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| Composition Setup |
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Bungee jumps involve elastic and gravitational potential energy. |
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(Photo courtesy Wikimedia Commons, uploaded by user Che010.) |
Bungee cords designed to U.S. Military specifications (DoD standard MIL-C-5651D,
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available
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at
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)
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are
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characterized
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by
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a
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force
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constant
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times
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unstretched
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length
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in
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the
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range
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kL
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~
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800-1500
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N.
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Jumpers
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using
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these
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cords
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intertwine
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three
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to
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five
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cords
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to
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make
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a
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thick
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rope
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that
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is
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strong
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enough
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to
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withstand
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the
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forces
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of
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the
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jump.
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Suppose
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that
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you
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are
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designing
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a
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bungee
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jump
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off
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of
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a
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bridge
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that
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is
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50.0
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m
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above
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the
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surface
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of
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a
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river
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running
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below.
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You
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have
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read
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that
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you
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should
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expect
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the
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cord
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to
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stretch
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(at
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peak
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extension)
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to
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about
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210%
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of
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its
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natural
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length.
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You
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have
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also
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read
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that
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you
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should
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use
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3
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cords
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together
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for
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jumpers
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with
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weights
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in
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the
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range
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100-150
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lbs,
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4
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cords
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for
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150-200
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lbs,
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and
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5
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cords
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for
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200-250
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lbs.
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Suppose
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you
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decide
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to
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use
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cords
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of
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length
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20
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m,
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which
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would
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seem
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to
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offer
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a
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safety
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zone
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of
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about
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8
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m
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(or
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really
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about
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6
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m
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if
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the
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cord
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is
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attached
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at
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the
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ankles).
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h3. Part A
Find the expected maximum length of the cords for a 200 lb person jumping with 4 cords, so that _kL_ =
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Part A
Find the expected maximum length of the cords for a 200 lb person jumping with 4 cords, so that kL = (4)(800
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N)
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=
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3200
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N.
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Since
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you
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are
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evaluating
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the
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safety
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factor,
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ignore
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any
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losses
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due
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to
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air
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resistance
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or
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dissipation
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in
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the
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cord.
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Ignore
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the
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mass
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of
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the
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rope.
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Solution
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| Cloak |
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jumper
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(treated
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as
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a
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)
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plus
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the
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earth
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(treated
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as
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an
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object
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of
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infinite
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mass)
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and
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the
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bungee
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cord
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and
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the
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bridge
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(also
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treated
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as
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an
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object
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of
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infinite
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mass).
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interact
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via
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,
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which
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contributes
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gravitational
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and
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via
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the
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of
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the
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cord
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which
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contributes
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elastic
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.
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External
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influences
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are
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assumed
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negligible.
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We begin with an initial-state
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and
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the
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corresponding
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Initial State | Final State |
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As indicated in the picture, we have chosen the zero point of the height to be the river's surface. With these pictures in mind, we can set up the Law of Change for our model:
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|Diagrams and Mechanical Energy]: |!bungeebadi.png!|!bungeebadf.png! ||Initial State||Final State|| As indicated in the picture, we have chosen the zero point of the height to be the river's surface. With these pictures in mind, we can set up the Law of Change for our model: {latex}\begin{large} \[ E_{\rm i} = mgh_{\rm i} = E_{\rm f} = mgh_{\rm f} + \frac{1}{2}k x_{f}^{2} \] \end{large}{latex}{note}Bungee cords provide an [elastic restoring force|Hooke's Law for elastic interactions] when stretched, but offer no resistance when |
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Bungee cords provide an elastic restoring force when stretched, but offer no resistance when "compressed", since they fold like an ordinary rope. Thus, the initial [ |Hooke's Law for elastic interactions]is zero in this case. {note} |
It
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seems
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that
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we
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have
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a
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problem
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here,
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because
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we
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do
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not
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know hf or xf. The easiest way to deal with this problem is to utilize our freedom to choose the zero point of the height axis. If we restructure our coordinate system to place h = 0 m at the point where the cord is stretched to its natural length L (as shown below).
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Initial State | Final State |
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This redefinition of the origin of the height in our coordinate system greatly simplifies the equation describing the evolution of the jumper's mechanical energy:
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_h{_}{~}f~ or _x{_}{~}f~. The easiest way to deal with this problem is to utilize our freedom to _choose_ the zero point of the height axis. If we restructure our [coordinate system] to place _h_ = 0 m at the point where the cord is stretched to its natural length _L_ (as shown below). |!bungeegoodi.png!|!bungeegoodf.png! ||Initial State||Final State|| This redefinition of the origin of the height in our [coordinate system] greatly simplifies the equation describing the evolution of the jumper's [mechanical energy]: {latex}\begin{large} \[ mgL = - mgx_{f} + \frac{1}{2} k x_{f}^{2} \] \end{large}{latex} {note}You can also solve using the initial coordinate system. You would simply have to substitute _h{_}{~}f~ = 50 m - _L_ \- _x_. After cancelling _mg_(50 m) from each side, you recover the same expression. {note} The discussion around the diagrammatic representation has left us with a quadratic equation in _x_~f~, which is solved to obtain: {latex} |
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You can also solve using the initial coordinate system. You would simply have to substitute hf = 50 m - L - x. After cancelling mg(50 m) from each side, you recover the same expression. |
The discussion around the diagrammatic representation has left us with a quadratic equation in xf, which is solved to obtain:
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\begin{large} \[ x_{f} = \frac{mg \pm \sqrt{(mg)^{2} + 2 k mg L}}{k} \]\end{large}{latex}
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It
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is
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not
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sensible
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that
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we
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should
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find
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a
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negative
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value
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for xf, so we must select the plus sign, giving:
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_x{_}{~}f~, so we must select the plus sign, giving: {latex}\begin{large} \[ x_{f} = \frac{mg}{k}\left(1+\sqrt{1+\frac{2kL}{mg}}\right) = 21.5\:{\rm m} \] \end{large}{latex} {tip}Does this bear out the original estimate that the cord should stretch to 210% of its initial length? {tip} {tip}Can you convince yourself by looking at the form of the symbolic answer that a 150 lb person attached to 3 cords and a 250 lb person attached to 5 cords would have the same _x{_}{~}f~ as the 200 lb person on 4 cords? {tip} {cloak:appA} {card:Part A} {card:label=Part B} h3. Part B Based on your analysis from Part A, what would be the peak force acting on a 200 lb person making the jump attached to 4 cords? {card:Part B} {deck:partdeck} |
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Does this bear out the original estimate that the cord should stretch to 210% of its initial length? |
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Can you convince yourself by looking at the form of the symbolic answer that a 150 lb person attached to 3 cords and a 250 lb person attached to 5 cords would have the same xf as the 200 lb person on 4 cords? |
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Part B
Based on your analysis from Part A, what would be the peak force acting on a 200 lb person making the jump attached to 4 cords?
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