For more information, please see full course syllabus of High School Physics

For more information, please see full course syllabus of High School Physics

### Power & Simple Machines

*Power*is a measure of how quickly we put work into a system. Mathematically,P = W t. - Since work is a transfer of energy (W=∆E), we can also formulate power as
P = ∆E t. - Finally, we can also formulate it as

[Note: this requires you to use theP = →F· →v__dot product__for vectors. If you aren't familiar with dot products, this is equivalent: P = |→F| |→v| cosθ.] - Power has the unit of joule per second ([J/S]). We call this a watt (W).
- Simple machines are based on the idea of conservation of energy. Instead of using a large force over a small distance, we can use a small force over a large distance and still put in an equivalent amount of work.

### Power & Simple Machines

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
- Introduction to Power & Simple Machines
- What's the Difference Between a Go-Kart, a Family Van, and a Racecar?
- Consider the Idea of Climbing a Flight of Stairs
- Power
- Alternate Formulas
- Units
- Block and Tackle, Redux
- Machines in General
- Example 1: Power of Force
- Example 2: Power &Lifting a Watermelon
- Example 3: Work and Instantaneous Power
- Example 4: Power and Acceleration of a Race car

- Intro 0:00
- Introduction to Power & Simple Machines 0:06
- What's the Difference Between a Go-Kart, a Family Van, and a Racecar?
- Consider the Idea of Climbing a Flight of Stairs
- Power 2:35
- P= W / t
- Alternate Formulas 2:59
- Alternate Formulas
- Units 4:24
- Units for Power: Watt, Horsepower, and Kilowatt-hour
- Block and Tackle, Redux 5:29
- Block and Tackle Systems
- Machines in General 9:44
- Levers
- Ramps
- Example 1: Power of Force 12:22
- Example 2: Power &Lifting a Watermelon 14:21
- Example 3: Work and Instantaneous Power 16:05
- Example 4: Power and Acceleration of a Race car 25:56

### High School Physics Online Course

### Transcription: Power & Simple Machines

*Hi, welcome back to educator.com, today we are going to be talking about power and simple machines.*0000

*Let is just start off with, what would you say is the difference between a go-kart, a family car and a race car is?*0006

*I will be honest, there are a lot of differences, but I would say their main difference is their top speed.*0013

*It is high fast they can go.*0020

*How quickly they can get in going a certain speed.*0022

*You are going to get a lot of difference in how fast a race car can get from 0 to 60, ad how fast the family van can get from 0 to 60.*0025

*The major issues here are, how fast they can go at their maximum, how much power can they put out.*0032

*What is the idea of power?*0037

*So far we have talked about speed and its connection to energy, but we have not talked about different rates of gaining that kinetic energy, we have just talked about it being there.*0040

*We have not talked about the difference between getting it to going fast quickly, we have only been talking about going fast versus going really fast.*0049

*There has been, the speed that you are going at, there has been no talk about how fast you can get to go in that speed, what is your acceleration has had no effect on this.*0057

*That is where power is going to come in, we are going to start talking about how quickly an object or system gains energy.*0066

*Consider the idea that we are climbing a flight of stairs.*0074

*Let us assume that we weigh 50 kg, for me 50 kg is fairly well under my weight.*0077

*The stairs are 5 m high.*0090

*There are two scenarios.*0092

*In one of them, you climb the stairs in 5 s you really hustle.*0093

*But in the other one, it take you 30 s.*0097

*In both these scenarios, we are going to have the exact same amount of energy at the end, the same amount of potential energy, (not develop the same power.)*0099

*We climb the same height, we are dealing with the same gravity, we have the same mass.*0108

*But, very different scenarios.*0113

*How fast you climb those stairs, that is something we should talk about, and care about.*0115

*In both cases, we have that same gain of energy of gravity, 50×9.8×5, so in both cases we have 2450 J, when we make it to the top of the stairs.*0120

*But they are clearly very different scenarios.*0129

*So we need a way to talk about the interaction between work and energy, and time.*0131

*Work and energy, and how fast we are able to put work and energy into a system.*0138

*How quickly we are able to change the work and energy in a system.*0142

*This is going to really matter for some applications.*0145

*For that race car, we want to be in a race car that can put massive quantity of energy into its system, really fast it can get off the starting line and win the race.*0147

*With this idea, we make a really simple creation to call this power, just, power = work/(amount of time it takes), work/time, that will give us a way to talk about how much work we are able to deal with in how much time.*0157

*Just like velocity was how much distance we have gone, divided by how much time to do it, we het a very similar idea with work/time for power.*0172

