Question

A bungee cord can stretch, but it is never compressed. When the distance between the two ends of the cord is less than its unstretched length L0, the cord folds and its tension is zero. A particular bungee cord has an unstretched length of L0 = 32 m; when it is stretched to L > L0, the cords tension obeys Hooke’s law with "spring" constant 50 N/m. To test the cords reliability, one end is tied to a high bridge of height 91 m (above the surface of a river) and the other end is tied to a steel ball of mass 93 kg, which is dropped off the bridge. The ball stops in the air a few meters before it hits the water — then the cord pulls it back up. Find the ball’s height above the water’s surface at this lowest point of its trajectory. The acceleration of gravity is 9.8 m/s 2 . For simplicity, neglect the cord’s own weight and inertia as well as any air drag on the ball and the cord. Answer in units of m.

Answer 1

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