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:

Ug=mgh

Kinetic energy:

KE=12mv2

Elastic potential energy for a stretched cord/spring:

Ue=12kx2

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)
View Complete Written Solution
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)

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Equations for Energy Conservation concept. If you need more Equations for Energy Conservation practice, you can also practice Equations for Energy Conservation practice problems.