Help please! This is a hard one! Please demonstrate work as well. A 97.0-kg bung

Question

Help please! This is a hard one! Please demonstrate work as well.

A 97.0-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 12.0 m. The jumper reaches reaches the bottom of her motion 36.0 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 12.0-m free-fall and a 24.0-m section of simple harmonic oscillation.
(a) For the free-fall part, what is the appropriate analysis model to describe her motion. Pick one from below:
i) particle under constant acceleration
ii) particle in simple harmonic motion
iii) particle under constant angular acceleration

(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 24.0 m?
s

(h) What is the total time interval for the entire 36.0-m drop?
s

Explanation / Answer

a) i) since only gravity activity so constant a b) we can use x = 1/2 g t^2 12=1/2 * 9.81 * t^2 so t= 1.564 s c) isolated since only earth and string act on the person d) since its isolated we can use energy initial energy = mgh final energy = 1/2 k x^2 so 97*9.81*36=1/2 * k * 24^2 so k =118.9 e) equillibrium happens when kx = mg so x= mg/k = 97*9.81/118.9=8 but this is the stretch of the string so distance below bridge is 12+8=20 f)we use the formula for a spring ?=v(k/m)=v(118.9/97)=1.107 rad/s g) we can find the period which is T= 2*pi/?=2*pi/1.107=5.68 sec since the equilibrium point is 8 m and is stretches to 24 m the amplitude is 16m so it goes from 8 below equillbrium to 16 above well from -A to A is half a period so this is -A/2 to A which is 3/4 of 1/2 T so 3/8 T = 3/8*5.68=2.13 s h) total time = 1.564 + 2.13 = 3.69 s

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