# Homework Help: Bungee Jumping (Finding distance and spring constant)

1. Oct 24, 2008

### jheld

1. The problem statement, all variables and given/known data
Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls.

Assume the following:

The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant .
Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward.
Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle.

Use g for the magnitude of the acceleration due to gravity.

How far below the bridge will Kate eventually be hanging, once she stops oscillating and comes finally to rest? Assume that she doesn't touch the water. solve d using the measurements that you were given.

And;
If Kate just touches the surface of the river on her first downward trip (i.e., before the first bounce), what is the spring constant ? Ignore all dissipative forces. (k in terms of L, h, m and g)

2. Relevant equations
F = ma
Vf^2 = Vi^2 + 2a(delta s)
1/2k(L^2)
spring equilibrium = 1/2k(delta s)^2

3. The attempt at a solution
i got d = sqrt(m*2*g*L/k), but that was wrong.

the answer is (mg +kL)/k, though I'm not sure why. and i'm not sure how to go about part 2.

2. Oct 24, 2008

### Staff: Mentor

For part 1, apply Hooke's law to find out how much the cord stretches.

For part 2, apply conservation of energy.

3. Oct 24, 2008

### jheld

i tried doing Fspring = -k(delta s)2 converting it into = -kL^2. i think that's the right way to do it. but, still not sure.

my final velocity squared will 2mgL. i bring that into the kinetic energy equation k = 1/2mv^2. so kinetic energy = 2/2*m(gL) ---> mgL. is that right?

then hooke's law is -ks^2. is my s = L ?.
how would i bring these two equations together?

4. Oct 24, 2008

### Staff: Mentor

That's not Hooke's law! Read this: Hooke's Law. Also, don't confuse the amount of stretch and L. (L is the unstretched length of the cord.)

I think you're confusing Hooke's law (F = -kx), which relates spring force with stretch, with the formula for spring potential energy (SPE = ½kx²), which also depends on the stretch.

5. Nov 15, 2008

### jheld

Coming back to this problem (now that I have some time on my hands).

Hooke's Law: F_spring = -k(delta s)

delta s = s - L (i think)
unfortunately in finding s, i get mg/k + L, which gives F_spring = mg, which i don't think is right.

conservation of energy:
K_i + U_gi + U_si = K_f + U_gf + U_sf
K_i = (1/2)(m)(v)^2

v = sqrt(2gL)
K_i = (1/2)(m)(sqrt(2gL))^2
= mgL. is this right? I think it is.

U_gi = mgh
h = L
U_gi = mgL. is this right? I think it is.

U_si = (1/2)k(delta s)
delta s = (L - L) = 0
U_si = 0. I know this is correct, and I understand why.

K_f = (1/2)mv^2
v = 0
K_f = 0. I think this is right.

U_gf = mg(delta s)
though from earlier, it seems my delta s (after bungee reaches length L and start stretching), is wrong, so I'm not sure how to handle this.

in the end (without completing everything):
K_i + U_gi = U_sf
2mgL = U_sf

Once I get the delta s correct, how do I find the length, though?

6. Nov 15, 2008

### jheld

I think I was thinking too deeply in this problem and didn't truly realize the things that I did have.

I already have part of the total distance, it is length of the unstretched rope, L.

And F_spring = -k(delta s).

delta s is the amount the rope will stretch from its unstretched length (technically the restoring force that holds the final length (from length L) with Kate of mass m, on it) to its final length.

ma = 0
0 = -k(delta s) + mg
k(delta s) = mg
delta s = (mg)/k.

therefore the total distance Kate will be at will be d = (mg)/k + L.

7. Nov 15, 2008

### Staff: Mentor

Good! Now it's time to tackle part 2.

8. Nov 16, 2008

### jheld

Okay.

So, I have chosen to make Ei to be before she 'jumps'. energy is conserved, it is an enclosed system. and I have chosen her to be standing at y = 0, to make my remaining calculations easier.
therefore,
(1/2)m(0)^2 + mg(0) + (1/2)k(0 - 0) = 0

Ef (touching the water at the bottom before springing up (so, elastic potential energy and gravitational energy are both at work)
Ef:
(1/2)m(0)^2 + mg(-h) + (1/2)k(h - L)^2

Ei = Ef
0 = -mgh + (1/2)k(h-L)^2
k = mgh(2)/(h-L)^2
Does this look right?

9. Nov 17, 2008

Looks good!