Question

A 62 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 12 m. She falls a total of 31 m. (A) Calculate the spring stiffness constant, k, of the bungee cord. Assume Hooke's law applies. (Hint: consider the jumper at three different points – at the beginning of the fall, at a fall distance equivalent to the length of the unstretched bungee cord and at the bottom of the fall when the cord is at maximum stretch. What are the different potential and kinetic energies at each point?)

Answer 1

Answer:

k = 104.46 N/m

Explanation:

Here we can use energy conservation

so we will have

initial gravitational potential energy = final total spring potential energy

as we know that she falls a total distance of 31 m

while the unstretched length of the string is 12 m

so the extension in the string is given as

$x = L - L_o$

$x = 31 - 12 = 19 m$

so we have

$mgH = \frac{1}{2}kx^2$

$62(9.81)(31) = \frac{1}{2}k (19^2)$

$k = 104.46 N/m$