bungee jump problem

[dopey9]

a bungee jumper stnd on a bridge of 100m above the floor of a valley.
She is attached to a bungee rope of length 25m and has a mass of 60kg. and i have taken g to be 10

Theres no: air resistance
damping in the bungee rope
and the weight of the bungee rope is negliable

i need to find the minimum required value for the spring constant k if she is to avoid hiting the valley floor

im using 1/2kx^2=mgh...but im getting confused in what im doing i dont know where to go next..i was wondering if i am doing it right n what i should do next ?....thankz

[Hootenanny]

im using 1/2kx^2=mgh...
Your on the right track, but what happens when the binjee jumper falls the first 25m? Will the rope be storing potential energy then? What is the maximum extension the rope can undergoe without the jumper hitting the floor?

[dopey9]

Your on the right track, but what happens when the binjee jumper falls the first 25m? Will the rope be storing potential energy then? What is the maximum extension the rope can undergoe without the jumper hitting the floor?

sorry im really confused?

[arildno]

Remember that for the first 25 meters of falling, the rope is SLACK and there is no tension in it.

[dopey9]

this is what i did

mgh= 0.5*k*(h-l)^2

60*10*100=0.5*k*(100-25)^2

k= 21.3

this is what i did and got k as ...have i done it right because im so confused

[arildno]

Okay, for clarity I'll take this step by step:
Consider the moment when the jumper has fallen h=25 meters.
Then, obviously, the kinetic energy gain equals the loss of potential energy:
\frac{1}{2}mv_{0}^{2}=mgh

We now look at conservation of mechanical energy during the phase where the rope stretches, and where L is the maximal length the rope stretches:
\frac{1}{2}mv_{0}^{2}=\frac{1}{2}kL^{2}-mgL
which can be rewritten as
mg(h+L)=\frac{1}{2}kL^{2}
with h=25, L=75, you get your own equation, so yes, you did the problem right!:smile:

[dopey9]

Okay, for clarity I'll take this step by step:
Consider the moment when the jumper has fallen h=25 meters.
Then, obviously, the kinetic energy gain equals the loss of potential energy:
\frac{1}{2}mv_{0}^{2}=mgh

We now look at conservation of mechanical energy during the phase where the rope stretches, and where L is the maximal length the rope stretches:
\frac{1}{2}mv_{0}^{2}=\frac{1}{2}kL^{2}-mgL
which can be rewritten as
mg(h+L)=\frac{1}{2}kL^{2}
with h=25, L=75, you get your own equation, so yes, you did the problem right!:smile:

Thankz .