Where does the potential energy go if a clamped spring is dissolved in acid?

AI Thread Summary
When a spring is compressed, it stores potential energy, which is described by the equation P.E = (1/2)*k*a^2. Upon dissolving the spring in acid, the potential energy is believed to convert into kinetic energy associated with the molecular motion of the resulting ions and hydrogen gas. The dissolution process involves the formation of a salt and gaseous hydrogen, indicating a transformation of energy states. The movement of ions from the solid state into solution signifies a release of energy as they interact with the acid. Ultimately, the potential energy stored in the spring is transformed into kinetic energy during the chemical reaction.
PrakashPhy
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Homework Statement


When a spring is compressed, potential energy is stored in it. If the spring is dissolved into acid; where does teh potential energy go.

Homework Equations



P.E = (1/2)*k*a^2

The Attempt at a Solution


I tried but got no idea about it.
 
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The potential energy of an elastically deformed body is the extra energy due to inter atomic interactions when the ions are moved from their equilibrium positions in the crystal lattice.

When a metal reacts with an acid, we get a salt and gaseous hydrogen in molecular form. The salt is usually hydrated and totally ionized in the solution. Therefore, the end effect of this reaction is that ions of the solid have moved in the solution and hydrogen ions have received some electrons and bonded in molecules. I would guess the potential energy has converted into extra kinetic energy of the molecular motion.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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