WEBVTT
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So here we're going to do an application problem thinking
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more specifically about line integral XYZ and the amount of
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work being done. Uh, so where we're going
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to start is saying that, um, we can
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start by remembering that the gravitational force is gonna be
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conservative, So thief force by the gravity on the
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man in the can is equal to the negative force
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by the man against the gravity. So to find
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the work done, we want to look at the
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magnitude of gravitational force acting on the man on the
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can of paint. So we have 160 plus m
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pounds, then the mass of the man is constant
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, and that's the 160. But the mass of
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the can is constantly decreasing because it's letting out all
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that paint. So the mass of the can is
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£9 less by the time that the top is reached
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. So based on that, we can say that
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em is equal to its initial£25. Minus nine
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are because it's losing those£9. Um, over
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S s is the height of the silo, and
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R is the distance covered, um, during some
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time, so based on that. We can look
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at the fact that f dot d r. Is
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now going to equal this portion right here. 1
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85 minus nine are over s. And the 1
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85 comes from the 1 60 plus the 25.
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And then that's gonna be D R. So the
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work done as we see is just gonna be this
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integral right here. So we take the line integral
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. We're gonna take the center, go here and
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it's from zero to s from the beginning of the
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time until we get to the top. So,
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since we know the height to be 90 ft s
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can become 90. So what we do when we
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evaluate this integral is 1 85 R minus nine R
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squared over two s evaluated at zero and s,
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or, in this case, zero and 90.
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So we evaluated at 90 and what we end up
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getting is that this is equal to 1 80 0.5
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s, which is equal toe 1 80.5 times 90
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. And that's going to give us our final answer
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of 16,245 ft pounds. That's gonna be our final
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answer for the work done