Answer:
Explanation:
Given that
g=9.8m/s²
The spring constant is
k=50N/m
The length of the bungee cord is
Lo=32m
Height of bridge which one end of the bungee is tied is 91m
A steel ball of mass 92kg is attached to the other end of the bungee.
The potential energy(Us) of the steel ball before dropped from the bridge is given as
P.E= mgh
P.E= 92×9.8×91
P.E= 82045.6 J
Us= 82045.6 J
Potential energy)(Uc) of the cord is given as
Uc= ½ke²
Where 'e' is the extension
Then the extension is final height extended by cord minus height of cord
e=hf - hi
e=hf - 32
Uc= ½×50×(hf-32)²
Uc=25(hf-32)²
Using conservation of energy,
Then,
The potential energy of free fall equals the potential energy in string
Uc=Us
25(hf-32)²=82045.6
(hf-32)² = 82045.6/25
(hf-32)²=3281.825
Take square root of both sides
√(hf-32)²=√(3281.825)
hf-32=57.29
hf=57.29+32
hf=89.29m
We neglect the negative sign of the root because the string cannot compressed
