(a) 35.0 N/m
The period of a simple harmonic motion is given by:
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
where
m is the mass attached to the spring
k is the spring constant
In this problem, we have:
T = 9.5 s is the period of the motion
m = 80.0 kg is the mass of the bungee jumper attached to the spring
Solving the equation for k, we find the effective spring constant of the cord:
[tex]k=m(\frac{2\pi}{T})^2=(80.0 kg)(\frac{2\pi}{9.5 s})^2=35.0 N/m[/tex]
(b) 17.6 m
When the jumper comes to rest, the cord is stretched with respect to its equilibrium position by a certain amount x. In this situation, the force responsible for the stretching of the cord is the weight of the bungee jumper:
[tex]F=mg=(80.0 kg)(9.8 m/s^2)=784 N[/tex]
Using Hooke's law:
[tex]F=kx[/tex]
and re-arranging for x, we find the stretching of the cord:
[tex]x=\frac{F}{k}=\frac{784 N}{35.0 N/m}=22.4 m[/tex]
And since the jumper comes to rest at a distance of 40.0 m below the jump point, this means that the unstretched length of the cord is
[tex]d=40.0 m-22.4 m=17.6 m[/tex]
