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The Ejection Seat at Lake Biwa Amusement Park in Japan (Figure 1) is an inverse bungee system. A seat with passengers of a total mass of 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.

Respuesta :

hamzaahmeds

This question involves the concept of the law of conservation of energy.

The spring constant of the cable is "915.6 N/m".

LAW OF CONSERVATION OF ENERGY:

According to the law of conservation of energy, energy can neither be created nor be destroyed. It can only change from one form to another form. In this scenario, the elastic potential energy stored by the cable will completely be transformed into the gravitational potential energy at the highest point.

[tex]Loss\ of\ Elastic\ Potential\ Energy=Gain\ in\ Gravitational\ Potential\ Energy[/tex]

[tex]\frac{1}{2}kx^2=mgh\\\\k= \frac{2mgh}{x^2}[/tex]

where,

  • k = spring constant = ?
  • m = mass = 160 kg
  • g = acceleration due to gravity = 9.81 m/s²
  • h = total height covered after launch = 12 m + 30 m = 42 m
  • x = stretched length = 12 m

Therefore,

[tex]k=\frac{2(160\ kg)(9.81\ m/s^2)(42\ m)}{(12\ m)^2}[/tex]

k = 915.6 N/m

Learn more about the law of conservation of energy here:

https://brainly.com/question/2264339

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