Question

A 90 kg student jumps off a bridge with a 10-m-long bungee cord tied to his feet. The massless bungee cord has a spring constant of 390 N/m. You can assume that the bungee cord exerts no force until it begins to stretch.

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

How far below the bridge is the student's resting position after the oscillations have been fully damped?

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the student = M = 90 kg

Length of the bungee-cord = L = 10 m

Spring constant of the bungee-cord = k = 390 N/m

Extension of the bungee cord when the student is at the lowest point = X₁

The lowest point of the student is when the student jumps off the bridge and comes to rest the first time that is the potential energy of the student's drop is converted into potential energy of the bungee-cord.

Mg(L + X₁) = kX₁²/2

(90)(9.81)(10 + X₁) = (390)X₁²/2

195X₁² - 882.9X₁ - 8829 = 0

X₁ = 9.363 m or -4.835 m

Distance cannot be negative.

X₁ = 9.363 m

Distance of the lowest point of the student below the bridge = D₁

D₁ = L + X₁

D₁ = 10 + 9.363

D₂ = 19.363 m

Extension of the bungee cord when the student is at rest = X₂

When the student is at rest the weight of the student is balanced by the force exerted by the bungee-cord.

Mg = kX₂

(90)(9.81) = (390)X₂

X₂ = 2.264 m

Distance of the resting point of the student from the bridge = D₂

D₂ = L + X₂

D₂ = 10 + 2.264

D₂ = 12.264

A) Distance of the student's lowest point from the bridge = 19.363 m

B) Distance of the student's resting point from the bridge = 12.264 m