Need the Answer for Part (D) only A 80.0 kg bungee jumper steps off a bridge wit

Question

Need the Answer for Part (D) only

A 80.0 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 13.0 m. She reaches the bottom of her motion 38.0 m below the bridge before bouncing back. Her motion can be separated into an 13.0 m free-fall and a 25.0 m section of simple harmonic oscillation. (a) For what time interval is she in free-fall? 1.62 (b) Use the principle of conservation of energy to find the spring constant of the bungee cord 95.33 /m (c) What is the location of the equilibrium point where the spring force balances the gravitational force acting on the jump Note that this point is taken as the origin in our mathematical description of simple harmonic oscillation m below the bridge 21.22 (d) What is the angular frequency of the oscillation? 16.78 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. rad/s (e) What time interval is required for the cord to stretch by 25.0 m? 1.91 (f) What is the total time interval for the entire 38.0 m drop? 3.53

Explanation / Answer

The spring constant K=95.33 N/m

Mass of body attached m=80 kg

This concept belong to oscillations of loaded spring

According to the force law

The angular frequency W=k/m

=96.33/80

=1.1 rad/s

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