Question

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 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

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

Part B

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

Answer 1

Part A Answer

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 Answer

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