### Video Transcript

In this lesson, we will learn how
to apply the conservation of energy principle to solve problems on moving
bodies. Let’s start by defining what
conservation of energy is.

Conservation of energy means that
the total energy of an isolated system remains constant. In equation form, we can say that
the total initial energy of the system, 𝐸 subscript 𝑖, is equal to the total final
energy of the system, 𝐸 subscript 𝑓. In this video, when we talk about
the total energy of the system, 𝐸, we are talking about the potential energy plus
the kinetic energy plus the work done against friction.

We should recall that the
gravitational potential energy of an object, 𝑃𝐸, is equal to 𝑚𝑔ℎ, where 𝑚 is
the mass of the object measured in kilograms, 𝑔 is the acceleration due to gravity
measured in meters per second squared, and ℎ is the height of the object above a
reference point measured in meters. We also need to recall the kinetic
energy of an object is equal to one-half 𝑚𝑣 squared, with 𝑚 once again
representing the mass measured in kilograms and 𝑣 being the speed of the object
measured in meters per second.

The work done against friction can
be expanded out to be the force of friction times the displacement, where the force
of friction is parallel to the displacement. The force of friction, 𝐹 subscript
𝑓, is measured in newtons and the displacement is measured in meters. A ball that is dropped will have
all of its initial energy as potential energy when it’s dropped and all of its final
energy as kinetic energy, just as it’s about to hit the ground. If air resistance is negligible,
then we can say the potential energy of the ball when it’s dropped is equal to the
kinetic energy of the ball just before it hits the ground.

Let’s apply the conservation of
energy principle to a few example problems, beginning with a body on a smooth
inclined plane.

A body started sliding down the
line of greatest slope of a smooth, inclined plane. When it was at the top of the
plane, its gravitational potential energy relative to the bottom of the plane was
1830.51 joules. When it reached the bottom of the
plane, its speed was 8.6 meters per second. Find the mass of the body.

When our object is at the top of
the incline, it has a potential energy of 1830.51 joules. When it gets to the bottom of the
inclined plane, it’s traveling at a speed of 8.6 meters per second. We can apply the principle of
conservation of energy to determine our unknown mass. In equation form, the principle
states that the total initial energy of a system is equal to the total final energy
of the system.

Looking back at our diagram, we can
see that our initial energy is all potential energy. And since the object is moving when
it reaches the bottom of the inclined plane, the total final energy is kinetic
energy. To find the mass, we need to expand
out the kinetic energy. Recall that kinetic energy is equal
to one-half 𝑚𝑣 squared, where 𝑚 is the mass of the object and 𝑣 is the speed of
the object. Substituting in the expanded form
for the kinetic energy, we now have the potential energy at the top of the incline
is equal to one-half 𝑚𝑣 squared of the object at the bottom of the incline.

Plugging in the values given to us
in the problem, we have a potential energy of 1830.51 and a speed of 8.6. To isolate the 𝑚, we multiply both
sides by two. This cancels out the one-half on
the right-hand side of the equation. Next, we divide both sides by 8.6
squared. This cancels out that term on the
right side of the equation. This leaves us with only the term
𝑚 on the right side of the equation. When we perform the calculations on
the left side of the equation, we come up with 49.5. We are solving for the mass, which
is measured in kilograms. The mass of the body that slid down
the inclined plane is 49.5 kilograms.

In the next problem, we will use
the principle of conservation of energy to calculate the work done against friction
on a rough inclined plane.

A body was projected up a rough
inclined plane from its bottom. Its initial kinetic energy was 242
joules. The body continued moving until it
reached its maximum height and then slid back down to the bottom. When it reached the bottom, its
kinetic energy was 186 joules. Find the work done against friction
𝑊 during the ascent and the gain in gravitational potential energy 𝑃 when the body
was at its maximum height.

Below, we have drawn a diagram to
represent the problem. In the diagram, we have labeled the
initial and final kinetic energy of the object at the bottom of the inclined
plane. We have also labeled the unknown
work done against friction for the ascent and the unknown gravitational potential
energy of the body at its maximum height. We have two unknown variables that
we will be solving for. Let’s start by solving for the work
done against friction during the ascent.

