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For more information, please see full course syllabus of AP Physics C: Mechanics
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Lecture Comments (20)

1 answer

Last reply by: Professor Dan Fullerton
Sat Dec 19, 2015 7:40 AM

Post by Jim Tang on December 18, 2015

for example 13 part D, shouldn't it be 1/2 mvT^2 - mgy?

2 answers

Last reply by: Jim Tang
Sat Dec 19, 2015 8:31 PM

Post by Jim Tang on December 18, 2015

on example 9, why did you choose that point for Kf and Uf? doesn't it still have some distance to the right to finish? was it just how the diagram was set up?

1 answer

Last reply by: Professor Dan Fullerton
Sat Dec 19, 2015 7:34 AM

Post by Jim Tang on December 18, 2015

on example 8, doesn't andy still have some gravitational potential energy after he slides down the mudslide? how do you know all of it is transferred to kinetic?

2 answers

Last reply by: Micheal Bingham
Tue Mar 17, 2015 8:50 PM

Post by Micheal Bingham on March 13, 2015

Hi for example XI AP Free Response 2003, I looked at the scoring guidelines, why is it not acceptable to place 2 photogates at x = 0 and x= 2 meters for the experiment? Why must 2 photogates be placed at x =2 ?

2 answers

Last reply by: Professor Dan Fullerton
Sun Jan 11, 2015 3:28 PM

Post by Carolyn Diamond on January 11, 2015

For example 11, why/how is kinetic energy being converted into spring potential energy? Also, is there spring potential energy at the beginning of the experiment, the end of the experiment, or both?

2 answers

Last reply by: Professor Dan Fullerton
Sun Dec 7, 2014 4:10 PM

Post by Thadeus McNamara on December 7, 2014

40:10 explain how the height is h + xsintheta.
I thought the height would just be h

1 answer

Last reply by: Professor Dan Fullerton
Fri Nov 21, 2014 6:29 AM

Post by Zhengpei Luo on November 20, 2014

In the  example of bungee jump, before certain point in the air, the rope is not taut at all I think.
But there is no way to calculate that point here in this problem.

1 answer

Last reply by: Professor Dan Fullerton
Fri Nov 21, 2014 6:28 AM

Post by Zhengpei Luo on November 20, 2014

on Ap physics C, should we still use mgh to represent potential energy? Or we use the other one

Conservation of Energy

  • For a conservative force doing work on a closed system, the work done provides a change in the system’s kinetic energy, and is equal to the opposite of the change in potential energy. The sum of the initial kinetic and potential energies therefore equals the sum of the final kinetic and potential energies.
  • Non-conservative forces change the total mechanical energy of a system, but not the total energy of a system. Work done by a non-conservative force is typically converted to internal (thermal) energy.

Conservation of Energy

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Objectives 0:09
  • Conservation of Mechanical Energy 0:32
    • Consider a Single Conservative Force Doing Work on a Closed System
  • Non-Conservative Forces 1:40
    • Non-Conservative Forces
    • Work Done by a Non-conservative Force
    • Formula: Total Energy
    • Formula: Total Mechanical Energy
  • Example I: Falling Mass 2:15
  • Example II: Law of Conservation of Energy 4:07
  • Example III: The Pendulum 6:34
  • Example IV: Cart Compressing a Spring 10:12
  • Example V: Cart Compressing a Spring 11:12
    • Example V: Part A - Potential Energy Stored in the Compressed Spring
    • Example V: Part B - Maximum Vertical Height
  • Example VI: Car Skidding to a Stop 13:05
  • Example VII: Block on Ramp 14:22
  • Example VIII: Energy Transfers 16:15
  • Example IX: Roller Coaster 20:04
  • Example X: Bungee Jumper 23:32
    • Example X: Part A - Speed of the Jumper at a Height of 15 Meters Above the Ground
    • Example X: Part B - Speed of the Jumper at a Height of 30 Meters Above the Ground
    • Example X: Part C - How Close Does the Jumper Get to the Ground?
  • Example XI: AP-C 2002 FR3 30:28
    • Example XI: Part A
    • Example XI: Part B
    • Example XI: Part C
    • Example XI: Part D & E
  • Example XII: AP-C 2007 FR3 35:24
    • Example XII: Part A
    • Example XII: Part B
    • Example XII: Part C
    • Example XII: Part D
  • Example XIII: AP-C 2010 FR1 41:07
    • Example XIII: Part A
    • Example XIII: Part B
    • Example XIII: Part C
    • Example XIII: Part D
  • Example XIV: AP-C 2013 FR1 48:25
    • Example XIV: Part A
    • Example XIV: Part B
    • Example XIV: Part C
    • Example XIV: Part D
    • Example XIV: Part E

