hit the far end is equal to our starting point 0.25 + 0.5 our velocity × T –
00:52:08.200 --> 00:52:16.200
you got to pullout that 0.79s because that is when we go into a constant acceleration.
00:52:16.200 --> 00:52:33.100
Solving this for T that means that subtract 0.25 from that side 1.75 is going to be equal to T /2 - we will distribute that 0.5, .395,
00:52:33.100 --> 00:52:47.000
which implies that 2.145 equals T /2 or T is double that or 4.29s.
00:52:47.000 --> 00:52:50.800
D, find the force constant of the spring.
00:52:50.800 --> 00:53:03.300
To do that, our potential energy stored in the spring must be our kinetic which implies that ½ KX² equals ½ MV².
00:53:03.300 --> 00:53:11.100
K is going to be equal to MV² /x² and we can substitute in our values 0.4kg.
00:53:11.100 --> 00:53:18.400
Our V was 0.5 m/s² and our x is .25m² .
00:53:18.400 --> 00:53:26.600
I did about 1.6N/m.
00:53:26.600 --> 00:53:34.900
Part E, the experiment is run again that this time the glider is attached to a spring rather than simply being pushed against it.
00:53:34.900 --> 00:53:40.000
When it is attached to a spring you are just going to go oscillating back and forth.
00:53:40.000 --> 00:53:43.300
We basically got a spring pendulum.
00:53:43.300 --> 00:53:46.200
Find the amplitude of the resulting periodic motion.
00:53:46.200 --> 00:53:49.100
The biggest amplitude you are going to have is when you have compressed that the most.
00:53:49.100 --> 00:53:52.900
It is not going to go past that because you cannot create energy.
00:53:52.900 --> 00:53:58.800
For part 1, that is going to be amplitude maximum is 0.25 m.
00:53:58.800 --> 00:54:01.000
For part 2, find the period of oscillation.
00:54:01.000 --> 00:54:06.100
We can use our formula for the period of a spring pendulum.
00:54:06.100 --> 00:54:21.400
Remember that from previous physics courses it is 2 π √M /k which is 2π √.4 / 1.6 or .4/ 1.6 that is going to be 1/4.
00:54:21.400 --> 00:54:29.200
√1/4 is ½ that just going to leave us with π s.
00:54:29.200 --> 00:54:33.000
If this part is a little fuzzy we have not got into oscillations in this course.
00:54:33.000 --> 00:54:38.600
It might be familiar because most folks had a prior physics course before taking APC mechanics.
00:54:38.600 --> 00:54:44.100
If you have not we are going to get there toward the end of the mechanics part of the course when we talk about oscillations.
00:54:44.100 --> 00:54:48.600
E might be a little fuzzy to you, do not worry about that.
00:54:48.600 --> 00:54:53.100
Hopefully, that gets you a great start with conservation of energy with a bunch of sample problems.
00:54:53.100 --> 00:54:56.000
Thank you so much for watching www.educator.com and make it a great day everybody.