Asked by Anonymous
A 72 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 13 m, and falls a total of 38 m. Calculate the spring stiffness constant "k" of the bungee cord, assuming Hooke's law applies. Calculate the maximum acceleration she experiences.
Answers
Answered by
bobpursley
mgh=1/2 k x^2
72(9.8)*38=1/2 k (38-13)^2
solve for k
Maximum acceleration? as she begins to fall, it is -9.8m/s^2
Now at the bottom, she is accelerating upward.
Forcebottom= -mg+k(38-13)
and Force= ma
solve for a.
72(9.8)*38=1/2 k (38-13)^2
solve for k
Maximum acceleration? as she begins to fall, it is -9.8m/s^2
Now at the bottom, she is accelerating upward.
Forcebottom= -mg+k(38-13)
and Force= ma
solve for a.
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