Transcription: Conservation of Energy

Hello, everyone, and welcome back to

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.0003

Our objectives include stating and applying the relation between work and mechanical energy.0011

Analyzing situations in which an object's mechanical energy is changed by external forces.0017

Applying conservation of energy and analyzing the motion of objects.0022

Solving problems that call for applications of both conservation of energy and Newton’s laws of motion.0026

Let us start by talking about conservation of mechanical energy.0032

That is considered a single conservative force doing work on a closed system.0036

In that case, we know that the work done by that force is that change in its kinetic energy by the work energy theorem.0041

We also know that the work done by the conservative force is the opposite of the change in its potential energy.0049

If those two are equal, we could then set the change in kinetic energy must be equal to0056

the opposite of the change + the change in potential energy.0061

Or rearranging it slightly, change in kinetic energy + change in potential energy must equal 0 for some close system.0068

If we expand this out a little bit or really saying is the initial energy must be equal to the final energy.0077

Or kinetic initial + potential initial must equal kinetic final + the final potential energy and conservation of mechanical energy.0084

What if you have non conservative forces?0099

Non conservative forces change the total mechanical energy of a system but not the total energy of the system.0102

The work done by the non conservative force is typically converted to some sort of internal or thermal energy.0108

If our total energy is kinetic + potential + the work done by non conservative forces0114

and the total mechanical energy we are going to define is just the kinetic + the potential.0125

Let us look at some examples here.0134

We have an object of mass M, it falls from some height h.0136

Find its speed prior to impact using conservation of mechanical energy and resistance.0140

You could do this with the kinematics you already know.0146

I recommend after we do this problem here with energy, take a minute and try it with kinematics.0149

Make sure you get the same answer.0154

We have our object there at B, it is going to fall some distance to a new position.0157

It has some kinetic + potential energy initial.0167

There it has some final amounts of kinetic energy + potential energy.0173

It changes its height by an amount H.0179

Writing our formula for conservation mechanical energy because we are only dealing with conservative forces here or neglecting air resistance.0186

The initial kinetic + the initial potential energy must equal the final kinetic + the final potential energy.0193

If it starts at rest of the initial kinetic to 0 and the initial potential energy is MGH assuming more uniform gravitational field.0203

Its final kinetic energy is ½ M × its final velocity².0212

Its final potential energy is 0 because it sets final point what we are calling 0.0217

We can simplify this and divide M both sides to say that gh=V² /2 or velocity² =2 gh.0225

If we want just the speed that is going to be √2gh.0237

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.0246

We can find its total energy as the sum of its gravitational potential energy and its kinetic will be MGH + ½ Mv²0258

or 20,000kg × 10 m/s² × 10,000m + ½ × its mass × squared of its velocity 250².0274

Complies that its total energy is 2.63 × 10⁹ joules.0294

That jet dives to an altitude of 2000m and drops in height 8000m, find the new velocity of the jet.0303

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.0314

Or V will be equal to 2 × our total energy - MGH all divided by mass.0336

We have to take the square root of that.0345

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/s².0348

Our height now 2000m all divided by the mass 20,000kg.0366

I come up with the speed of about 472 m/s.0378

The jet trading in the altitude for speed keeping the same total mechanical energy.0385

