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

### Work

- Work is the process of moving an object by applying a force. The object must move for work to be done, and the force must cause the movement.
- Work is a scalar quantity. The units of work are joules.
- Work is calculated as the integral of scalar product of the force vector with the differential of displacement. In one dimension, work is the area under the force vs. displacement graph.
- The force from a spring is opposite the direction of its displacement from equilibrium, therefore it is a restoring force.
- Fspring=-kx
- The slope of a graph of the force from a spring against its displacement from equilibrium gives you the spring’s “spring constant.”
- The work done on an object changes the object’s energy.

### Work

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
- What is Work?
- Work in One Dimension
- Examples of Work
- Stuntman in a Jet Pack
- A Girl Struggles to Push Her Stalled Car
- A Child in a Ghost Costume Carries a Bag of Halloween Candy Across the Yard
- Example I: Moving a Refrigerator
- Example II: Liberating a Car
- Example III: Lifting Box
- Example IV: Pulling a Wagon
- Example V: Ranking Work on Carts
- Non-Constant Forces
- Force vs. Displacement Graphs
- Hooke's Law
- Determining the Spring Constant
- Work Done in Compressing the Spring
- Example VI: Finding Spring Constant
- Example VII: Calculating Spring Constant
- Example VIII: Hooke's Law
- Example IX: Non-Linear Spring
- Work in Multiple Dimensions
- Work-Energy Theorem
- Example X: Work-Energy Theorem
- Example XI: Work Done on Moving Carts
- Example XII: Velocity from an F-d Graph

- Intro 0:00
- Objectives 0:07
- What is Work? 0:36
- What is Work?
- Units of Work
- Work in One Dimension 1:31
- Work in One Dimension
- Examples of Work 2:19
- Stuntman in a Jet Pack
- A Girl Struggles to Push Her Stalled Car
- A Child in a Ghost Costume Carries a Bag of Halloween Candy Across the Yard
- Example I: Moving a Refrigerator 4:03
- Example II: Liberating a Car 4:53
- Example III: Lifting Box 5:30
- Example IV: Pulling a Wagon 6:13
- Example V: Ranking Work on Carts 7:13
- Non-Constant Forces 12:21
- Non-Constant Forces
- Force vs. Displacement Graphs 13:49
- Force vs. Displacement Graphs
- Hooke's Law 14:41
- Hooke's Law
- Determining the Spring Constant 15:38
- Slope of the Graph Gives the Spring Constant, k
- Work Done in Compressing the Spring 16:34
- Find the Work Done in Compressing the String
- Example VI: Finding Spring Constant 17:21
- Example VII: Calculating Spring Constant 19:48
- Example VIII: Hooke's Law 20:30
- Example IX: Non-Linear Spring 22:18
- Work in Multiple Dimensions 23:52
- Work in Multiple Dimensions
- Work-Energy Theorem 25:25
- Work-Energy Theorem
- Example X: Work-Energy Theorem 28:35
- Example XI: Work Done on Moving Carts 30:46
- Example XII: Velocity from an F-d Graph 35:01

### 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: Work

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

*I am Dan Fullerton and in this lesson we are going to talk about work.*0003

*Our objectives include calculating the work done by a force of an object undergoing a displacement that displacement is going to be very important.*0008

*Relating the work done to the area under graph of forces a function of position.*0016

*Using integration to calculate the work performed by a non constant force on an object undergoing a displacement.*0021

*Using a dot product to calculate the work performed by a specified constant force on an object undergoing displacement on a plane.*0028

*Suppose we are going to start by talking about what is work?*0036

*Work is the process of moving object by applying the force.*0040

*There are keys there the object has to move for work to be done.*0045

*You can push and push and push on something if it does not move you are not doing any work.*0049

*The force also must cause the movement.*0055

*If you apply a force up to something just a holdup there you keep holding it you are applying a force that is not moving *0059

*or it is moving but it is not because of the force your applying you are not the one doing the work.*0065

*Work is a scalar quantity and its units are joules.*0070

*You also can have positive and negative work.*0074

*For positive would be the work you are doing.*0077

*Negative would be work done on you but that does not have a direction north south east west.*0079

