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.
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.
3 answers
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
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
NA IS 897
1 answer