Kate's Bungee Jump: Calculating Distance Below Bridge

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SUMMARY

The discussion centers on calculating the distance below a bridge that Kate, a bungee jumper, will hang after jumping. The problem utilizes the conservation of energy principle, where the gravitational potential energy (mgd) equals the elastic potential energy of the bungee cord (1/2 k(d-L)²). The final answer for the distance d is derived as d = mg/k + L, indicating that the bungee cord behaves as an ideal spring with a spring constant k. The conversation also touches on the implications of energy loss due to friction when determining k for the scenario where Kate just reaches the water.

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  • Understanding of conservation of energy principles
  • Knowledge of spring mechanics and Hooke's Law
  • Familiarity with gravitational force calculations
  • Basic algebra for solving equations
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  • Study the derivation of Hooke's Law and its applications in spring mechanics
  • Learn about energy conservation in mechanical systems
  • Explore the effects of friction on energy calculations in physics
  • Investigate the dynamics of oscillatory motion in bungee jumping scenarios
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Homework Statement


Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls.

Assume the following:

The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k.
Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward.
Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle.

Use g for the magnitude of the acceleration due to gravity.

How far below the bridge will Kate eventually be hanging, once she stops oscillating and comes finally to rest? Assume that she doesn't touch the water.

Homework Equations


conservation of energy


The Attempt at a Solution


the problem states to use only terms introduced in the problem, so don't use X in kx

mgd = \frac{1}{2}k(d-L)^{2}

\frac{2mgd}{k} = (d-L)^{2}

\sqrt{\frac{2mgd}{k}} = d - L

\sqrt{\frac{2mgd}{k}} + L = d

\sqrt{d} * \sqrt{\frac{2mg}{k}} + L = d

\sqrt{\frac{2mg}{k}} + \frac{L}{\sqrt{d}} = \frac{d}{\sqrt{d}} = \sqrt{d}

(\sqrt{\frac{2mg}{k}} + \frac{L}{\sqrt{d}})^{2} = d

im stuck here since i have to solve for d, and i don't even think the result will be similar to the answer

the answer is \frac{mg}{k} + L

where did i go wrong?
 
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Why do you think those energies should be equal? When she is at rest her weight mg is equal to the force of the spring pulling up, k*(d-L). The energy approach is used if you want to find the value of d where she is closest to hitting the water.
 
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wow that was simple, hm

well the next part of the problem asks to find the k constant if she jumps and JUST reachs the water,, is the energy approach, then correct?

BTW how did you know to set the forces equal to each other and not the energys?

thanks
 
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Because of the phrase "once she stops oscillating and comes finally to rest". That implies that frictional forces have been acting on her and dissipating energy. The energy calc wouldn't account for that lost energy. On the other hand, for the 'just reaches the water' case, it's reasonable to assume that there hasn't been that much energy lost to friction, and since KE is zero you can equate the PE's. Now you tell me why I can't just balance the forces in that case?
 
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