Question

A 75 kg bungee jumper leaps from a bridge. When he is 30 meters above the water, and moving at a speed of 20 m/s, the bungee cord begins to stretch. The cord stops his motion exactly at the water surface. Determine the spring constant of the bungee cord. Assume no thermal losses.

Answer 1

k = 52.2 N / m

Step-by-step explanation:

For this exercise we are going to use the conservation of mechanical energy.

Starting point. When it is 30 m high

Em₀ = K + U = ½ m v² + m g h

Final point. Right when you hit the water

`Em_(f)` = K_{e} = ½ k x²

in this case the distance the bungee is stretched is 30 m

x = h

as they indicate that there are no losses, energy is conserved

Em₀ = Em_{f}

½ m v² + m g h = ½ k h²

k =
`(m (v^(2) + 2 g h))/(h^(2) )`

let's calculate

k =
`(75 \ ( 20^(2) + 2 \ 9.8 \ 30))/(30^(2) )`

k = 52.2 N / m