Spring Constant - Bungee Jumping

AI Thread Summary
The discussion revolves around calculating the spring constant of a bungee cord used in a jump scenario. The initial attempt incorrectly used a mass of 60 kg instead of the correct 65 kg and misapplied the equations for gravitational and elastic potential energy. The equilibrium point indicates that the gravitational force equals the spring force, leading to a more straightforward calculation of the spring constant. The correct spring constant, derived from the equilibrium condition, is 637 N/m, highlighting the importance of accurately defining variables and understanding the context of equilibrium in physics problems. Clarifying the relationship between energy and force is crucial for accurate problem-solving in this scenario.
Speedking96
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Homework Statement



A person with a mass of 65 kg goes bungee jumping. At the lowest point, he is located 30 m below his starting point. If, at the equilibrium point, the bungee cord measures 15m, what is its spring constant?

Homework Equations



F = k*x

Ep = (1/2)(k)(x^2)

The Attempt at a Solution



At the bottom, all the gravitational potential energy is converted to elastic potential energy.

mgh = (1/2)(k)(x^2)
(60 kg)(9.8 m/s^2)(30 m) = (1/2)(k)(30^2)

k = 39.2 N/m

I am not sure if this is correct, why did they give the equilibrium information? What does that mean?
 
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What is x in your expression for the bungee cord's potential energy measured relative to?
 
Just because two relevant equations use the same symbol (x here), doesn't mean they refer to the same quantity. Whenever you quote an equation, you ought to state what each symbol represents for that equation.
 
Speedking96 said:

Homework Statement



A person with a mass of 65 kg goes bungee jumping. At the lowest point, he is located 30 m below his starting point. If, at the equilibrium point, the bungee cord measures 15m, what is its spring constant?

Homework Equations



F = k*x

Ep = (1/2)(k)(x^2)

The Attempt at a Solution



At the bottom, all the gravitational potential energy is converted to elastic potential energy.

mgh = (1/2)(k)(x^2)
(60 kg)(9.8 m/s^2)(30 m) = (1/2)(k)(30^2)

k = 39.2 N/m

I am not sure if this is correct, why did they give the equilibrium information? What does that mean?
First thing, a typo somewhere: in the problem the mass is 65 kg, but you plugged in 60 kg at the end.

Second, to address your question: in this case equilibrium means that the forces are in balance. That should point you to a slightly easier method of solution.,
 
At equilibrium point, the gravitational force is equal to the force in the spring:

mgh = k x
(65 kg)(9.8 m/s^2)(15 m) = (k)(15 m)

k = 637 N/m
 
Speedking96 said:
At the bottom, all the gravitational potential energy is converted to elastic potential energy.

mgh = (1/2)(k)(x^2)
(60 kg)(9.8 m/s^2)(30 m) = (1/2)(k)(30^2)

k = 39.2 N/m

I am not sure if this is correct, why did they give the equilibrium information? What does that mean?

You have used incorrect mass .Mass is 65 kg .While using (1/2)kx2 ,you need to be careful in the sense that 'x' represents the displacement from the unstretched length.

30m is the distance between the topmost and lowest point ,not between the lowest point and the equilibrium position .

Speedking96 said:
At equilibrium point, the gravitational force is equal to the force in the spring:

mgh = k x
(65 kg)(9.8 m/s^2)(15 m) = (k)(15 m)

k = 637 N/m

Wrong.

How can you equate mgh i.e energy with 'kx' i.e force ?
 
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