WEBVTT
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This is College Physics
Answers with Shaun Dychko.
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So the box begins in
this corner of the rug
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and the options are to move it along the
smooth concrete this distance *d one*
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and then again along distance *d
two* to get to this position,
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or we can move it with --
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by pushing it with a greater
force across the rough rug
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but go the shorter distance
which is the hypotenuse
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directly from corner to corner.
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So let's consider the case across
the concrete first of all.
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So the total distance traveled
would be *d one* plus *d two*.
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So that's three meters plus four
meters which is seven meters
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and the work done to
push along the concrete
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would be the force applied on the
concrete which is 14 Newtons,
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times the total distance on the concrete,
seven meters, for a total of 280 joules.
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Then for the case where
we go across the rug,
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we're going to push with this force,
the rug force which is 55 Newtons
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multiplied by this
distance across the rug
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which is the square root of *d
one* squared plus *d two* squared.
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So that's 55 Newtons times square root
of three squared plus four squared
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which gives 275 joules.
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So it turns out that there is
less work to go across the rug,
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or I guess to put it in terms that
work with these options here,
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we should say that across the floor is
5 joules extra compared to the rug.
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So option B is the answer.