# Problem: Your adventurous friend Lola goes bungee jumping. She leaps from a bridge that is 100 m above a river. Her bungee cord has an un-stretched length of 50 m and a spring constant k = 600 N/m. Lola has a mass of 48 kg. What is her height above the river at her lowest point? (Hint: use energy for this, assume no energy is lost, and remember that the energy stored in a stretched bungee cord [or spring] is (1/2)k(Δx)2, where Δx is the amount of stretch)

###### FREE Expert Solution

Gravitational potential energy:

$\overline{){{\mathbf{U}}}_{{\mathbf{g}}}{\mathbf{=}}{\mathbf{m}}{\mathbf{g}}{\mathbf{h}}}$

Kinetic energy:

$\overline{){\mathbf{K}}{\mathbf{E}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}{{\mathbf{v}}}^{{\mathbf{2}}}}$

Elastic potential energy for a stretched cord/spring:

$\overline{){{\mathbf{U}}}_{{\mathbf{e}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{k}}{{\mathbf{x}}}^{{\mathbf{2}}}}$

Lona comes to rest at hf = (50 + Δx) m (length of unstretched bungee cord + stretching)

The gravitational potential energy at the jumping point, Ui,g = mghf

Final gravitational potential, Uf,g = 0 (since reference height is zero)

Kinetic energy at the jumping point, KE0 = Kinetic energy at the lowest point, KEf = 0 (velocity is zero at those points)

90% (175 ratings) ###### Problem Details

Your adventurous friend Lola goes bungee jumping. She leaps from a bridge that is 100 m above a river. Her bungee cord has an un-stretched length of 50 m and a spring constant k = 600 N/m. Lola has a mass of 48 kg.

What is her height above the river at her lowest point? (Hint: use energy for this, assume no energy is lost, and remember that the energy stored in a stretched bungee cord [or spring] is (1/2)k(Δx)2, where Δx is the amount of stretch)