Solving Spring Constant: Find k with Hooke's Law

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Homework Help Overview

The problem involves determining the spring constant using Hooke's Law in the context of a ball dropped from a height that compresses a spring upon impact. The scenario includes gravitational potential energy and elastic potential energy considerations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using conservation of energy to relate gravitational potential energy to elastic potential energy in the spring. Questions arise about how to set up the equations correctly and the implications of the initial and final states of the system.

Discussion Status

Several participants have provided insights into the energy transformations involved, and there is an ongoing exploration of the correct formulation of the equations. Some attempts to calculate the spring constant have been made, but there is no explicit consensus on the final value yet.

Contextual Notes

There is a note of clarification regarding the distance the ball falls, indicating it falls slightly more than the initial height due to spring compression. This detail may affect the calculations being discussed.

lking226
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Homework Statement


A ball is dropped from rest at a height of 50.0m above a spring. After the ball hits, it compresses the spring .340m. Ignoring any non-conservative forces, find the spring constant if the mass of the ball is 4.50 kg.


Homework Equations


Hooke's law: k = mg/x


The Attempt at a Solution


I'm not sure what to do.
k = 4.50 kg x 9.8 m/s / 0.340 m = 129.71
That doesn't seem right.
Help?
 
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Conservation of energy would work very well here.
 
but how do you use conservation of energy with springs?
 
lking226 said:
but how do you use conservation of energy with springs?

The potential energy stored in a spring is given by 1/2kx2. So what type of energy is being converted into the elastic potential energy in the spring?
 
the kinetic energy of the ball?
 
lking226 said:
the kinetic energy of the ball?

The energy is possesses at rest 50m above the spring.
 
so the gravitational potential energy of the ball at 50m converts to the elastic potential energy in the spring?
 
lking226 said:
so the gravitational potential energy of the ball at 50m converts to the elastic potential energy in the spring?

That would be correct. Can you now form an equation and solve for the spring constant 'k'?
 
mghi + kxi^2 = mghf + kxf^2 ??
 
  • #10
lking226 said:
mghi + kxi^2 = mghf + kxf^2 ??

Initially the spring is not extended, so what does the left side reduce to? At the final stage, the final height is zero. So the entire equation simplifies to?
 
  • #11
so then just

mghi = kxf^2
 
  • #12
lking226 said:
so then just

mghi = kxf^2

Right, so what is k equal to now?
 
  • #13
4.50 kg x 9.8 m/s x 50.0 m = k x 0.340m
k = 6485.3 ?
 
  • #14
lking226 said:
4.50 kg x 9.8 m/s x 50.0 m = k x 0.340m
k = 6485.3 ?

That should be

[tex]4.5 kg \times 9.81 m/s^2 \times 50m = \frac{1}{2}k(0.34m)^2[/tex]


Solve again for k
 
  • #15
38187.71?
 
  • #16
lking226 said:
38187.71?

That looks correct to me
 
  • #17
Thank you so much!
 
  • #18
To nit-pick, the ball doesn't fall 50.0 meters, it falls 50.34 m after compressing the spring.
 

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