AP Physics C: Mechanics 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 moving0059

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 force0152

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 gravity0343

so that is going to be MG for force and the displacement is going to be your height 2m0349

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 moved0435

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 information0460

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 rectangles0774

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 x0803

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 5m0834

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 that0861

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 run0941

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 above1111

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 again1286

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 integral1360

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 work1453

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 one1495

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 be1666

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