Let us take a look at our pendulum example.0394

The pendulum comprises the light string of length L swings mass M back and forth.0397

We have look at this previously but let us highlight a couple things here.0403

There is the length of our pendulum L.0407

If we went here, we already calculated this in our last lesson and we said that this must be L cos θ.0409

This amount h= L - L cos θ or L 1 – cos θ.0418

We have done that derivation before.0429

As this mass swings back and forth in the pendulum, it is transferring kinetic and potential energy.0432

Its highest point it has the maximum potential energy.0439

Its lowest point, its potential energies at a minimum and it is all kinetic energy.0443

And swings to the other side transfers that kinetic energy back to potential energy.0449

For a split second that stops but it is at its highest point.0453

We can write an equation here for the maximum potential energy is MGH which is going to be MG × our height L1 - cos θ.0459

Which means assuming you are in a system we are not dealing0474

with any non conservative forces when we get to this point it is all kinetic energy.0477

Our maximum kinetic energy which is ½ Mv² has to be equal to MG L 1 - cos θ.0482

We can solve for the velocity of our pendulum at that lowest point when we got the maximum kinetic energy.0495

Let us take a second to do that.0501

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.0503

Which implies then the V² =2g L × 1 – cos θ or V is just going to be equal to0519

and let us make sure we know that that is our maximum velocity.0528

It is going to be equal to 2 gL 1 - cos θ square root.0531

If they wanted to make a graph of this as we look at the displacement in the X vs. Energy in the Y.0539

Our total energy must remain constant by law of conservation of energy.0547

There is our total.0552

We have our highest potential energies and where the maximum displacements.0554

We will draw our little points in here and at our lowest point the middle point no displacement0560

we have maximum kinetic energy and minimum 0 potential energy.0566

We will put our kinetic energy the maximum here.0571

The furthest displacements that is not moving so it is kinetic energy to 0 out here.0574

Our kinetic energy graph might look something like this.0579

As best as I can eyeball that in their.0584

There would be our kinetic energy and our potential energy might look something like that.0587

At any point, wherever we happened to be the total sum of the potential + the kinetic has to equal our total.0597

We have that maintain that total because we are not dealing with any non conservative forces.0605

Take a look at another example.0611

The diagram below shows a cart possessing 16 joules of kinetic energy0614

traveling on a frictionless horizontal surface to a horizontal spring.0618

If the cart comes to rest after compressing the spring a distance of 1m what is the spring constant of the spring ?0622

The initial kinetic energy must be equal to the final spring elastic potential energy0630

because when the kinetic energy become 0 we put all of energy and compressing the spring.0638

16 joules must equal to ½ KX² assuming it follows Hooke’s law.0644

K must be 32 joules / X² which is going to be 32 joules / 1m² or 32 N/m.0652

There is a spring constant.0669

Lets do another example.0672

A popup toy has a mass of 0.02kg and the spring constant of 150 N/m.0674

A force is applied to the toy to compress the spring 0.05m.0681

Calculate the potential energy stored in the compressed spring.0686

Alright potential energy in the compressed spring.0691

Make it follow Hooke’s law is ½ KX² which would be ½ × our spring constant 150 N/ m × our spring compression 0.05²0695

which is going to be 0.1875 joules.0711

Let us take this one step further.0720

The toy was activated all the compressed springs potential energy is converted a gravitational potential energy.0722

As it pops up it goes up into the air.0727

Find the maximum height to which the toy is propelled.0730

Our initial potential energy must equal our final potential energy.0735

No kinetic in there because it at rest that both its initial and its final positions.0742

Our initial potential energy the spring potential energy was 0.1875 joules that must be equal to its final gravitational potential energy mgh.0746

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.0758

Taking a look at another one.0786

A car initially travels 30 m/s and that slows uniformly as it skids to stop after the brakes are applied.0787

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.0794

Let us make a couple axis here.0804

We have kinetic energy on the Y and we have the work done by the force of friction on the X.0815

