Question

The Ejection Seat at Lake Biwa Amusement Park in Japan is an inverse bungee system. A seat with passengers of total mass 160 kg is connected to elastic cables on the sides. The seat is pulled down 12.0 m, stretching the cables. When released, the stretched cables launch the passengers upward above the towers to a height of about 30.0 m above their starting position at the ground. Assume that the cables are vertical
Determine the spring constant of the cable.

Answer 1

Answer:

Approximately $6.5 \times 10^{2}\; {\rm N\cdot m^{-1}}$ (assuming that $g = 9.8\; {\rm m\cdot s^{-2}}$ and that the spring is ideal.)

Explanation:

If the elastic cable is ideal, Hooke's Law would apply. Let $k$ denote the spring constant of the cable. At a displacement of $x$ from the equilibrium position of the spring, the elastic potential energy stored in the cables would be $(1/2)\, k\, x^{2}$.

In this question, the initial displacement from the equilibrium position was $x = 30.0\; {\rm m}$. Assuming that there is no energy loss, all the elastic energy stored in the cables would be turned into gravitational potential energy (${\rm GPE}$)

Let $m$ denote the mass of the passenger and the chair, combined. Let $g$ denote the gravitational field strength. If the altitude gain of the passenger and the chair is $h$, the ${\rm GPE}$ of the passenger and the chair combined would be $m\, g\, h$.

The entirety of the ${\rm GPE}$ of the passenger and the chair was converted from the elastic potential energy initially stored in the cables. There was no energy lost. Therefore, the following would be an equality:

$\displaystyle \frac{1}{2}\, k \, x^{2} = m\, g\, h$.

It is given that $x = 12.0\; {\rm m}$, $m = 1.6 \times 10^{2}\; {\rm kg}$, and $h = 30.0\; {\rm m}$ (relative to the starting position of the passenger and the chair, with the cables already stretched.) Assuming that $g = 9.8\; {\rm m\cdot s^{-2}}$, substitute in the values and solve for the spring constant $k$:

$\begin{aligned}k &= \frac{2\, m\, g\, h}{x^{2}} \\ &= \frac{2 \times 1.6\times 10^{2}\; {\rm kg} \times 9.8\; {\rm m\cdot s^{-2}} \times 30.0\; {\rm m}}{(12.0\; {\rm m})^{2}} \\ &\approx 6.5 \times 10^{2}\; {\rm kg \cdot s^{-2}} \\ &= 6.5 \times 10^{2}\; {\rm (kg \cdot m \cdot s^{-2}) \cdot m^{-1}} \\ &= 6.5\times 10^{2}\; {\rm N \cdot m^{-1}}\end{aligned}$.