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For more information, please see full course syllabus of AP Physics C: Mechanics
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  • Power is the rate at which work is performed, or the rate at which a force does work.
  • Units of power are Joules/second, known as Watts.


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 0:06
  • Defining Power 0:20
    • Definition of Power
    • Units of Power
    • Average Power
  • Instantaneous Power 1:03
    • Instantaneous Power
  • Example I: Horizontal Box 2:07
  • Example II: Accelerating Truck 4:48
  • Example III: Motors Delivering Power 6:00
  • Example IV: Power Up a Ramp 7:00
  • Example V: Power from Position Function 8:51
  • Example VI: Motorcycle Stopping 10:48
  • Example VII: AP-C 2003 FR1 11:52
    • Example VII: Part A
    • Example VII: Part B
    • Example VII: Part C
    • Example VII: Part D

Transcription: Power

Hello, everyone, and welcome back to educator.com0000

I am Dan Fullerton and in this lesson we are going to talk about power.0003

Our objectives include calculating average and instantaneous power.0007

Calculating the power required to maintain the motion of an object and calculating the work performed by a force applying constant power.0011

Let us start by defining power.0019

Power is the rate at which work is done or it is the rate at which a force does work.0022

Units of power are joules/second which we also know as watts.0028

Oftentimes, given the symbol capital W.0032

Be careful capital W and watts can look like capital W for works0034

You got to know what you are talking about whether it is a unit or whether you are talking about the quantity work.0038

Power, if we wanted to find average power, is the change in work and amount of work done in some amount of time.0044

Units as we said are joules /second or watts.0054

Instantaneous power, we can find by looking at the average power over a very small interval and telling me that time interval infinitesimally small.0063

Power is the time rate of change of work with respect to time but we also stated that the differential of work is F.DR so we could then write that Power= F.DR t.0073

We also know that drdt is our definition of velocity.0102

We can write Power = Force dotted with velocity.0111

A couple different ways to find power.0121

Let us make a couple examples.0126

Bob pushes a box across a horizontal surface at a constant speed of 1 m /second.0129

If the box has a mass if 30 kg, find the power Bob applies given the coefficient of kinetic friction is 0.3.0134

Let us start with the free body diagram.0143

There is box, we have the normal force acting on it, we have its weight down,0147

we have some applied force, the force of Bob on the box and we must have some amount of friction and it is kinetic frictions we will call the FK.0154

Writing Newton’s second law equation in the x direction, net force in the x direction is going to be equal0164

to the force of Bob minus the kinetic frictional force which is equal to Max.0171

We have got these keywords in the problem, constant speed which means at Ax = 0.0180

We now know that the force of Bob must equal the force of kinetic friction, by the way friction is fun.0187

It is μ K × Fn.0197

Let us take a look at Newton’s 2nd law.0203

In the y direction, net force in the y direction is going to be our normal force - MG and again no acceleration that is equal to 0,0205

which implies that the normal force equals MG.0217

We can take that and we can plug Ng in there, for the normal force.0220

Going back to that equation, force of Bob = μ K × MG which is going to be 0.3 or coefficient of kinetic friction ×0228

our mass 30 kg × the acceleration due to gravity 10 m /second squared so that is 300 × 0.3 or 90 N.0244

For after power though, power is force with velocity, they are in the same direction so this is just going to be FV cos θ0257

which is FV or our 90 N × the velocity 1 m /s means that Bob is applying 90 watts of power.0267

Let us take a look at another example, a 9000 kg truck accelerates uniformly from rest to a final speed of 36 m /s in 12 s.0287

What is the average power required to accomplish this?0298

We are looking for average power, that is going to be the average force dotted with the average velocity.0303

Butt force is MA, Newton’s 2nd law so this is MA × our average velocity that is going to be 9000 kg our mass, the acceleration we can find by Δ V/T.0313

Change in speed is 36 m /s so that will be 36 -0 over time 12 s and our average velocity all of its constant acceleration you go from 0 to 36.0328

Remember that the average is halfway between those two or 18m /s.0339

That is going to be 486000 W or 486 kilowatts.0346

Let us take a look at a couple motors delivering power.0358

Motor A, list of 5000 N steel crossbar upward and a constant 2 m /s.0362

Motor B, list a 4000 N steel support upward and the constant 3 m /s.0368

Which motors applying more power?0373

Let us figure the power for A first, that is going to be F × V is 5000 FN × V 2 m /s or 10,000W or 10kw.0376

Taking a look at motor B, motor B, we can use the same formula force × velocity but now it is a 4000 N force0395

at 3 m /second which is 12,000W or 12kw.0405

Which motor applies more power?0412

Got to be B.0416

A little bit trickier problem.0420

The box of mass M is pushed up a ramp at constant velocity V to maximum height H in time T by force F as shown in the diagram.0422

The ramp makes an angle of θ with a horizontal as shown in the diagram here.0430

What is the power applied by the force?0434

Let us take a look here.0439

We have, as I look at this, we have a force, a mass, an angle, an H, and a bunch of different choices here.0441