Our initial velocity was 30 m/s, our final velocity is 0.0825

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.0831

It is going to have some final point 0.0837

Because we are looking at the work done by friction on the x, all the work done by friction is what slowing it down.0843

Therefore, our graph is just going to look like that.0851

Let us do a block on the ramp problem.0862

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.0865

Find the total work done by friction on the block as it slides down the ramp.0872

Let us make ourselves a diagram here.0877

Give ourselves a nice little ramp to play with.0884

On the ramp, we have a 2kg block and the height of our ramp 3m.0894

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.0907

At the bottom, it has kinetic energy at the bottom + we have whatever work was done by friction.0917

This implies that the work done by friction must be the gravitational potential energy at the top - the kinetic energy at the bottom0926

which is going to MG h at the top - the kinetic energy at the bottom which is given as 50 joules.0936

The work done by friction is its mass 2 kg × acceleration due to gravity 10 m/s² ×0947

the height 3m -50 joules or 20 × 360 -50 is just going to 10 joules.0958

10 joules must be the total work done by friction on the block as it slides down the ramp.0968

Let us do a multipart problem here.0975

Andy, the adventurous adventurer while running from evil bad guys in the Amazonian rain forest0977

saving the world from impending doom of course, trips, falls, and slides down the frictionless mudslide of height 20m is depicted here.0983

Once he reaches the bottom of a mudslide however he flies horizontally off a 50m cliff.0993

How far from the base of the cliff Andy land?0999

A multipart problem.1003

Let us see.1004

Here at the top he has all gravitational potential energy.1006

If we then convert then do is kinetic energy here we can find the velocity with which he goes the side of a cliff1010

and then this becomes a projectile motion problem for the object launch horizontally off the cliff with the height of 15m.1016

We can break it up into two parts, the mudslide piece and the cliff piece.1024

Let us do this where we have ourselves a little bit more room.1029

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 mv².1034

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.1048

He is going to go off horizontally with 20 m/s how far does he land there?1067

That is a kinematics problem.1073

Let us take a look.1076

Horizontally, when we look at our kinematics problem we have some horizontal velocity 20 m/s1078

that is not going to change for neglecting air resistance.1084

We do not know how long he is in the air so we cannot figure out yet how far he travels.1088

We will call that D.1093

We have to analyze the vertical motion in order to find out how long is in the air.1096

We will call down the positive y direction.1101

The initial vertically 0.1104

The final vertically we do not know and frankly do not care a lot.1107

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.1112

Our acceleration is 10 m/s² positive because we call down the positive y direction and we are looking for T.1119

Finding a kinematic equation I probably look at δ Y=V initial T + ½ ay T².1129

We have this nice little helper that V initial 0.1139

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.1143

√30/10² is about 1.73s.1158

The time we can now filling up here horizontally 1.73 s.1163

To figure out how far it goes the δ X or I think we call that D1169

is just going to be horizontal velocity × time or 20 m/s × 1.73s should give us about 34.6m.1176

Distance here 34.6m.1193

A multipart problem where we had to pull in conservation of energy.1199

A rollercoaster problem.1205

A rollercoaster car begins at height H above the ground and completes a loop along its path.1209

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.1215

I'm going to recognize here first that this is a kinetic initial + potential initial energy.1224

Kinetics going to be 0.1232

The potential energy there is going to be all gravitational.1234

When we get to the highest point here, we have kinetic final + potential final because that is what we are worried about.1237

This distance is r and that distances is r.1245

2r is our diameter.1249

As I look at this, our initial gravitational potential energy MGH must be equal to our kinetic and gravitational here.1255

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.1266

MG × 2r this implies then as we multiply through here let us get rid of our M we can divide all those out1279

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.1290

Or solving for V final² that is just going to be 2g × (h- 2r).1304

There is V final².1317

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.1320

Its centripetal acceleration has to be greater than the acceleration due to gravity otherwise it is going to fly off.1326

