Bungie jumper jumps of cliff. (Energy)

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SUMMARY

The discussion focuses on calculating the spring constant and maximum acceleration experienced by a bungee jumper using Hooke's Law. The bungee jumper, weighing 55 kg, leaps from a height of 31 meters with a bungee cord length of 11 meters. The calculated spring constant of the bungee cord is 84 N/m, derived from energy conservation principles. Additionally, the maximum acceleration experienced by the jumper at the bottom of the jump exceeds 9.8 m/s² due to deceleration forces acting on her.

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Homework Statement


A 55 bungee jumper leaps from a bridge. She is tied to a bungee cord that is 11 long when unstretched, and falls a total of 31 .
a)Calculate the spring constant of the bungee cord assuming Hooke's law applies.
b)Calculate the maximum acceleration she experiences.

The Attempt at a Solution


first i drew out the problem in 3 sections. the first (A) the jumper on a cliff 31 m above h=0, then i drew (B) the jumper 11 m below the cliff because that is the length of the rope, and the third (C) the jumper at h=0, all the way at the bottom.

a)= 84 N/m
-to get this i solved for velocity at B by setting the energy at A = energy at B. mgh(at A)=1/2mv^2+mgh(at B) and got v=14.7 m/s. then i used Energy at B=Energy at C, mgh(at B)+1/2mv^2=1/2kx^2 and got k=83.5 N/m.

b)i know the point of greatest acceleration can't be at A, at B its simply acceleration due to gravity, and at C the jumper is at h=0 and has a moment of no movement, so no acceleration. however the answer isn't 9.8.
 
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For part B), don't forget that DECELERATION is also acceleration, simply in the opposite direction of velocity. The acceleration required when she is stopped at the bottom of the cliff is much greater than 9.8.
 
tal444 said:
For part B), don't forget that DECELERATION is also acceleration, simply in the opposite direction of velocity. The acceleration required when she is stopped at the bottom of the cliff is much greater than 9.8.

true. ima give that idea a try
 

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