We can apply the conservation of
energy principle, which states that the total initial energy of a system is equal to
the total final energy of the system. Initially, the only energy the
object has is kinetic energy. So we can replace the total initial
energy of the system with the initial kinetic energy. The total final energy of the
system is gonna be the kinetic energy of the object at the bottom of the incline
plus the total work done against friction. The total work done against
friction is going to be twice the work done against friction on the ascent because
our object goes up and then back down the inclined plane. This is why we use two 𝑊 to
represent the total work down against friction.

Substituting in our values for
initial kinetic energy and final kinetic energy, we are now ready to solve for
𝑊. We start by subtracting 186 from
both sides of the equation, canceling out that term on the right-hand side. The left side of the equation
becomes 56, and the right side of the equation becomes two 𝑊. Next, we divide both sides of the
equation by two, canceling out the term on the right-hand side of the equation. Dividing 56 by two, we get 28. And work is measured in joules. The work done against friction
during the ascent is 28 joules.

We can once again use the
conservation of energy principle to determine the gravitational potential energy of
the body at its maximum height. The total initial energy of the
system is still the initial kinetic energy of the object. But the final total energy of the
system at the maximum height is the gravitational potential energy 𝑃 plus the work
done against friction on the ascent 𝑊. Substituting in the values for the
total initial kinetic energy and the work done against friction during the ascent,
we are ready to solve for 𝑃.

To isolate 𝑃, we must subtract 28
from both sides of the equation, thereby canceling out the 28 on the right-hand
side, giving us a value for the gravitational potential energy of 214 joules. The gain in gravitational potential
energy 𝑃 when the body is at its maximum height is 214 joules.

For our next example, we will apply
the conservation of energy to calculate the change in energy of a body in free
fall.

A body of mass nine kilograms fell
vertically from a point 3.4 meters above the ground. At a certain moment, the speed of
the body was 3.9 meters per second. Determine the change in the body’s
gravitational potential energy from this point until it reached a point 68
centimeters above the ground. Take 𝑔 equals 9.8 meters per
second squared.

Our object was dropped from a
height of 3.4 meters. We are looking for the change in
gravitational potential energy between when our object is at a speed of 3.9 meters
per second and when our object is at a height of 68 centimeters above the
ground. Before we can determine the change
in gravitational potential energy of our object, we must first find its height above
the ground when its speed is 3.9 meters per second. We can use the conservation of
energy principle, the total initial energy of the system equals the total final
energy of the system, to determine the unknown height.

The total initial energy is
potential energy as our object is dropped from a height above the ground. Our total final energy is made up
of potential energy, as our object is at some unknown height above the ground, and
also kinetic energy, as our object is now moving. We should recall that the potential
energy of an object is equal to the mass of the object times the acceleration due to
gravity times the height of the object above the ground.

To differentiate between our two
heights, we can call our initial height ℎ subscript 𝑖 and our final height ℎ. To substitute in for the kinetic
energy of the object, we need to remember that the kinetic energy is equal to
one-half times the mass times the speed of the object squared. Looking at the expanded equation,
we can see there’s an 𝑚 in each one of our terms. Since our 𝑚 is nonzero in this
problem, we can cancel it out of each of the terms.

Substituting in the values from our
problem, we have 9.8 for 𝑔, 3.4 for initial height, and 3.9 for our speed. The left side of the equation
becomes 33.32 and the right side of the equation becomes 9.8 times ℎ plus 7.605. To isolate ℎ, we can subtract 7.605
from both sides of the equation, cancelling out the term on the right-hand side. Subtracting the term from the left
side of the equation, we get 25.715.

The final step to solve for ℎ is
dividing both sides by 9.8, canceling out the term on the right-hand side. The height of the object above the
ground when its speed is 3.9 meters per second is 2.62398 meters. Now that we know the height, we can
find the change in potential energy from when the object’s at 2.62398 meters and at
68 centimeters above the ground. The change in potential energy,
Δ𝑃𝐸, is equal to the final potential energy minus the initial potential
energy. Gravitational potential energy is
equal to 𝑚𝑔ℎ. So we can replace 𝑃𝐸 final with
𝑚𝑔ℎ final and 𝑃𝐸 initial with 𝑚𝑔ℎ initial.

