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ActivPhysics%20OnLine

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(c) the woman's kinetic energy is maximized when she is near the middle of an oscillation. ... y2 = 4.0 m at the top ... of 5.0 N/m. Notice on the simulation if ... – PowerPoint PPT presentation

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Title: ActivPhysics%20OnLine


1
  • ActivPhysics OnLine
  • Problem 5.4 Inverse Bungee Jumper
  • Choose xy coordinates in the usual way. y
  • x
  • upwards y
  • downwards - y
  • In this problem we are assuming there is no
    friction.


2
  • Problem 5.4 Question 1
  • Do the simulation.
  • Notice (a) the womans gravitational potential
    energy increases as her height above the floor
    increases (b) the springs potential energy
    increases as the spring is stretched
  • (c) the womans kinetic energy is maximized when
    she is near the middle of an oscillation.

3
  • Problem 5.4 Question 2 Construct an energy
    equation for the womans motion. What is the
    force constant of the spring, k, if her head just
    touches the overhead support without injury?
  • The systems total energy E is the sum of
    kinetic energy K, grav. potential energy Ug, and
    the springs potential energy Us.
  • E K Ug Us


4
  • (This problem is similar to Randall Knights
    Physics for Scientists and Engineers, Example
    10.7 on pages 285-286.)
  • E K Ug Us
  • Let ye the equilibrium position of the spring,
    i.e. the position at which there is no restoring
    force.
  • E m v2/2 mgy
  • k (y - ye)2 /2


5
  • Since total energy is conserved, the energy of
    the system when she is at the top position y2
    energy of the system when she is at the bottom
    position y1.
  • mv22/2 mgy2 k (y2 - ye)2/2
  • mv12/2 mgy1 k (y1 -ye)2/2
  • Her kinetic energy should be 0 at the top and is
    0 also at the bottom where she is at rest.
  • mgy2 k (y2 - ye)2/2
  • mgy1 k (y1 -ye)2/2


6
  • The gravitational potential energy also must be 0
    at the bottom where y 0.
  • mgy2 k (y2 - ye)2/2
  • k (y1 -ye)2/2
  • k (y1 -ye)2/2 - k (y2 - ye)2/2 mgy2
  • k (y1 -ye)2/2 - (y2 - ye)2/2 mgy2
  • Substitute in values.
  • m 50 kg
  • g 10 m/s2


7
  • Since the y values are zero at the floor
  • y2 4.0 m at the top
  • y1 - ye - 3.5 m when the spring is stretched
    and her feet touch the floor.
  • At the top the spring is slightly compressed and
  • y2 - ye 0.5 m.
  • k (-3.5m)2 (0.5m)2
  • (2)(50 kg)(4.0m)(10m/s2)
  • k 4000 N/ 12 m2


8
  • k 333 N/ m
  • The simulation only proceeds in steps of 5.0
    N/m. Notice on the simulation if
  • k 330 N/ m, her head does not quite reach the
    top, but if k 335 N/ m, then she hits her head
    and is injured.

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