### Video Transcript

A body, starting at rest, slides
down a rough plane inclined at an angle of 45 degrees to the horizontal. The coefficient of friction between
the body and the plane is three-quarters. Let π‘ one be the time required to
traverse a certain distance down the slope and π‘ two be the time required for the
same body to travel the same distance down a smooth slope at the same angle of
inclination. Express π‘ one in terms of π‘
two.

Okay, in this situation, we have a
plane thatβs inclined at 45 degrees to the horizontal. We then imagine a body on the plane
that starts at rest. Maybe we set it on the plane with
our hand, then starting to slide downward. This motion is resisted by the
roughness of the plane, creating an opposing frictional force. Nonetheless, the body does slide
downhill, and after some amount of time, π‘ one, itβs moved a distance weβll call
π. Weβre then told to imagine an
identical scenario with the same kind of body and the same kind of slope except that
as the body slides down this time, itβs not resisted by friction. That is, the slope is smooth, so
the coefficient of friction between the body and the plane is zero.

Just like before, this body too
moves through a distance π, but now it takes a time π‘ two. This time will be different from π‘
one because slope one is rough, while slope two, we can call it, is smooth. Our goal is to develop a
mathematical equation relating π‘ one to π‘ two. As we get started, letβs record the
fact that the coefficient of friction, weβll call it π, for our first plane is
three-quarters. And then clearing some space on
screen, weβll begin by solving for this time π‘ one. To figure this out, weβll need to
know how our body moves down this rough slope. We know that, initially, itβs at
rest and that over time it picks up speed, so this body must be accelerating as it
slides downhill.

To solve for that acceleration,
letβs consider the net force that acts on this body sliding down a rough slope. Newtonβs second law tells us that
this sum of forces is equal to the bodyβs mass times its acceleration. Say that this is an up-close view
of our body as it slides down the rough slope. In terms of the forces acting on
it, we know thereβs a weight force, π times π, a reaction or normal force, π
. And then also because the slope is
rough, thereβs a frictional force. Since our body is moving down the
incline and friction always opposes the direction of motion, that force, which weβll
label πΉ, points up the incline.

In this sketch, letβs establish the
downward direction along the slope as the positive π₯-direction. Likewise, the positive π¦-direction
will be perpendicularly away from the incline. If we consider the forces acting in
the π₯-direction then, we see that there is a component of the weight force acting
in the positive π₯-direction and then a frictional force which acts purely in the
negative. Considering the right triangle made
by the components of our weight force, we know that this angle here in that triangle
is equal to 45 degrees. It always matches the angle of
inclination of our plane. Therefore, the π₯-direction
component of the weight force is π times π times the sin of 45 degrees. And knowing that the sin of 45
degrees equals the square root of two over two, we can write this overall force as
the square root of two over two times π times π.

From this, we subtract the
frictional force acting on our body. The reason the frictional force is
negative is because it acts up the incline, whereas positive π₯ is down the
incline. For a moving object experiencing
friction, that force is equal to whatβs called the coefficient of friction times the
reaction force experienced by the object. In the case of our body sliding
down this incline, the magnitude of the reaction force is equal to the magnitude of
this component of the weight force. That component equals π times π
times the cos of 45 degrees. So, overall, the frictional force
is equal to the coefficient of friction π times π times π times the cos of 45
degrees. But then cos of 45 degrees, like
sin of 45 degrees, also equals the square root of two over two. And we now have, in simplified
form, all of the forces acting on our body on the rough plane in the
π₯-direction. By Newtonβs second law, this sum is
equal to the bodyβs mass times its acceleration in this dimension.

Notice, on the left-hand side, that
we can factor out square root of two over two times π times π and that, on both
sides of the equation, we have the mass π of our body. Therefore, we can divide both sides
by this mass and cancel this term out completely. And lastly, we can substitute in
the value we know for π. Itβs given as three-quarters. So the acceleration of our body
sliding down the rough incline equals the square root of two over two times π times
one-quarter, or simply the square root of two over eight π. In order to distinguish this
acceleration from the acceleration of our body as it slides down the smooth incline,
letβs call it π sub one and record it at the top of our screen.

Knowing this, we move on to our
smooth incline, and now we want to calculate the acceleration of this body. We can start doing this by
revisiting our free-body diagram. All of these forces still apply
except one, the frictional force πΉ. This force is now gone because our
block is smooth. And so if we go back to our
original Newtonβs second law force balance equation, we can say that now this whole
term here is zero because thereβs no coefficient of friction between our smooth
incline and the body. We simply have then that the square
root of two over two times π times π is equal to π times π sub π₯, the
acceleration of the body on the smooth incline in what weβve called the
π₯-direction.

Note that once more the mass of our
body cancels from this equation. π sub π₯, in this case, equals the
square root of two over two times π. Weβll record this value as π sub
two, and note that itβs four times greater than π sub one. However, itβs a relationship
between π‘ one and π‘ two that we really want to solve for, so letβs keep going. Itβs possible to relate the
accelerations weβve solved for with these times. We do it using whatβs called an
equation of motion. These are equations that describe
object motion whenever they experience uniform or constant acceleration. Thatβs true for both of these
bodies, and so weβre free to use any equation of motion that helps us.

A useful one for our purposes tells
us that the displacement of a body is equal to the original velocity of that body
times the time elapsed plus one-half the bodyβs acceleration times that time elapsed
squared. If we apply this relationship to
the situation with the rough plane, we can say that π, the distance that our body
moves through, is equal to its initial velocity times π‘ one plus one-half π one
times π‘ one squared. Since our body started from rest,
π£ zero is zero. And so if we rearrange this
equation to solve for π‘ one, we find itβs equal to the square root of two times π
over π one. If we then substitute in the value
for π one, square root of two over eight times π, we find that π‘ one equals the
square root of 16π over root two π or, recognizing that 16 is four squared, four
times the square root of π over the square root of two times π. This then is the value weβll use
for π‘ one.

Now letβs apply the same equation
of motion, but this time to solve for π‘ two. This is equal to the square root of
two times π over π two. And if we then substitute in the
square root of two over two times π for π two, π‘ two we find is equal to the
square root of four times π over root two times π or two times the square root of
π over the square root of two times π. This is the value weβll use for the
time π‘ two. And notice how this time value
compares with the time π‘ one. Both have this factor of the square
root of π over the square root of two times π. But in π‘ one, that factor is
multiplied by four, while in π‘ two, itβs multiplied only by two.

We can say then that π‘ one is
twice as big as π‘ two or π‘ one equals two times π‘ two. This is our answer. And now letβs check if it makes
physical sense that π‘ one is greater than π‘ two. π‘ one is the time it takes our
body to slide across a distance π on a rough surface. π‘ two is the time required to
slide across that same distance on a smooth surface. We would expect then that π‘ one is
greater than π‘ two because the body in that case is opposed by friction. So our finding that π‘ two is less
than π‘ one does make physical sense and gives us greater confidence in our
result.