Energy Minimisation

Problem Statement

A spring of stiffness 100 N/m is stretched by a 1 N force. What is the minimum total energy of the system?

Mathematical Model

The total energy of a spring-force system consists of:

• The potential energy stored in the spring
• Minus the work done by the external force

For a spring with stiffness $k$ stretched by a distance $x$ under a force $F$:

Total energy: $E=\frac{1}{2}k{x}^2-Fx$

Where:

• $k$ is the spring stiffness (100 N/m in this problem)
• $x$ is the extension of the spring
• $F$ is the applied force (1 N in this problem)

Physical Interpretation

The first term $\frac{1}{2}k{x}^2$ represents the elastic potential energy stored in the spring.

The second term $Fx$ represents the work done by the external force, which is subtracted because this energy is supplied to the system rather than stored within it.

Optimisation Setup

Design Variable

• $x$ = extension of the spring (scalar, 1-dimensional problem)

Objective Function

We want to minimize the total energy:

$L(x):=E=\frac{1}{2}k{x}^2-Fx$

Solution Approach

To find the extension that minimizes the total energy, we differentiate the energy function with respect to $x$ and set it to zero:

$\frac{dE}{dx}=kx-F=0$

Solving for $x$:

$x^{\ast }=\frac{F}{k}$

For our specific problem with $k=100$ N/m and $F=1$ N:

$x^{\ast }=\frac{1}{100}=0.01\text{ m}=1\text{ cm}$

Second Derivative Test

To verify this is indeed a minimum, we check the second derivative:

$\frac{d^{2}E}{d{x}^2}=k=100>0$

Since the second derivative is positive, $x^{\ast }=0.01$ m is indeed a minimum.

Minimum Energy Value

The minimum energy is:

$E_{\min }=\frac{1}{2}k{x}^{\ast 2}-F{x}^{\ast }=\frac{1}{2}\cdot 100\cdot (0.01{)}^2-1\cdot 0.01=0.005-0.01=-0.005\text{J}$

The negative energy value indicates that the system configuration at equilibrium is at a lower energy state than the reference configuration (unstretched spring).

Static Equilibrium Connection

This result is consistent with the principle of static equilibrium, where the spring force equals the applied force:

$kx=F$

At equilibrium, the net force is zero, and the system is in a minimum energy state - a fundamental principle in mechanics.

Alternative Derivation

We can also approach this problem from the perspective of virtual work. At equilibrium, the virtual work done by all forces for any virtual displacement must be zero.

For a small displacement $\delta x$:

• Work done by spring force: $-kx\cdot \delta x$
• Work done by external force: $F\cdot \delta x$

Setting the total virtual work to zero: $(-kx+F)\cdot \delta x=0$

Since $\delta x$ can be arbitrary, we need $-kx+F=0$, giving the same result as before.

Generalization: Potential Energy Minimization

This problem illustrates a fundamental principle in mechanics: a system in stable equilibrium minimizes its potential energy.

For a general mechanical system with potential energy $V(q)$ where $q$ represents generalized coordinates:

1. Equilibrium positions occur at critical points: $\nabla V(q)=0$
2. Stable equilibrium occurs at local minima: ${\nabla }^{2}V(q)$ is positive definite

Practical Applications

This principle of energy minimization has wide applications:

• Structural analysis and design
• Molecular configuration in chemistry
• Position control in robotics
• Shape optimization in engineering design

Key Insights

1. The minimum energy principle provides an elegant way to find equilibrium configurations
2. The solution has a clear physical interpretation: the spring extends until its restoring force balances the applied force
3. The negative energy value indicates work has been done on the system
4. This approach can be extended to more complex mechanical systems with multiple degrees of freedom
5. The energy minimization approach is often more straightforward than force balance for complex systems