Conservation of Elastic and Gravitation Energy - 2

[PeachBanana]

Homework Statement

A 70 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 37 m .

Calculate the spring stiffness constant of the bungee cord, assuming Hooke's law applies.

88 N/M

Calculate the maximum acceleration she experiences.

Homework Equations

Spring Force = -k * x
F = m * a

The Attempt at a Solution

Spring Force = (88 N/M)(24 m)

I used 24 m because that was the maximum stretch of the cord.
Spring Force = 2,112 N

a = F / m
a = (2112 N) / (70 kg)
a = 88 m/s^2
I added 9.8 m/s^2 because she was in free fall.
a = 97.8 m/s^2

Could someone explain why this is incorrect?

 

[Delphi51]

a = (2112 N) / (70 kg)
a = 88 m/s^2

I get 30, not 88.
Why add 9.8? My thinking is that
F = ma
kx - mg = ma
a = kx/m - g
Maximum (upward) acceleration is when x is at its maximum but
9.8 would be subtracted rather than added.

 

[PeachBanana]

I made a typo on the acceleration, oops. I added 9.8 m/s^2 because I thought the jumper was moving in the same direction as gravity as she jumped. Could you explain why you subtracted?

 

[Delphi51]

kx - mg = ma
The spring force kx is upward. Gravity mg is downward.