WEBVTT
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We will now like to explore a
property of work done by forces
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called path independence.
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So recall that our
work is defined
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to be an integral of
F.ds from some initial
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to some final point.
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Let's look at an example.
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The force that we're
going to look at
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is the gravitational force.
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And let's draw a
coordinate system.
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Let's take an initial
point and a final point,
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and let's choose plus x
and plus y and our unit
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vectors i-hat and j-hat.
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And in this example
we're going to consider
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the gravitational force near the
surface of the earth, which is
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a constant pointing downwards.
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And what I'd like to do
is consider two paths.
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I'd like to consider a path,
that first path goes straight
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up and over.
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So this will be path 1.
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It has two legs.
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And the second path, path 2
will go horizontal and vertical.
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So that's going to be path 2,
and it has two different legs.
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And we'd like to evaluate
this integral on both paths.
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Now the way we do that is, first
off, let's start with path 1.
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We'll call this y initial
and we'll call that y final.
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And we'll draw the gravitational
force mg, mg on both legs.
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And now we see that
when we do the integral
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along this first leg of path
1, the gravitational force
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is opposite the direction
we're moving the displacement.
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So the interval being negative,
the force is constant.
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And so the work done is
simply minus mg times
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delta y, which is y final minus
y initial on that first part.
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Now on the second part,
this is a right angle.
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And because it's a right
angle the dot product is 0.
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So that's the total
work done on path 1.
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Now on path 2, we'll again,
let's just draw our forces.
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We have gravity down and
we have gravity down.
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And we can see that, once again,
on the first horizontal leg
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of path 2 this integral
is 0 because it's
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moving perpendicular to the
force, the direction of just
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along the path.
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And on this leg,
just as on that leg
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there, the gravitational
force is down,
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the integral is negative, we
get exactly the same result.
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Now what we'd like to do is
consider a more general path
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rather than these two
horizontal and vertical legs.
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So let's draw a
coordinate system again.
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Let's introduce i, and let's
introduce f plus y and plus x.
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And now let's consider a path
which is moving like that.
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Now on this path,
what we want to do
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is break it down into
horizontal and vertical pieces,
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horizontal, vertical,
horizontal, vertical,
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horizontal, vertical,
horizontal, vertical,
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horizontal, vertical.
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And let's focus on
one of these pieces,
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which has a displacement,
ds, going from one
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point to another.
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Now if we blow that up, so
we have our displacement,
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ds, then what we have here
is a displacement, dx,
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in the horizontal direction and
dy in the vertical direction.
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And so we see again that
when we compute this work,
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and our gravitational
force is downward, then
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the horizontal display-- part
of this displacement, the dot
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product in the
horizontal direction, 0,
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only the vertical part cancels.
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And so when we add up the
work along these horizontal
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and vertical pieces, only
the vertical piece counts.
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We're just adding
up the dy's until we
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get y final minus y initial.
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Now let's be a little more
analytic here and write out
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our ds as a vector dx
i-hat plus dy j-hat
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and our gravitational
force as minus mg j-hat.
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Now the work is the
integral from the initial
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to the final place of fg.ds.
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And so when we do that
we have minus mg j-hat
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dot dx i-hat plus dy j-hat.
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And from her initial
to the final place.
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And notice that we
have j-hat dot i-hat.
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They are perpendicular
so that dot product is 0.
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And j-hat dot j-hat is 1.
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And so what we're left
with here in the integral
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is simply now we're
integrating from y initial
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to y final because we
only have minus mgdy.
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And that's what we said before,
only the vertical parts are
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adding up.
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And so when you do this
integral, mg is a constant.
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We get mg times y
final minus y initial,
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and that agrees with
what we had before.
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So this type of force
is an illustration
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of a force where
the work done is
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independent of the path
we choose from the initial
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to the final place.