*Work is a scalar units are joules.*0085

*Let us take a look at work in one dimension.*0090

*Only the force in the direction of the displacement contributes to the work done.*0093

*If we have a box with the force at some angle θ moving through some displacement Δ x,*0096

*if we want to find the work done we need to know the component of the force that actually causing the displacement.*0102

*This vertical component is not helping you cause the displacement.*0108

*What is really doing that it is the horizontal component of that force which is going to be F cos θ.*0112

*If we go to find the work done our work is going to be F cos θ.*0121

*Δ x or F dotted with Δ x, it is a dot product there.*0129

*Let us look at some examples.*0139

*Let us assume a stuntman in a jet pack blast through the atmosphere accelerating the higher and higher speeds.*0141

*Sounds exciting, perhaps a touch of danger.*0147

*The jet pack is applying to force causing it to move.*0149

*How is expanding gasses are pushed a backward of a jet pack with a reactionary force *0152

*by Newton’s third law that pushes the jet pack forward causing a displacement?*0157

*The expanding exhaust gas is doing work on the jet pack causing that displacement.*0162

*As a second example of a girl struggles to push her stalled cart but cannot make it move.*0170

*Pushes and pushes and pushes just tired out and turning red she expands tons of effort but the cart does not move.*0176

*If the cart is not moving no work is done.*0184

*Do not equate the amount of work you do with how much energy you are exerting. *0187

*How hard you are trying for example.*0196

*In the physics sense work requires a displacement.*0199

*Another example, a child in a ghost costume carries a bag of Halloween candy across the yard.*0204

*The child applies a force upward on the bag but the bag is moving horizontally.*0210

*The forces of the child arms on the bag are not the one that is causing the displacement so no work is being done by the child's arms on the bag.*0216

*If he wanted a little more complicated the child's pushing with the legs, the legs are causing a forward displacement.*0225

*At some point the child is doing some work on it.*0232

*As far as the arms are just holding up that is not really doing any work because it is not in the direction of the displacement.*0234

*Let us do some examples.*0243

*An appliance salesman pushes a refrigerator 2m across the floor by applying the force of 200 N, find the work done.*0245

*Our displacement Δ X is going to be 2m, the applied force is 200 N, the angle between them is 0°.*0254

*The forces in the direction of the displacement and cos θ if cos 0 which is just 1.*0264

*Work which is force dotted with Δ X or F cos θ, Δ X is just going to be 200 N × 2m which is 400 joules.*0272

*My straightforward simple application.*0290

*On the other hand, your buddy’s car is stuck on the ice.*0295

*You push down on the car to provide more friction for the tires because you are increasing the normal force by pushing down on the car.*0299

*Get more friction allowing the car's tires to propel it forward 5 m on the west with the ground.*0307

*How much work did you do?*0312

*The key here you are pushing down and although that is allowing the car to move and you are not the one doing work.*0315

*The car is the one that is providing the force the cause of the displacement.*0321

*You do a 0 work on the car.*0324

*Example 3, how much work is done the lifting of 8 kg box from the floor to a height of 2m above the floor?*0331

*Work must F dotted with Δ x.*0339

*Our force, the force that you need to lift that box has to at least equal the force of gravity *0343

*so that is going to be MG for force and the displacement is going to be your height 2m *0349

*so that is going to be 8kg × acceleration due to gravity 10 m / s² × height 2m or 160 joules.*0356

*Let us take a look at the pulling wagon example.*0373

*Barry and Sydney pull a 30 kg wagon with the force of 500 N the distance of 20 m.*0377

*The force acts at an angle 30° from horizontal, find the work done.*0383

*Work is F dotted with Δ x but all of that force is in the direction of the displacement.*0391

*Only the x component is which is going to be F cos θ or F cos 30°.*0398

*That is going to be equals 500 N the total force × the cos 30° to give you the component in the direction the displacement × 20 m.*0406

*All that together and I come up with about 8660 joules.*0421

*Let us do a ranking test.*0432

*4 carts initially at rest on a flat surface are subjected to vary in forces is the carts moved *0435

