Question

A bungee-jumping company operates on a bridge 200 m above the ground. They use bungee cords that are 100 m when they are unstretched; these bungee cords have a spring constant of 25Nm. What could the mass of their largest customer be before hitting the ground becomes an issue?

Answer 1

Answer:

m = 63.7 kg

Explanation:

As we know that when mass connected to the bungee cord stretch the string then the gravitational potential energy of the person will convert into potential energy of the string at the end

now here we know that when person jump from the top and reach at the end then loss in gravitational potential energy is given as

$U = mgH$

$U = m(9.81)(200)$

$U = 1962 m$

now when it is at the end of the motion stretch in the string will be

$x = 200 - 100 = 100 m$

now potential energy of string is given as

$U_{spring} = \frac{1}{2}kx^2$

$U_{spring} = \frac{1}{2}(25)(100^2)$

now by energy conservation we have

$1962 m = \frac{1}{2}(25)(100^2)$

$m = 63.7 kg$