Understanding Power in Physics

Power are going to be the topic of this lesson in my brand new physics playlist, which eventually will cover a full year of university algebra-based physics. Now in this lesson in power, we're going to go through the basic definition, the formula, the SI units, and we're going to work a handful of calculations. My name is Chad, and welcome to Chad's Prep, where my goal is to take the stress out of learning science. Now if you're new to the channel, we've got comprehensive playlists for general chemistry, organic chemistry, general physics, and high school chemistry. And on chadsprep.com you'll find premium master courses for the same that include study guides and a ton of practice. You'll also find comprehensive prep courses for the DAT, the MCAT, and the OAT. Okay, so power, it's the rate at which work can be delivered, if you will. And so as such, it's equal to work over the change in time here. So, and if we take a look at those SI units, it turns out we measure it in what we call the watt. And the abbreviation for the watt is just a simple W. And if you look at kind of what that is, well, the SI unit for work here, W, is joules in time seconds. And so a watt is equal to a joule per second. And if you kind of break that up further, well, a joule, if you take a look at work, work equals force times displacement. And so you can see that the joule here is going to equal the Newton meter. We said that in an earlier lesson in this chapter. And so you could also see this written as the Newton meter per second. So, but that Newton using... F equals mA in the last chapter, we saw that you could express the Newton as the kilogram meter per second squared, and we could substitute this in right here for that Newton, and rearrange some terms, combine the meter here and the seconds here, and overall you'd see this is going to equal the kilogram meter squared per second cubed. Cool. You're most likely going to see it, if at all, written as the joule per second, but definitely could be asked. It's not, it's totally fair game. to have the watt expressed in all SI base units. Okay, so let's talk a little more about power for a second. So, if we take a look at this relationship here, we can see that power is proportional to work. So, if you double the amount of work done, you're gonna double the power. They're directly proportional. But you see there's an inverse relationship with the amount of time. So inversely proportional, power is inverse proportional to one... or I should say to time. So, it's proportional to one over time. So... in this case it's about the rate at which work can be delivered. So if you want to have a faster rate, i.e. more power, you actually want a shorter duration of time. And so if you actually, like we said, double the work, you double the power. But if you double the amount of time in which that work is delivered, you're actually cutting the power in half. There's an inverse relationship. All right. Now we also said that work is force times displacement, and we also wrote that out like force times delta x, and we did it in the x component. We're going to look at power almost universally just in one dimension. In fact, you'll probably not even see that they even mention dimensionality in any way, shape, or form. But it turns out there's a little bit of a trouble trying to take a look at what it is in more than one dimension, so we just kind of don't. So you're only ever going to look at it in one dimension, and probably won't even be overtly addressed in any way, shape, or form. But if you take a look at rewriting this just a little bit, we say that it's equal to force times displacement. So we now can see that in addition to power being proportional to work, power is also proportional to the force and power is also proportional to the displacement. So if you're applying more force, double the force, you're doubling the power. So if you're moving an object further, if you double the displacement, then you double the power. So just another, breaking it down, different way of looking at it. Now one other thing this provides for us though is we can get one other definition out of this. So if you look at delta x over delta t, that is the definition of velocity. And so one alternate form for the equation for power is force times velocity. And so we really got two equations for power. Now this top one, that's the definitional version of it. So for that formula, but this one comes up on occasion, not quite as commonly, but on occasion as well. And so you've really got a couple different options. Can you calculate the work and the time, or do you know some sort of constant force or average force being applied and the velocity of the object that is in motion? Then you can go this route instead. So you should kind of keep these both in the back of your mind when you think of power. Again, the top one is more common and is the definition, but this one will come up a time or two as well. Let's work some problems. Alright, so the first question is an elevator question. It says a loaded elevator weighs 2,000.0 kilograms. What is the average power required to raise the elevator 30.0 meters in 10.0 seconds? So take a look. We're gonna have this 2,000 point O kilogram elevator and we want to raise it 30 meters so we can look at that as a vertical displacement which I'll