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 400 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

(a) Write the equation oflaw of conservation of momentum. 1 kx=mgx 2 2 (1) Here, k is spring constant, x is the lowest point, m is themass of the student and g is the acceleration due to gravity Substitute 400 Nm fork, 90Kg form ,(8-x)m for x and 9.8m/s2for g in (1) x(400N/m) (8-x)m=(90Kg)(9.8m/s2 )(x) 100x2 2041x+6400 = 0 Solving for x x-3.87m Hence, the student's lowest point far below the bridge is 3.87m (b) After the oscillations have been fully damped, the system is in equilibrium Write the equation of forces in equilibrium system kx2 mg (2) Here, xis the distance ofresting position Substitute 400 N/m for k. (x-8)m for x.90Kg form and 9.8m/s for g in (2) (400 N/m(x-8)m)=(90Kg)(9.8m/s2) (x-8)m 2.205 m x-10.205m Hence, the distance ofresting position is 10.205m