# Lesson Video: Power Mathematics

In this video, we will learn how to find the power of a constant force using the relation 𝑃 = 𝐹 × 𝑣.

17:40

### Video Transcript

In this video, we’re talking about power, specifically mechanical power that arises from a force acting on an object to give it a constant velocity. Along the way, we’ll learn a few different units in which power can be expressed.

Let’s begin by reminding ourselves of the connection between power and work. The work done on some object by a force 𝐹 is equal to the product of that force times the displacement the object experiences. So given this mass 𝑚 at rest on a flat surface, we could do work on this mass by lifting it up through a displacement 𝑑. The force we would apply to do this is equal to the weight force 𝑚 times 𝑔, the acceleration due to gravity, acting on the mass.

The relationship between work and power is that power is an amount of work that’s done over some amount of time. In other words, if we divided the work we did on this box by the time it took to elevate the box through this displacement, then that would equal the power exerted on the box. Power then is work over time. And if we focus on this part of the equality, notice that on the left-hand side we have a displacement divided by a time. This is equal to an object’s velocity, the displacement an object experiences divided by the time taken to be displaced that far. This means we can write power as force times velocity.

And note that in this equation, we’re assuming that the force 𝐹 and the velocity 𝑣 are constant. In other words, in the context of elevating this box, for example, the force we apply exactly matches the magnitude of the weight force on the box, which means that as our mass is elevated, it doesn’t accelerate but rather maintains a constant velocity. This relationship then assumes that objects are in equilibrium, not accelerating.

To see another example of this, say that we’ve got a mass on an incline. This mass is attached to a rope that runs over a pulley, and then we’re pulling on the other end. Say we exert a force 𝐹 and that this force is equal to the sum of all the forces that tend to make this mass move down the incline. That is, it’s equal to the gravitational force in this direction along with the force of friction resisting the mass’s motion. Due to this balance of forces on the mass, the mass is moving up the incline, we’ll say, at a speed 𝑣. We would say then that the power applied to this mass is 𝐹 times 𝑣.

Even though in general we will limit ourselves to cases where 𝐹 and 𝑣 are constant, it’s certainly physically possible that one or both of them might vary. For example, going back to our case of this mass at rest on a flat surface, imagine that instead of exerting a vertical force, we start to push on the mass horizontally. If we were to make a plot of the horizontal speed of our mass against the applied force, we would find that if our applied force was, say, zero, then the speed would also be zero. And if we assume that there is a nonzero frictional force between our mass and the surface it might move across, then that means that even as we increase the applied force 𝐹 up to a point, our mass won’t move. Eventually, though, with a strong-enough push, our box will be set in motion.

If we wanted to calculate the total power exerted on the mass over this time, we can see that it’s not as simple as multiplying one force by one speed. After all, these quantities vary. Graphically, though, if we calculate the area under this curve, then that is equal to the total power applied. All this to say, even when 𝐹 and 𝑣 are not constant, it’s still may be possible to calculate the power.

Now, before we get to some example exercises, it’s worth saying a word about the units of the values in this equation. First, considering force, we know that the SI base unit of force is the newton, abbreviated capital N. And then the SI base unit for power 𝑃 is the watt, abbreviated capital W. For velocity, we use the base units of meters and seconds. A meter per second is the base unit of velocity. All this means that if we multiply one newton of force by one meter per second of speed, then we get one watt of power.

But these aren’t the only units in which we might find forces, speeds, and powers expressed. For example, another unit of force is called the kilogram-weight, and its abbreviated like this. Another unit of power is the horsepower, specifically the metric horsepower, abbreviated hp. And we might find a speed or velocity given to us in units of kilometers per hour.

Something to keep in mind with these units is that if we were to multiply one kilogram-weight by one kilometer per hour, we would not get one metric horsepower. The values work out a bit differently in a less straightforward way than the SI base units. For this reason, we’ll tend to work in the SI units, but it’s nonetheless useful in knowing how to convert between, say, newtons and kilogram-weight or watts and metric horsepower.

So here those conversions are for force, power, and speed. One kilogram-weight is equal to 9.8 newtons. One metric horsepower is equal to 735 watts. Note that there’s also a unit called mechanical horsepower, which works out to a different number of watts, so we need to be careful. And one meter per second is equal to 3.6 kilometers per hour. Knowing all this, let’s look now at an example exercise.

Given that a car’s maximum speed is 270 kilometers per hour and its engine generates a force of 96 kilograms-weight, determine the power of its engine.

