# Elastic potential energy and springs

The formula for elastic potential energy(for a spring) has been derived by assuming the following things:

1.Work done by a deforming force on a spring from a relaxed state (state where spring is not deformed) to another point would be equal to the elastic potential energy gained by the spring from the relaxed state to that point.

Using this assumption we derived elastic potential energy as: U = $$1/2$$ Kx$$^2$$

Now let us say that we apply a constant deforming force on a block connected to the spring, now there will be a varying spring force being acted on the block to counter the constant deforming force. Once this deforming force gets greater then the constant deforming force then the deformation of the spring would stop and then the spring will gradually regain its shape.

Now generally it is assumed that the work done by the deforming force from the moment it starts to act to the moment that the deformation of the spring stops is equal to the elastic potential energy gained by the spring.

that is, Initial kinetic energy of spring = final elastic potential energy of spring


But unless the restoring spring force gets larger in magnitude then the constant deforming force the kinetic energy of the spring would keep on increasing and also due to deformation some elastic potential energy will be increasing. Let a point 'A' describe this situation

It is said that the total mechanical energy is conserved in such a system, but in such a situation mechanical energy is not being conserved.

because it is assumed that the kinetic energy when a constant deforming force is just applied is equal to the final elastic potential energy therefore energy is conserved in these two situations but when we compare the total mechanical energy in either of these two situations with the situation at point A then the total energy is not constant(or conserved).

Therefore how can we say that mechanical energy is conserved in such a system? and if it isn't conserved then how we define elastic potential energy to be = $$1/2$$ Kx$$^2$$

• Why would you expect a system's mechanical energy to be conserved when an external force is applied? Conservation of mechanical energy only applies in the absence of external forces. For a trivial counterexample, consider a block being pushed by a constant external force across a frictionless table. The block's kinetic energy increases while its potential energy stays the same. – probably_someone Dec 7 '18 at 15:09

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.

The net force acted on the block is: $$F(x)=-kx+F_0$$ Then work done $$∆W=\int {F(x)dx}$$ $$=\frac {1}{2} k {x}^2+F_0x$$ But we can still say that elastic potential energy is $$\frac {1}{2} k x^2$$.because the only central force acted towards the center is $$-kx$$.

So the change of the mechanical energy (extra change of the kinetic energy) of the system is therefore $$∆E=F_0x$$. That is ,total energy is conserved,but the energy of the block spring system is increasing. You can't expect the total mechanical energy of the block-spring system to be conserved because you are applying an external force on the system.

So , first of all you need to understand that only "internal" conservative forces changes potential energy as: $$-W_{internal} = ΔU \tag{1}$$ Note:- I am considering non-conservative internal forces to be zero.

Note when we write Work energy theorem , the $$W_{net}$$ includes both $$W_{internal}$$ and $$W_{external}$$. $$W_{internal} + W_{external} = ΔKE$$

By using $$(1)$$: $$W_{external} = ΔU + ΔKE \tag{2}$$

If $$W_{external}$$ is zero then we say total energy of system is conserved.

Now let's consider spring as our system in which let $$W_{internal}$$ mean work done by internal particles of spring on each other due to elastic forces and the block applies the force $$kx$$ on the spring which is $$W_{external}$$. As spring is massless this implies $$ΔKE$$ = 0 ,So we could write work energy theorem for the spring as: $$W_{internal} + W_{external} = 0$$ Using (1) $$W_{external} = ΔU_{spring}$$

By calculation , $$W_{external}$$ turn outs to be $$kx^2/2$$. $$ΔU_{spring} = kx^2/2$$

Now you are asking why mechanical energy is not conserved , Can you tell why it need to be conserved as external force $$kx$$ is acting on the spring and this could be verifed from $$(2)$$.