WEBVTT
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Let's now explore an example of
a force in which the work done
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it is not path independent.
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And the classic example
is the friction force.
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So let's consider
the following setup.
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Suppose we have a horizontal
surface with friction
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and we have an object.
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And we're moving this object.
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So let's choose an origin.
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We'll call this plus
x, our i hat direction,
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it's all going to be
one-dimensional motion.
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And we're going to move
this object from an initial
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to a final state.
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And were going to
move it directly
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in a straight line from the
initial to the final state.
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And this will be our path 1.
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And in our second
case, what we'd
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like to contrast with
that, is that we'd
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like to move the object
out to a point xa
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and then back to
the final point.
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So this is our path 2.
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Has two legs.
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And we'd like to compare the
work done on these two paths.
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So for path 1 we'll begin by
calculating-- our force here
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is the kinetic friction force.
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And the kinetic friction force
remember is, in this case,
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it's going to oppose the motion.
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So we have force
kinetic for path 1,
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and that is minus mu k mg
in the i-hat direction.
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And so when we do the
integral for the work
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from x initial to x
final, this is path 1,
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then we have minus mu k
mg i-hat dotted into-- Now
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what is the ds for this path?
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It's simply dx
i-hat, so dx i-hat.
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Notice we're not putting
any sign into dx.
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The sign will show up
in terms of our end
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points of our integral.
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So we do the dot
product here, we
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have i-hat dot i-hat, that's 1.
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And so this interval, we can
pull out all the constants,
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mu k mg.
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We're just integrating dx
from x initial to x final.
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And so we get mu k mg times
x final minus x initial.
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Now for path 2 we have
two separate integrals.
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So for path 2 we'll
just show the first part
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where we're going
from x initial to xa.
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Then the friction force
is opposing the motion.
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And we always just write dx in
terms of the coordinate system,
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dx i-hat because you'll see that
the signs show up in the end
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points of the integral.
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And then when
we're coming back--
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I'll put that in a different
color and I'll put it below it.
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So when we come back,
notice the friction force
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is going to change direction.
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ds will still be
written that way
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but pay close attention to the
end points of the integral.
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So now what we have
is two integrals.
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So W is the integral
from x initial to xa.
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And now I'm going to take the
dot products here directly.
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It's the same friction
force, we still
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have this same integral,
which is minus mu k mg dx.
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Now here's where it's
a little bit tricky.
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Notice on this path fk
is plus mu k mg i-hat.
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And so when we dotted into
dx we have a plus sign,
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we'll just continue
that integration here,
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of mu k mg dx from
xa to x final.
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Both of these integrals are
straightforward integrals
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to do.
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This is minus mu k mg
xa minus x initial.
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And over here, we have a plus
mu k mg x final minus x initial.
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Notice x final minus
xa, rather, is negative.
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And so both of these integrals
are negative, as we expect.
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And so what we see here is
that we have two pieces,
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so minus 2 mu k mg xa.
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And then we have that
other piece, mu k mg
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x final minus x initial.
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Hang on, this is
actually a plus sign.
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So our answer is very
different because the
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displaced the amount that
we've traveled is different.
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So what we see here is an
example of a force which
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is the work done is
not path independent
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but depends on the path
taken from the initial
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to the final states.