*With power defined as work/time, we can easily create a few equivalent formulae.*0180

*First, since work is a measure of how much energy is being shifted around, we know work = change in energy, always.*0185

*Another formula is, power = Δenergy/time.*0191

*There is another interesting formula we can create.*0199

*Alternately, we can look back to how we originally formulated work.*0201

*Work = F.d = Fdcosθ, in this case it is going to help us to use that dot product.*0206

*This allows us to use velocity.*0222

*Power = work/time = F.d/time.*0224

*We pull off that force, and we get, F.d/time, but distance/time = velocity, so, we get, F.v, so, Power = F.v, which also, if we do not want to use the dot product, Fv×cosθ.*0229

*That same idea that worked with work, works with power.*0253

*So, F.v, or ΔEnergy/time, or work/time, these are all same way to say power.*0257

*What unit is power in?*0266

*Work and energy are both in joules, time is in seconds, so, power = work/time, implies, J/s is the unit for power.*0268

*For ease, we can call 1 J/s a watt, watt is in honour of James Watt who has done a lot of work with energy in 1800's.*0287

*A watt is a measure of weight.*0295

*Just like 1 m/s is the rate that you are moving at, 1 J/s, 1 watt is the rate that you are putting energy into a system.*0298

*One moment, you could have a totally different power, the next moment, just like your velocity can change.*0307

*watt gives us an instantaneous measure of how much energy is going into a system, that is the definition of power.*0314

*In other unit systems, there is also the horse power, you probably heard cars referred to in terms of horse power, and the kilowatt hour, another way of saying watts×time, watt×1000×time.*0324

*Horse power, kilowatt hour, all ways of saying power, that is why energy bills involve these things, cars involve these things, you see these things any time you want to talk about how quickly we can get energy into, or out of the system.*0343

*We are going to make a little tangent here.*0356

*This does not directly have to do with power, but I know, it has been driving you crazy that we have not discussed block and tackle system more.*0359

*I know you remember block and tackle problems extremely well, we talked about them in advanced uses of Newton's second law, and they seemed like magic, they seemed absolutely incredible, and it has been blowing your mind, you keep thinking about it, how is it possible, Physics must be lying, fret no more, I am going to make it better for you.*0368

*Finally we have the understanding of work and energy to see how those systems make perfect sense.*0391

*There is nothing magical about them, there is nothing insane about them, the world is not coming apart as it seems, it makes perfect sense when you look at it in terms of energy.*0394

*I know you love thinking about it, but once again, let us talk about it briefly, quick reminder about how the block and tackle system works.*0401

*Let us say we have got the force of gravity pulling down on this block, Fg, it is pulling down on this block.*0411

*If we want to keep it still, or move it up, say we want to keep it still, we are going to have to pull with the force of gravity here, put that much tension into it, so we have got canceling it out, the force of gravity over here.*0417

*But over here, something weird happens.*0430

*If we pull with a certain tension, then that tension is going to get pulled in here, but it is also going to get pulled in here.*0433

*So we still got that same mg, that same force of gravity, but over here, we are going to have the tension = (1/2)Fg, because if this is (1/2)Fg, and this is (1/2)Fg, then when we put them together, we are going to wind up combining two, one whole force of gravity.*0440

*With the block and tackle system, we are able to distribute our forces over multiple pulleys.*0460

*We are able to distribute the same tension force in multiple places.*0466

*It seems crazy, we are able to get more force for the same original cost.*0470

*How is this possible! This seems like madness.*0474

*The thing to notice here, is that the block and the single pulley system, say we wanted to raise this block 1 m.*0479

*If we wanted to raise the block by 1 m, how much rope would we have to pull?*0488

*We would have to pull 1 m of rope.*0491

*We have to pull, say some force F.*0494

*Over here, we know that we only have to pull at half of that force, F.*0497

*To be able to get that raising happening.*0501

*But, if we want this block to raise up 1 m, we do not just have to get this move by 1 m, if this moves 1 m, we would be lopsided, we have to get this move 1 m as well.*0503

*Both sides of our rope system have to move up a metre, that means we have to get 2 m of motion in our rope.*0514

*Over on the left side, we are able to use 1 m of motion for 1 m of motion.*0524

*Over here, if we want to get that 1 m of lift, we have to put 2 m of distance into our rope.*0530

*Even if we can use half the force, we have to pull double the distance.*0536

*So, our work, the amount of energy in our system is preserved, the force×distance.*0542

