### 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.