# Elastic strength: what is it, formulas and exercises

The **spring force** exerts force on an object to resist a change in its shape. It manifests itself in an object that tends to regain its shape when under the action of a deforming force.

Elastic force is also called restoring force because it opposes deformation to return objects to their equilibrium position. The transfer of elastic force takes place through the particles that make up objects.

For example, when a metal spring is compressed, a force is exerted that pushes the particles in the spring, reducing the separation between them; at the same time, the particles resist being pushed by exerting a force against compression.

If, instead of compressing the spring, it is pulled, stretching, the particles that make it up are further apart. Likewise, the particles resist separation by exerting a counter-elongation force.

Objects that have the property of recovering their original shape, opposing the force of deformation, are called elastic objects. Springs, elastics and elastic cords are examples of elastic objects.

__What is the elastic force?__

__What is the elastic force?__

The elastic force ( *F ** _{k}* ) is the force that an object exerts to recover its natural equilibrium state, being affected by an external force.

To analyze the elastic force, the ideal spring mass system will be taken into account, which consists of a spring placed horizontally attached from one end of the wall to the other and a block of negligible mass. The other forces acting on the system, such as the friction force or the force of gravity, will not be taken into account.

If a horizontal force is exerted on the mass, directed towards the wall, it is transferred to the spring, compressing it. The spring moves from its balanced position to a new position. As the object tends to remain in balance, the elastic force manifests itself in the spring that opposes the applied force.

The displacement indicates how much the spring deforms and the elastic force is proportional to this displacement. As the spring is compressed, the variation in position increases and the elastic force increases accordingly.

The more the spring is compressed, the more opposing force it exerts until it reaches a point where the applied force and the elastic force are balanced; consequently, the spring mass system stops moving. When you stop applying force, the only force that acts is elastic force. This force accelerates the spring in the opposite direction to the deformation until the equilibrium state is restored.

Likewise, it occurs when stretching the spring, pulling the mass horizontally. The spring is stretched and immediately exerts a force proportional to the displacement opposite the elongation.

__formulas__

__formulas__

The elastic force formula is expressed by Hooke’s Law. This law establishes that the linear elastic force exerted by an object is proportional to the displacement.

*F *_{k}* = -k.Δ s* [1]

*F *_{k}* =* tensile strength

*k* = constant proportionality

*Δ s* = displacement

When the object moves horizontally, as in the case of a spring attached to the wall, the displacement is *Δ x* and the expression of Hooke’s Law is written:

*F *_{k}* = -k.Δ x* [2]

The negative sign in the equation indicates that the spring force is in the opposite direction to the force that caused the displacement. The proportionality constant *k* is a constant that depends on the type of material from which the spring is made. The unit of the constant *k* is *N/m* .

Elastic objects have an elastic limit that will depend on the deformation constant. If it exceeds the elastic limit, it permanently deforms.

Equations [1] and [2] apply to small spring displacements. When the displacements are larger, terms with a greater power of *Δ ***x *** are added* .

**Kinetic energy and potential energy referred to an elastic force**

Elastic force does the work on the spring, moving it into its balanced position. During this process, the potential energy of the spring’s mass system increases. The potential energy due to the work done by the elastic force is expressed in equation [3].

*U = ½ k. **Δx *^{2}* *[3]

Potential energy is expressed in Joules (J).

When deformation force is not applied, the spring accelerates toward the equilibrium position, reducing potential energy and increasing kinetic energy.

The kinetic energy of the spring’s mass system, when it reaches the equilibrium position, is determined by equation [4].

*E *_{k}* = ½ mv *^{2}* *[4]

*m* = mass

*v* = spring speed

To solve the spring mass system, Newton’s second law is applied, taking into account that the elastic force is a variable force.

__Practical examples of exercises__

__Practical examples of exercises__

**getting deformation force**

How much force is needed to apply a spring to stretch 5 cm if the spring constant is 35N / m?

As the application force is opposite to the spring force, *F ** _{k}* is determined by assuming that the spring is stretched horizontally. The result does not require the minus sign, as only the application force is needed.

Hooke’s Law

*F *_{k}* = -k.Δx*

The spring constant *k* is *35N/m.*

*Δ **x = 5cm = 0.05m*

*F *_{k}* = -35N / m. 0.05m*

*F *_{k}* = – 1.75N = – F*

It takes *1.75 N* of force to deform the *5 cm* spring .

**Getting the deformation constant**

What is the deformation constant of a spring that extends *20 cm* under the action of a force of *60N* ?

*Δx = 20cm = 0.2m*

*F* = 60N

*F *_{k}* = -60N = – F*

*k = – F *_{k}* / Δx*

*= – (- 60N) / 0.2 m*

*k = 300 N / m*

The spring constant is *300N / m*

**Obtaining potential energy**

What is the potential energy related to the work done by the elastic force of a spring that compresses *10cm* and its strain constant is *20N/m?*

*Δ **x = 10 cm = 0.1 m*

*k = 20 N / m*

*F *_{k}* = -20N / m. 0.1m*

*F *_{k}* = -200N*

Spring tensile strength is *-200N.*

This force does the work on the spring to move it into its balanced position. Doing this work increases the potential energy of the system.

Potential energy is calculated with equation [3]

*U = ½ k. **Δx *^{2}

*U = 1/2 (20N / m). (0.1m) *^{2}

*U = 0.1 Jules*