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Lecture Comments (27)

1 answer

Last reply by: Professor Dan Fullerton
Tue Oct 10, 2017 5:55 AM

Post by Sahitya Senapathy on October 10, 2017

For example 15, would it be -1000N because Ff is done in the opposite direction of Fbob so you multiply by cos(180)? That's what you did for example 13, so I'm a bit confused.

1 answer

Last reply by: Professor Dan Fullerton
Mon May 15, 2017 7:25 AM

Post by Parsa Abadi on May 14, 2017

for example 5 shouldnt gravity be negative? since up is positive.

1 answer

Last reply by: Professor Dan Fullerton
Mon Mar 6, 2017 6:14 AM

Post by Woong Ryeol Yoo on March 4, 2017

Hi mr Fullerton

I have a question on 5.34 Q (A) in your AP Physics 1 Essentials 2nd edition.

Since it's asking for the potential energy, in addition to the elastic potential energy, do we have to take the gravitational potential energy into account as well since the object has height?

1 answer

Last reply by: Sarmad Khokhar
Sat Dec 10, 2016 11:01 AM

Post by Sarmad Khokhar on December 10, 2016

Is Example 6 answer wrong  because  the purple triangle has area of 6 not 7.5

1 answer

Last reply by: Professor Dan Fullerton
Thu Aug 6, 2015 8:46 AM

Post by Anh Dang on August 4, 2015

in example 13, where'd you get the cos(180) from?  How did you get it and why?  I'm a bit confused.

2 answers

Last reply by: Abhishek Raj
Fri Dec 12, 2014 1:47 AM

Post by Abhishek Raj on December 11, 2014

If possible provide me solution of the following question.
I've taken a pic of this question please solve and let me know.

1 answer

Last reply by: Daniel Fullerton
Thu Oct 30, 2014 6:11 AM

Post by Foaad Zaid on October 29, 2014

Hello Professor, for the work done by friction example, how come it wouldn't be appropriate to plug the normal force (calculated in the y-component of newton's 2nd law) and multiply that by the constant to obtain the force of kinetic friction? Followed by multiplying that by the displacement of 10m? Thank you in advance.

1 answer

Last reply by: Professor Dan Fullerton
Mon Sep 29, 2014 9:10 PM

Post by Max Starr on September 29, 2014

on example 8 why isn't the spring constant negative?

1 answer

Last reply by: Professor Dan Fullerton
Tue May 7, 2013 6:18 AM

Post by Nawaphan Jedjomnongkit on May 7, 2013

From the example of work done by friction that you use friction force = Force apply in x , so what about in y ? If think about f=uN and from free body diagram that you draw will give N = 800 - 150 which is 650 and the friction force will be 650x0.3 N which give f=195N ???

7 answers

Last reply by: Professor Dan Fullerton
Thu Jul 7, 2016 5:38 PM

Post by natasha plantak on April 15, 2013

At 12:38 (example #6) wouldn't the first triangle be 1/2 x 4 x 3= 6 instead of equaling 7.5, since the height of the triangle is 4 not 5?

Work & Power

  • Work is the process of moving an object by applying a force. W=Frcosθ.
  • Only the component of force in the direction of the object's displacement contributes to the work done.
  • The area under a force vs. displacement graph is the work done by a force.
  • Hooke's Law is an empirical law describing the restoring force from a stretched or compressed spring. F=-kx, where k is the spring constant (in N/m), and x is the spring's displacement from its equilibrium position.
  • Power is the rate at which a force does work. P=W/t=Fvcosθ.

