Friend pulls the spring without you looking

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SUMMARY

The discussion revolves around a physics problem involving a 0.500 kg mass attached to a spring with a spring constant of 45 N/m. When the mass is pulled and released, it reaches a speed of 3.375 m/s at the equilibrium position. The solution employs the law of conservation of energy, equating elastic potential energy (Ee) and kinetic energy (Ek) to find the displacement (x) from the equilibrium position using the formula 1/2(k)(x)^2 = 1/2(m)(v)^2.

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Homework Statement


0.500kg mass resting on frictionless surface is attached to a horizontal spring with a spring constant of 45N/m. When you are not looking, your lab partner pulls the mass to one side and then releases it. When it passes the equlibrium position, its speed is 3.375m/s. How far from the equilibrium position did your lab partner pull the mass before releasing it?


Homework Equations


Elastic Potential Energy
Kinetic Energy


The Attempt at a Solution


What is confusing me is how I am going to apply the law of conservation of energy here. Could I say that whilst he is pulling the mass, only Ee is present and not Ek?

I was thinking I would do something like this to end up getting the "x" value needed:

1/2(k)(x)^2 = 1/2(m)(v)^2 <-- then solve for x, which is the amount of stretch or compression..
 
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aeromat said:
What is confusing me is how I am going to apply the law of conservation of energy here. Could I say that whilst he is pulling the mass, only Ee is present and not Ek?
Assume that when he releases the mass it starts from rest.

I was thinking I would do something like this to end up getting the "x" value needed:

1/2(k)(x)^2 = 1/2(m)(v)^2 <-- then solve for x, which is the amount of stretch or compression..
Sounds good to me.
 

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