a 62.0kg bungee jumper jumps from a bridge. she is tied to a bungee cord whose unstretched length is 12.0 m and falls a total of 31m a) calculate the spring constant k of the bungee cord and b) calculate the maximum acceleration she experiences.

Answer 1

loss of potential energy = m g (31)
= 62 * 9.81 * 31
= 18,855 Joules

gain in potential energy of spring = (1/2) k x^2
=.5 * k (31-12)^2
= 180.5 k
so
k = 18855/180.5 = 104 N/m

F = m a

force down = m g = 62*9.81 = 608 Newtons
force up = kx
when x = 31-12 = 19
F up maximum = 104*19 = 1976 Newtons
net force up at max of spring = 1976-608
= 1368 Newtons up max
a = F/m = 1368/62 = 22 m/s^2 up