call Delta Y 30 point O meters and we want to do this in 10.0 seconds There's our basic setup here. We want to calculate that power, and if we recall, power is equal to work over time. Now the time we have, that's 10 seconds of supply, but we've got to get that work. In this case, we've got to remember that work equals force times displacement, because we've got the displacement, we'll have to figure out what the force is from a free body diagram here. But that's the approach we're going to take here. So force times displacement all over delta T. And that displacement, I wrote it as D. It could be delta X, but in this case, because it's in the Y direction, it's probably more properly delta Y as this. I'll just write D, and we can use it in any single dimension, which is typical for power problems. Okay, so if we look at a free body diagram of this thing, we definitely have its weight. pointing down and there's gonna have to be some upward applied force on this elevator to overcome that and Again when an elevator here starts moving it's accelerating instantaneously, but then after that it typically moves up at constant velocity Which is what I'm gonna kind of assume here. We're just calculating average power anyways And so in this case we can see that some of the forces in the y direction equals zero. Since we're assuming constant velocity, F minus mg equals zero, and we can see that this upward applied force must be equal to its weight, which in this case is 2,000 kilograms times 9.8 meters per second, which I guess we can write out. Okay, somewhere in the ballpark of 20,000 Newtons, just shy of that, right? Cool, and that's what's gonna get plugged in right here for this force. We've got the displacement of 30 meters, the time of 10 seconds. We're ready to do a little bit of plugging and chugging here. Okay, let's get that force first, so 2,000 times 9.8. It's gonna get us 19,600. Newtons, displacement of 30 meters, time 10 seconds. And based on the numbers we're given, we're limited to 3 sig figs. And so we'll take 19,600 times 30 divided by 10. And we're going to get 58,800. It'll end up at 3 sig figs anyways. And again, that is in units of watts, which again is a joule per second. You can see that kind of works out, the newton meter per second, with a newton meter being the same thing as a joule. All right, the second one is related to this first one. In fact, we'll kind of find out that it's the same problem. A loaded elevator weighs 2,000.0 kilograms. What is the average power required to raise the elevator at constant speed if its standard operating speed is... is 3.0 meters per second. Okay, so in this case we're given a velocity. In the first one we're given both the displacement and the time, which we could have used to get a velocity, technically. So, but in this one we're actually given the velocity. such if you're given a velocity it's probably going to be more convenient to use this form of the power formula instead and so in this case in fact let's just plug it right in so we know the force again of a constant velocity here is going to have to be that 2000 kilograms times 9.8 which we figured out was the 19 600 newtons so and now we're told the 3.0 meters per second So we're given, yeah, just 3.0, two sig figs this time. So, and it's just force times velocity, and we should get the same answer. And if you notice, if we just took the 30 meters over 10 seconds, that would be 3 meters per second velocity anyways, right? so 19,600 times 3 is indeed 58,800 but based on the numbers given this time we've got to round it to two sig figs so in this case that'll round up to 59,000 watts Alright, so this last question is going to involve finding the horsepower of a motorcycle engine. In this case, the horsepower is actually pretty common in the automotive industry and stuff for any kind of engine, right? So rather than using the SI unit of watts, it's much more common. But we're going to find the conversion is provided to get us from... watts to horsepower in this problem. So a motorcycle engine delivers a 1500 Newton force while maintaining a velocity of 25 meters per second. How much horsepower is the engine providing and one horsepower equals 746 watts. Okay. So we're given the force and the velocity, which makes the latter expression for power here, the convenient one to use in this problem. Again, that force was given as 1500 newtons at a velocity of 25 meters per second. So now we can do this all in one big step. We could actually, a newton meter per second is a watt. We can now convert watts to horsepower. So, but I think that's a little bit inherently confusing. So I'm going to get the answer out in watts first and then convert it at the end. All right, so 1500 times 25 is 37,500. And that again is in watts. And now we want to convert those watts to horsepower. And that conversion again is given as one horsepower equals 746 watts. All right, so 37,500 divided by 746 and we get 50.268. So in this case we're going to round that to two sig figs. And so that's just going to round down to 50, that's a problem, right? So 50 horsepower, but that's only got one sig fig. And again, my preference is not to put a line over the zero like that, technically you can do that. My preference in such cases is to write this in scientific notation instead and write 5.0 times 10 to the first horsepower.

