A 55.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her

Question

A 55.0-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 37.0 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 23.0-m section of simple harmonic oscillation.

For what time interval is she in free-fall?

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 or 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?

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

(g) What time interval is required for the cord to stretch by 23.0 m?
s

(h) What is the total time interval for the entire 37.0-m drop?
s

Explanation / Answer

a) 14=.5*9.81*t^2
solve for t
1.690 s

b)
55*9.81*37=.5*k*23^2
solve for k
k=2*55*9.81*37/23^2
75.47 N/m

c)
75.47*x=55*9.81
x=55*9.81/75.47
x=7.15
this is 37-7.15 m above the max stretch
d)
?=sqrt(k/m)
1.17 rad/s
e)
T=2*?/?
5.37 s

f) 1.690+5.37
7.06 s

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