Video Transcript
In this video, we’re going to talk
about the conservation of energy. We’ll learn what this principle
says, what its limits are. And we’ll get some practical
experience working with energy conservation.
To start out, imagine that you and
a friend are busy shooting a music video. Standing on a single-story height
by a driveway where a car is parked, you see it would make a great shot if you
jumped off this height onto the roof of the car and then from there onto the
driveway. The only possible issue you see, as
this is a low-budget video production, is that you might damage the car and incur
expenses.
To better understand whether
getting this shot is a good idea, we’ll want to know more about the conservation of
energy. The conservation of energy means
that, within an isolated system, the total energy at any one time equals the total
energy at any other time. Written as an equation, we can say
that the initial system energy is equal to the final system energy. This principle is simple to
remember, as long as we keep in mind the limits under which it applies.
The main thing to keep in mind is
that energy conservation requires an isolated system, one where no energy enters or
leaves. This idea of a system can best be
explained through an example. Let’s say you hold a ball in your
hand and drop it from some height above ground. As we normally think about this
scenario, in our system, we might include the ball, your hand, and the earth. And in that case, we could show
that the gravitational potential energy of the ball when it’s in your hand is equal
to the kinetic energy of the ball just as it hits the ground. In other words, we could show that
the energy of the ball is conserved. But what if our system was not
isolated?
Imagine that there was a vertically
downward moving wind that, as soon as the ball was released, helped push it towards
the ground. In that case, the ball would speed
up more than its initial potential energy will lead us to think. When it hit the ground, its kinetic
energy would exceed its initial potential energy. It would seem as though energy is
not conserved. But that’s only because we violated
a condition of conservation of energy. Our system, in the case of the wind
coming in, is not an isolated system. We’ve added energy.
When we look at the expression for
energy conservation, we realize this covers all kinds of energy. Broadly speaking, energy comes in
two main types: kinetic or energy of motion and potential energy. We can expand our energy
conservation expression to include these two main types. So we can say that the initial
kinetic and potential energy of a system together is equal to the final kinetic and
potential energy of that system. And here initial and final are any
two times in the evolution of the system. Let’s get some practice applying
this conservation principle through a few examples.
A wagon sits at the top of a
hill. The wagon is given a push that
increases its speed negligibly but is just sufficient to set the wagon in motion
down a straight slope. The wagon rolls 53.9 meters down
the slope, which is inclined 16.5 degrees below the horizontal, and reaches the
bottom of the hill. If friction is negligible, what
speed is the wagon moving when it reaches the bottom of the hill?
We can call the speed of the wagon
when it reaches the bottom of the hill 𝑣 sub 𝑓. And we’ll start on our solution by
drawing a diagram of the situation. A wagon sits perched at the top of
a long flat hill whose length we’ve called 𝐿. It’s 53.9 meters. The hill is inclined at an angle
below the horizontal — we’ve called 𝜃 — given as 16.5 degrees. We’re told the wagon is given a
very slight push to set it in motion and then rolls down the hill. And when it reaches the bottom of
the hill, we want to solve for its speed, 𝑣 sub 𝑓.
In this scenario, our system
consists of the wagon and the hill. And since energy is neither added
to nor taken away from this system, we can say that it’s conserved. Applying the general statement of
energy conservation to this scenario, we’ll use our freedom to decide when the
initial moment and when the final moments are. Choosing the initial moment to be
just when the cart is pushed at the top of the hill, and the final moment when it
just reaches the bottom of the hill.
We can expand this conservation
expression to say that the kinetic energy plus potential energy initially is equal
to the kinetic energy plus potential energy finally. At the outset, the wagon is not in
motion. So its initial kinetic energy is
zero. And if we choose the height at the
bottom of the hill to be our reference level of zero, we can say that the final
potential energy of the cart is also zero, since it’s at zero height. So the wagon’s initial potential
energy is equal to its final kinetic energy.
The wagon’s potential energy is
entirely gravitational. And we recall that gravitational
potential energy is equal to 𝑚 times 𝑔 times ℎ. Applying that relationship, we
write 𝑚 times 𝑔 times the initial height of the wagon equals KE sub 𝑓. We also recall that an object’s
kinetic energy equals half its mass times its speed squared. We can write this in for our KE sub
𝑓 expression.
Looking at this equality, we see
that the mass of the wagon appears on both sides. So it cancels out. We want to solve for the final
speed of the wagon. So we rearrange this equation and
see the wagon’s final speed is equal to the square root of two times 𝑔 times ℎ sub
𝑖. The acceleration due to gravity,
𝑔, we take to be exactly 9.8 meters per second squared.
Looking at our diagram, we see that
the height of the hill, ℎ sub 𝑖, relates to the hill length, 𝐿, and the angle of
inclination, 𝜃. In particular, we can write that ℎ
sub 𝑖 is equal to 𝐿 times the sin of 𝜃. So we substitute 𝐿 sin 𝜃 in for ℎ
sub 𝑖 in our expression for 𝑣 sub 𝑓.
