WEBVTT physics/ap-physics-c-mechanics/fullerton
00:00:00.000 --> 00:00:03.600
Hello, everyone, and welcome back to educator.com
00:00:03.600 --> 00:00:07.600
I am Dan Fullerton and in this lesson we are going to talk about power.
00:00:07.600 --> 00:00:11.600
Our objectives include calculating average and instantaneous power.
00:00:11.600 --> 00:00:19.100
Calculating the power required to maintain the motion of an object and calculating the work performed by a force applying constant power.
00:00:19.100 --> 00:00:22.000
Let us start by defining power.
00:00:22.000 --> 00:00:28.100
Power is the rate at which work is done or it is the rate at which a force does work.
00:00:28.100 --> 00:00:31.900
Units of power are joules/second which we also know as watts.
00:00:31.900 --> 00:00:34.600
Oftentimes, given the symbol capital W.
00:00:34.600 --> 00:00:38.400
Be careful capital W and watts can look like capital W for works
00:00:38.400 --> 00:00:44.300
You got to know what you are talking about whether it is a unit or whether you are talking about the quantity work.
00:00:44.300 --> 00:00:54.400
Power, if we wanted to find average power, is the change in work and amount of work done in some amount of time.
00:00:54.400 --> 00:01:02.700
Units as we said are joules /second or watts.
00:01:02.700 --> 00:01:12.800
Instantaneous power, we can find by looking at the average power over a very small interval and telling me that time interval infinitesimally small.
00:01:12.800 --> 00:01:42.400
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.
00:01:42.400 --> 00:01:51.000
We also know that drdt is our definition of velocity.
00:01:51.000 --> 00:02:00.800
We can write Power = Force dotted with velocity.
00:02:00.800 --> 00:02:06.000
A couple different ways to find power.
00:02:06.000 --> 00:02:08.800
Let us make a couple examples.
00:02:08.800 --> 00:02:14.200
Bob pushes a box across a horizontal surface at a constant speed of 1 m /second.
00:02:14.200 --> 00:02:23.100
If the box has a mass if 30 kg, find the power Bob applies given the coefficient of kinetic friction is 0.3.
00:02:23.100 --> 00:02:27.200
Let us start with the free body diagram.
00:02:27.200 --> 00:02:33.700
There is box, we have the normal force acting on it, we have its weight down,
00:02:33.700 --> 00:02:44.300
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.
00:02:44.300 --> 00:02:51.300
Writing Newton’s second law equation in the x direction, net force in the x direction is going to be equal
00:02:51.300 --> 00:03:00.400
to the force of Bob minus the kinetic frictional force which is equal to Max.
00:03:00.400 --> 00:03:07.300
We have got these keywords in the problem, constant speed which means at Ax = 0.
00:03:07.300 --> 00:03:17.600
We now know that the force of Bob must equal the force of kinetic friction, by the way friction is fun.
00:03:17.600 --> 00:03:23.100
It is μ K × Fn.
00:03:23.100 --> 00:03:25.100
Let us take a look at Newton’s 2nd law.
00:03:25.100 --> 00:03:37.300
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,
00:03:37.300 --> 00:03:40.100
which implies that the normal force equals MG.
00:03:40.100 --> 00:03:47.700
We can take that and we can plug Ng in there, for the normal force.
00:03:47.700 --> 00:04:04.200
Going back to that equation, force of Bob = μ K × MG which is going to be 0.3 or coefficient of kinetic friction ×
00:04:04.200 --> 00:04:16.900
our mass 30 kg × the acceleration due to gravity 10 m /second squared so that is 300 × 0.3 or 90 N.
00:04:16.900 --> 00:04:27.500
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
00:04:27.500 --> 00:04:46.800
which is FV or our 90 N × the velocity 1 m /s means that Bob is applying 90 watts of power.
00:04:46.800 --> 00:04:58.000
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.
00:04:58.000 --> 00:05:03.500
What is the average power required to accomplish this?
00:05:03.500 --> 00:05:13.000
We are looking for average power, that is going to be the average force dotted with the average velocity.