*This is force × 1 = F, for work.*0547

*Over here though, we have got, work = (1/2)F × 2 = F, so they wind up being the exact same things, checks out.*0555

*For us to be able to manipulate how the system works, we still have to maintain that conservation of energy, that conservation of work.*0568

*It is equivalent work, because the change in the system, the real change in the system, is how high up we are able to change that block's height.*0576

*If you want to do that, we are going to have to put in the same amount of work, no matter how we go about it.*0585

*Work that goes in, is equal to F in both cases, because you have to pull double the rope.*0590

*And if we had a multiply pulley system, where we were able to have four pulleys, we only have to pull with a quarter of force, we would wind up having to pull 4 times the distance.*0597

*Everything works out, there is nothing magical about it, it makes perfect sense.*0605

*It is the same idea in place with all of our machines.*0610

*Let us look at ramps and levers.*0613

*First we will look at levers.*0617

*If we want to get some object to move up here, traditionally you have a lever, you stick it under, you got a fulcrum, you got a long lever arm, and you pull, you pry.*0618

*You put a low pressure here, and you get a really strong force here.*0627

*High pressure here, low pressure here.*0632

*How is it being done, it is being done based on work.*0634

*You got that small force over here, small force, but it covers a really long distance.*0637

*On the other side, we get this small distance covered, which means, to be that work to be preserved, it is going to be have to put up a massive force.*0643

*The reason why a lever works, is that the energy is conserved.*0651

*If you put a slight force over a long distance, and the other side has a slight distance, then it is going to need a big force to compensate.*0654

*Force×distance has to be equal in any case.*0663

*Over the lever, if you put that fulcrum really close to one side, we will be able to get a giant force with little distance.*0667

*So, the amount of work is preserved, it makes perfect sense.*0674

*You see the exact same thing with a ramp.*0677

*If we want to get this box up this ramp, then we can push with this little force over this really big distance.*0679

*But, if we want to get this box directly up, then we will have to lift a whole lot harder, but we wind up going a smaller distance.*0685

*The ramp works, because we are able to get a small force over a long distance, whereas if you just want to lift it up with brute strength, we need a really powerful force, but we will be able to save some distance.*0694

*The same idea, force×distance, they are always equal.*0706

*The way that you are going to distribute, how you are going to put it in, what ratio you want to put it in, that is up to you.*0709

*But, it is going to have to come out equal when you multiply the two.*0715

*However you put it in, work is going to be conserved, energy is going to be conserved, the energy that goes into it is going to be the same however you do it.*0719

*Machines do not allow us to break the rules of Physics, they just allow us to take advantage if the resources that we have on hand.*0726

*They allow us to use the rules of Physics on our side.*0732

*If we have a little force, but a lot of distance or time, we can figure out an alternative rather than needing a really big force.*0736

*Like with the ramp, like with the lever, like with the pulley system, you can have that slight force, and then figure out the way to multiply by using more distance, or more force, or both , so you can take advantage of what you have, by being able to use the same amount of energy.*0742

*Same amount of energy will go into the system, same work, but it is up to us to figure how to get that work into it, and that is where the cleverness of machines comes into being, at least simple machines.*0757

*Now we are ready for our examples.*0769

*How much work is going to be involved here?*0771

*50 kg block is pushed along a horizontal surface at a constant velocity by a parallel force 47 N.*0773

*It covers 10 m in 5 s.*0779

*What is the power of the force?*0781

*Lets us draw a quick diagram.*0783

*Do we have to care about the mass?*0801

*We do not have to care about the mass.*0802

*Our power formula is, work/time.*0804

*We can figure out the work, work = F.d = Fdcosθ = Fd (since parallel).*0807

*So, 47×10 = 470 J of work.*0820

*What is the time? 5 s, so , Power = 470 J / 5s = 94 J/s = 94 W, that is the power of the force.*0828

*It does not change, the power remains the same, because we got a constant velocity.*0850

*Remember, we could have also used, F.v.*0855

*If you wanted to, we could figure out, it travels 10 m in 5 s, means that we got, 2 m/s = v, and F = 47 N, so, power = 47×2 = 94 W, two different ways.*0861

*Example 2: 2 kg watermelon starts at rest, and it is lifted vertically, 9 m.*0887

*It takes time 20 s, and ends at rest.*0898

*Over that lifting, what was the power developed in lifting that watermelon?*0900

*You started at some height, you reached another height, how do we deal with that?*0905