Work & Power

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

  • Intro 0:00
  • Objectives 0:09
  • What Is Work? 0:31
    • Power Output
    • Transfer Energy
    • Work is the Process of Moving an Object by Applying a Force
  • Examples of Work 0:56
  • Calculating Work 2:16
    • Only the Force in the Direction of the Displacement Counts
    • Formula for Work
  • Example 1: Moving a Refrigerator 3:16
  • Example 2: Liberating a Car 3:59
  • Example 3: Crate on a Ramp 5:20
  • Example 4: Lifting a Box 7:11
  • Example 5: Pulling a Wagon 8:38
  • Force vs. Displacement Graphs 9:33
    • The Area Under a Force vs. Displacement Graph is the Work Done by the Force
    • Find the Work Done
  • Example 6: Work From a Varying Force 11:00
  • Hooke's Law 12:42
    • The More You Stretch or Compress a Spring, The Greater the Force of the Spring
    • The Spring's Force is Opposite the Direction of Its Displacement from Equilibrium
  • Determining the Spring Constant 14:21
  • Work Done in Compressing the Spring 15:27
  • Example 7: Finding Spring Constant 16:21
  • Example 8: Calculating Spring Constant 17:58
  • Power 18:43
    • Work
    • Power
  • Example 9: Moving a Sofa 19:26
  • Calculating Power 20:41
  • Example 10: Motors Delivering Power 21:27
  • Example 11: Force on a Cyclist 22:40
  • Example 12: Work on a Spinning Mass 23:52
  • Example 13: Work Done by Friction 25:05
  • Example 14: Units of Power 28:38
  • Example 15: Frictional Force on a Sled 29:43

Transcription: Work & Power

Hi everyone. I am Dan Fullerton and I am thrilled to welcome you back to

Today we are going to talk about work and power.0004

Now, our objectives are going to be, first, to define work, to calculate the work done by a force, to utilize Hooke's Law to describe the force that you get from compressing or stretching a spring, recognizing power as the rate at which work is done...0007

...and finally, calculating the power supplied for a variety of situations.0023

So with that, why don't we dive in and talk about what is work.0027

Well, you do work on an object when you move it and the rate at which you do work is your power output.0031

When you do work on an object you transfer energy from one object to another.0038

That is a key point here -- work transfers energy.0043

So work is the process of moving an object by applying a force.0046

An object must be moving when you apply a force, therefore you do work.0051

Now to give some examples of work -- A stunt man in a jet pack blasts through the atmosphere accelerating to higher and higher speeds.0056

We have a force causing an object to move.0063

The jet pack is applying a force causing it to move, the hot expanding gases are pushed backwards out of the jet pack, and the reactionary force -- Newton's Third Law of the Gas -- is pushing the jet pack forward causing a displacement.0066

You need to have that displacement for work.0078

And the expanding exhaust gas, therefore is doing work on the jet pack.0081

Let us take another example -- A girl struggles to push her stalled car, but cannot make it move.0088

She expends a lot of effort; she is sweating; she is feeling like she is doing a lot of work, but from a physics' perspective, no work is being done since the car is not moving.0092

Very different definition -- every day work, compared to the physics' definition of work.0104

Another example -- We have a child in a ghost costume at Halloween, carrying a bag of candy across the yard.0109

If the child applies a force horizontally upward on the bag, but the bag is moving horizontally, the forces of the child's arms on the bag are not what causes the displacement.0115

The force of the child's hands on the bag is up -- the displacement is horizontal, therefore, no work is being done by the child's arms due to the force that is upwards.0124

When we want to calculate work quantitatively, we will use the formula -- work is equal to the force times the object's displacement.0136

And the units of work are going to be Newton-meters (N-m), force times distance, or joules (J).0144

Only the force in the direction of the displacement counts for calculating quantitatively the work.0154

When the force and displacement are not in the same direction, you must take the component of the force that is in the direction of the displacement.0161

So you could write work as F cos(θ), where θ is the angle between the object's displacement vector and the force vector times that displacement vector, so F cos(θ) times δr or F(δr)cos(θ).0167

And of course, if the force and the displacement are in the same direction, θ is 0, cos(θ) is 1, and that term is just going to cancel out -- you will just have F(δr).0183

So let us take a look at an example of moving a refrigerator.0196

An appliance salesman pushes a refrigerator 2 m across the floor by applying a force of 200N.0200

Let us find the work done.0206

Well let us start off with our formula -- work equals force times displacement (δr), which is going to be 200N and our displacement is 2 m or 400N-m.0208

And as we just discussed 400N-m is also known as a 400 J, so our answer there would be 400 J.0225

How about liberating a car? A friend's car is stuck on the ice.0239

You push down on the car to provide more friction for the tires by way of increasing the normal force -- remember back from dynamics, the frictional force is μ times the normal force.0243

If you push down, you will get more normal force up, which means you are going to get more frictional force.0254

However, that allows the car's tires to propel it forward 5 m on the less slippery ground.0258

How much work did you do?0263

Well, this is kind of a tricky question because the force you are applying is in the downward direction and the car's displacement is horizontal.0265