Looking at this expression, we’re
given the length 𝐿, the angle 𝜃, and 𝑔 is a known constant. So we’re ready to plug in and solve
for 𝑣 sub 𝑓. Entering these values on our
calculator, we find that, to three significant figures, 𝑣 sub 𝑓 is 17.3 meters per
second. That’s the speed of the wagon when
it reaches the bottom of the hill.
Now let’s look at an example where,
in our initial condition, we have both potential and kinetic energy.
A hiker being chased by a bear runs
to the edge of a cliff and jumps off. The hiker is running horizontally
at 4.5 meters per second when he reaches the cliff edge. The hiker’s speed when he hits the
ground is 16 meters per second. What vertical downward distance did
the hiker fall? If the hiker stepped forward off
the cliff edge with negligible horizontal velocity, what speed would they hit the
ground at?
So we want to solve for the
distance the hiker falls. And we also wanna solve for the
speed the hiker would have if the hiker left the cliff edge with negligible
horizontal velocity. We can call this distance ℎ. And we can label the speed they hit
the ground at 𝑣 sub 𝑔. We’ll start by drawing a diagram of
the situation.
In the first scenario, our hiker is
running away from the bear at a speed of 4.5 meters per second in the horizontal
direction, before leaping off the edge of the cliff. We know that, under these
conditions, the hiker’s overall speed when he hits the ground is 16 meters per
second. And we want to use that information
to solve for the height of the cliff, ℎ.
If we approach this from an energy
perspective, we can say that the hiker’s energy, kinetic plus potential, just as he
leaves the edge of the cliff is equal to his energy just as he lands. Those are our initial and final
moments. In this system of the hiker and the
cliff, energy is conserved. So we can write that 𝐸 sub 𝑖 is
equal to 𝐸 sub 𝑓 and expand that expression to include kinetic and potential
energies.
If we set our zero value for height
at the base of the cliff, that means that when the hiker lands, he has no potential
energy due to gravity. So that term goes to zero. It turns out that’s the only term
we can eliminate because, initially, as the hiker leaves the cliff’s edge, he has
both gravitational potential energy and kinetic energy due to his motion.
Recalling the equation forms of
gravitational potential energy, 𝑚 times 𝑔 times ℎ, and kinetic energy, one-half
𝑚𝑣 squared, we can expand our energy balance equation. So it reads one-half the hiker’s
mass times 𝑣 sub 𝑥 squared, the horizontal speed as the hiker leaves the
cliff. Plus the hiker’s mass times 𝑔
times the height of the cliff is equal to one-half the hiker’s mass times 𝑣
squared. Where 𝑣 is the hiker’s overall
speed when he lands. We see that the mass of the hiker
appears in every term. So we can cancel it out. And since we want to solve for the
height of the cliff, ℎ, we rearrange this equation to do so. And we find that ℎ is equal to 𝑣
squared minus 𝑣𝑥 squared all over two 𝑔, where we treat the acceleration due to
gravity as exactly 9.8 meters per second squared.
Since we know 𝑣 and 𝑣 sub 𝑥,
we’re ready to plug in and solve for ℎ. When we do and enter these terms on
our calculator, we find that, to two significant figures, ℎ is 12 meters. That’s the height of the cliff from
which the hiker jumps.
Next, we picture a slightly
different scenario where, in this one, instead of the hiker running off the cliff’s
edge at speed, the hiker jumps off from rest. We want to solve for the speed of
the hiker when he hits the ground, 𝑣 sub 𝑔. And once again, our initial point
will be when the hiker jumps. And the final point will be when
the hiker lands.
Just like before, in this system,
energy is conserved. But unlike before, the initial
kinetic energy of our hiker is zero. In this case, our final potential
energy will again be zero since we let our height zero reference point be at the
ground level. Our equation, therefore, reduces to
PE sub 𝑖 equals KE sub 𝑓, where the potential energy is entirely due to
gravitation.
Referring once again to our
gravitational potential energy and kinetic energy expressions, we write that 𝑚𝑔ℎ
is equal to one-half 𝑚𝑣 sub 𝑔 squared. Once again, the mass of the hiker
cancels out from both sides. And rearranging to solve for 𝑣 sub
𝑔, we find it’s equal to the square root of two times 𝑔 times ℎ, the height of the
cliff. We know 𝑔 and we’ve solved for ℎ
in the earlier part of the problem. So we’re ready to plug in and solve
for 𝑣 sub 𝑔. When we enter these terms on our
calculator, to two significant figures, it’s 15 meters per second. That’s the hiker’s speed when he
reaches the ground after jumping from rest.
Let’s summarize what we’ve learned
about the conservation of energy. We’ve seen that energy conservation
means that, within an isolated system, energy is neither created nor destroyed. Written as an equation, we’ve said
that the initial energy in a closed system is equal to the final energy in that
system. We’ve also seen that system energy
can consist of both kinetic or moving energy as well as potential energy. This led us to expand our
expression for energy conservation to say that initial kinetic plus potential energy
is equal to final kinetic plus potential energy. And lastly, we’ve seen that an
isolated system is one that neither gains nor loses energy. It’s within this sort of system
that energy is conserved.