*to the right to set distance is depicted here in the diagram below.*0440

*Rank the 4 carts from least to greatness in terms of the work done by the applied force on the carts.*0444

*Their inertia and the normal force applied by the surface to the cart.*0451

*Alright, let us give ourselves some more room here.*0457

*We had just blown up and the way I do a ranking test like this is probably make a table of information *0460

*especially when you have multiple questions around the same setup.*0467

*You can do them all at once.*0470

*We are going to need to know items for cart A, B, C and D.*0472

*The things we are going to want to know include the force, angle, the force acts with compared to the displacement.*0479

*The displacement we will call that D, the work done W.*0487

*We need to know the mass that is the measure of inertia and the normal force.*0492

*Starting with the force we can read that right off the base we have 20 N, 30 N, 25 N, and 50 N.*0499

*The angle we can read write off the diagrams as well.*0510

*We have 30° for A, 60° for B, 30° for C, and 0° for D and their displacements can come right off here as well.*0513

*We have 8m, 12m, 8m, and 6 m.*0528

*To find the work done, that is just going to be a component of the force in the direction of the displacement F cos θ × that displaced in itself.*0535

*Our work done here F cos θ 20 Cos 30 × 8 =139 joules going down 180 joules for B, 173 joules for C and 90 joules for D.*0546

*We can read their mass the measure of their inertia right off the graphs as well.*0563

*Our mass is going to be 10 kg, 20 kg, 15 kg, 12 kg.*0567

*For the normal force, it might be helpful draw free body diagrams here.*0577

*Let us go up to A and realize that we have weight down.*0581

*We also have the vertical component of the force which is going to be F sin θ that is going to be F sin 30° in the normal force acting up.*0587

*These have to be in balance because our car is accelerating up off the plane.*0601

*The normal force must equal MG + F sin 30.*0606

*Normal force = MG + f sin 30 I come up with 110 N.*0610

*Same idea over here for B but now we have our force acting upwards of the vertical component is going to be F sin 60°.*0618

*We have the normal force and we have MG.*0628

*The normal force + F sin 60 have to equal MG.*0632

*The normal force is MG –f sin 60 is going to give us 174 N.*0637

*Looking at C, we have free body diagram.*0646

*We have F sin 30°, we have our normal force up, we have our weight M.*0655

*Once again, the same basic design as we had in B but with F sin 30 in the different force.*0670

*For C, I come up with about 137.5 N for the normal force.*0676

*D is a nice straightforward we have normal force up, MG down.*0683

*The normal force equals the force of gravity is 120 N/ our normal force 120.*0688

*To do our ranking, for work we are going to rank from smallest to greatest.*0698

*We have D, A, C, and B.*0707

*For inertia, that is just mass is a measure of inertia so we start off with A, D, C, B.*0714

*For our normal force, we have let us see A, D, C, and B from smallest to greatest.*0727

*What happens when you have a force that varies? A non constant force.*0740

*Here, we are showing a graph of force that varies as a function of the displacement.*0744

*How would you deal with something like that?*0750

*Here, we can go to some of our calculus skills and work done because the area under the force vs. displacement graph.*0753

*If we wanted the entire work done we can take the entire area under that but how we can to find something like that?*0763

*Here is where integration is going to come in so handy.*0771

*What we can do is we can break up our very detailed curve into a bunch of tiny little rectangles *0774

*that we are going to approximate as rectangles that make those skinnier until infinitesimally thin.*0784

*Then we add up all of those to give us the total work.*0791

*We will call that with dx and the height is just F(x) at that point dx.*0795

*Our total work done is going to be the integral the sum of all these little rectangles from some initial value of x *0803

*to some final value of x, F(x) the height of our rectangle × width dx, that infinitesimally small piece of width.*0813

*Let us do that with some examples.*0828

*The area under force displacement graph is the work done by the force.*0830

*Consider the situation of a block pull across a table with the constant 5 N force over displacement 5m *0834

*and then net force tapers off over the next 5m, find the work done.*0841

*We do not actually have to use any calculus here.*0846

*We can use common sense and say the area under the graph here that is a rectangle that is going to be 5 N × 5m or 25 joules length × width.*0849