The force is not in the direction of the displacement, therefore no work is done.0274

Or if you wanted to do that mathematically -- if there is our force vector -- here we have our displacement vector, δr and the angle between them is 90 degrees.0285

So if work is F(δr) cos(θ) -- well since θ equals 90 degrees, and cos(90 degrees) is 0...0294

...then you could say that work equals 0 in this instance.0311

Let us take a look at another example.0320

Let us say that we push a crate up a ramp with a force of 10N.0322

Despite our pushing however, the crate slides down the ramp at a distance of 4 m.0326

How much work did you do?0331

And here is where we are going to have to re-define or maybe clarify that definition of work a little bit.0332

Let us draw a ramp to begin with. There is our ramp on here.0339

Let us put our crate -- and what is going to happen is despite all of our efforts, it is going to move some displacement of δr = 4 m.0345

And as we do this, we are going to apply a force on the box of 10N up the ramp.0360

How much work did we do?0369

Well let us go back to our mathematical definition -- Work = F(δr) cos(θ).0371

Our force is 10N, δr was 4 m and the angle between them -- well if the force is going up the ramp and the displacement is going down the ramp, our angle is going to be 180 degrees.0383

Cos(180 degrees) is -1, so we are going to get an answer of 10 × 4 = 40 × -1 or -40 J.0404

We have done negative work on the box. What does that mean?0413

Well that means that the force was in the opposite direction of the displacement.0419

So we are kind of re-defining that initial definition of work or clarifying that definition.0422

All right, let us take a look at lifting a box.0432

Now we want to find out how much work is done in lifting an 8 kg box from the floor to a height of 2 m above the floor.0434

Let us start with our box -- there it is -- We are going to apply some force (F) in order to make it move, a displacement of about 2 m.0444

Well what force do we have to apply to lift that box off the ground?0455

We have to overcome the force of gravity.0459

So the force we apply has to be equal to mass times the acceleration due to gravity here on the surface of the earth (mg).0461

The work then is going to be F(δr) cos(θ) or in this case, (F) is mg, so we have mg(δr).0470

Force and our displacement are in the same direction, so cos(0 degrees), therefore is 1.0483

The cosine term goes away and we just have mg(δr), or this implies then that work is equal to our mass 8 kg times the acceleration due to gravity.0489

Let us round that off and say that that is roughly 10 m/s2 times the displacement of 2 m -- 8 × 10 = 80 × 2 = 160 J of work.0500

Let us take a look at an example now where we are applying a force that is not specifically in the direction of the displacement.0519

Barry and Sidney pull a 30 kg wagon with a force of 500N, a distance of 20 m.0525

The force acts at an angle of 30 degrees here above the horizontal. Calculate the work done.0531

We will go back to our definition again.0539

Work equals F(δr) times the cosine of the angle between those vectors (θ), so that is going to be 500N, our applied force, times our displacement (20 meters) cos(30 degrees).0541

So we have 500 × 20 × cos(30) is about 8660 J.0561

Let us take a look at force vs. displacement graphs.0574

The area under a force vs. displacement graph is the work done by the force.0577

So if you have a force vs. displacement graph -- if you want to know the work done, just take the area underneath it.0582

Let us consider the situation of a block being pulled across a table with a constant force of 5N for a displacement of 5 m -- so that part of the graph -- and then, over the next 5 m, that force tapers off to 0 in a linear fashion.0589

Find the work done.0604

To do that, all we have to do is take the area of these two sections of our graph.0605

Over here in this section, we have a rectangle -- so the area is going to be the base times the height -- 5 m × 5N = 25 J.0610

Over here we have a triangle.0624

The area of a triangle is one-half base height, or 1/2 × 5 × 5 -- 1/2 of 25 will be 12.5 J for our area here.0626

So the total work done is going to be the area of the first part of our graph, 25 J plus the area of the second part of our graph, 12.5 J, therefore, our work must be 37.5 J.0639

Let us take a look at work from a varying force with an example.0660

A box is wheeled to the right with a varying horizontal force.0664

The graph below represents the relationship between the applied force and the distance the box moves.0667

What is the total work done in moving the box that displacement of 10 m?0673

We have to find the area under the graph in order to find the total work done.0678

And there are a lot of different ways we could break this up, but I like to find nice, simple shapes myself.0682

