For more information, please see full course syllabus of AP Physics C: Mechanics

For more information, please see full course syllabus of AP Physics C: Mechanics

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### 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
- Objectives
- Conservation of Mechanical Energy
- Non-Conservative Forces
- Non-Conservative Forces
- Work Done by a Non-conservative Force
- Formula: Total Energy
- Formula: Total Mechanical Energy
- Example I: Falling Mass
- Example II: Law of Conservation of Energy
- Example III: The Pendulum
- Example IV: Cart Compressing a Spring
- Example V: Cart Compressing a Spring
- 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
- Example VII: Block on Ramp
- Example VIII: Energy Transfers
- Example IX: Roller Coaster
- Example X: Bungee Jumper
- 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
- Example XII: AP-C 2007 FR3
- Example XIII: AP-C 2010 FR1
- Example XIV: AP-C 2013 FR1

- 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

### AP Physics C: Mechanics Online Course

I. Introduction | ||
---|---|---|

What is Physics? | 7:12 | |

Math Review | 1:00:51 | |

II. Kinematics | ||

Describing Motion I | 23:47 | |

Describing Motion II | 36:47 | |

Projectile Motion | 30:34 | |

Circular & Relative Motion | 30:24 | |

III. Dynamics | ||

Newton's First Law & Free Body Diagrams | 23:57 | |

Newton's Second & Third Laws of Motion | 23:57 | |

Friction | 20:41 | |

Retarding & Drag Forces | 32:10 | |

Ramps & Inclines | 20:31 | |

Atwood Machines | 24:58 | |

IV. Work, Energy, & Power | ||

Work | 37:34 | |

Energy & Conservative Forces | 28:04 | |

Conservation of Energy | 54:56 | |

Power | 16:44 | |

V. Momentum | ||

Momentum & Impulse | 13:09 | |

Conservation of Linear Momentum | 46:30 | |

Center of Mass | 28:26 | |

VI. Uniform Circular Motion | ||

Uniform Circular Motion | 21:36 | |

VII. Rotational Motion | ||

Rotational Kinematics | 32:52 | |

Moment of Inertia | 24:00 | |

Torque | 26:09 | |

Rotational Dynamics | 56:58 | |

Angular Momentum | 33:02 | |

VIII. Oscillations | ||

Oscillations | 1:01:12 | |

IX. Gravity & Orbits | ||

Gravity & Orbits | 34:59 | |

X. Sample AP Exam | ||

1998 AP Practice Exam: Multiple Choice | 28:11 | |

1998 AP Practice Exam: Free Response Questions (FRQ) | 28:11 |

### Transcription: Conservation of Energy

*Hello, everyone, and welcome back to www.educator.com.*0000

*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 to*0056

*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 forces*0114

*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 Mv²*0258

*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 dealing*0474

*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 to*0519

*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 displacement*0560

*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 energy*0614

*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 energy*0630

*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 0.05²*0695

*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 bottom*0926

*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 forest*0977

*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 cliff*1010

*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/s*1078

*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 D*1169

*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 out*1279

*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 energy*1486

*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/s²*1556

*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 energy*1634

*at B - the gravitational potential energy at B - the spring potential energy at B*1642

*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 own*1839

*as you pause the video, give it a minute or 2 then come back here and check your answers*1844

*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 + 2⁻¹*1934

*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 and*1981

*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 about*2057

*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 travel*2073

*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 and*2245

*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 h*2378

*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 energy*2394

*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 to*2735

*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 E*2841

*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 boxes*2999

*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 www.educator.com and make it a great day everybody.*3293

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