*Here we have a triangle ½ base height or by observation you can probably look at that *0861

*and say that is half of what we have over there 12.5 joules.*0866

*Add those up and find the work done is to 37.5 joules.*0872

*Let us take a look at Hooke’s law talking about springs.*0882

*The more you stretch or compress a spring the greater the force of the spring.*0885

*It always wants to restore.*0889

*If you squish it, it wants to go back to where it was.*0890

*If you pull it, it wants to go back to where it was.*0893

*We call it a restoring force.*0895

*The spring’s forces opposite the direction of its displacement.*0897

*We are going to model this as a linear relationship where the force applied by the spring is equal to some constant.*0901

*We will call that k the spring constant multiplied by the springs displacement from its equilibrium or rest position.*0907

*We call this relationship Hooke’s law where the force of the spring is equal – KX.*0914

*Assuming that we have that when your relationship between force and the displacement from the equilibrium position.*0920

*Not all objects follow Hooke’s law.*0928

*It is an empirical law not a pure law of physics that is usually a pretty good approximation.*0931

*Let us take a look at how we might determine the spring constant for a spring.*0937

*If we make a graph of force vs. Displacement the slope of the graph is going to be our spring constant K rise over run *0941

*which is going to be our change in force over change in displacement.*0952

*For something like this, that is going to change let us pick 2 points on our line.*0957

*There are a couple easy ones.*0961

*20 N -0 N / 0.1m -0 m = 20/0.1 or 200 N/m would be our spring constant.*0963

*If the spring was there for we have a higher spring constant.*0977

*If it was when shares had less force per unit extension would have a smaller spring constant.*0981

*K = 200 N/m we found by taking the slope of the force vs. Displacement graph.*0986

*How about compressing the spring?*0994

*Find the work done as we compress a spring from 0 to .1 meter?*0996

*Well, the work done remember is the area under force vs. Displacement graph.*1001

*We could take the area under the graph in this case makes a nice happy triangle.*1006

*Remember the area of the triangle is ½ base height.*1013

*The work is our area is ½ base × height is going to be ½ the base of our triangle is 0.1m.*1017

*The height is 20 N just going to give us it takes 1 joule.*1028

*It takes 1 joule of work in order to compress that spring 0.1m.*1034

*Take a look at another example where we are trying to find the spring constant.*1041

*A spring is subjected to a varying force and its elongation is measured.*1045

*Determine the spring constant of the spring.*1049

*First thing we are given some forces in elongation and very popular in AP exams is to have the plot the data.*1052

*It is probably worth to take a minute in doing that here.*1058

*We have a point at 00, we have a point at a force of 1 N, elongation of 0.3 that is going to be right around here.*1061

*We have 8.67m, 3 N force, so that is going to be somewhere right in here at 1m a force of 4 N.*1072

*At 1.3m, we have 5 N so that is going to be right around here and finally at 1.5m we have 6 N.*1087

*We are going to do is draw our best fit line, we are not connecting the dots.*1102

*We are drawing a line that has best fits our data.*1107

*There are lots of different ways to do this but typically if you target roughly the same amount of points above *1111

*and below the line by the same distance you are in the ballpark of what we are after.*1116

*But we are not connecting the dots.*1120

*Our actual best fit line actually does not have to go through any of the data points.*1122

*We will see that looks like it goes through a couple there when we find the slope.*1127

*To find the spring constant, the key is we do not use data points instead we use points on our best fit line.*1135

*If they happen to overlap the data points that is great but it does not have to be the case.*1142

*As I look at my line I see a bunch of places I could go.*1147

*Let us see.*1151

*It looks like we have got a pretty good one right there that is right in a grid match and that should be easy to see.*1153

*And 00 there with line looks pretty easy.*1160

*Our slope is rise over run between these two we have a rise of 5 N and our run is from 0 to 1.25m gives us 4 N/m.*1163

*That must be our spring constant.*1185

*Another example at 10 N force compresses a spring ¼ m from its equilibrium position.*1190

