Does a Compressed Vertical Spring Only Store Elastic Potential Energy?

AI Thread Summary
A compressed vertical spring stores elastic potential energy, which is calculated using the formula Ee = 0.5kx^2. When the spring is released, this energy converts into kinetic energy, but gravity acts against this kinetic energy as the object moves upward. To determine how high an object will rise after being released from the spring, the gravitational potential energy equation, E = mgh, can be used, where h represents height. The relationship between the spring's potential energy and the gravitational potential energy is established by equating them, leading to the formula h = 0.2J/mg. The discussion confirms that while the equations represent different forms of energy, they are interconnected in the context of the spring's release.
bullroar_86
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if a vertical spring is being compressed down, is the Elastic potential energy the only energy?

Ee = .5kx^2

is gravity included in that? or does it not play a role?

and how could I tell how far an object would fly up in the air once the spring is released?


thanks
 
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Thats the energy with regards to the spring's compression.
The spring's potential energy will be translated to kinetic energy, how does gravity affect this energy/
 
well i know that the force of gravity will work against the kinetic energy...


I have an elastic potential energy of .2 J, so the kinetic energy will also be .2 J (up) ..

I know the mass of the item, and g, so I know that

Fg = mg


just not sure how to put the two together and get a height out of it.
 
express the gravitation in terms of energy, the .2J would result in a change of gravitational potential energy.
 
whozum said:
express the gravitation in terms of energy, the .2J would result in a change of gravitational potential energy.


ahh ok so,

.2J = mgh

and

h = .2J/mg
 
whozum said:
Thats the energy with regards to the spring's compression.
The spring's potential energy will be translated to kinetic energy, how does gravity affect this energy/

well I'm still a little confused, the equation E = mgh is still a potential energy equation, not the kinetic energy equation (E = .5mv^2)


is the above solution correct? (previous post)
 
bullroar_86 said:
well I'm still a little confused, the equation E = mgh is still a potential energy equation, not the kinetic energy equation (E = .5mv^2)


is the above solution correct? (previous post)

What's confusing about it?
The solution looks fine to me.
 
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