For more information, please see full course syllabus of AP Physics C: Mechanics

For more information, please see full course syllabus of AP Physics C: Mechanics

### Power

- 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.

### 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 0:06
- 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

### AP Physics C: Mechanics Online Course

I. Introduction | ||
---|---|---|

What is Physics? | 7:12 | |

Math Review | 1:00:51 | |

II. Kinematics | ||

Describing Motion I | 23:47 | |

Describing Motion II | 36:47 | |

Projectile Motion | 30:34 | |

Circular & Relative Motion | 30:24 | |

III. Dynamics | ||

Newton's First Law & Free Body Diagrams | 23:57 | |

Newton's Second & Third Laws of Motion | 23:57 | |

Friction | 20:41 | |

Retarding & Drag Forces | 32:10 | |

Ramps & Inclines | 20:31 | |

Atwood Machines | 24:58 | |

IV. Work, Energy, & Power | ||

Work | 37:34 | |

Energy & Conservative Forces | 28:04 | |

Conservation of Energy | 54:56 | |

Power | 16:44 | |

V. Momentum | ||

Momentum & Impulse | 13:09 | |

Conservation of Linear Momentum | 46:30 | |

Center of Mass | 28:26 | |

VI. Uniform Circular Motion | ||

Uniform Circular Motion | 21:36 | |

VII. Rotational Motion | ||

Rotational Kinematics | 32:52 | |

Moment of Inertia | 24:00 | |

Torque | 26:09 | |

Rotational Dynamics | 56:58 | |

Angular Momentum | 33:02 | |

VIII. Oscillations | ||

Oscillations | 1:01:12 | |

IX. Gravity & Orbits | ||

Gravity & Orbits | 34:59 | |

X. Sample AP Exam | ||

1998 AP Practice Exam: Multiple Choice | 28:11 | |

1998 AP Practice Exam: Free Response Questions (FRQ) | 28:11 |

### Transcription: Power

*Hello, everyone, and welcome back to educator.com*0000

*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 works*0034

*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 equal*0164

*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 θ*0257

*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 force*0395

*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 here*0593

*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 T⁴*0907

*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 www.educator.com.*0999

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

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