Let us see if we can solve this.0450

I'm going to start by looking at our sin of θ to see how far this is going.0452

Sin of θ is the opposite over the hypotenuse so that is going to be H/ D.0458

Which implies that the distance up that, the hypotenuse is going to be H /sin θ.0465

The velocity as you go up the ramp is just the distance travel divided by × that will be H over T sin θ.0475

If I wanted power, that is force × velocity that is just going to be FH/ T sin θ.0484

I would say that C works and D also works, force × velocity of course.0497

Looking at other choices MGH /T, that is the energy /amount of time but it does not take into account any possible friction.0505

And same here, we are pulling that sin θ so those pieces all would only be, if we consider those if we want a frictionless environment.0514

We are not frictionless so those are not going to work.0522

I would say that C and D here our best answers.0524

Let us see if we can find of power from a position function.0532

Find the power delivered by the net force to a 10 kilogram mass at time T = 4 seconds, given the position of the mass as 4T³ - 2t.0535

Let us start by finding the velocity as a function of time that is just the first derivative a position which is going to be 12 T² – 2.0546

If we wanted acceleration, why we are here?0559

Acceleration is the derivative of velocity or the second derivative of position that is just going to be 24 T.0563

Finding the power delivered, bunch of different ways we could do that but let us start by finding the net force.0572

That is mass × acceleration which is going to be 10 kg × 24 T our acceleration or 240T.0578

Power then is force × velocity which is going to be FV cos θ which implies then that power here0593

is going to be 240T × velocity 12T² - 2 which implies then that power = 240 × 12.0606

That is 28 ADT³ - 4 ADT and plug into our time of 4 seconds, I come up with the power of about 182,400W or 182.4kw.0620

How about a motorcycle, a 400 kg motorcycle travels along a highway at 30 m /s?0649

If the motorcycle breaks with an acceleration of 3 m /s², what is the average power required to bring it to a full stop?0655

Power is force × average velocity, it is going to be mass × acceleration, our force × our average velocity.0666

Our mass is 400 kg, our acceleration 3 m /s², we are going to worry about the magnitude since we are after the power.0676

And it does all that and the average velocity of, it started at 30 it goes to 0,0686

a constant acceleration, the average velocity is halfway between 0 and 30 or 15 m /s.0691

That is going to give us 18000W or 18kw.0699

Let us finish up by looking in an old AP free response problem.0709

We will take a look at the 2003 exam Mechanics free response 1.0713

Take a minute, go to the web site there, and download it.0717

If you cannot find it that way, google it, take a minute, print it out, give it a try, and come back and hit play again.0720

As we look at part A, given a function of X we are asked to find the speed of the box of × T = 0.0735

If x is 0.5 T³ + 2T that means the velocity which is the derivative of X with respect to T must be 1.5 T² +2.0742

Since you want to know this, 1T=0 that just means V at time T=0 must be 2 m /s.0758

There is part A.0769

For part B, we are asked to determine the following as function of time.0772

The kinetic energy of the box and net force on the box and the power being delivered to the box.0778

Let us take a look first at the kinetic energy of the box, that is ½ MV² which be ½ M × 1.5 T² +2².0785

Or 50 our mass 100 × 1.5 T² +2².0800

For part 2, we are asked to find the net force.0813

Net force equals mass × acceleration which is MDVDT, which is M × the derivative with respect to T of 1.5 T² + 2 which is going to be M × 3T or 300 T.0817

B3, a power being delivered to the box.0843

The power is force × velocity which is going to be our force 300 T.0848

We already did our velocity 1.5 T² + 2,0856

multiplying that through that is going to be 450 T³ + 600T.0861

There is part B.0874

Moving on to the C, let us give ourselves some more room here.0877

Calculate the net work done on the box from 0 to 2 seconds.0883

What we can do, the net work will be the integral of the power with respect to time from T = 0 to 2 seconds,0887

which will be the integral from 0 to 2 of 450 T³ + 600 T DT which we just determined.0895

Or integrating that is 450 T⁴/4 + 600 T²/2 all evaluated from 0 to 2 which is going to be 452.5/2 T⁴0907

225 T⁴/2 + 300 T² evaluated from 0 to 2, which will be 225 × 2⁴ ÷ 2 + 300 × 2² or 3000 joules.0926

Part D, indicate below whether the work done on the box by the student from 0 to 2s would be greater than, less than, or equal to the answer in part C.0955

It is got to be greater than.0966

Why? The student’s word has to be greater than the net work because the student had to work against friction.0967

The student has to do more work compared to what you just have in the box.0975

The net work is a student minus friction, the work by the student minus the work than friction.0978

The work done by the student has to cover the net work + the work done by friction.0983

Explain that in word somehow to justify your answer.0993

Hopefully, that gets you a good start on power.0997

Thank you so much for watching

We will see you in the next lesson and make it a great day everybody.1002