How High Will a Mass Fly on a Vertical Spring?

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A vertical spring with a constant of 200 N/m compresses by 0.0245 m when a 500-g mass is placed on it. When the mass is pushed down an additional 0.0755 m and released, the discussion revolves around how high the mass will fly. The potential energy stored in the spring converts to gravitational potential energy as the mass rises. The mass reaches its highest point when its kinetic energy is zero, meaning all energy is now gravitational potential energy. The equilibrium position is identified as the point of departure for the mass.
azizlwl
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A vertical spring with constant 200N/m has a light platform on its top. When a 500-g mass is set on the platform, the spring compresses 0.0245m. The mass is now pushed down 0.0755m farther and released. How far above this latter position will the mass fly?

The answer from the book.
If it does, Us at start=Ug at end, where zero Ug is at its lowest position.
200(0.10)2/2=0.5(9.8)h
......
I'm not sure if i intepret it correctly.
Initial spring energ=highest level it goes.

But if the KE energy is transformed to PE, what additional energy to make it 'fly'?

Thank You.
 
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The spring PE is converted to KE and the KE is converted into gravitational PE. Since the KE is 0 at the beginning and end you can just ignore it and consider a straight transfer of spring PE to gravitational PE.
 
azizlwl said:
But if the KE energy is transformed to PE, what additional energy to make it 'fly'?
When it reaches the highest point, its KE is zero so everything is gravitational PE. Of course, at the intermediate points where the mass is rising the mass has KE, all of which can be traced to the original spring PE.

Dale beat me to it!
 
Yes I really interpret it wrongly.
I thought the question is the height at which the object leave the spring.

Can we calculate the point/height of departure?
 
Is it because it is at the highest velocity, equilibrium position or of any other reasons?
 
It is the equilibrium position.
 
Thanks.
 
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