A 65.0-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 11.0 m. The jumper 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 11.0-m free fall and a 25.0-m section of simple harmonic oscillation.
(a) For the free-fall part, what is the appropriate analysis model to describe her motion
(b) For what time interval is she in free fall?
(c) For the simple harmonic oscillation part of the plunge, is the system of the bungee jumper, the spring, and the Earth isolated or non-isolated?
(d) From your response in part (c) find the spring constant of the bungee cord.
Answer:
a) Generally the appropriate analysis model to describe is the Simple harmonic motiom(SHM)
b) [tex]t=1.4983[/tex]
c) For the simple harmonic oscillation part of the plunge, the system of the bungee jumper, the spring, and the Earth ISOLATED
d )[tex]k=73.3824N/m[/tex]
Explanation:
From the question we are told that:
Mass of Bungee jumper [tex]M=65kg[/tex]
Un-stretched length of the cord is [tex]l_{un} 11.0 m[/tex]
Bottom of motion distance [tex]d_m=36.0m[/tex]
Free fall distance [tex]d_f=11.0-m[/tex]
Simple harmonic oscillation[tex]d_{shm}=25m[/tex]
a)
Generally the appropriate analysis model to describe is the Simple harmonic motiom(SHM)
b)
Generally the Newtons's equation for motion is mathematically given by
s=ut+1/2at^2
Since she fell downward
[tex]-d_f=-\frac{1}{2}(9.8)t^2\\\\ -11=-\frac{1}{2}(9.8)t^2\\\\ t=\sqrt{\frac{22}{9.8}}\\\\[/tex]
[tex]t=1.4983[/tex]
(c)
For the simple harmonic oscillation part of the plunge, the system of the bungee jumper, the spring, and the Earth ISOLATED
d)
Generally the equation for potential energy is mathematically given by
[tex]P.E=mgh[/tex]
and Gravitational potential energy
[tex]P.E_G=\frac{1}{2}k(h-L)^2[/tex]
Since
[tex]P.E=P.E_G\\\\mgh =\frac{1}{2}k(h-L)^2\\\\k=\frac{2(mgh)}{(h-L)^2}\\\\k=\frac{2(65*9.8*36)}{(36-11)^2}\\\\[/tex]
[tex]k=73.3824N/m[/tex]
