**Bungee Jumping**

Part A = d = mg/k + L

Part B = k = 2mgh / (h – L)^2

**Solution Below:**

- The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k.
- 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 for the magnitude of the acceleration due to gravity.

Part A

Express the distance in terms of quantities given in the problem introduction.

Remember that the force exerted by a spring is F = -kx. Since the cord is being stretched downwards (and will spring upwards), we can remove the negative sign since the force will be acting in the positive y direction. And x = (d – L), where d is the total (stretched) length of the bungee cord. Just set this equal to the gravitational force:

F(spring) = k(d – L)

F(gravity) = mg

k(d – L) = mg

(d – L) = mg/k

d = mg/k + L

d = mg/k + L

Part B

Express k in terms of L, h, m, and g.

When Kate jumps, her gravitational potential energy will be converted to spring potential energy, as the cord stretches. Remember that gravitational potential energy is U = mgh, and that spring potential energy is E = 1/2kx^2. Just set the two equal to each other. Also remember from Part A that x = (d – L), except that in this part, ‘d’ is replaced by ‘h’, since the cord is extended all the way down to the river. So for x, you have: x = (h – L).

U = mgh

E = 1/2k(h – L)^2

mgh = 1/2k(h – L)^2

2mgh = k(h – L)^2

k = 2mgh / (h – L)^2

k = 2mgh / (h – L)^2

Instead of “(d – L)^2″ is (h – L)^2 for the second part

Good catch. Sorry for the confusion!