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.1332

Rearranging this, V² then must be greater than equal to gr but V² we just said was 2g × (h-2r).1343

We could write then that 2g × (h-2r) has to be greater than or equal to gr.1355

Which implies then h -2r must be greater than or equal to r/2 factoring out the g.1366

Therefore, getting h all by itself h has to be greater than or equal to 5r / 2.1380

As long as h is more than 2 ½ × r you are in great shape.1388

This is assuming you are frictionless so I would not build it right there.1395

You do not want to put some leeway in there if you are a roller-coaster designer.1397

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.1401

H is greater than or equal to 5 r/2.1409

Let us do the problem with the bungee jumper.1412

Alicia a 60 kg bungee jumper steps off the 40m high bridge.1415

The bungee cord behaves like a spring with K =40 N / m.1420

Assume there is no slack in the cord.1424

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.1427

I'm going to make a quick diagram here of where we are going to know our speeds.1437

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.1445

We want to know the speed of the height of 30m above the ground we will call that B.1457

This is another 15m there.1463

He is stepping off the 40m high bridge so there is our starting point.1467

There is another 10m there and we want to know how close she gets to the ground down here as well.1473

Let us start on part A, find the speed of the jumper the height of 15m above the ground or here at A.1479

Give ourselves a little bit of room, we can take a look and say that the gravitational potential energy1486

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.1493

Which implies after speed we want kinetic energy all by itself.1505

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.1509

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 Y².1523

Which implies then the VA² equals we will multiply by 2/ M.1550

The right hand side we get 2g δ Y - K / M δ Y² which implies that va² = 2 × 9.8 m/s²1556

because we do not want here running into the ground in your calculations.1575

× 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/s².1581

Or the velocity at A must be 8.6 m/s.1599

We found the first part the speed of Alicia at the height of 15m above the ground.1608

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.1613

She has fallen 10m at this point.1621

Gravitational potential energy at the top equals gravitational potential energy at B + the spring potential energy at B +1625

the kinetic energy at B which implies then that the kinetic energy B equals the gravitational potential energy1634

at B - the gravitational potential energy at B - the spring potential energy at B1642

which implies again that ½ M V at B² = MG × change in position δ y -1/2 K δ Y².1653

Or VB² = 2g δ y - K / M δ y² which implies then that vb²-= 2 × 9.8.1667

Our δ y at 10m - K / M still 40/60 × δ y 10m² or 129m² /s².1685

Taking the square root of that velocity at B must be 11.4 m/s.1698

Finally, how close does she get to the ground?1709

This occurs when the velocity and kinetic energy are equal to 0.1713

Potential energy at the top equals a gravitational potential energy at the bottom + the spring potential energy at the bottom.1719

Which implies that our change in gravitational potential energy must equal the stored potential energy and the spring.1732

Or MG δ Y = ½ K δ Y².1743

Which implies then, let us get rid of that two at about ½ by multiplying by two.1752

We can also divide by that by δ Y 2Mg = K δ Y or δ Y = 2 Mg / K.1757

Which is 2 × Alicia is 60kg g 9.8 m/s² /spring constant 14 N/m which is 29.4m.1774

Be careful here, that means if δ Y =29.4m the jumper must be 40m - that from the ground that are lowest points.1788

40 -29.4 is going to be equal to 10.6m from the ground.1801

I think to the point where we can do a couple simple problem.1818

Let us do a couple of old AP free response problems to finish up this unit here.1821

Let us start by taking a look at the 2002 Mechanics free response number 3 problem.1828

I will give you a minute, you can go download the problem at the address above or google search it.1833

As we look at that one, I want you to take a minute, look here and see if you cannot solve it at your own1839

as you pause the video, give it a minute or 2 then come back here and check your answers1844

or if you get stuck, use this to help you get past that sticking point.1849

As we look at 2002 free response 3.1855

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.1859

We will create a graph for these sorts of things by now.1870

Of course, you will graph it very carefully plugging the points.1876

I am going to try and give a rough idea of where we are at.1881

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 1.1884

We are somewhere here.1893

Our graph is going to look something like that.1895

You got that general shape.1908

Now B says, determine the force associated with the potential energy function given above.1910

