In this video, we’re going to learn
about conservative and non-conservative forces. We’ll see what these two force
types are and how they’re different from one another.
To start out, imagine that you are
the captain of a gigantic, inflatable blimp. After a long day of riding, you’re
running low on fuel and you want to make it back to the landing pad using as little
fuel as possible. Of all the many paths that can get
you to the landing pad, which one will be the most fuel efficient? To better understand this question,
we’ll want to learn something about conservative and non-conservative forces.
At the heart of this question of
conservative and non-conservative forces is the work done by these forces. Say that we have a spring, and it’s
unstressed and extended to its natural length. If we then compress that spring
using a force, then the energy used in doing that work is not lost energy, rather
it’s stored as potential energy in the spring. In this way, we can say that the
spring force conserves energy. It is a conservative force.
Or think of another example. Say we take a mass 𝑚 and we move
it on top of a table of height ℎ. The work it takes to do this is
equal to the force of gravity 𝑚 times 𝑔 multiplied by the height against that
force we moved this mass. Considering the mass on the top of
the table, we know that that energy isn’t lost. It’s stored as gravitational
potential energy. So, gravity, like the spring force,
conserves the energy used to do work.
We can say that a force is a
conservative force if the work it does results in potential energy. In other words, the energy that
goes into doing the work is conserved rather than dissipated, or lost. There’s something else that’s true
though about conservative forces. And we can use our mass in this
example of gravity to show it.
One question we might have when we
see the mass in its starting position and then in its ending position is, how did it
move between those two points. For example, maybe we picked up the
mass and then walked it over to the table top and placed it down. Or maybe we slid it along the
ground with our foot then picked it up when it got to the edge of the table and
placed it on top. Or perhaps we picked the box up as
we were moving towards the table in one smooth motion.
Considering these different
possible pathways for getting the box from its start to its end point, we realize
that from the perspective of gravitational potential energy the path makes no
difference. Given the mass of the box, all that
matters is the height difference between where it started and where it ended. This leads us to a second way of
describing a conservative force. A force is conservative if the work
it does does not depend on the path taken. That is, it’s path-independent.
We’ve seen that two examples of
conservative forces are gravity and the spring force. Now, what about non-conservative
forces? We can say that a force is
non-conservative if the work it does does depend on the path taken. Imagine we had a large heavy crate
sitting on a rough concrete floor. And say we wanted to slide the
crate along the floor to the opposite corner in the room.
We can see that the particular path
we choose affects how much work it takes to move the box from its start to its end
point. Not only that, but for all the
energy we input to do that work to move the box, when the box reaches its final
location, we don’t have any potential energy to show for it. That energy has been dissipated as
heat energy through friction. This is why another name for a
non-conservative force is a dissipative force. It’s not easy to recover the energy
that goes into the work done by a non-conservative force.
We’ve seen that one example of a
non-conservative force is friction. Another example is air resistance,
which at the molecular level is also due to friction. Let’s practice using these ideas of
conservative and non-conservative forces through a couple of examples.
You are in a room in a basement
with a smooth concrete floor and a nice rug. The rug is three meters wide and
four meters long. You have to push a very heavy box
from one corner of the rug to its opposite corner. The magnitude of friction between
the box and the rug is 55 newtons, but the magnitude of friction between the box and
the concrete floor is only 40 newtons. Will you do more work against
friction going around the floor or across the rug? How much extra work would it
This is an exercise involving
non-conservative forces. And to start out, let’s draw a
diagram. In this situation, we have a very
heavy crate on the corner of a three-by-four-meter rug. We want to move the crate to the
opposite corner. And we’re told that if we move the
crate straight across the rug, the force of friction is 55 newtons. While if we choose instead to move
the crate on the smooth concrete floor around the rug, the overall force of friction
is 40 newtons. We want to figure out which of
these two paths will take more work.
Recalling that work is equal to
force times distance, we can write that the work required to move the crate across
the rug equals 𝐹 sub 𝑟, given as 55 newtons, times the distance across the rug the
crate moves. Since the rug is a rectangle, that
distance 𝑑 sub 𝑟 is equal to the square root of three meters squared plus four
meters squared, or five meters. Plugging in these values and
calculating 𝑊 sub 𝑟, it’s equal to 275 joules.
