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
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
To solve this problem, we can start by analyzing the forces acting on the student during the bungee jump.
1. First, let's calculate the total energy (TE) of the system, which is equal to the potential energy (PE) at the highest point. We can use the formula PE = mgh, where m is the mass of the student, g is the acceleration due to gravity, and h is the height of the bridge above the river.
PE = (91.0 kg)(9.8 m/s^2)(36.7 m + 1.05 m) = 33665.45 J
2. Next, let's calculate the work done (W) by the spring to bring the student to the lowest point of the jump. We can use the formula W = TE, where TE is the total energy.
W = 33665.45 J
3. The work done by the spring is equal to the force exerted by the spring (F) multiplied by the displacement (d). In this case, the displacement will be twice the length of the bungee cord (2 L), as the student will fall to the lowest point and then rebound back up.
W = Fd
F(2 L) = 33665.45 J
4. The force exerted by the spring is given by Hooke's Law: F = kx, where k is the spring constant and x is the displacement from the equilibrium position.
F = k(2 L) = 33665.45 J
5. Finally, we can rearrange the equation to solve for the spring constant (k).
k = F / (2 L)
k = 33665.45 J / (2 * 25.3 m)
k ≈ 667.2 N/m
Therefore, the minimum spring constant required for the bungee cord is approximately 667.2 N/m to allow the student to get as close as possible to the water without getting wet.