00:05:13.000 --> 00:05:28.100
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.
00:05:28.100 --> 00:05:39.500
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.
00:05:39.500 --> 00:05:46.100
Remember that the average is halfway between those two or 18m /s.
00:05:46.100 --> 00:05:57.800
That is going to be 486000 W or 486 kilowatts.
00:05:57.800 --> 00:06:02.000
Let us take a look at a couple motors delivering power.
00:06:02.000 --> 00:06:07.900
Motor A, list of 5000 N steel crossbar upward and a constant 2 m /s.
00:06:07.900 --> 00:06:13.000
Motor B, list a 4000 N steel support upward and the constant 3 m /s.
00:06:13.000 --> 00:06:16.300
Which motors applying more power?
00:06:16.300 --> 00:06:35.300
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.
00:06:35.300 --> 00:06:44.900
Taking a look at motor B, motor B, we can use the same formula force × velocity but now it is a 4000 N force
00:06:44.900 --> 00:06:52.500
at 3 m /second which is 12,000W or 12kw.
00:06:52.500 --> 00:06:56.000
Which motor applies more power?
00:06:56.000 --> 00:06:59.900
Got to be B.
00:06:59.900 --> 00:07:02.200
A little bit trickier problem.
00:07:02.200 --> 00:07:10.400
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.
00:07:10.400 --> 00:07:14.600
The ramp makes an angle of θ with a horizontal as shown in the diagram here.
00:07:14.600 --> 00:07:19.100
What is the power applied by the force?
00:07:19.100 --> 00:07:21.000
Let us take a look here.
00:07:21.000 --> 00:07:29.800
We have, as I look at this, we have a force, a mass, an angle, an H, and a bunch of different choices here.
00:07:29.800 --> 00:07:32.500
Let us see if we can solve this.
00:07:32.500 --> 00:07:37.900
I'm going to start by looking at our sin of θ to see how far this is going.
00:07:37.900 --> 00:07:44.700
Sin of θ is the opposite over the hypotenuse so that is going to be H/ D.
00:07:44.700 --> 00:07:55.300
Which implies that the distance up that, the hypotenuse is going to be H /sin θ.
00:07:55.300 --> 00:08:04.300
The velocity as you go up the ramp is just the distance travel divided by × that will be H over T sin θ.
00:08:04.300 --> 00:08:16.800
If I wanted power, that is force × velocity that is just going to be FH/ T sin θ.
00:08:16.800 --> 00:08:25.200
I would say that C works and D also works, force × velocity of course.
00:08:25.200 --> 00:08:33.900
Looking at other choices MGH /T, that is the energy /amount of time but it does not take into account any possible friction.
00:08:33.900 --> 00:08:42.000
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.
00:08:42.000 --> 00:08:44.600
We are not frictionless so those are not going to work.
00:08:44.600 --> 00:08:52.100
I would say that C and D here our best answers.
00:08:52.100 --> 00:08:55.300
Let us see if we can find of power from a position function.
00:08:55.300 --> 00:09:06.300
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.
00:09:06.300 --> 00:09:19.200
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.
00:09:19.200 --> 00:09:22.800
If we wanted acceleration, why we are here?
00:09:22.800 --> 00:09:32.300
Acceleration is the derivative of velocity or the second derivative of position that is just going to be 24 T.
00:09:32.300 --> 00:09:38.000
Finding the power delivered, bunch of different ways we could do that but let us start by finding the net force.
00:09:38.000 --> 00:09:53.200
That is mass × acceleration which is going to be 10 kg × 24 T our acceleration or 240T.
00:09:53.200 --> 00:10:06.400
Power then is force × velocity which is going to be FV cos θ which implies then that power here
00:10:06.400 --> 00:10:19.700
is going to be 240T × velocity 12T² - 2 which implies then that power = 240 × 12.
00:10:19.700 --> 00:10:49.500
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.
00:10:49.500 --> 00:10:55.200
How about a motorcycle, a 400 kg motorcycle travels along a highway at 30 m /s?
00:10:55.200 --> 00:11:05.900
If the motorcycle breaks with an acceleration of 3 m /s², what is the average power required to bring it to a full stop?
