You can calculate the additional mass yourself and then consider whether it may be measurable.RobbyQ said:TL;DR Summary: Mass Energy and a compressed spring
Reference: https://www.physicsforums.com/forums/high-energy-nuclear-particle-physics.65/post-thread
If I take a spring with clamps and I weight that system accurately. Then I compress the spring and clamp it thus giving it potential energy. If I now weigh the clamped spring I should see an increase in mass because of the added energy. Is this the case and something that could be proved in the lab ?
Yes understandable PeroK. Hence why I wondered if this could be measured even. But from a mass/energy equivalence is it conceptually correct i.e if I were to convert both the uncompressed and compressed systems into pure energy would I observe the added PE difference?PeroK said:PS even tiny variations in the surface gravity over time may be more significant.
There's no such thing as "pure energy". Energy is a property of a system.RobbyQ said:Yes understandable PeroK. Hence why I wondered if this could be measured even. But from a mass/energy equivalence is it conceptually correct i.e if I were to convert both the uncompressed and compressed systems into pure energy would I observe the added PE difference?
And the mass deficit is the binding energy? So I cannot think of the spring system in this way?PeroK said:There's no such thing as "pure energy". Energy is a property of a system.
The proof of "mass-energy equivalence" is that the rest mass of the Hydrogen atom is less than the rest mass of the proton plus the rest mass of the electron. This has been measured to a very high degree.
The formation of a hydrogen atom involves loss of PE. Hence, a reduction in rest mass.RobbyQ said:And the mass deficit is the binding energy?
Yes, but it's too small to measure relative to the overall mass of the system in a macroscopic environment.RobbyQ said:So I cannot think of the spring system in this way?
What experimental setup do we use to measure PE loss in Hydrogen atom formation. Or, in fact, the PE needed to separate the electron and proton until they reach their rest mass.PeroK said:The formation of a hydrogen atom involves loss of PE. Hence, a reduction in rest mass.
Yes, but it's too small to measure relative to the overall mass of the system in a macroscopic environment.
If you jump off the ground, then technically the centre of mass of the Earth moves a tiny amount in the opposite direction. But, that's too small to measure, especially given that lots of other things are happening all the time on the Earth.
Google is your friend, as they say.RobbyQ said:What experimental setup do we use to measure PE loss in Hydrogen atom formation. Or, in fact, the PE needed to separate the electron and proton until they reach their rest mass.
Is it an implementation of the Einstein's Box (which I understand was a thought experiment)?
It's true that you have more energy available in a compressed spring. Try using a compressed spring to launch a toy car.RobbyQ said:But from a mass/energy equivalence is it conceptually correct i.e if I were to convert both the uncompressed and compressed systems into pure energy would I observe the added PE difference?
The relationship between mass and energy in a compressed spring is described by Hooke's Law, which states that the potential energy stored in a spring is directly proportional to the square of its displacement from its equilibrium position. This means that as the mass of the object attached to the spring increases, so does the potential energy stored in the spring.
The compression of a spring increases its potential energy. This is because as the spring is compressed, it stores potential energy in the form of elastic potential energy. The more the spring is compressed, the more potential energy it can store.
Yes, the mass of an object can affect the compression of a spring. According to Hooke's Law, the amount of compression in a spring is directly proportional to the force applied to it. Therefore, a heavier object will cause more compression in the spring compared to a lighter object.
The compression of a spring increases its stiffness. This is because the more a spring is compressed, the more force it exerts to return to its original length. This increase in force is known as the spring's stiffness or spring constant, and it is directly proportional to the amount of compression.
The formula for calculating the potential energy of a compressed spring is PE = 1/2 * k * x^2, where PE is the potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position. This formula is derived from Hooke's Law and is applicable as long as the spring remains within its elastic limit.