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 h_{f} = (50 + Δx) m (length of unstretched bungee cord + stretching)

The gravitational potential energy at the jumping point, U_{i,g} = mgh_{f}

Final gravitational potential, U_{f,g} = 0 (since reference height is zero)

Kinetic energy at the jumping point, KE_{0} = Kinetic energy at the lowest point, KE_{f} = 0 (velocity is zero at those points)

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)

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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.