Next, we wanna calculate the work
done if we move the crate across the floor instead of the rug. This is equal to 𝐹 sub 𝑓, which
is 40 newtons, times 𝑑 sub 𝑓, which is the distance the crate would move. This distance, skirting the edge of
the rug, is equal to four plus three, or seven meters. Plugging in for these two values,
when we calculate 𝑊 sub 𝑓, we find it’s 280 joules. So, comparing the work done on the
rug to the work done on the floor, we see that more work is required to move the
crate across the floor by five joules.
Now, let’s look at an example
involving a conservative force.
A particle with a mass 𝑚 is
suspended from a string of negligible mass and a length of 1.0 meters, as shown in
the diagram. The particle is displaced to a
position where the taut string is at an angle of 30 degrees from the vertical. And the particle is released from
rest at that position. The particle moves through an arc,
where the lowest point of the arc is the point 𝑃. What is the instantaneous speed of
the particle at point 𝑃? What is the vertically upward
displacement of the particle from point 𝑃 when its instantaneous speed is 0.81
meters per second?
We can call this instantaneous
particle speed at point 𝑃 𝑣. And the vertically upward
displacement of the particle from 𝑃, we’ll name 𝑑. We’re told that point 𝑃 on our
diagram is the location of where the mass would be when the string is hanging
straight down. We start off solving for the speed
of the mass when it’s at point 𝑃. Because gravity is a conservative
force, we know that the potential energy of the mass 𝑚 when it’s in its original
position can be converted into kinetic energy of the mass when it’s at point 𝑃.
We can write our energy balance
equation to say that the initial kinetic plus potential energy is equal to the final
kinetic plus potential energy of the mass. At the outset, the speed of the
mass is zero. So, its initial kinetic energy is
zero. And if we set the altitude of the
mass at point 𝑃 to be zero, then its final potential energy will be zero as
well. So, we can say the initial
potential energy of the mass is equal to its final kinetic energy.
Recalling that gravitational
potential energy equals mass times the acceleration due to gravity times height and
that kinetic energy equals one-half an object’s mass times its speed squared, we can
write that 𝑚 times 𝑔 times ℎ is equal to one-half 𝑚𝑣 squared. We see that the mass of this object
cancels out. And rearranging to solve for the
speed 𝑣, we find it’s equal to the square root of two times 𝑔 times ℎ.
𝑔, the acceleration due to
gravity, we treat as exactly 9.8 meters per second squared. And we see on our diagram that ℎ is
the height difference between the point 𝑃 and the original location of the
mass. That height difference is equal to
1.0 meters, the length of the string, minus the length of the string times the cos
of 30 degrees. Plugging this value for ℎ and the
known value for 𝑔 into our expression for 𝑣, to two significant figures, 𝑣 is 1.6
meters per second. That’s how fast the mass is moving
when it’s at its lowest point 𝑃.
In part two, we imagine that the
speed of our mass, we’ll call it 𝑣 sub 𝑑, is 0.81 meters per second. We want to solve for the vertical
distance 𝑑 of our mass at that speed above point 𝑃. In this case, when we write out our
energy balance equation, we’re only able to eliminate the initial potential energy
of our mass because we imagine its initial point is at point 𝑃.
At point 𝑃, the mass has kinetic
energy. And at the point where its speed is
0.81 meters per second, it will have both kinetic and potential energy. Using the variables we’ve decided
on, we can write that one-half 𝑚𝑣 squared, where 𝑣 is the speed we solved for in
part one, is equal to one-half 𝑚𝑣 sub 𝑑 squared plus 𝑚 times 𝑔 times 𝑑, the
distance we want to solve for. Again, the object’s mass cancels
out. And we can rearrange to solve for
𝑑. It’s 𝑣 squared minus 𝑣 sub 𝑑
squared all over two times 𝑔. When we plug in for these values
and calculate 𝑑, to two significant figures, we find it’s 3.3 centimeters. That’s the vertical displacement of
the mass above point 𝑃.
Let’s summarize what we’ve learnt
so far about conservative and non-conservative forces. We’ve seen that a force is
conservative if the work it does does not depend on the path taken. That is, it’s path-independent. And a force is non-conservative if
it does. Another way to say this is that a
force is conservative if the work it does results in potential energy and
non-conservative if the work it does dissipates energy. And finally, by way of example of
these forces, we’ve seen that gravity and the spring force are examples of
conservative forces, while friction and air resistance are examples of