Question

A person of mass m will bungee-jump from a bridge over a river where the height from the bridge to the river is h. The bungee cord has an un-stretched length of 23h and when it stretches beyond its equilibrium length it behaves as a spring with spring constant k. Find an expression for the minimum spring constant k that will stop the person just before hitting the river (ignore the height of the person), in terms of m, g, and h.

Answer 1

Answer:$k_{min}=\frac{18mg}{h}$

Explanation:

Given

mass of person is m

Distance between bridge and river is h

chord has an un-stretched length of $\frac{2h}{3}$

Let spring constant be k

Person will just stop before hitting the river

Conserve energy i.e. Potential Energy of Person is converted in to elastic energy of chord

$mgh=\frac{kx^2}{2}$

$x=h-\frac{2h}{3}=\frac{h}{3}$

$mgh=\frac{kh^2}{18}$

$k=\frac{18mg}{h}$

Thus $k_{min}=\frac{18mg}{h}$