Question

we Tal trie max at

Two men are standing on a bridge, getting ready to bungee jump off using the same bungee at the same time. The masses of the men are 100 kg and 70kg. The k of the cord is 3 kN/m.

What is the maximum L can be so that at full length, it will just barely reach the water 100 meters below?

After the men reach the water, the heavier one legs go and stays in the water, launching the lighter man up from the water with the cord.

How high will the lighter student be launched up from the water?

What is the max acceleration reached?

It's a work-energy problem

Answer 1

The spring constant of the cord is

$k=3\ kN/m=3000\ N/m$

The length of the stretched cord is

$l=100\ m$

The length of the unstretched cord is taken as L.

So, the length of the stretched part is

$\Delta L=100-L$

The total weight of the two men while jumping is

$W=\left (100+70 \right )9.8=1666\ N$

At the bottom, this force balances the spring force. i.e

$F=W=k\Delta L=1666\ N$

$\Rightarrow 3000\times\left ( 100-L \right )=1666\ N$

$\Rightarrow 300000-3000L =1666$

$\Rightarrow L =\frac{300000-1666}{3000}=99.44\ m$

So, the maximum length is 99.44 m.

Now, the spring potential energy at the bottom is

$E_s=\frac{1}{2}k\Delta L^2$

$E_s=\frac{1}{2}3000 \left ( 100 - 99.44 \right )^2=462.59\ J$

Now, only the 70 kg man is attached to the cord.

Now let us assume the maximum height attain by the 70 kg man is h. At this height the potential energy of the man is

$E_p=mgh=70\times 9.8h=686h$

Conservation of energy gives us

$E_p=E_s=686h=462.59$

$\Rightarrow h=\frac{462.59}{686}=0.674\ m$

So, the lighter student will attain a height of 0.674 m from the water.

The maximum acceleration is given by

$a=\frac{k\Delta L}{m}-g=\frac{3000\times (100-L)}{70}-9.8=14\ m/s^2$

So, the maximum acceleration is 14 m/s².