A 58.0 kg bungee jumper steps off a bridge with a light bungee cord tied to her

Question

A 58.0 kg bungee jumper steps off a bridge with a light bungee cord tied to her body and to the bridge. The unstretched length of the cord is 12.9 m. The jumper reaches reaches the bottom of her motion 37.3 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 12.9 m free fall and a 24.4 m section of simple harmonic oscillation. (Use the exact values you enter in previous answer(s) to make later calculation(s).)

(a) For the free-fall part, what is the appropriate analysis model to describe her motion. provide explanation for your choice
(i)particle in simple harmonic motion
(ii)particle under constant angular acceleration
(iii)particle under constant 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?
Explain with reasoning.

(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 24.4 m?
______________s

(h) What is the total time interval for the entire 37.3 m drop?
______________ s

Explanation / Answer

b) 12.9=.5*9.81*t^2
solve for t
1.6225 s

d)
58*9.81*37.3=.5*k*24.4^2
solve for k
k=2*58*9.81*37.3/24.4^2
71.3 N/m

e)
71.3*x=58*9.81
x=58*9.81/71.3
x=7.98
this is 37.3 - 7.98 m above the max stretch
f)
?=sqrt(k/m)
1.1087 rad/s
g)
T=2*?/?
5.667 s

h) 1.6225+5.667
7.29 s

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