Potential Energy bungee jump question?
The Royal Gorge bridge over the Arkansas River is 310.0 m above the river. A 60.0-kg bungee jumper has an elastic cord with an unstressed length of 64.0 m attached to her feet. Assume that, like an ideal spring, the cord is massless and provides a linear restoring force when stretched. The jumper leaps, and at at her lowest point she barely touches the water. After numerous ascents and descents, she comes to rest at a height h above the water. Model the jumper as a point particle and assume that any effects of air resistance are negligible.
Please help! Each time I do this problem I keep getting the wrong answer.
Thanks!
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Well... it is kind of a strange question...
"Potential Energy bungee jump"
and
"Model the jumper as a point particle"
So what exactly do you want???
I can tell you this (and I hope it will help you to find the correct answer):
H = height of the bridge = 310 m
L = unstreched length of the cord = 64 m
m = mass of the jumper = 60 kg
g = acceleration by gravity = 9.8 m/s²
k = elastic constant of the cord = ?
h = point of equilibrium = ?
At the top of the bridge, the jumper has a gravitational potential energy of:
PEg = m × g × H
PEg = (60 kg) × (9.8 m/s²) × (310 m)
PEg = 182280 J
That must also be the elastic potential energy stored in the cord at the lowest point.
The stretching of the cord at the lowest point is:
s = H - L
s = (310 m) - (64 m)
s = 246 m
Now you can find the elastic constant of the cord:
PEg = PEe = k × s² / 2
(182280 J) = k × (246 m)² / 2
k = 6.024 N/m
Now you can find the stretching of the cord when at rest.
F = k × d
m × g = k × d
(60 kg)×(9.8 m/s²) = (6.024 N/m) × d
d = 97.6 m
So the length of the cord when at rest is:
S = L + d
S = (64 m) + (97.6 m)
S = 161.6 m
The height above the ground of that point is:
h = H - S
h = (310 m) - (161.6 m)
h = 148.4 m
The angular velocity of the SHM is:
ω = √( k / m )
ω = √( (6.024 N/m) / (60 kg) )
ω = 0.3169 rad/s
The frequency of the SHM is:
ω = 2π f
(0.3169 rad/s) = 2π × f
f = 0.0504 Hz
The period of the SHM is:
T = 1 / f
T = 1 / (0.0504 Hz)
T = 19.8 s
Choosing time t = 0 s at the moment the jumper reaches the lowest point for the first time, the equation for vertical position as a function of time is:
y = -h×cos(ω×t)
y = (-148.4 m)×cos((0.3169 rad/s)×t) + (148.4 m)
The equation for velocity (using the same time t = 0 s) is:
v = h×ω×sin(ω×t)
v = (47.03 m/s)×sin((0.3169 rad/s)×t)
The equation for acceleration (using the same time t = 0 s) is:
a = h×ω²×sin(ω×t)
a = (14.9 m/s²)×sin((0.3169 rad/s)×t)
So... that's about all I think...
I hope this is useful, and feel free to contact me if needed ;-)
1
PE=mgh
mass*acceleration of gravity*height.