We are given potential energy we want force.1917

Force is - du DL which in this case is going to be - D / Dx of our equation 4/2 + X.1920

We can pull the 4 out that is a constant.1931

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 + 2⁻¹1934

or the force is going to be equal 24 × x + 2⁻² which is 4/ x + 2².1952

There is part B.1968

Taking a look here at C, suppose the object is released from the origin, find the speed at x = 2m.1971

That sounds like a conservation of energy problem.1978

We know its change in potential energy is its potential energy at 0 - its potential energy at 0.2 and1981

all that has to be equal be turned into the change in energy be turned into its kinetic energy ½ mv².1988

Which implies that U of 0 is 2, U of 2 is 1.1996

2 -1 is going to be 1 joule = ½ f MV².2000

We can solve that for V that means 2/ M is equal to V².2008

V is going to be equal to V² is 2/m which is 2/0.5 or 4.2016

V is √4 or 2m/s.2024

That did not seem so bad.2031

Let us move on and take a look at part D.2033

In a lab you are given a glider of a mass ½ kg on the track,2039

the gliders acted on by the force determined in B and your goals to determine the validity of your theoretical calculation.2042

We can get to pick some equipment to help us do this.2048

What we are going to do is pick the equipment and outline a procedure we will use.2053

They are probably a bunch of different ways you can do this but what I'm thinking about2057

is I would probably take one of those photo D timers at probably two of them.2062

And set the photo gate a small distance apart right here that x =2m.2068

If you measure the distance between the gates with the meter stick and you have obtained that time for the gliders travel2073

between the gates you could then use V = D/T to determine its velocity and confirm your previous findings.2078

You would probably need a photo gate timer.2086

We need a meter stick, I think it will be is set for equipment if you follow that procedure.2093

The trick is to make sure that you measure the distance between gates when you really close to that the 2m mark.2100

Determine that the time and then you can get the velocity.2106

Make sure you are in the ballpark to confirm your findings.2110

That is D and E.2114

In E, make sure you explain that in a complete sentence fairly clearly.2115

Let us go on to our next free response question from 2007.2122

Let us look at free response 3, a spring and glider question with the photo gate again.2127

It looks familiar.2133

The apparatus they show you here is used to study conservation mechanical energy.2136

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.2140

They say assuming no energy is lost, write the equation for conservation and mechanical energy that would apply here.2147

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 KX².2154

But it tells us in the problem that K is 40 N /m.2167

We can write that ½ Mv² is equal to ½ × 40 X² or 20 x².2172

There is an equation of conservation of energy that would apply to this problem.2184

For part B, it looks like we are getting into some graphing again.2189

Plot V² vs. X² label the axis, units, and scale.2192

Plotting points should be easy points on the AP exam.2198

We will give ourselves some axis and you have got to label the scale and things like that.2205

As by I did this and have X² and m² on the x, I have v² and m² for s² on the Y.2214

We have 0.005, 0.01, 0.015, and so one for the x.2227

In the y, I did 1, 2, 3, 4, 5, 6 something like that.2235

I am not going to the plot the points myself point by point but when we do this you plot the points and2245

draw a best fit line should get something that looks fairly linear and like it goes through 00.2250

My line looks something like that.2257

We are not connecting points.2259

We are drawing the best fit line.2260

We plotted points labeled the axis including units and scale.2263

And moving on the part C, draw the best fit line and use the best fit line to obtain the mass and the glider.2268

For C, to determine the mass and the glider I am first going to recognize here that the slope is going to K/M.2277

As I look at that, because as we do our plots here let see we have our equation ½ MV² = ½ 40x².2284