And on chadsprep.com you'll find premium master courses for the same that include study guides and a ton of practice. You'll also find comprehensive prep courses for the DAT, the MCAT, and the OAT. Okay, so power, it's the rate at which work can be delivered, if you will.

And so as such, it's equal to work over the change in time here. So, and if we take a look at those SI units, it turns out we measure it in what we call the watt. And the abbreviation for the watt is just a simple W. And if you look at kind of what that is, well, the SI unit for work here, W, is joules in time seconds.

And so a watt is equal to a joule per second. And if you kind of break that up further, well, a joule, if you take a look at work, work equals force times displacement. And so you can see that the joule here is going to equal the Newton meter.

We said that in an earlier lesson in this chapter. And so you could also see this written as the Newton meter per second. So, but that Newton using...

F equals mA in the last chapter, we saw that you could express the Newton as the kilogram meter per second squared, and we could substitute this in right here for that Newton, and rearrange some terms, combine the meter here and the seconds here, and overall you'd see this is going to equal the kilogram meter squared per second cubed. Cool. You're most likely going to see it, if at all, written as the joule per second, but definitely could be asked. It's not, it's totally fair game.

to have the watt expressed in all SI base units. Okay, so let's talk a little more about power for a second. So, if we take a look at this relationship here, we can see that power is proportional to work.

So, if you double the amount of work done, you're gonna double the power. They're directly proportional. But you see there's an inverse relationship with the amount of time.

So inversely proportional, power is inverse proportional to one... or I should say to time. So, it's proportional to one over time. So... in this case it's about the rate at which work can be delivered.

So if you want to have a faster rate, i.e. more power, you actually want a shorter duration of time. And so if you actually, like we said, double the work, you double the power. But if you double the amount of time in which that work is delivered, you're actually cutting the power in half.

There's an inverse relationship. All right. Now we also said that work is force times displacement, and we also wrote that out like force times delta x, and we did it in the x component. We're going to look at power almost universally just in one dimension. In fact, you'll probably not even see that they even mention dimensionality in any way, shape, or form.

But it turns out there's a little bit of a trouble trying to take a look at what it is in more than one dimension, so we just kind of don't. So you're only ever going to look at it in one dimension, and probably won't even be overtly addressed in any way, shape, or form. But if you take a look at rewriting this just a little bit, we say that it's equal to force times displacement. So we now can see that in addition to power being proportional to work, power is also proportional to the force and power is also proportional to the displacement.

So if you're applying more force, double the force, you're doubling the power. So if you're moving an object further, if you double the displacement, then you double the power. So just another, breaking it down, different way of looking at it. Now one other thing this provides for us though is we can get one other definition out of this.

So if you look at delta x over delta t, that is the definition of velocity. And so one alternate form for the equation for power is force times velocity. And so we really got two equations for power. Now this top one, that's the definitional version of it.

So for that formula, but this one comes up on occasion, not quite as commonly, but on occasion as well. And so you've really got a couple different options. Can you calculate the work and the time, or do you know some sort of constant force or average force being applied and the velocity of the object that is in motion? Then you can go this route instead. So you should kind of keep these both in the back of your mind when you think of power.

Again, the top one is more common and is the definition, but this one will come up a time or two as well. Let's work some problems. Alright, so the first question is an elevator question. It says a loaded elevator weighs 2,000.0 kilograms.

What is the average power required to raise the elevator 30.0 meters in 10.0 seconds? So take a look. We're gonna have this 2,000 point O kilogram elevator and we want to raise it 30 meters so we can look at that as a vertical displacement which I'll call Delta Y 30 point O meters and we want to do this in 10.0 seconds There's our basic setup here.

We want to calculate that power, and if we recall, power is equal to work over time. Now the time we have, that's 10 seconds of supply, but we've got to get that work. In this case, we've got to remember that work equals force times displacement, because we've got the displacement, we'll have to figure out what the force is from a free body diagram here.

