Question

numerade 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 = 104.5 N/m

Explanation:

When she reaches the lowest point the velocity of the jumper is 0. This means all her kinetic energy has been absorbed by the cord.

The energy absorbed by a spring (a cord can be treated as a spring when it's being stretched) is:

$Ee = \frac{1}{2}*k*\Delta x^{2}$

Where:

Ee = elastic potential energy

k = elastic constant

Δx = lenght the spring was stretched

When a body falls, it's potential gravitational energy is converted to kinetic energy. So the kinetic energy of a body can be calculated as the gravitational potential energy it lost.

The gravitational potential energy is:

$Ep = m * g * \Delta h$

Where:

Ep: potential gravitational energy

m: mass

g: acceleration of gravity

Δh: change of height (final - initial)

When she has fallen 31 m she has lost

62 kg * 9.81 m/s^2 * 31 m = 18854 J

Since her velocity at the bottom is 0, her kinetic energy is 0 J.

All her kinetic energy has been absorbed by the cord and transformed into elastic potential energy. So we can equate these two energies.

Ep = Ee

Then:

$Ep = \frac{1}{2}*k*\Delta x^{2}$

Δx = 31 m - 12 m = 19 m

From here the only unknown is k:

$k = 2 * Ep / (\Delta x ^{2})$

$k = 2 * 18854 J / ((19 m) ^{2}) = 104.5 N/m$