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Let's apply the work
energy theorem principle

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to the motion of a block
sliding down an inclined plane.

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And here is an inclined
plane at an angle, theta.

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And lets choose a
coordinate system.

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We'll choose x equals 0
up here, or i-hat here.

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And here is our
coordinate function.

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And suppose that the
object starts at xi,

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that it ends at x final.

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If we want to calculate
the work energy then

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what we're going to learn
is that this theorem

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is two different sides.

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Calculating the change
in kinetic energy

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is simply a description.

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This is 1/2 M v final squared
minus 1/2 v initial squared.

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That's a property,
parameters of the system,

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at the initial state, the
speed, or the velocity's speed,

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which is speed squared.

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And the same property v
final in the final state.

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Now over here we have
two types of forces

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acting on this object.

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So here we need a free-body
force diagram first.

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And this side is where the
physics [INAUDIBLE] lie.

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And so we want to draw
our force diagram.

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So if we have our block,
we have the friction force,

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we have the normal
force, we have mg.

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Recall, if the plane
is inclined theta

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that's also the angle theta.

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If we chose i-hat,
j-hat unit vectors,

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I just want to repeat that
on my free-body diagram.

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Now we can think of
this integral as just

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one-dimensional motion
in the x direction.

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And so we have two
different forces

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that we have to calculate.

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The friction force is
in the minus x direction

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so we're integrating minus the
friction force with respect

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to displacement from
x initial to x final.

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And what about the
gravitational force?

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Well the gravitational force
has a component, mg sine

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theta in the x direction.

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So when we integrate
that x component

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we have now plus because
it's in the same direction.

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Remember, we're displacing a
little bit dx down the inclined

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plane so we're going from x
initial x final of mg sine

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theta, which is a constant, dx.

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And so when you're applying
the work energy theorem

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you need to
integrate your forces

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and actually calculate the work.

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Now again, if you looked
in the j-hat direction,

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and we applied Newton's second
law, n minus mg cosine theta 0,

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and our rule for friction
is its mu k times

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n or mu k mg cosine
theta, then what

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we have is in both instances
we have a constant force.

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So it's just force
times displacement.

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So we have minus mu k mg cosine
theta times the displacement,

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which is x initial x final.

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Over here we have mg sine theta
times x final minus x initial.

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And now we've calculated
separately both sides

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of our work kinetic
energy principle.

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As in all our physical
laws the equal sign

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means the work is equal to
the change in kinetic energy.

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I'll emphasize that by now
placing the equal sign because

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of our physical law.

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And so equaling 1/2 M v final
squared minus v initial mean

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squared.

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And now I have a relationship
between the parameters

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of the initial state, which
I'm calling x initial and v

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initial, and the parameters
that describe the final state, x

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final, v final.

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And depending on which set
of these parameters are given

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I can conceivably solve
for the other ones.

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One thing I do want
to point out when

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we do this example is
we've described work

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as a dot product from A to
B. Take the friction force.

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Well in this instance, if
we wrote this out explicitly

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it would be minus fk i-hat
dot dx i-hat from the initial

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to the final.

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i-had dot i-hat is 1.

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And so you see, we recover from
x initial to x final of fk dx.

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And that was the first piece.

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The second piece, the
gravitational force,

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dotted into ds from
x initial to x final.

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Well, if you wrote down the
gravitational force, mg,

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as a i-hat component
and a negative mg cosine

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theta j-hat component, then when
we take the dot product again

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where our ds-- here, we'll
write ds as dx i-hat--

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then when you dot product
mg dot ds, mg dot ds,

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we have i-hat dot
i-hat, which is one.

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But j-hat dot i-hat, they're
perpendicular so that's 0,

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so the only piece that survives
in the scalar product mg

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dot ds is these two pieces.

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And so we get the integral
from x initial to x

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final of mg sine theta times dx.

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And that's precisely
our second piece here.

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So here's the simple application
of the work energy theorem.