The belly button of the jumper is 2.10/2 = 1.05 meters above bridge level when he jumps.
The belly button of the jumper is at the water when he stops if he falls face down.
Therefore his center of gravity moved down 36.7 + 1.05 meters.
His change in potential energy is m g h
= 91*9.8*37.75=33665 Joules (check with you)
the bungee cord stretched from 25.3 to 36.7 meters or
11.4 meters stretch
The energy absorbed by that stretch =(1/2)k x^2
so
(1/2) k (11.4)^2 = 33665
k = 518 n/m
A tall bald student (height 2.1 meters and mass 91.0 kg) decides to bungee jump off a bridge 36.7 meters above the river. The bungee cord is 25.3 meters long as measured from the attachment at the bridge to the foot of the jumper. Treat the bungee as an ideal spring and the student as a 2.1 meter rod with all the mass at the midpoint. This particular student desires to stay dry. What is the minimum spring constant (N/m) of the bungee that will allow the student to get as close as possible to the water but still stay dry? Assume that he beings at a standing position and "falls" from the bridge.
This is what I did but it turned out to be incorrect. Can anyone correct my errors? Thank you so much
25.3m + 27.4 m = 27.4m
TE=mgh
(91.0kg)(9.8 m/s^2)(36.7 + 1.05)
= 33665.45
work = Fd
33665.45/9.3 = 3619.94 N
F=kx
3619.94/9.3m = 389N/m
2 answers
have you ever heard of or seen a bungee jump? your math is wrong and you have no idea what your talking about as far as the bungee concept.
1 answer