Oscillation:Mass dropped on Vertical Spring

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SUMMARY

The discussion focuses on a bungee-jumping scenario involving a woman with a mass of 60 kg attached to an elastic rope with a spring constant of 220 N/m and a natural length of 15 m. To determine how far she falls before coming to rest, the conservation of energy principle is applied, considering both gravitational potential energy and elastic potential energy. The relevant equations include the angular frequency formula, ω² = k/m, and energy equations for kinetic and potential energy. The correct approach involves calculating the total potential energy from both forces acting on the jumper.

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Homework Statement



A woman bungee-jumper of mass 60 kg is attached to an elastic rope of natural length 15m. The rope behaves like a spring of spring constant k= 220Nm. The other end of the spring is attached to a high bridge. The woman jumps from the bridge.
a. Determine how far below the bridge she falls, before she instantaneously comes to rest.

Homework Equations



a= -(omega)Acos((omega)t)
omega^2=k/m
E=Ep+Ek
Ep=1/2mv^2
Ek=1/2 mv^2

The Attempt at a Solution


I'm not sure anymore, please help
 
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You can use conservation of energy but find the correct form of the potential energy. You have two forces, gravity and the elastic force of the cord, both of them contribute to the potential energy of the woman. ehild
 

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