But that's the approach we're going to take here. So force times displacement all over delta T. And that displacement, I wrote it as D. It could be delta X, but in this case, because it's in the Y direction, it's probably more properly delta Y as this. I'll just write D, and we can use it in any single dimension, which is typical for power problems.

Okay, so if we look at a free body diagram of this thing, we definitely have its weight. pointing down and there's gonna have to be some upward applied force on this elevator to overcome that and Again when an elevator here starts moving it's accelerating instantaneously, but then after that it typically moves up at constant velocity Which is what I'm gonna kind of assume here. We're just calculating average power anyways And so in this case we can see that some of the forces in the y direction equals zero. Since we're assuming constant velocity, F minus mg equals zero, and we can see that this upward applied force must be equal to its weight, which in this case is 2,000 kilograms times 9.8 meters per second, which I guess we can write out. Okay, somewhere in the ballpark of 20,000 Newtons, just shy of that, right?

Cool, and that's what's gonna get plugged in right here for this force. We've got the displacement of 30 meters, the time of 10 seconds. We're ready to do a little bit of plugging and chugging here. Okay, let's get that force first, so 2,000 times 9.8.

It's gonna get us 19,600. Newtons, displacement of 30 meters, time 10 seconds. And based on the numbers we're given, we're limited to 3 sig figs. And so we'll take 19,600 times 30 divided by 10. And we're going to get 58,800. It'll end up at 3 sig figs anyways.

And again, that is in units of watts, which again is a joule per second. You can see that kind of works out, the newton meter per second, with a newton meter being the same thing as a joule. All right, the second one is related to this first one. In fact, we'll kind of find out that it's the same problem. A loaded elevator weighs 2,000.0 kilograms.

What is the average power required to raise the elevator at constant speed if its standard operating speed is... is 3.0 meters per second. Okay, so in this case we're given a velocity.

In the first one we're given both the displacement and the time, which we could have used to get a velocity, technically. So, but in this one we're actually given the velocity. such if you're given a velocity it's probably going to be more convenient to use this form of the power formula instead and so in this case in fact let's just plug it right in so we know the force again of a constant velocity here is going to have to be that 2000 kilograms times 9.8 which we figured out was the 19 600 newtons so and now we're told the 3.0 meters per second So we're given, yeah, just 3.0, two sig figs this time. So, and it's just force times velocity, and we should get the same answer.

And if you notice, if we just took the 30 meters over 10 seconds, that would be 3 meters per second velocity anyways, right? so 19,600 times 3 is indeed 58,800 but based on the numbers given this time we've got to round it to two sig figs so in this case that'll round up to 59,000 watts Alright, so this last question is going to involve finding the horsepower of a motorcycle engine. In this case, the horsepower is actually pretty common in the automotive industry and stuff for any kind of engine, right? So rather than using the SI unit of watts, it's much more common. But we're going to find the conversion is provided to get us from...

watts to horsepower in this problem. So a motorcycle engine delivers a 1500 Newton force while maintaining a velocity of 25 meters per second. How much horsepower is the engine providing and one horsepower equals 746 watts.

Okay. So we're given the force and the velocity, which makes the latter expression for power here, the convenient one to use in this problem. Again, that force was given as 1500 newtons at a velocity of 25 meters per second. So now we can do this all in one big step.

We could actually, a newton meter per second is a watt. We can now convert watts to horsepower. So, but I think that's a little bit inherently confusing.

So I'm going to get the answer out in watts first and then convert it at the end. All right, so 1500 times 25 is 37,500. And that again is in watts.

And now we want to convert those watts to horsepower. And that conversion again is given as one horsepower equals 746 watts. All right, so 37,500 divided by 746 and we get 50.268. So in this case we're going to round that to two sig figs. And so that's just going to round down to 50, that's a problem, right?

So 50 horsepower, but that's only got one sig fig. And again, my preference is not to put a line over the zero like that, technically you can do that. My preference in such cases is to write this in scientific notation instead and write 5.0 times 10 to the first horsepower.

Transcript for:Understanding Power in Physics

Transcript for:
Understanding Power in Physics