*Potential energy.*0909

*What is the change in energy?*0911

*ΔE = mgΔh = 2×9.8×9 = 176.4 J of work.*0913

*We know that, power = ΔE/t = 176.4 J/20 s = 8.82 W.*0933

*There you go, the change in energy divided by how much time it takes to put it i n there, just like velocity, just like acceleration, it is what you have got already, distance or speed divided by how much it is altered, how much it is being changed by, how much it is being increased by, that tells us how much the power being developed is.*0952

*The power of the system is the change of energy, how much work is going into that system.*0970

*Just like velocity is how much distance is going into an object, whereas the acceleration is how much velocity is going into an object, just a way of thinking how much velocity is going into an object, a way of thinking how much you are putting into a thing.*0978

*S0, 8.82 W is what is put in, in that 20 s.*0987

*Example 3: 20 kg block is initially at rest on a flat frictionless surface.*0992

*Parallel force of 10 N acts on the block.*0997

*What is the work done on the block in (A) the first second (B) the second second (C) the third second (D) the instantaneous power at the end of 3rd second.*0999

*First thing to think about, work = F.d.*1009

*Is this object accelerating?*1015

*It is on a flat frictionless surface, it has got a force acting on it, of course it is accelerating.*1016

*The amount of distance it is going to cover is going to change the entire time.*1020

*It is also going to have a change in velocity.*1023

*Now we have got two different ways of looking at this.*1025

*We can approach this by wither thinking about the distance that it has changed, it is going to be able to give us our work, and from the work, we will be able to get our power.*1027

*But we can also think that the velocity that it has at each time, would be a way to tell us what is the change in energy, and from the change in energy, we can get the amount of power.*1040

*These are both perfectly good ways to do it, and we will do both of them just to be able to understand two different ways to approach this problem.*1055

*First way, we are going to go with distance.*1060

*What is the formula for distance?*1066

*We got, F = ma, 10 N = 20 kg×a, a = (1/2) m/s/s.*1068

*What is the other formula for distance?*1082

*From basic kinematics, d(t) = (1/2)at ^{2} + v_{0}t + d(0) = (1/2)at^{2} = (1/4)t^{2}, is our distance, (d(0) and v_{0} are zero).*1084

*Now we have got a distance formula.*1107

*Now we want to find out where is it at 1 s, 2 s and 3 s.*1112

*Plug things in, d(0) = 0, d(1) = (1/4), d(2) = (1/4)×2 ^{2} = 1, d(3) = (1/4)×3^{2} = (9/4).*1122

*If you wanted to see what the distance covered in that period of time is, that change in distance, we can say what is the distance between 0 and 1?*1149

*That is clearly (1/4).*1157

*What is the distance between 1 and 2? That is (3/4).*1159

*What is the distance between 2 and 3? That is (5/4).*1165

*These three changes in distance, if we want to figure out what the work involved is, we know, work = Fd.*1171

*Then, work that happened from 0 to 1st second is, 10×(1/4) = 2.5 J of work.*1180

*What was the work done between 1 and 2? That is, (3/4)×10 = 7.5 J.*1200

*What is the work from 2 to 3 s? That is, (5/4)×10 = 12.5 J.*1209

*Now, if we want to know how much power, what the average power was, over each one of these, we just go, 2.5 J/1 s = 2.5 W, was the average power in that first second.*1218

*In the 2nd second, 7.5 J/1 s = 7.5 W.*1228

*In the 3rd second, 12.5 J/1 s = 12.5 W.*1237

*Remember, these are going to be the average powers, because this is not going to give us the instantaneous.*1244

*Clearly, the amount of power is changing the faster it goes, because it is getting the chance to cover more distance, and that is how that force being applied more and more, since work = Fd.*1248

*If we want to know what the instantaneous power is, we are going to need to know what its speed is at a given moment.*1257

*Remember, power = F.v, v(3 s) = at = (1/2)×3 = 3/2 is the velocity at the 3rd second.*1264

*If we want to figure out the instantaneous power, Power at the third second, power (3 s) = 10 N × (3/2) = 15 W. (Dot product becomes multiplication because of one dimension, in dot product we are dealing with more dimensions, in more dimensions, we will have to know how to use the dot product, we will have to multiply the first components together, add them to the second components, multiply and add them to the third components, so on and so forth, for as many components you have. You probably will do with only 2 or 3 since you dealing with Physics, but it will work with any.)*1304

*So, the instantaneous power is 15 W for that third second, which makes sense, we see that our average, 2.5, 7.5, 12.5, it continues to go up, so the at the very end of the third second, we got 15 W of instantaneous power.*1344

*Remember, the instantaneous power can change just like your instantaneous velocity can change.*1358