So what I would probably do is look at something like -- looks like we have a triangle over here and it should be easy to find the area of that purple triangle.0687

Looks like we have another triangle over here.0697

If we find the area of that green triangle, then that will just leave us down here a rather long red rectangle and that will give us the entire area under the graph.0701

The area over here of this triangle -- 1/2 base height.0714

So we have 3 m × 5 = 15 and 1/2 of that will be 7.5 J there.0718

Over here in our green triangle, we have a base of 5 m and we have a height of 3N, so 5 × 3 = 15 and we have 7.5J here again.0725

And our red rectangle has a height of 1 and a length of 10, so that is going to be 10 J there, so our total work is going to be -- our total area or 10 + 7.5 + 7.5 or 25 J.0739

Let us talk a little bit about springs.0763

The more you stretch or compress a spring, the greater the force of the spring.0765

The more you push on it -- the more you compress it, the harder it pushes back or the more you stretch it, the more it wants to return to its equilibrium or its happy position.0771

The spring's force then, is opposite the direction of its displacement from its equilibrium.0779

And we can model this as a linear relationship where the force applied by the spring is equal to some constant, which we will call the spring constant -- kind of how strong the spring is...0783

...multiplied by the spring's displacement from its equilibrium, or rest or happy position -- whatever you want to call it.0794

This is known as Hooke's Law.0800

The force on the spring is equal to the opposite of the spring constant (K) -- how strong it is times its displacement and that negative sign just means it is restoring force.0802

If you pull it this way, the force wants to go back.0813

If you compress it this way, the force wants to push it back to where it started.0815

That displacement is always from its equilibrium position.0819

The negative tells you it is a restoring force.0823

So as an example, if we have a nice spring here -- there it is -- and we will start with an axis and call that distance its happy or equilibrium position -- we will call that x = 0.0826

If we then go and we try and stretch our spring out -- now at this point we have an (x) that is significantly greater than 0, so the force of the spring is going to be in the opposite direction.0841

There is the negative sign again -- why that is a restoring force.0855

So how do you find the spring constant of a spring?0862

Well the easiest way is probably to look it up on the box when you go buy a spring.0865

But assuming you do not have the box anymore, make a graph of the force required to stretch the spring against its displacement from its equilibrium position.0871

This is not the length of the spring here, this is how far you have stretched it from its happy position.0881

So force vs. displacement graph -- the slope is going to give you the spring constant (K) in newtons per meter (N/m).0886

So slope, which is rise over run -- for something like this, let us pick a couple of points.0892

These are easy points to pick, so let us pick that point right there, and that point right there.0902

So the rise is going to be going from 0 to 20N and that will be 20N and the run, we will go from 0 to 0.1 meters, so over 0.1 meters will give us a spring constant of 200N/m.0905

The bigger that (K) value, the stronger the spring.0921

So let us take a look at the work done in compressing a spring.0928

Here we have a force vs. displacement graph.0931

If we want the work done in compressing the spring -- well notice a force vs. displacement graph -- we have here an area.0934

The area under the force vs. displacement graph, still works; it still gives us the total work done.0942

So in this case, our work is going to be the area of our triangle -- nice big triangle there.0947

Work done will be 1/2 base times height or 1/2 × 0.1 m × the height (20N), or 2 × 1/2 = 1 J.0957

So let us do an example where we are finding the spring constant.0978

A spring is subjected to a varying force and its elongation is measured.0985

Determine the spring constant of the spring.0990

We have a bunch of points here to plot so let us start with that.0993

We have (0,0), we have an elongation of 0.3 with the force of 1, so somewhere right around there.0997

We have 0.67 with a force of 3 so that will be somewhere right around here -- make another point.1006

At 1 m, we have a force of 4N, at 1.3 m we have a force of about 5N and finally at about 1.5 m we have a force of 6N.1015

So the first thing I will do is use a straight edge to draw a best fit line here -- something like that -- use a straight edge yourself.1031

And when we do that now we have to find the slope of that line.1040

What we will do is pick a couple of points that are on that line and let us say we have a point right there -- is an exact point on the line and (0,0) is there, so that will make it pretty easy.1044

Our slope is our rise over our run or 6N/1.5 m or 4N/m.1056

Pretty easy to find the spring constant, just by taking our graph.1074

Let us take another example where we are calculating the spring constant again.1078

