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Lecture Comments (2)

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?

Power & Simple Machines

  • Power is a measure of how quickly we put work into a system. Mathematically,
    P = W

  • Since work is a transfer of energy (W=∆E), we can also formulate power as
    P =∆E

  • Finally, we can also formulate it as
    P =


    [Note: this requires you to use the 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 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

Transcription: Power & Simple Machines

Hi, welcome back to, 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)at2 + v0t + d(0) = (1/2)at2 = (1/4)t2, is our distance, (d(0) and v0 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)×22 = 1, d(3) = (1/4)×32 = (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)mv2 = 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 ~ 5.22×105 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)mv2/power = (1/2)×1500×(50)2/(5.22×105) = 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