Energy problem on bungee jumper

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SUMMARY

The discussion centers on calculating the speed of a bungee jumper at a height of 12.5 m above the water using conservation of energy principles. The jumper has a mass of 64.5 kg, and the bungee cord has a force constant of 65.5 N/m with an unstretched length of 10.1 m. The correct approach involves determining the stretch of the cord (delta X) when the jumper is at 12.5 m above the water, which is calculated using the equation mgh = 0.5kx². The accurate calculation yields a speed of 6.37 m/s, contrasting with the incorrect initial calculation of 9.29 m/s.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Familiarity with elastic potential energy calculations (E = 0.5 kx²)
  • Knowledge of gravitational potential energy (mgh)
  • Basic algebra for solving equations
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  • Review the principles of conservation of energy in mechanical systems
  • Learn how to calculate elastic potential energy in spring systems
  • Study the relationship between gravitational potential energy and height
  • Practice solving problems involving energy conservation with varying mass and spring constants
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decamij
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I'm having trouble solving this problem. can n e one help me?

A bungee jumper of mass 64.5 kg (including safety gear) is
standing on a platform 48.0 m above a river. The length of
the unstretched bungee cord is 10.1 m. The force constant
of the cord is 65.5 N/m. The jumper falls from rest and just
touches the water at a speed of zero. The cord acts like an
ideal spring. Use conservation of energy to determine the
jumper’s speed at a height of 12.5 m above the water on
the first fall.

I used the work equation to find the total energy, then subtracted the elastic potential energy and gravitational potential energy at 12.5m above ground. Then i solved for Ek, and used THAT equation to solve for speed.
i got 9.29m/s, but the answer is 6.37m/s.

What did i do wrong?
 
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What were the numbers that you put into the term for elastic potential energy?
 
elastic potential energy is E = 0.5 kx^2, right?

I put: k = 65.5, x = 35.5 FIRST, but then i ended up getting a negative answer, so i used x = 12.5.

But is what i tried to do correct, did i miss anything? I'm starting to think that i calculated the total energy wrong. What did you get for total energy?
 
decamij said:
elastic potential energy is E = 0.5 kx^2, right?
Correct.
I put: k = 65.5, x = 35.5 FIRST, but then i ended up getting a negative answer, so i used x = 12.5.
Both x = 35.5 m and x = 12.5 m are wrong. How much is the cord stretched (beyond its unstretched length) when the jumper is 12.5 m above the water?
 
you should find the "delta X", by this equation: mgh=.5kx^2, I got 30.45 m for delta X,
then you should put this value in the equation: mgh=.5kx^2+mgh`+.5mv^2 and find the velocity..the initial X, 10.1, is extra in the given problem.
 

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