91Ó°ÊÓ

The given data can be listed below,

• The length of the bungee cord is,$l=30\text{m}$
• The height of the platform is,$y_1=45\text{m}$
• The maximum height the cord will fall is,$y_2=41\text{m}$
• The mass of the person is,$m=95\text{kg}$
• The force on the bungee cord is$F=380\text{N}$

The energy an item or body possesses as a result of its motion is known as kinetic energy. Every atom that is engaged in some sort of motion possesses it.

The stretched amount of cord is given by,

$\begin{array}{rcl}d& =& 41\text{m}-30\text{m}\\ & =& 11.0\text{m}\\ & & \end{array}$

The cord is initially at equilibrium so the initial gravitational potential energy which can be given by,

$\begin{array}{rcl}{U}_{grav,1}& =& mg{y}_1\\ & =& 0\\ & & \end{array}$

The elastic potential energy is given by,

$U_{el,2}=\frac{1}{2}k{x}^2$

Here, kis the force constant and x is the stretched distance of cord.

Substitute all the values in the above,

$\begin{array}{rcl}{U}_{el,2}& =& \frac{1}{2}k{\left(11\text{m}\right)}^2\\ & =& 60.5k\\ & & \end{array}$

The kinetic energy of the person is zero as it will eventually come at rest.

So, from the law of conservation of energy the force constant is given by,

$\begin{array}{rcl}{K}_1+U_{grav,1}+U_{el}& =& K_2+U_{grav,2}+U_{el}\\ 0+0+0& =& 0-mg{y}_2+60.5k\\ k& =& \frac{mg{y}_2}{60.5}\\ & & \end{array}$

Here, m is the mass of the person, g is the acceleration due to gravity, and $y_2$is the maximum fallen distance.

Substitute all the values in the above,

$\begin{array}{rcl}k& =& \frac{-\left(95\text{kg}\right)\left(9.8{\text{m/s}}^2\right)\left(-41\text{m}\right)}{60.5}\\ & =& 630.9\text{N/m}\\ & & \end{array}$

The restoring force is given by,

$F=kx$

The stretched length of the cord is calculated by substitute all the values in the above,

$\begin{array}{rcl}380.0\text{N}& =& 630.9x\\ x& =& 0.602\text{m}\\ & & \end{array}$

Thus, the stretched length of the bungee cord is $0.602\text{m}$.