For plotting V² vs x² we have ½ MV² = 20x².2304

Multiply that by 2 mv² = 40x² or if V² = 40/M or k/Mx².2316

It fits the form Y = MX where X is x² our y is V².2326

Our slope must be K/M as 40/M.2332

We find our slope and when I find the slope I came up with about slope is K/M so m is k/slope.2335

My slope was right around 201/s².2345

Our mass which is going to K/slope becomes 40 N/m / 201/s² or 0.2kg.2351

That covers C1 and 2.2369

It looks like we move on the part D.2372

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 h2378

above the photo gate and the experiment is repeated.2384

Assume no energy is lost write the new equation for conservation of mechanical energy that would apply.2387

Potential energy in the spring + gravitational potential energy equals kinetic energy or ½ KX² + gravitational potential energy2394

is going to be MGH where height is going to be h + X sin θ factor must equal ½ mv².2405

Part B2, with a graph of V² vs. X² for this new experiment be a straight line.2419

It looks now like V² is a function of both X and x².2426

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 X².2437

This from the ½ KX² factored this from our potential energy factor.2455

To that finishes up to 2007 question.2462

Let us take a look at the 2010 exam free response 1.2465

A coffee filter question which is a very popular lab and physics courses for looking at retarding and drag forces.2471

Here we are dropping some coffee filters we are measuring their terminal velocities.2481

We have got a graph here in A of the mass of the filters vs. their terminal speed.2486

Let us say what they want us to do.2492

In A derive an expression relating the terminal speed to the mass.2495

Going back to our lecture our lesson on retarding and drag forces.2500

Draw a coffee filter free body diagram we have a weight down and it looks like the retarding forces proportional CV².2506

At terminal velocity it is no longer accelerating, it is a constant speed that means the net force in the y direction must be 0.2514

That means CV² must equal MG.2523

CV² = MG and that V² that is V terminal².2526

V term is just the square root of MG/your constant C.2535

What do we do next?2545

Assuming that function relationship, use the grid to plot a linear graph as a function of M to verify the relationship.2547

Use empty boxes in the table to record calculated values you are graphing.2554

Label the vertical axis as appropriate and place numbers on the axis.2558

The first thing I would do is we know that is going to be proportional to V².2563

I will add a row for V² vs M if we want to get a linear graph.2568

To the table under terminal speed I would add a row for vt² and go calculate that.2572

Your values should run from something near about 0.26 to about 1.12 as you fill that in.2579

We are going to make our graph.2587

Let us go to the next page to give ourselves lots of room here.2589

We have VT² for Y, 10 m²/s² we will have a mass on our x in kg.2609

I will leave it to you to choose appropriate scales I think 1 × 10⁻³, 2 × 10⁻³, and so on.2622

In the Y 0.2, .4, .6 something like that.2628

When you plot your points and then draw your best fit line you should get something that looks fairly linear.2634

Do not connect your points draw the best fit line.2640

It looks like we have done that there.2646

We shown that that is linear if we have done everything correctly.2648

For part B2, let us see what we got.2652

Use your graph to calculate C.2655

As I look at our graph, let us say we have VT² vs. M.2659

It looks like our slope is going to be VT²/M.2671

Our slope is VT²/M that is going to be g/C.2675

If I find my slope and take a couple points on the line not θ points, find your slope.2685

I found my slope I came up with the slope of about 215.9 m² /kg s².2691

Hopefully, you are something remotely close to that must equal g/C.2702

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 B2.2707

That is how I would go about that piece.2723

It looks like we are moving on to part C.2726

Sketch and approximate graph of speed vs. Time from the time the filters are released up to2735

the time T equals capital T when the fall in some distance Y.2740

We have got V here and we have got time here.2754

We know they are going to start at 0 and they are going to approach some asymptote.2758

Our terminal velocity VT and that happens right about we get right about when we get to time T.2766

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.2773

How can you use the graph to approximate the distance Y?2780

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.2784

That is going to be area under graph from T equal 0 to T equals capital T.2792

Finding all of this area in here we will not even sketch that in a little bit.2803