*If you are driving in a car that is accelerating, every instant that you move along, your velocity is getting larger and larger.*1364

*So, if the car is accelerating, you got a larger velocity instant by instant.*1370

*In this case, we got a larger and larger power instant by instant.*1374

*If we have got an alternate method, let us start using it, how will that alternate method work?*1378

*So, we know, we did distance already, we can figure this out using distance, but we can also use velocity to tell us changes in energy.*1382

*Now we are going to do this using velocity.*1390

*If we want to see what its velocity is at 1 s, at 2 s, at 3 s, what is the formula for velocity?*1394

*We know, v(t) = at = (1/2)t, (since previous values still work), makes sense.*1403

*In this case, what will be velocity at 0 s? Zero, it is still.*1424

*v(1 s) = (1/2) m/s, v(2 s) = 1 m/s, and v(3 s) = (3/2) m/s.*1431

*If you want to see what the energy in its movement at that time is, we know, E(0) = (1/2)mv ^{2} = 0, E(1 s) = (1/2)×20×(1/2)^{2} = 2.5.*1443

*E(2 s) = (1/2)×20×1 =10, and E(3 s) = (1/2)×20×(3/2) ^{2} = 22.5.*1473

*So, we have got, 2.5 J, 10 J, 22.5 J.*1495

*If we want to figure out what the change in energy is, because we are working towards figuring out what the work is in each of these seconds, change in energy = work, work = ΔE.*1500

*ΔE(0 to 1) = 2.5 - 0 = 2.5 J, ΔE(1 to 2) = 10 - 2.5 = 7.5 J, ΔE(2 to 3) = 22.5 - 10 = 12.5 J, it is the exact same thing that we saw by doing it the other way.*1514

*Figuring out through the work, figuring out through the change in energy, of course they are going to give the same answer because they are the same thing.*1543

*If we want to figure what the power developed at the 3rd second was, we just do the exact same thing we did previously, so we can skip that, because we are just figuring out instantaneous power, and we discussed that on the last slide.*1548

*But, it is kind of cool to be able to see that we have got two different ways of approaching it.*1559

*So, whichever way that makes more sense to you, is the way you want to do it.*1563

*The important thing to think is, "Okay, great! I have got lots of methods, I have got lots of ways I can attempt a problem." Figure out what is the best for you, what is the best possible way to approach a problem, and then do it.*1567

*There are lots of tools for any given job, and it is up to you to figure out what tool to use.*1577

*Last example: This one is a fun one.*1583

*We have got a race car of mass 1500 kg, and it has an engine capable of putting out 700 hp ~ 5.22×10 ^{5} W of power.*1585

*Neglecting air friction, and friction on the ground, the air drag, the car begins at rest, and we assume that the car puts out its maximum amount of power, how long will it take the car to accelerate to 50 m/s on flat ground?*1595

*In this case, what do we want to use, what power formula are we going to use?*1608

*Do we know what the work is? Do we know what the forces involved are?*1612

*We do not really.*1615

*Do we know what the change in energy is?*1616

*We do know what the change in energy is.*1618

*It starts at a stop, it stops going at 50 m/s.*1619

*Do we know what its instantaneous velocity is?*1623

*No, because that is going to change depending.*1625

*So, we do not really want to go with that one, because that will give us the force, and the force is not really useful, so the best choice for this one is, power = ΔE/t.*1629

*We know what the time is, can we figure out what is the change in energy?*1639

*We do not know what the time is, we are solving for the time.*1681

*But we do know what the change in energy is, we do know what the power is, so we are good to go.*1686

*That gives us, t= ΔE/power = (1/2)mv ^{2}/power = (1/2)×1500×(50)^{2}/(5.22×10^{5}) = 3.59 s, that tells us how long it will take us for that car to accelerate form a dead stop, to going at 50 m/s, and that is equivalent to 110 miles/h.*1702

*3.59 s to get to 110 miles/h, or 50 m/s, that is pretty darn good, that explains why race cars are so powerful.*1725

*Hope you enjoyed this lesson, hope power made sense, just think of it as the change in work, the change in energy, and how long it took to get there.*1732

*There is a great analog between speed and velocity and energy and power, it is the same thing.*1739

*How fast are we changing, how much are we changing from moment to moment.*1752

*Hope you enjoyed this.*1756

1 answer

Last reply by: Professor Selhorst-Jones

Mon Dec 10, 2012 1:23 PM

Post by Abdelrahman Megahed on November 30, 2012

Example 4:

How come when you use other method not COE you end up with 7.4s?