Question

A 90 kg student jumps off a bridge with a 8-m-long bungee cord tied to his feet. The massless bungee cord has a spring constant of 390 N/m.

A)How far below the bridge is the student's lowest point?

B)How long does it take the student to reach his lowest point? You can assume that the bungee cord exerts no force until it begins to stretch

Answer 1

A)

x = stretch in the cord

L = length of cord = 8 m

h = height above the lowest point = L + x

m = mass = 90 kg

k = spring constant = 390 N/m

using conservation of energy

spring potential energy = gravitational potential energy

(0.5) k x² = mg h

(0.5) k x² = mg (L + x)

(0.5) (390) x² = (90) (9.8) (8 + x)

x = 8.7 m

depth below bridge = L + x = 8 + 8.7 = 16.7 m

B)

consider the free fall motion :

t = time for free fall

d = distance travelled = 8 m

using the formula

d = (0.5) g t²

8 = (0.5) (9.8) t²

t = 1.3 sec

considor the motion during stretch of the cord :

t' = time taken = Time period /4 = (0.25) (2) sqrt(m/k) = (0.25) (2 x 3.14) sqrt(90/390) = 0.75 sec

total time = 1.3 + 0.75 = 2.05 sec