Question

A bungee jumper who has a mass of 80.0 kg leaps off a very high platform. A crowd excitedly watches as the jumper free-falls, reaches the end of the bungee cord, then gets “yanked” up by the elastic cord, again and again. One observer measures the time between the low points for the jumper to be 9.5 s. Another observer realizes that simple harmonic motion can be used to describe the process because several of the subsequent bounces for the jumper require 9.5 s also. Finally, the jumper comes to rest a distance of 40.0 m below the jump point. Calculate (a) the effective spring constant for the elastic bungee cord and (b) its unstretched length.

Answer 1

(a) 35.0 N/m

Step-by-step explanation:

Answer 2

(a) 35.0 N/m

The period of a simple harmonic motion is given by:

`T=2\pi \sqrt{(m)/(k)}`

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:

`k=m((2\pi)/(T))^2=(80.0 kg)((2\pi)/(9.5 s))^2=35.0 N/m`

(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:

`F=mg=(80.0 kg)(9.8 m/s^2)=784 N`

Using Hooke's law:

`F=kx`

and re-arranging for x, we find the stretching of the cord:

`x=(F)/(k)=(784 N)/(35.0 N/m)=22.4 m`

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

`d=40.0 m-22.4 m=17.6 m`