Question

A 74.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her and to the bridge. The unstretched length of the cord is 14.0 m. The jumper reaches reaches the bottom of her motion 40.0 m below the bridge before bouncing back. We wish to find the time interval between her leaving the bridge and her arriving at the bottom of her motion. Her overall motion can be separated into an 14.0-m free-fall and a 26.0-m section of simple harmonic oscillation.

(a) For the free-fall part, what is the appropriate analysis model to describe her motion.

-particle under constant angular acceleration

-particle under constant acceleration    

-particle in simple harmonic motion

(b) For what time interval is she in free-fall?s

(c) For the the simple harmonic oscillation part of the plunge, is the system of the bungee jumper, the spring, and the Earth isolated or non-isolated?

-isolated

-non-isolated    

(d) From your response in part (c) find the spring constant of the bungee cord.N/m

(e) What is the location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper?
m below the bridge

(f) What is the angular frequency of the oscillation?rad/s

(g) What time interval is required for the cord to stretch by 26.0 m?s

(h) What is the total time interval for the entire 40.0-m drop? s

Answer 1

(a) For the free fall part we can model the jumper as a particle under constant acceleration. Since during that time, the particle has only constant acceleration equal to acceleration due to gravity.

(b) The jumper will have free fall until when the cord remains unstretched. Therefore the jumper will be under free fall till it travels a distance of 14m. Time taken to reach 14 m, can be calculated as,

Substituting values we get,

(c) The whole system is isolated since there is no transfer of energy from any other external systems.

(d) Velocity of the particle after travelling 14 m will be,

After traveling 14 m in free fall the particle will have a kinetic energy and potential energy. After travelling the rest of the 40m under the influence of spring, the particle will only have the potential energy. The rest of the energy would have been converted to spring potential energy. The Kinetic energy of the particle after travelling 14 m will be equal to the potential energy lost. Hence we can write the initial energy as,

We will consider this energy as negative (we are taking potential energy at the bridge is 0, because we don't know the distance of person from the ground.

Now at a distance of 40 m. The energy can be written as (the spring extended 26m).

By conservation of energy we can write,