When deforming any spring the deforming force is always greater than the restoring force until equilibrium is reached. So, if a constant deforming force caused an extension in any spring the restoring force will increase until it becomes equal to the restoring force and equilibrium is reached. However, I do not understand how the restoring force even increases with extension. I imagine it is related to the fact that the more we deform a spring the more potential energy it has, but I do not know how exactly this affects the restoring force. An explanation which includes the behaviour of the bonds and molecules in any spring during the deformation is appreciated.
Forces – Why is the Restoring Force Directly Proportional to Extension in Springs and Elasticity?
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It is possible because the velocity goes from a negative velocity to a positive velocity (depending on how you chose the axis). The object has a velocity towards the ground, but due to a force in the opposite direction, the object decelerates to zero. At zero velocity, there is still a net force acting on the object due to the spring, so the object will accelerate in the opposite direction.
Try to compare it with this: you thrown an object vertically upwards with a velocity $v_{0}$, so the only force acting on the object is gravity downwards with an acceleration $-g$. At a certain point, the object will slow down to a velocity of $0$ and then fall back downwards to earth. In this whole process, the acceleration is constant, but the velocity is still zero at its highest point.
More mathematically: $a = -g = \frac{dv}{dt}$. This is a simple differential equation and can be solved easily with integrals: $-gdt = dv$ so $\int_{t_0}^{t}-gdt = -g\int_{t_0}^{t}dt = -g\cdot(t-t_0) = -g\cdot t$ if we take $t_0 = 0s$ and also $\int_{v_0}^vdv = v(t)-v_0$, and so we get $-g\cdot t = v(t)-v_0$ or $v(t) = v_0 - g\cdot t$. Acceleration $a$ is constant and thus $a(t) = -g$. Now $v(t)$ will be equal to zero for $t = \frac{v_0}{g}$, so $v(\frac{v_0}{g}) = 0$ but also $a(\frac{v_0}{g}) = -g \neq 0$.
A mathematical equation like this can also be deducted for your spring problem, and then you'll see that the velocity will in fact be zero when the acceleration is maximum, and that the velocity will be maximum for zero (net) acceleration. This is known as a harmonic oscillator but requires some knowledge of differential equations.
A simple example of a constant external force being applied to a spring-mass system is the force of attraction $mg$ on a mass $m$ is a gravitational field of strength $g$.
Release the mass at the end of an unstretched spring, then when the spring has been stretched by an amount $x$ the work done by the external force (gravitational attraction) is $mgx$.
The elastic potential energy stored in the spring is $\frac 12 kx^2$ where $k$ is the spring constant.
The difference between these two quantities, $mgx - \frac 12 kx^2$, is the increase in kinetic energy of the mass.
Eventually the constant external force will be smaller than the force exerted by the spring and the mass will slow down and finally stop.
This will happen when $mgx_{\rm stop} = \frac 12 k x_{\rm stop}^2$
At this position all the work done by the external force is stored as elastic potential energy.
This example is no more than the oscillation of a mass at the end of a spring but noting that $x$ is the total extension of the spring and not the extension of the spring from the static equilibrium position.
Best Answer
You can think of two atoms in a solid joined together via a bond.
The graph of potential (pe) energy against separation of the atoms is shown below as is a graph of the force between atoms and their separation.
The equilibrium separation $r_o$ is when the potential energy is a minimum and the force between the atoms is zero. In reality since quantum harmonic oscillators have non-zero amplitude even in ground state due to a finite zero point energy the equilibrium is a bit beyond $r_0$ due to the asymmetry of the potential, but we can ignore that effect here.
Now it so happens that for a lot of bonds the graph of force against separation is a straight line about the equilibrium position as shown in mauve in the lower graph. This is equivalent to saying that a parabola fits quite well to the shape of the upper plot near equilibrium.
So in this region you can think of the atoms being connected by springs which obey Hooke's law with the spring unextended when the separation of the atoms is $r_o$.
Trying to increase the separation (stretching the solid) stretches the bond and results in an attractive force between the atoms which is proportional to the increase in the separation from their equilibrium position.
This attractive force summed over all the bonds will then become equal and opposite to the external force which caused the bonds to stretch.
A similar thing happens when the solid is compressed but now there is a force of repulsion which might be thought of as the negative electron shells between adjacent atoms repelling one another.
In terms of energy the external force stretching a solid does work extending the bond and that energy is stored as potential energy within the bond.
When the external force is removed the atoms move towards the minimum of potential energy, their equilibrium separation.