We have a 10N force -- F = 10N -- compressing a spring 0.25 m from its equilibrium position, So x = 0.25m.1082

Find the spring constant (K).1092

We will start off by writing Hooke's Law, and let us just worry about the magnitude for now.1095

We will worry about direction later.1100

So F = Kx, therefore the spring constant (K) is F/x or 10N/0.25 m for a spring constant of 40N/m.1102

Let us talk about power for a couple of minutes.1122

If work is the process of moving an object by applying a force, power is the rate at which that force does work.1126

Power is the rate at which work is done.1132

The units of power are joules per second (J/s), which we also call a watt (W).1134

Now you have to be careful using watts as your units because 'work' is capital W, and now we have the unit watts as capital W.1140

So you have to be careful and understand what you are doing when we write these.1148

Our formula for power is going to be work over time, the rate at which work is done.1152

And since power is the rate at which work is done, it is possible to have the same amount of work done but with a different supplied power if it has two different time intervals.1157

For example, Robin Pete move a sofa 3 m across the floor by applying a combined force of 200N horizontally.1167

If it takes them 6 s to move the sofa, what amount of power did they supply?1174

Well the power supplied is going to be the work done divided by the time it took, which is going to be F(δr) -- the displacement and force are in the same direction, so we do not have to worry about that cos(θ) term -- divided by t.1180

So we have 200N as our force, they moved it 3 meters, our displacement in a time of 6 s, so 200 × 3/6 is just going to be 100 J/s or 100 W.1193

At the same time though Kevin pushes another sofa 3 m across the floor by applying a force of 200N.1209

Kevin, however, takes 12 s to push the sofa.1215

What amount of power did Kevin supply?1218

Well the same formula -- Power will be F(δr)/T, which is going to 200 × 3/12 s this time or 50 W.1221

So same amount of work done, took him twice as much time so he had half the power output.1234

When we are calculating power, there are a couple of different ways we can do this.1242

We already talked about power as being the work done divided by the time, but that is also F(δr) cos(θ) divided by time.1246

But take a look, we have δr over (t) here, displacement over time.1260

That looks like velocity.1266

Velocity is δr/t, so we could rewrite this as power is equal to force times velocity times the cosine of the angle between those.1269

Another version of that same formula, another way to write it, another way to calculate it.1282

Let us take a look at that with an example.1286

Motor A lifts a 5,000N steel crossbar upward at a constant velocity of 2 m/s.1291

Motor B lifts a 4,000N steel support upward at a constant 3 m/s.1297

Which motor supplies more power? Let us figure out the power from each one.1301

The power from motor A is going to be the force times velocity, no cosine-theta term needed because they are in the same direction again.1307

That is 5,000N, our force, times our velocity of 2 m/s or 10,000 W, which we could write as 10 kilowatts (kW).1315

The power for motor B on the other hand, we calculate the same way, but now we have a force of 4,000N and we are doing this at a velocity of 3 m/s for 12,000 W or 12 kW.1329

Which motor supplies more power? Well obviously it must be motor B.1350

Let us take a look at an example with a cyclist.1361

A 70-kg cyclist develops 210 W of power while pedalling at a constant velocity of 7 m/s East.1363

What average force is exerted eastward on the bicycle to maintain this constant speed?1371

Let us start with our givens.1376

We know the mass is 70 kg; we know that the power is 210 W; our velocity is 7 m/s in eastward direction and we are trying to find an average force.1378

Power is force times velocity, therefore if we want just the force we will rearrange this as power over velocity or 210 W divided by 7 m/s.1403

It should give us a force of 30N, and of course, that is going to be in the eastward direction as well if we make that a vector and we are going to track our direction -- 30N East.1419

Let us take a look at work on a spinning mass.1433

A 5 kg ball is spun by a chain in a horizontal circle of radius 2 m at a speed of 3 m/s.1436

So a horizontal circle being spun pretty quickly. What is the work done on the ball by the chain?1443

First thing, let us draw a graph of this, let us draw a diagram.1449

If we look at it from the top, our horizontal circle, that is my best attempt at a circle.1452

Any point in time -- there is our object -- it has some velocity tangent to the circle.1458

The force is always toward the center of the circle because it is a centripetal force.1463

The force is always perpendicular to the displacement in the velocity.1469

Because of that, no work is done on the ball by the chain.1475

You cannot do any work because the force is toward the center of the circle.1479

The velocity, the displacement, at any instantaneous point in time is always 90 degrees from that, it is always perpendicular.1482