All right, so let’s say that this is our car moving along at its maximum speed of 270 kilometers per hour. The car does this, we’re told, when its engine generates a force of 96 kilograms-weight. That force, we’ll call it 𝐹, pushes the car forward, which we might think would make the car accelerate. But actually, the car maintains this constant speed because there are opposing forces to the car’s motion. We don’t need to consider them in detail, but in general these will be frictional forces.

Our goal is to determine the power of this car’s engine. And we can start doing this by recalling that power is equal to force multiplied by speed. This suggests that we can calculate the power of our car’s engine by multiplying the given force by the given speed. The challenge to this, though, is that the units involved are not on a consistent basis. Ultimately, we would like to express our engine’s power in horsepower. But if we multiply these values as is, we’ll get a result in kilograms-weight kilometers per hour. To clarify the meaning of the units involved, we can convert them to more familiar SI base units.

One kilogram-weight is equal to 9.8 newtons of force. Likewise, one kilometer per hour is equal to one over 3.6 meters per second. This means that if we take the value 96 and we multiply it by 9.8, then we’ll have a total force in units of newtons. And then if we work with 270, dividing this by 3.6, we’ll have a speed in units of meters per second. All this is helpful because it means if we were to calculate this power now, we would get an answer in the SI base unit of power, watts.

As we mentioned, though, we would like to give our final answer in units of metric horsepower. One metric horsepower is equal to 735 watts, which means if we divide the right-hand side of our equation by 735 watts per horsepower, where, just like with our other conversions, we’re effectively multiplying by one, which is why we can apply this operation to just one side of our equation, then the units of newton meters per second cancel with watts in the denominator and our units of horsepower come up top so that finally we have an expression for power in the units of interest.

When we enter this expression on our calculator, we get a result of exactly 96 horsepower. This is the power of our car’s engine.

Let’s now look at an example where, instead of calculating power, we calculate speed.

A tractor has an engine of 187 horsepower and it is pulling against a force of 374 kilograms-weight. Find its maximum speed.

All right, so let’s say that this is our tractor. And we’re told that it has an engine with a power of 187 horsepower. It exerts this power to pull against a force of 374 kilograms-weight. This is a way of telling us that the force opposing the tractor’s forward motion is 374 kilograms-weight. For the tractor to move forward as designed, it needs to match this force in the opposite direction. Its engine supplies the power to do that. The tractor will move at a constant speed when the forward force 𝐹 matches 374 kilograms-weight in magnitude.

And when all 187 horsepower of its engine are used, it will be moving at its maximum speed. These three quantities — power, force, and speed — are related by this equation. Since we want to solve for speed rather than power, we can divide both sides by the force 𝐹, cancelling that out on the right. This leaves us with the result that 𝑃 over 𝐹 equals 𝑣. In theory, we could solve for the maximum speed of the tractor by inputing our power and our force to this equation.

The trouble, though, is we would end up with units of horsepower per kilograms weight, an unfamiliar unit for speed. To express the maximum speed in more familiar units, we can convert horsepower and kilograms-weight into their respective SI base unit equivalents. One metric horsepower is equal to 735 watts. And one kilogram-weight equals 9.8 newtons. So if we multiply the numerator of our fraction by 735 watts per horsepower, then in terms of the units, horsepower will cancel out and we’ll be left with units of watts. And if we likewise multiply our denominator by 9.8 newtons per kilogram-weight, then kilogram-weight will cancel out, leaving us with units of newtons.

Note that in each case, as we multiplied by these conversion factors, we were effectively multiplying by one. That’s what makes these operations allowable algebraically. When all the dust settles then, we end up with units of watts per newton. And as we can infer from our equation, power divided by force is equal to speed, a watt per newton is equal to a meter per second.

We’re almost ready to calculate our maximum speed 𝑣. Before we do, there’s one last change to make. We could report 𝑣 in meters per second, but given the units we started with, horsepower and kilograms-weight, it’s more common to report speeds in kilometers per hour. Remembering that one meter per second is equal to 3.6 kilometers per hour, if we multiply our fraction by 3.6 kilometers per hour per meter per second, effectively multiplying by one, then we see the units of meters per second cancel from numerator and denominator.

Finally, then, we have an expression for 𝑣 in the units of interest. Computing the speed, we find a result of 135 kilometers per hour. This is the maximum speed the tractor can attain.

Let’s look now at one last example where our vehicle is moving on an inclined surface.

A vehicle of mass three metric tons was moving at 51 kilometers per hour along a horizontal section of road. When it reached the bottom of a hill inclined to the horizontal at an angle whose sine is 0.5, it continued moving at the same speed up the road. Given that the resistance of the two sections of road is constant, determine the increase in the vehicle’s power to the nearest horsepower. Take the acceleration due to gravity 𝑔 to equal 9.8 meters per second squared.