To get a little bit better feel for what we are talking about there.2810

From T =0 to T = T and that is the distance traveled Y.2815

You can do that by integrating under that.2827

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.2829

But really it is taking the area as the final important answer.2835

In D, determine an expression for the approximate amount of mechanical energy dissipated δ E2841

and air resistance during the time the stack falls at distance Y or Y is greater than the capital Y.2848

Express your answer and those terms of those variables have already given us.2854

We can use conservation of energy to do that initial potential + initial kinetic energy must equal final potential + final kinetic energy.2859

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.2870

That is just going to be our MGY at this Y point – ½ MV terminal².2880

There would be our expression for that energy.2890

I think that finishes up 2010 question.2897

Let us do one more here, we have got the 2013 exam mechanics question 1.2901

Amazingly, we have another glider on the track with the spring.2911

Maybe you are seeing a trend here.2915

Take a minute and download the question look it over and give it a shot and comeback here.2920

We have got our data what is the first thing that is going to ask us to do.2925

On the axis below, plot the data points and draw a smooth curve that best fits the data.2931

Once again, more graphing, it is very important stuff.2938

There is my graph.2949

We have velocity in m/s vs. Time in seconds.2951

My plot when I go to plot the points looks kind of like that.2957

Something close and label your scales on the axis and all that important stuff.2967

For part B, that says a student wishes to use the data to plot position.2972

Describe a method the student can use to do this.2976

It is very similar from our last question.2980

I plot the area under the VT curve as a function of time.2983

Lots of ways you can do that, you can break this up into a bunch of little boxes and figure out how many boxes2999

you have covered at each point in time and plot those data points.3004

That might be the easiest way to graphically do it.3008

You could fit a curve to it and actually integrate.3010

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.3014

On the axis below now, sketch of position as a function of time for the glider labeling in the intercepts.3022

Assuming we did this we are going to draw with that should look like.3028

Let us go here and give ourselves another graph.3032

We are plotting position as a function of time, position in meters vs. time.3046

When I do this, my graph is going to look something like this.3053

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.3057

0.79s it becomes fairly linear.3071

It is concave upward and then we go in a linear fashion.3073

We no longer have any force on it.3076

You can plot the data points yourselves that come up with something like that.3082

Then we move on to part C, find the time at which the glider makes contact with the bumper at the far right.3088

Let us give ourselves a little room here for this one.3095

For part C, we can use our kinematics where X= X initial + VT.3098

We know V is our slope or 0.5 m/s.3107

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 –3116

you got to pullout that 0.79s because that is when we go into a constant acceleration.3128

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, .395,3136

which implies that 2.145 equals T /2 or T is double that or 4.29s.3153

D, find the force constant of the spring.3167

To do that, our potential energy stored in the spring must be our kinetic which implies that ½ KX² equals ½ MV².3171

K is going to be equal to MV² /x² and we can substitute in our values 0.4kg.3183

Our V was 0.5 m/s² and our x is .25m² .3191

I did about 1.6N/m.3198

Part E, the experiment is run again that this time the glider is attached to a spring rather than simply being pushed against it.3206

When it is attached to a spring you are just going to go oscillating back and forth.3215

We basically got a spring pendulum.3220

Find the amplitude of the resulting periodic motion.3223

The biggest amplitude you are going to have is when you have compressed that the most.3226

It is not going to go past that because you cannot create energy.3229

For part 1, that is going to be amplitude maximum is 0.25 m.3233

For part 2, find the period of oscillation.3239

We can use our formula for the period of a spring pendulum.3241

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 1/4.3246

√1/4 is ½ that just going to leave us with π s.3261

If this part is a little fuzzy we have not got into oscillations in this course.3269

It might be familiar because most folks had a prior physics course before taking APC mechanics.3273

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.3278

E might be a little fuzzy to you, do not worry about that.3284

Hopefully, that gets you a great start with conservation of energy with a bunch of sample problems.3288

Thank you so much for watching and make it a great day everybody.3293