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

Answer:

k = 52.2 N / m

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 = $\frac{m (v^{2} + 2 g h)}{h^{2} }$

let's calculate

       k = $\frac{75 \ ( 20^{2} + 2 \ 9.8 \ 30)}{30^{2} }$

       k = 52.2 N / m