All right, so in this example, we have this vehicle moving originally along a horizontal section of road. Its speed at that point is 51 kilometers per hour. And we’re told that eventually this vehicle reaches the base of a hill and then starts to move up the hill, maintaining the same speed. If we call the angle that this incline makes with the horizontal 𝜃, we’re told that the sin of 𝜃 equals 0.5 or one-half.

We’re also told that the resistance of the horizontal stretch of road to the vehicle’s forward motion is equal to the resistance of the inclined stretch of road. In other words, the frictional forces opposing the vehicle’s motion are the same in either case. What is different is that once the vehicle is moving uphill, it now has to work against the force of gravity. To maintain its original speed, the vehicle will need to increase the power its engine generates. That’s the increase we want to calculate here.

Before we clear some space on screen to work, let’s recall the fact that the mass of our vehicle, we’ll call it 𝑚, is given to us as three metric tons. We’ll abbreviate that three t. To get started on our solution, we can recall that when an object is subject to a force causing it to move at a speed 𝑣, the product of that force and 𝑣 is equal to the power applied to it. In our case, though, we really want to solve for an increase in power, we’ll call it Δ𝑃, that the car’s engine needs to supply. This increase in power is to supply an increase in force, Δ𝐹, needed for the car to maintain its same speed, 𝑣, as it moves onto this inclined plane.

To get a clearer sense for the forces involved here, let’s look at this up-close sketch and draw a free body diagram. Recall that this involves sketching in all the forces that are acting on our vehicle. We know that our vehicle is subject to a gravitational force. Its magnitude is the vehicle’s mass times the acceleration due to gravity. It’s also acted on by a normal force and by a force that resists its motion across this road surface. Lastly, there’s the force applied to the vehicle, thanks to the power from its engine. This is what allows it to move up the incline.

Our problem statement told us that what we’ve called 𝐹 sub r, the resistive force, hasn’t changed from when our vehicle was traveling on flat ground to this incline. What has changed is that we now need to work against gravity. Specifically, if we break the gravitational force into its two perpendicular components, we’re now opposed by this component of that force. To solve for the magnitude of this component, we recognize that this triangle we’ve sketched is a right triangle where this angle here, just like the angle of our incline, has a measure of 𝜃. That tells us that the component of the gravitational force we now need to work against is 𝑚𝑔 times the sin of 𝜃.

In magnitude, this is equal to the increase in force that needs to be exerted on our vehicle to move it uphill. So we can replace Δ𝐹 in our equation with 𝑚 times 𝑔 times the sin of 𝜃. And note that we know 𝑚, that’s given to us, as is the acceleration due to gravity 𝑔, as is the sin of 𝜃 and the speed 𝑣. If we plug in all these values, though, we see that the units involved don’t match one another. That is, they don’t all come from the same system of units. And even those that do aren’t expressed in quite the same way. For example, our acceleration due to gravity has distance expressed in meters, while our speed has distance expressed in kilometers.

To make sense of our calculation for Δ𝑃, we’ll want the units on the right-hand side to all be on the same basis. To do this, we can recall that one metric ton is equal to 1000 kilograms and that one kilometer per hour is equal to one over 3.6 meters per second. Since one metric ton equals 1000 kilograms, three metric tons is equal to 3000 kilograms. And then considering this unit conversion here, we find that if we divide 51 by 3.6, then we have effectively changed the units of this value to meters per second.

Now all of our units are SI base units, kilograms for mass, meters for distance, and seconds for time. If we were to calculate Δ𝑃 now, we would get a result in units of watts, the SI base unit of power. Our question, though, wanted us to state this increase in power in units of horsepower. One metric horsepower is equal to 735 watts. So if we divide the right-hand side of our equation by 735 watts per horsepower, what we’ll find, and we can check it if we want, is that all of the SI base units will cancel out, while the units of horsepower will move into the numerator. All this to say, we’ll get the units that we’re after, horsepower.

When we do go ahead and calculate this fraction, rounding to the nearest horsepower, we get a result of 283. This is the increase in power necessary from this vehicle’s engine to keep it moving at the same speed as it moves up the incline.

As we finish up, let’s review some key points from this lesson. We’ve seen here that mechanical power is the rate at which a force does work. Written as an equation, we would say that the force applied to an object to make it move at a constant speed 𝑣 when multiplied by that speed is equal to the power applied. This equation assumes a balance of forces acting on whatever object of interest we’re considering. In other words, the acceleration of that object is zero. And lastly, we saw that the units in which 𝑃, 𝐹, and 𝑣 are expressed can vary.

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