*Find the spring constant of the spring.*1196

*Our force is 10 N and our displacement from the equilibrium or happy position of the spring is 0.25 m.*1199

*Our spring constant that is going to be the force divided by the displacement by Hooke's law which is 10 N / 0.25m or 40 N/m.*1210

*Taking a step further we have a spring that obeys Hooke’s law where F =kx.*1230

*We are looking at the magnitude of the force here.*1235

*The negative on the Hooke’s law just means it is a restoring force.*1237

*How much work is done and compressing the spring from equilibrium to some point x?*1241

*The work done and it is going to be the integral of F • dx.*1248

*We are going to integrate from some X =0 and its equilibrium position to some final point x.*1255

*The force is KX and we have our dx.*1261

*K is a spring constant.*1269

*It can be pulled out of the integral sign.*1271

*K integral from 0 to x of x and dx complies the net work done is going to K the integral of x is x² / 2.*1273

*Our limits of integration we have to evaluate that from 0 to x which is going to K × what these mean again *1286

*we are going to take this value and plug it into our variable.*1294

*We got x² /2 - and then we take our bottom variable 0 and plug it in.*1298

*0² /2 is 0 so we end up with ½ K c².*1305

*If the work that we doing compressing the spring that amount is ½ KX² where did that energy go?*1311

*It is stored potential energy in the spring so what this is really saying is that the potential energy stored in the spring is ½ KX².*1318

*We just found a formula for the potential energy of a spring that is Hooke’s law.*1331

*We said not all springs obey Hooke’s law.*1338

*What if you have a nonlinear spring?*1340

*When it does not obey Hooke’s law?*1342

*Here we have an example where the force required to extend this spring is half k² that is the force not the store potential energy.*1344

*How much work is done in compressing this spring from equilibrium to some point x?*1353

*We can follow the same basic procedure.*1358

*Work is going to be the integral from x equal 0 to some final value x of F • dx which will be the integral *1360

*from 0 to x but now our force is ½ KX² dotted with dx.*1371

*We will pull our constant out of the integral.*1379

*That will be k/2 can come out integral from 0 to x of x² Dx.*1381

*Which can b k/2 the integral of x² is x³/3.*1389

*We will evaluate that from 0 to X.*1395

*This implies them that the work done is K/2 and we are going to take X and plug in for a variable.*1398

*X³/3 – the bottom value 0 plug in for a variable -0³/3 which is just 0.*1408

*Therefore we, find that we have kx³/6 that is the work required to compress this nonlinear spring.*1419

*Let us talk about now work in multiple dimensions.*1432

*Assume we have an object moving along some path here and we are going to break up the path of the little tiny bits or going to call Dr.*1435

*There it is at this point.*1443

*The force does not have to be completely in that direction.*1445

*The force at this little bit of piece of our path on this dr is in that direction.*1448

*There is F and the way they can find the work for the entire path is to find just little tiny bits of work *1453

*we do over each a little bit of the path dr or sometimes we will write that as DL.*1461

*Add all of those little bits of work do all the little pieces in the path.*1467

*Our little tiny bit of work that we get for that little piece is going to be our force dotted with our little bit of displacement along the path Dr.*1472

*Our total work is just going to be the integral of all these little bit of dw.*1483

*Add all of these infinite tests with small pieces of tiny bits of work to get the entire work.*1489

*Or we can say that is going to be the integral from some initial position vector or one *1495

*which takes you to over here to some final position vector over here r2 of F dotted with Dr.*1500

*That is going to be a more general formula that we can use for work.*1510

*Work = integral of f • dr.*1514

*There is a generalized version we can use that covers one dimension and multiple dimensions.*1519

*From this so we can actually derive the work energy theorem.*1526

*Work is the integral from some initial X value to some final x value of F(x) dx.*1530

*We also know that force is mass × acceleration.*1541

*Acceleration is the derivative of velocity with respect to time.*1547

*Or we can write we know velocity is the derivative of the position or dx dt therefore DX = VDT even when we rearrange them.*1553

*We could then write work is equal to the integral.*1566

*We are going to replace F with mdv dt and we are going to replace Dx with our vdt.*1571