So you cannot do any work on that spinning mass, not by that force.1488

That force is changing its direction, keeping it moving in a circle, but it is not doing any work on the object; it is not causing that displacement.1494

Let us take a look at one where we are talking about work done by friction now as we again explore that definition of work in the force having to cause that displacement and how we are just going to massage that a little bit.1505

We have an 80 kg wooden box pulled 10 m horizontally across a wood floor at a constant velocity by a 250N force at an angle of 37 degrees above the horizontal.1519

If the coefficient of friction between the floor and the box is 0.3, find the work done by friction.1531

Wow! There is a lot there. Let us start by exploring this problem a little bit more.1537

First of all we know it is being pulled at a constant velocity.1542

The moment I see that, right away I think, 'You know we must have 0 acceleration.'1546

The net force must be 0 by Newton's Second Law.1553

Let us draw our box here -- it is an 80 kg box with a 250N force that is being applied in an angle of 37 degrees above the horizontal.1560

It is going to be pulled at a displacement of 10 m, and we know the coefficient of friction between the box and the floor is 0.3.1578

Let us start off with a free body diagram (FBD) here because we have a lot going on.1594

There is my box. I must have its weight (mg) down, a normal force opposing that.1597

Our applied force of 250N at an angle of 37 degrees, and we must have our frictional force opposing that motion, Ff.1606

Now I am going to make my pseudo free body diagram (P-FBD) and get all my forces to line up with an axis.1618

So I am going to break this 250N up into components and when I do that, we will have (mg) down still.1624

I still have my normal force up -- 250 times the sine of 37 to give me the vertical component is going to give me 150N up.1633

Its horizontal component 250(cos37) is going to be 200N and I have the frictional force opposing that.1645

Let us write Newton's Second Law in the (x) direction.1655

Net force in the (x) direction and I look at my P-FBD.1660

I have 200N to the right, minus the frictional force to the left and that must all be equal to 0 because the acceleration is 0; it is moving at a constant velocity, therefore, the frictional force must be 200N.1665

The work done by friction then must be that frictional force times the displacement.1680

Frictional force is going to be opposite in direction to the displacement, so I could write that as 200N, displacement (10 m), but I have to bring in my cosine-theta term -- Cos(180 degrees) which will be -1 or -2000 J of work done by friction.1686

Why a negative? Because the box's displacement is in one direction, the force of friction is in the opposite direction.1709

Let us explore the units of power a little bit.1718

Determine the unit of power in terms of fundamental units -- Kilograms (kg), meters (m), and seconds (s).1720

Let us start by using our definition of power.1725

Power is work over time, which is going to be force times displacement divided by time.1728

And force, by the way, Newton's Second Law, is mass times acceleration.1736

So we have broken this down into some more detailed -- a different definition based on fundamental units, let us find the units of these.1743

Units of mass are kilograms, acceleration is meters per second squared (m/s2), displacement will be meters -- we will have a squared there -- and time, well, we have seconds down here again.1752

So our total unit must be kg × m 2/s3 and that all must be equal to a watt.1766

What are the units for power? Watt.1778

One last example problem -- the frictional force on a sled.1782

Bob supplies 2000 W of power, P = 2000 W, in pushing a heavy sled across a frozen lake at a constant speed of 2 m/s.1787

So, constant speed right away, I think, acceleration = 0, it must be at equilibrium and that speed is 2 m/s.1796

Find the frictional force acting on the sled.1803

Let us take a look at the FBD for this case.1806

We have the force of Bob acting in one direction.1810

We have the frictional force acting in the opposite direction.1815

We know they must balance out because it is a constant speed of 2 m/s and of course we must have also here the normal force and the gravitational force, the object's weight.1819

Because it is moving at a constant speed, the force of Bob must be equal to the frictional force, since the acceleration up here is 0.1832

And if power is force times velocity, then we could say that the force of Bob must be equal to the power over the velocity, or 2000 W over 2 m/s, which is going to give us 1000N as the force of Bob.1841

And since the force of Bob equals the frictional force, we could then say that the frictional force is also equal to 1000N -- it is just in the opposite direction.1861

Hopefully that gets you a good start on work and power.1872

Thanks for watching We will talk to you soon. Make it a great day.1876