Energy and Simple Harmonic Motion of spring

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SUMMARY

The discussion focuses on the energy dynamics of a 2.00-kg object suspended from a vertical spring with a spring constant of 50.0 N/m. The object is pulled 0.200 m downward and released, prompting calculations for translational kinetic energy, gravitational potential energy, elastic potential energy, and total mechanical energy at various heights. The correct approach involves using the conservation of mechanical energy principle, specifically the equations K = 0.5mv², Ug = mgh, and Us = 0.5kx². The elastic potential energy at the initial position requires accounting for the spring's equilibrium stretch before adding the additional displacement.

PREREQUISITES
  • Understanding of simple harmonic motion principles
  • Familiarity with energy conservation laws in physics
  • Knowledge of spring constants and Hooke's Law
  • Ability to perform calculations involving kinetic and potential energy
NEXT STEPS
  • Calculate the equilibrium stretch of the spring using mg = kΔLe
  • Explore the concept of total mechanical energy in oscillatory systems
  • Learn about the relationship between kinetic energy and velocity in simple harmonic motion
  • Investigate the effects of varying spring constants on oscillation behavior
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators looking for practical examples of simple harmonic motion and energy calculations.

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Homework Statement



A 2.00-kg object is hanging from the end of a vertical spring. The spring constant is 50.0 N/m. The object is pulled 0.200m downward and released from rest. Complete the table below by calculating the translational kinetic energy, the gravitational potential energy, the elastic potential energy, and the total mechanical energy E for each of the vertical positions indicated. The vertical positions h indicate distance above the point of release, where h = 0.

h (meters) KE PE(gravity) PE(elastic) E
0
0.200
0.400

Homework Equations



Ef = E0 (Conservation of mechanical energy)

K = .5mv2 Us = .5kx2 Ug = mgh

The Attempt at a Solution



I got all of the gravitational potential energies ( 0 , 3.92J , 7.84J) using Ug = mgh and KE at 0, where v = 0. For the elastic potential energy at 0, I thought that you would be able to use .5kx2 with x = 0.200m, but that doesn't give the right answer (which is 8.76 J). I think the answer has something to do with the equation for conservation of mechanical energy, but what I tried didn't yield a correct answer or even make much sense. I'm probably missing something obvious, but I would appreciate some help.
 
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Taciturn Type said:

Homework Statement



A 2.00-kg object is hanging from the end of a vertical spring. The spring constant is 50.0 N/m. The object is pulled 0.200m downward and released from rest.


For the elastic potential energy at 0, I thought that you would be able to use .5kx2 with x = 0.200m,

The elastic potential energy is kΔL2/2, where ΔL is the length the spring is stretched with. The spring was stretched by the hanging object already, and stretched further by 0.2 m. Calculate the equilibrium ΔLe from mg=kΔLe, and add 0.2 m to get the initial ΔL.


ehild
 

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