After earning an 'A' in PHY 183 you land a job with ACME Bungee Jump company. They need to know what spring stiffness k_S to make the cords so that a jumper of mass 200kg will only fall 30m. The standard length of an un-stretched cord is just 10m.

Jumper of mass 200kg

Jumper will fall 30m

Standard length of un-stretched cord is 10m

Initial State: Jumper motionless at top

Final State: Jumper motionless 30m below

k_s

Jumper is motionless before they start

No energy lost to the surroundings

System: Jumper + Cord + Earth

Surroundings: Nothing

course_planning

$$\Delta E_{system} = W_{surroundings}$$

$$\Delta K_{jumper} + \Delta U_{grav} + \Delta U_{spring} = W_{surroundings} = 0$$

$$\Delta K_{jumper}$$ is 0 because jumper is motionless before + after jump

$W_{surroundings}$ no work

$$\Delta U_{grav} = -\Delta U_{spring}$$

$$mg(y_{f}-y_{i}) = \dfrac{1}{2}ks({s_{f}}^2 - {s_{i}}^2)$$

${s_{i}}^2 = 0$ (no stretch)

$$m_{j}(y_{f}-y_{i}) = -\dfrac{1}{2}ks({s_{f}}^2)$$

$$k_{s} = \dfrac{-2m_{j}g(y_{f}-y_{i})}{{s_{f}}^2}$$

$$k_{s} = \dfrac{-2(200kg)(9.81m/s^2)(-30m)}{{(30m-10m)}^2}$$

$$k_{s} = 294N/m$$