*Why we do that?*1584

*We have a nice little simplification here DT / DT makes a ratio of 1 and we end up with the integral of MV dv.*1587

*Our variable of integral now is velocity.*1595

*We are going to integrate over the limits from some V equals V initial to some final velocity of M DV × V.*1598

*This implies then that their work is the integral from our initial to our final velocity of MV dv.*1609

*Mass in this case should be a constant we can pull that out of the integral sign.*1621

*Which is going to be M integral from the initial to V final V DV which is going to be M the integral V is V² /2 evaluate it from V initial to V final.*1626

*Which is going to be M × we will take our top variable plug it in for V final²/2 - our bottom value for variable V initial²/2.*1641

*Which implies then kinetic energy is ½ mv².*1658

*What we have here is that work is equal to all we have ½ V final² – ½ V initial² that is going to be *1666

*final kinetic energy - initial kinetic energy which is our change in kinetic energy.*1678

*Work equals a change in kinetic energy.*1686

*There is the work energy theorem.*1689

*You are on something like a horizontal surface and dealing only with conservative forces we are neglecting friction things like that.*1692

*The work that you do on the object becomes the change in its kinetic energy.*1697

*When you do work on something you give an energy.*1703

*When the object does work on something else it gives it energy.*1706

*Work is a transfer of energy of sorts.*1710

*Let us take a look at an example what is the work energy theorem.*1714

*A pickup truck with a mass of 1000 kg is traveling at 30 m /s, the driver sees a dog on the road 31m ahead.*1717

*What force must the brakes exert in order to stop the truck?*1728

*It is going to come to rest in the distance of 30 m assuming constant acceleration.*1732

*What do we know?*1738

*Mass is 1000 kg, our initial velocity is 30 m/s, our final velocity is going to be 0.*1740

*We need all this to happen in a displacement of 30m.*1755

*Our net work is going to be the change in kinetic energy to be the final kinetic energy - the initial kinetic energy.*1760

*The final we want to be 0 since we want it to be at rest and that is going to be 0 -1/2 M 1000 × its initial velocity squared.*1769

*30² 900 × 1000 × - ½ is -450,000 joules.*1781

*But we also know that the net work is the force × Δ x.*1791

*Assuming we have that constant force here.*1798

*Therefore force is our net work divided by Δ x or -450,000 joules /30 m.*1802

*We get a force of -15,000 N.*1818

*There is our answer.*1823

*What is that negative mean?*1826

*All that just means that the force you are applying is opposite the direction we initially called positive which is the direction of the initial velocity.*1828

*The force is opposite direction and velocity which you need in order to slow it down to bring to a stop.*1836

*Let us take a look at the work done on moving carts.*1843

*Another ranking test.*1846

*In the following diagrams of force F x on a cart in motion on a fictional surface to change its velocity.*1848

*The initial and final velocity of each cart is shown on the left, final on the right.*1854

*You do not know how far in which direction the car traveled?*1859

*The magnitude of the energy change of each cart from greatest to least.*1862

*With that these different carts let us give ourselves a little bit more room on the next page to do some work here.*1868

*Our ranking leaves from the greatest to least.*1875

*As we go to our next page with a little more room we are going to make a table again to help me out here.*1878

*Our carts, we will list A, B, C, and D.*1885

*Things are going to want to know will probably the mass of the cart, the initial velocity, the final velocity, *1889

*the initial kinetic energy, final kinetic energy, and we will call that the magnitude of the work done or energy change.*1898

*That one is going to get a little tricky.*1907

*Let us go through this.*1909

*For A, our mass is 2kg B3 C5 and D4.*1911

*Our initial velocity, 5, 3, 5, -1.*1917

*For final velocity, 2, -3, 6, and 2.*1925

*Our initial kinetic energy ½ mv² we have 25, we have ½ mv², here is going to be 13.5, we have 62.5.*1933

*For the final, ½ mv² is just going to be 2.*1947

*For the final kinetic energy is ½ mv final² is going to give us 4, 13.5, 90, 8.*1953

