A 55 kg Bungee Jumper Has Fallen Far Enough
Introduction
A bungee jumper with a mass of 55 kg has fallen far enough that the bungee cord is now fully stretched. The cord has a spring constant of 120 N/m.
What is the jumper’s speed just before he begins to rebound?
Solution
The gravitational potential energy of the jumper at the top of the fall is:
$$PE_g = mgh$$
where:
* $m$ is the jumper’s mass (55 kg)
* $g$ is the acceleration due to gravity (9.8 m/s2)
* $h$ is the height of the fall
We don’t know the height of the fall, but we can use the fact that the bungee cord is now fully stretched to find it. The elastic potential energy stored in the cord is:
$$PE_e = frac{1}{2}kx^2$$
where:
* $k$ is the spring constant of the cord (120 N/m)
* $x$ is the stretch of the cord
The stretch of the cord is equal to the height of the fall, so we have:
$$PE_e = frac{1}{2}kh^2$$
The total mechanical energy of the jumper is conserved, so we have:
$$PE_g = PE_e$$
Substituting the expressions for $PE_g$ and $PE_e$, we get:
$$mgh = frac{1}{2}kh^2$$
Solving for $h$, we get:
$$h = sqrt{frac{2mgh}{k}}$$
Substituting the given values, we get:
$$h = sqrt{frac{2(55 text{ kg})(9.8 text{ m/s}^2)(h)}{120 text{ N/m}}}$$
$$h = 22.4 text{ m}$$
Now that we know the height of the fall, we can find the jumper’s speed just before he begins to rebound using the conservation of energy:
$$PE_g = KE$$
where $KE$ is the jumper’s kinetic energy. Substituting the expressions for $PE_g$ and $KE$, we get:
$$mgh = frac{1}{2}mv^2$$
Solving for $v$, we get:
$$v = sqrt{2gh}$$
Substituting the given values, we get:
$$v = sqrt{2(9.8 text{ m/s}^2)(22.4 text{ m})}$$
**Therefore, the jumper’s speed just before he begins to rebound is 20.4 m/s.**