Question

a bungee jumping company wants to set up a bungee jumping location on the top of a bridge that is 300 m above the ground. for safety reasons, the company wants to select a bungee cord spring constant such that a jumper with a mass of 115 kg will reach his lowest point - that is, the point when the change in gravitational potential energy equals the amount of energy stored in the bungee cord at the bottom of the jump - at 50 m above the ground. what is the approximate spring constant the company should choose, assuming that air resistance, friction, and the weight of the cord can be ignored and that the cord immediately begins to stretch as soon as the jumper begins to fall? (recall that ( g = 9.8 , m/s^{2} )) a. 9 n/m b. 23 n/m c. 4 n/m d. 15 n/m

Explanation:

Step1: Calculate the change in gravitational potential energy

The change in height \(h = 300 - 50=250\space m\).
The formula for gravitational potential energy is \(U = mgh\), where \(m = 115\space kg\), \(g = 9.8\space m/s^{2}\), and \(h = 250\space m\).
So, \(U=115\times9.8\times250\).

$$U = 115\times2450=281750\space J$$

Step2: Relate gravitational potential energy to elastic potential energy

The elastic potential energy of a spring is \(U_s=\frac{1}{2}kx^{2}\). Here, the extension of the spring \(x = 250\space m\) (since the cord stretches as soon as the jumper falls).
Since \(U = U_s\), we have \(281750=\frac{1}{2}k\times(250)^{2}\).
First, simplify \(\frac{1}{2}\times(250)^{2}=\frac{1}{2}\times62500 = 31250\).
Then, solve for \(k\): \(k=\frac{2\times281750}{62500}\).

$$k=\frac{563500}{62500}\approx9\space N/m$$

Answer:

A. \(9\space N/m\)