Conservation of energy spring problem

AI Thread Summary
The problem involves a block of mass 2.00 kg attached to a spring with a force constant of 500 N/m, pulled to 0.05 m and released from rest. The total mechanical energy of the system remains constant due to the absence of friction, expressed as E = 1/2mv^2 + 1/2kx^2. To find the speed of the block as it passes the equilibrium position, the equation simplifies to 1/2mv^2 = 1/2k(x_o)^2, where x_o is the initial stretch of the spring. The elastic potential energy is considered at the initial position, and the kinetic energy is calculated at the equilibrium position. This approach confirms that the conservation of energy principle is applied correctly in solving the problem.
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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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