Conservation of energy spring problem

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SUMMARY

The conservation of energy spring problem involves a block of mass m=2.00 kg attached to a spring with a force constant κ=500 N/m, initially pulled to a position of xi=0.05 m and released from rest. The speed of the block as it passes through the equilibrium position can be determined using the equation 1/2mv^2 = 1/2k(x_o)^2, where x_o represents the initial stretch of the spring. The total mechanical energy remains constant in a frictionless system, allowing for the calculation of speed without needing to separately account for elastic potential energy at equilibrium.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Familiarity with the principles of conservation of energy
  • Knowledge of kinetic and potential energy equations
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of the conservation of mechanical energy in oscillatory systems
  • Learn about the effects of friction on spring systems
  • Explore advanced topics in harmonic motion and energy transfer
  • Investigate real-world applications of spring dynamics in engineering
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Students in physics courses, educators teaching mechanics, and anyone interested in understanding the principles of energy conservation in spring systems.

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Homework Statement


A block of mass m=2.00 kg is attached to a spring of force constant κ=500 N/M. the block is pulled to a position xi=.05m and released from rest. Find the speed the block has as it has passed equilibrium. Assume frictionless.

Homework Equations


ΔEsystem=ƩT(transfer)
Δk+ΔU=W

The Attempt at a Solution


I am assuming that this is a non isolated system.So I am having trouble setting up this equation I know work done by spring=1/2kX^2 and I know what Δk is, but Do I need to include the difference in elastic potential? I have seen this worked out without the elastic potential. why is it not included? and they just solved 1/2mv^2=1/2kx^2?
 
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You know that when the mass vibrates and there is no friction the total energy of the system remains constant. The total energy E is,

E = 1/2mv^2 + 1/2kx^2 = constant = 1/2k(x_o)^2 where x_o is the initial amount the spring is stretched.

1/2mv^2 + 1/2kx^2 = 1/2k(x_o)^2 but we want to know v when x = 0 so,

1/2mv^2 = 1/2k(x_o)^2

So use the last equation you wrote but remember x is the initial amount the spring was stretched.
 

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