We are looking for the change in
gravitational potential energy from when the object’s at a height of 2.62398 meters
and when the object is at a height of 68 centimeters above the ground. Therefore, 2.62398 meters will be
our initial height and 68 centimeters will be our final height. But there is one problem. They’re measured in different
units, meters versus centimeters. We will, therefore, need to convert
68 centimeters into meters, using the conversion factor that one meter is equal to
100 centimeters. When we convert 68 centimeters to
meters by dividing by 100, we get 0.68 meters.

Both of the terms on the right side
of the equation have an 𝑚 times 𝑔 in it. Therefore, we can factor out the
term 𝑚 times 𝑔 and put it out front. Substituting the values from our
problem with the mass of nine, 𝑔 of 9.8, final height of 0.68, and initial height
of 2.62398, we calculate the change of potential energy to be negative 171.459
joules, where the negative tells us that this is a loss in energy.

The problem asked us to find the
change of potential energy and does not require us to tell them whether it’s a gain
or a loss. So we can apply our absolute value
to both sides of the equation. The change in gravitational
potential energy between the two positions is 171.459 joules.

In our next example problem, we
will be using the conservation of energy principle where we need to account for
friction.

The figure shows a body of mass
one-quarter kilograms before it started to slide along the surface. The two surfaces 𝐴𝐵 and 𝐶𝐷 are
smooth. However, the horizontal plane 𝐵𝐶
is rough, and its coefficient of kinetic friction is seven-tenths. If the body started moving from
rest, find the distance that the body covered on 𝐵𝐶 until it came to rest. Consider the acceleration due to
gravity to be 𝑔 equals 9.8 meters per second squared.

We can add the labels of mass is
equal one-quarter kilogram and 𝜇, the coefficient of kinetic friction, is equal to
seven-tenths to the figure. We can apply the principle of
conservation of energy, where the total initial energy of the system is equal to the
total final energy of the system, to our problem.

Initially, at position 𝐴, the
object is at rest, so the only form of energy that the object has is gravitational
potential energy. The object comes to rest somewhere
along the plane of 𝐵𝐶. Therefore, the final energy does
not have any potential energy or kinetic energy, only the work done against
friction. We should recall that the
gravitational potential energy of an object is equal to the mass of the object times
acceleration due to gravity times the height of the object above the ground.

We can substitute in 𝑚𝑔ℎ for the
potential energy in our formula. We should also recall that the work
done by friction is equal to the force of friction times the distance traveled,
which allows us to substitute in force of friction times distance for the work. The problem does not give us the
force of friction, but it does give us the coefficient of kinetic friction. So we must use the definition of
force of friction, which is the coefficient of kinetic friction times the normal
reaction force.

The normal reaction force is the
force that a surface puts on an object. When an object is traveling along a
horizontal surface as, it does in this problem, without any other vertical forces
besides the force of gravity and the normal reaction force acting on it, then we can
say the normal reaction force is equal to the force of gravity, where the force of
gravity is the mass of the object times the acceleration due to gravity.

Looking at our expanded equation,
we can see that there’s an 𝑚 and a 𝑔 on both sides of the equation. Since these are both nonzero
numbers, we can cancel them out. Now we can plug in the values from
our problem, four for the height and seven-tenths for the coefficient of kinetic
friction. To isolate 𝑑, we can multiply both
sides of the equation by 10 over seven. This will cancel out the
seven-tenths on the right side. The left side of the equation
multiplies out to be 40 over seven. The distance the body covered on
𝐵𝐶 until it came to rest is 40 over seven meters.

Let’s review the key points of the
lesson.

Key Points

The conservation of energy
principle, total energy of an isolated system remains constant. In equation form, we can say the
conservation of energy principle is 𝐸 subscript 𝑖, the total initial energy of the
system, is equal to 𝐸 subscript 𝑓, the total final energy of the system. When friction is present, the work
done against friction must be included in the calculation.