*To get this piece remember that the work done is the change in kinetic energy.*1967

*It is pretty easy on some of these to see what is going on.*1973

*Final - initial or initial – final.*1975

*We are looking at the absolute value that will be 21.*1977

*Here we have 13.5 and notice it is going one direction initially in the opposite direction when we are done .*1981

*This is where you get into some tricky.*1990

*If work is changing kinetic energy to kinetic energy change but direction change.*1992

*The kinetic energy is not a scalar.*1998

*It gets to be kind of a tricky question and really has a lot to do with semantics.*2001

*You are doing work on the cart in one direction it is going the opposite way.*2005

*Are you doing positive work or negative work?*2009

*Which all you can get into an argument in semantics.*2012

*What is really going to be useful in that?*2016

*Probably what you want to know was how much energy it is going to take in order to change its velocity that much.*2018

*And that should be pretty straightforward.*2024

*It takes 13.5 joules to get to 0 and another 13.5 to get back to 3 m /s going the opposite direction.*2026

*We are going to interpret that as the energy needed which is why the question is worded the way it is and say that is going to be 27 joules.*2035

*We did a little bit of thinking there about what we are really after and what the questions asking.*2044

*You can get lost in semantics on these and not worth spending all that time figuring out exactly what your quick question means.*2049

*Instead what are you trying to accomplish with the question?*2057

*Take that answer.*2059

*Down here for C we go from 5 to 6 that change in kinetic energy is going to be 27.5.*2062

*That is the energy we have made.*2070

*For the last one, we are going from -1 to 2.*2072

*We are going from one direction to the other the amount of energy would take for us to do that is going to be 10 joules.*2077

*That is how we are going to interpret and use this question.*2083

*Therefore, if we are going from greatest to least the order that we would do this in would C, B, A, and D.*2086

*Let us take a look at one last question here.*2099

*Given a force versus displacement graph below for net force applied horizontally to an object of mass M at initially at rest on the frictionless surface.*2103

*Determine the objects final speed in terms of F net, these r1 and r2 and r3 points and mass.*2112

*You may assume the force does not change its direction.*2119

*We got a force vs. displacement graph.*2123

*If we want the work done and we could take the area that and if the work is the change in kinetic energy we can go from there to solve for velocity.*2126

*Work is going to be the area under a graph which is ½ base × height + width of the second rectangle + ½ base height for this third triangle.*2137

*The second triangle our third section of the curve.*2153

*This implies then we have got ½ r1 × F max + this distance here is going to be r2 – r1 × height F max + ½.*2158

*Our base here is going to be r3 – r2 F max and all of that has to equal ½ mv² by the work energy theorem.*2178

*This implies then that velocity squared we can multiply everything else by 2/M is going to be let us factor out f max / M.*2191

*I would get r3, -r2, +r2.*2204

*We will have + r2 – r1 which implies then that our velocity is going to be the square root of all of that which will be f max /m × (r3 + r2 – r1).*2215

*We can get the velocity even from the area under our force displacement graph.*2240

*Hopefully that gets you a good start on work.*2247

*Thank you so much for joining us at www.educator.com.*2249

*I hope to see you again real soon and make it a great day everyone.*2251

3 answers

Last reply by: Professor Dan Fullerton

Sun Aug 21, 2016 9:39 AM

Post by Cathy Zhao on August 20, 2016

On Example X, why you got -15,000 N instead of 15000N? How come you got a negative number? Thanks!

2 answers

Last reply by: Parth Shorey

Tue Sep 29, 2015 7:38 PM

Post by Parth Shorey on September 29, 2015

I still don't understand how you got 1/2(1000) on the work-energy theorem problem ?

3 answers

Last reply by: Professor Dan Fullerton

Wed Sep 30, 2015 6:36 AM

Post by Parth Shorey on September 29, 2015

Does the inegral represent the initial starting point and final on work-energy theorem?

1 answer

Last reply by: Professor Dan Fullerton

Mon Mar 9, 2015 6:50 PM

Post by Nike Oyedokun on March 9, 2015

Which Lecture is equivalent to College introduction to Physics