In the dangerous "sport" of bungee-jumping, a daring student jumps from a balloon with a specially designed elastic cord attached to his ankles, as shown in Figure P5.69. The unstretched length of the cord is 28.0 m, the student weighs 680 N, and the balloon is 36.0 m above the surface of a river below. Calculate the required force constant of the cord if the student is to stop safely 4.00 m above the river.
ok what i like to do is list what i know.
the elastic cord stretches from 28m to (36-4)m
so x=36-4-28=4m
m=680N
h1=36
h2=4m
the question is asking for k
ok lets start
mgh1 = mgh2 + .5kx^2
.5kx^2 = mg(h1-h2)
k = 2mg(h1-h2)/x^2
k=2720
All of the gravitational PE the idiot has must translated to elastic PE
mgh=1/2 k x^2
mg*32=1/2 k x^2 so my question is x. How long is the stretched cord? It is not clear to me where it is attached.
NA IS 897
To solve this problem, we need to use the concept of potential energy.
The gravitational potential energy of the student at the beginning is given by:
PE_initial = m * g * h_initial,
where m is the mass of the student, g is the acceleration due to gravity, and h_initial is the initial height of the student.
The elastic potential energy of the cord when it stretches to 4.00 m is given by:
PE_elastic = (1/2) * k * (x^2),
where k is the force constant of the cord and x is the amount the cord has stretched.
At the stopping point, the student is at a height of 4.00 m above the river. Since he is stopped, his velocity is zero and therefore his kinetic energy is also zero.
Therefore, the total mechanical energy of the system is conserved. So we can write the equation:
PE_initial + PE_elastic = PE_final + KE_final,
where PE_final is the gravitational potential energy of the student at the stopping point and KE_final is his kinetic energy at the stopping point.
Since the student is stopped, KE_final is zero. Therefore, the equation simplifies to:
PE_initial + PE_elastic = PE_final.
Now we can plug in the values given in the problem:
PE_initial = m * g * h_initial,
PE_final = m * g * h_final,
where h_final is the final height of the student at the stopping point.
Given values:
m = 680 N / g (mass can be calculated as weight divided by the acceleration due to gravity),
g = 9.8 m/s^2,
h_initial = 36.0 m,
h_final = 4.00 m,
x = h_initial - h_final.
We can rearrange the equation to solve for k:
PE_elastic = PE_final - PE_initial,
(1/2) * k * (x^2) = m * g * (h_final - h_initial),
k = (2 * m * g * (h_final - h_initial)) / (x^2).
Plug in the values and calculate k.