Bungee jumper(Conservation of mechanical energy)

  • Thread starter Raziel2701
  • Start date
1. The problem statement, all variables and given/known data
The Royal Gorge bridge over the Arkansas River is 310 m above the river. A 60-kg bungee jumper has an elastic cord with an unstressed length of 50 m attached to her feet. Assume that, like an ideal spring, the cord is massless and provides a linear restoring force when stretched. The jumper leaps, and at at her lowest point she barely touches the water. After numerous ascents and descents, she comes to rest at a height h above the water. Model the jumper as a point particle and assume that any effects of air resistance are negligible.

2. Relevant equations
1/2kx^2 is the elastic potential
mgh is grav potential
1/2mv^2 is KE

3. The attempt at a solution
I set the river to be my zero potential height, so before the jumper goes, she has an initial gravitational potential of mgh where h is 310m. At the bottom, when she barely touches the water, she has an elastic potential of 1/2kx^2, where x is 310-50=260. So I solved for k. getting 5.392899 N/m

Now when she has oscillated a bit and has finally come to a stop, she has gravitational and elastic potential. So this is what I solved for:

[tex]mgh_{initial}=\frac{1}{2}k(h-260)^2 +mgh[/tex] The h in the left hand side of the eqn is 310, I tried to solve for h in the right hand side but I got back to zero as an answer. So am I setting up wrong?

Also I need to find the max velocity.
Assuming an ideal spring with no damping, h=0 (and h=310) would be the only point where the sum of gravitational potential energy and spring energy equal the initial gravitational potential energy. That is because those are the only two points where there is no kinetic energy! You need to include a term for kinetic energy in the right side of your equation to complete describe the particle.

And assuming conservation of energy, kinetic energy would be greatest when the sum of gravitational potential energy and spring energy is smallest.

Are you supposed to solve h for when she comes to rest? Because assuming there is no wind resistance, it doesn't seem like there would be any damping to include in the equation, and she never actually would come to rest.

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