00:11:05.900 --> 00:11:16.100
Power is force × average velocity, it is going to be mass × acceleration, our force × our average velocity.
00:11:16.100 --> 00:11:26.400
Our mass is 400 kg, our acceleration 3 m /s², we are going to worry about the magnitude since we are after the power.
00:11:26.400 --> 00:11:31.500
And it does all that and the average velocity of, it started at 30 it goes to 0,
00:11:31.500 --> 00:11:38.700
a constant acceleration, the average velocity is halfway between 0 and 30 or 15 m /s.
00:11:38.700 --> 00:11:49.600
That is going to give us 18000W or 18kw.
00:11:49.600 --> 00:11:53.200
Let us finish up by looking in an old AP free response problem.
00:11:53.200 --> 00:11:57.300
We will take a look at the 2003 exam Mechanics free response 1.
00:11:57.300 --> 00:12:00.400
Take a minute, go to the web site there, and download it.
00:12:00.400 --> 00:12:14.800
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.
00:12:14.800 --> 00:12:21.800
As we look at part A, given a function of X we are asked to find the speed of the box of × T = 0.
00:12:21.800 --> 00:12:38.100
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.
00:12:38.100 --> 00:12:48.700
Since you want to know this, 1T=0 that just means V at time T=0 must be 2 m /s.
00:12:48.700 --> 00:12:51.700
There is part A.
00:12:51.700 --> 00:12:58.000
For part B, we are asked to determine the following as function of time.
00:12:58.000 --> 00:13:05.200
The kinetic energy of the box and net force on the box and the power being delivered to the box.
00:13:05.200 --> 00:13:19.800
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.
00:13:19.800 --> 00:13:33.600
Or 50 our mass 100 × 1.5 T² +2².
00:13:33.600 --> 00:13:37.300
For part 2, we are asked to find the net force.
00:13:37.300 --> 00:14:03.200
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.
00:14:03.200 --> 00:14:08.600
B3, a power being delivered to the box.
00:14:08.600 --> 00:14:16.200
The power is force × velocity which is going to be our force 300 T.
00:14:16.200 --> 00:14:20.900
We already did our velocity 1.5 T² + 2,
00:14:20.900 --> 00:14:34.300
multiplying that through that is going to be 450 T³ + 600T.
00:14:34.300 --> 00:14:37.300
There is part B.
00:14:37.300 --> 00:14:43.000
Moving on to the C, let us give ourselves some more room here.
00:14:43.000 --> 00:14:47.000
Calculate the net work done on the box from 0 to 2 seconds.
00:14:47.000 --> 00:14:55.300
What we can do, the net work will be the integral of the power with respect to time from T = 0 to 2 seconds,
00:14:55.300 --> 00:15:07.400
which will be the integral from 0 to 2 of 450 T³ + 600 T DT which we just determined.
00:15:07.400 --> 00:15:26.600
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
00:15:26.600 --> 00:15:55.300
225 T⁴/2 + 300 T² evaluated from 0 to 2, which will be 225 × 2⁴ ÷ 2 + 300 × 2² or 3000 joules.
00:15:55.300 --> 00:16:05.700
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.
00:16:05.700 --> 00:16:07.300
It is got to be greater than.
00:16:07.300 --> 00:16:15.300
Why? The student’s word has to be greater than the net work because the student had to work against friction.
00:16:15.300 --> 00:16:18.500
The student has to do more work compared to what you just have in the box.
00:16:18.500 --> 00:16:23.600
The net work is a student minus friction, the work by the student minus the work than friction.
00:16:23.600 --> 00:16:33.500
The work done by the student has to cover the net work + the work done by friction.
00:16:33.500 --> 00:16:36.900
Explain that in word somehow to justify your answer.
00:16:36.900 --> 00:16:39.300
Hopefully, that gets you a good start on power.
00:16:39.300 --> 00:16:42.000
Thank you so much for watching www.educator.com.
00:16:42.000 --> 00:16:44.000
We will see you in the next lesson and make it a great day everybody.