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Power: Example

by Jared Rovny
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    00:01 Let's now look at a quick example of how you could look for work and power in a given system.

    00:07 For this simple system again, we have an apple and we're starting it at a height and we have a particular mass, m.

    00:12 For these again, just variables, we don't have any numbers here which we're getting more and more used to.

    00:17 We could ask how much work was done by gravity as the apple fell a height, h.

    00:21 We could ask what was the average power exerted by gravity over this entire distance, and we could also ask what was the instantaneous power, the power right at that particular moment, halfway through the fall.

    00:32 So I would recommend you try this based on the definitions that we've introduced so far and give it a shot and see how it goes and then we'll give it a try here as well.

    00:39 If you've given this problem a shot, hopefully it will look something like this.

    00:44 For the first part, we're gonna see how much work was done by gravity.

    00:47 We resort back to our definition of work.

    00:50 Work is a force times a distance times the cosine of the angle between the force and distance.

    00:56 In this case, our object is falling so the distance here is downwards.

    01:00 The force of gravity is also acting downwards and so the angle again between the force and the distance is zero degrees and so they're acting on the same direction.

    01:09 So, we only need force times distance.

    01:12 The force of gravity is mg, the distance here is h and so the work done by gravity is simply mgh.

    01:25 So, this first part was quite simple, just a reiteration of work, but now we can use this quantity for work that we just found to find the average power exerted by gravity through the entire time.

    01:36 We again found that work was equal to mgh and we know that power is equal to the work per unit time.

    01:46 What we're going to do here is find the average power, which is just like average velocity, means that we're going to take the total work done over the total change in time.

    01:57 So, the work is not so bad, we've already found this, so we have m times g times h, so we just need to divide by the time, the amount of time this fall took to take place.

    02:06 So this means that we're going to have to resort one more time to our equations of motion.

    02:11 We have our position is equal to our initial position, plus our initial velocity times the time minus 1/2gt squared which just comes from putting in minus g for acceleration.

    02:23 I recommend you always review the equations of motion and how we use this so that you can call them to mind later as we're doing here to find variables that you might need to find like the time in this case.

    02:33 So here what we're going to do is say that there was no initial velocity and solve for the time here.

    02:39 So we have minus 1/2gt squared equals x minus x0.

    02:46 So this is just final minus initial.

    02:48 We can take the square root after multiplying both sides by this minus 2 over g and see that this is equal to minus 2 x minus x0 over g.

    03:02 Now again, we have to be careful with this x minus x0.

    03:06 The final position x is equal to zero. The initial position x nought is equal to h.

    03:13 So final minus initial is zero minus h which is equal to a negative number, negative h, which is good for our square root because we don't wanna try to take a square root of a negative.

    03:24 We put in the minus h and we see that the minuses cancel and we get the square root of 2h over g.

    03:31 So, this was all a quick aside to figure out what the time was.

    03:36 So that was the question we were answering here is what is the time, which we just found.

    03:39 So we'll put this time in to our equation.

    03:42 We get the square root of 2h over g and because we don't have any variables given to us in this problem, there's really not much more we can do with what we've been given.

    03:52 We simply have that our power is equal to mgh over the entire time of the whole fall which was 2h over g.

    04:00 So, now we found the work done by gravity, we found the average power that gravity had to do in order to bring this apple through a particular amount of work and a particular amount of time.

    04:11 We can do one more thing which is instead of asking about the average power, so I'm gonna write average for the power here, we can ask about the power at a particular point since the power will be changing depending on how much work was done at a given amount of time.

    04:24 Doing that, it might be easier to take this particular point, which is halfway through the fall, and use our other definition of power, that power is a force times a velocity.

    04:36 So in this case, what we have is a velocity at this height h over 2, since we're halfway through the fall, and we still know that the force is the force of gravity mgh.

    04:48 So our only question is, can we find this velocity halfway through the fall.

    04:51 So power equals mgh, and then we need v, so let's try to find this velocity.

    04:57 Well, this is something that we know how to do quite well now as we've practiced many times with the energy example, is the energy at the top of the path is just the gravitational potential energy, mgh, because there is no motion, no kinetic energy, and then the energy halfway through the path is partly kinetic and partly potential.

    05:16 So, this is mg times its height, which is h over 2, that's the potential energy, plus and then we have the kinetic energy because it's in motion, so 1/2 mass times velocity squared.

    05:29 So, now we have 2 energies and we can find the velocity here.

    05:32 So this is all a brief caveat again to find velocity, so what is the velocity? By equating these 2 energies we can see E1 equals E2 which means mgh equals mgh over 2 plus 1/2mv squared.

    05:51 So now we can solve for velocity very easily here, we have 1/2 mv squared equals mgh minus a half mgh which is 1/2mgh.

    06:02 We can multiply both sides by 2, divide both sides by m and see that the velocity in this case is equal to the square root of g times h.

    06:11 So, this is it, we have our velocity.

    06:13 This whole thing was just to find this quantity, so we can plug that back in up here and have the force mgh times the velocity which we just found is the square root of gh.

    06:25 And again, so this is the power halfway through the fall, maybe we can write it like that, again, because we don't have any numbers given in this problem, you could always put in numbers for a problem you were given.

    06:35 We just have an expression just in terms of the variables that we solved for in this particular problem.

    06:42 So, this is it. This is a good summary of how power works, how power works with the idea of the physical quantity called work and how to find time to get the power or how to find the velocity to find the power.

    06:53 Now that we've wrapped up a basic summary of how work and power work together with force and energy, we're ready to move to our last mechanics topic which is momentum, and we're gonna do that next to wrap up the mechanics section of this whole course.

    07:06 Thanks for listening.


    About the Lecture

    The lecture Power: Example by Jared Rovny is from the course Work.


    Included Quiz Questions

    1. 485W.
    2. 455W.
    3. 433W.
    4. 499W.
    5. 480W.
    1. 2.30 x 103 J.
    2. 4.3 x 103 J.
    3. 3.20x 103 J.
    4. 2.9 x 103 J.
    5. 1.73 x 103 J.

    Author of lecture Power: Example

     Jared Rovny

    Jared Rovny


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