For more information, please see full course syllabus of AP Physics 1 & 2

For more information, please see full course syllabus of AP Physics 1 & 2

### Work & Power

- Work is the process of moving an object by applying a force. W=FrcosÎ¸.
- 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=FvcosÎ¸.

### 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
- Objectives
- What Is Work?
- Examples of Work
- Calculating Work
- Example 1: Moving a Refrigerator
- Example 2: Liberating a Car
- Example 3: Crate on a Ramp
- Example 4: Lifting a Box
- Example 5: Pulling a Wagon
- Force vs. Displacement Graphs
- Example 6: Work From a Varying Force
- Hooke's Law
- 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
- Work Done in Compressing the Spring
- Example 7: Finding Spring Constant
- Example 8: Calculating Spring Constant
- Power
- Example 9: Moving a Sofa
- Calculating Power
- Example 10: Motors Delivering Power
- Example 11: Force on a Cyclist
- Example 12: Work on a Spinning Mass
- Example 13: Work Done by Friction
- Example 14: Units of Power
- Example 15: Frictional Force on a Sled

- 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

### AP Physics 1 & 2 Exam Online Course

### Transcription: Work & Power

*Hi everyone. I am Dan Fullerton and I am thrilled to welcome you back to Educator.com.*0000

*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/s ^{2} 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/s ^{2}), 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}/s^{3} 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 Educator.com. We will talk to you soon. Make it a great day.*1876

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

Sir,

If possible provide me solution of the following question.

https://i.imgur.com/o8tFSqD.jpg

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?

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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?