Question

A bungee jumper of mass 75kg is attached to a bungee cord of length L=35m. She walks off a platform (with no initial speed), reaches a height of 72m below the platform, and continues to oscillate. While air resistance eventually slows her to a stop, assume there is no air resistance for these calculations.
1. What is the spring constant of her bungee cord?
2. What is her speed when she is 35m below the platform (i.e., just before the cord starts to stretch)?
3. If she had instead jumped vertically off the platform, would the maximum displacement of the bungee cord increase, decrease, or stay the same?

Answer 1

given
m = 75 kg
Lo = 35 m
h = 72 m
extension of the spring, x = 72 - 35

= 37 m
1) Let k is the spring cosnatnt.

Apply conservation of energy

(1/2)*k*x^2 = m*g*h

k = 2*m*g*h/x^2

= 2*75*9.8*72/37^2

= 77.3 N/m <<<<<<<<--------------------Answer

2) again apply conservation of energy

(1/2)*m*v^2 = m*g*Lo

v^2 = 2*g*Lo

v = sqrt(2*g*Lo)

= sqrt(2*9.8*35)

= 26.2 m/s <<<<<<<<--------------------Answer

3) stay the same (if initial vertical speed is zero)

increase (if she had some initial vertical speed)