Question

A 65 kg bungee jumper leaps from a bridge. She is tied to a bungee cord that is 12 m long when unstretched, and falls a total of 33 m.
Requried:
a. Calculate the spring constant of the bungee cord assuming Hooke's law applies.
b. Calculate the maximum acceleration she experiences.

Answer 1

Answer:

(a) k = 30.33 N/m

(b) a = 9.8 m/s²

Explanation:

First, we need to find the force acting on the bungee jumper. Since, this is a free fall motion. Therefore, the force must be equal to the weight of jumper:

F = W = mg

F = (65 kg)(9.8 m/s²)

F = 637 N

(a)

Now applying Hooke's Law:

F = k Δx

where,

k = spring constant = ?

Δx = change in length of bungee cord = 33 m - 12 m = 21 m

Therefore,

637 N = k(21 m)

k = 637 N/21 m

k = 30.33 N/m

(b)

Since, this is free fall motion. Thus, the maximum acceleration will be the acceleration due to gravity.

a = g

a = 9.8 m/s²