Have you ever noticed that the spring regains its normal shape despite the force you put while compressing or stretching it? Why do you need to exert extra stress to change a springs position? The secret is the stored Spring potential energy. The physics behind the work, energy, and force of elastic substances like springs! Let’s understand the Spring potential energy.

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**Spring potential energy**

When you compress or stretch a spring, as soon as the stress is relieved, the spring attains its normal shape instantly. Its Elastic potential energy helps it do so. Generally, these elastic substances follow the Hooke’s law.

### Hooke’s Law

Before unveiling the mechanism of spring potential energy, we need to understand the Hook’s Law. According to this Law, the force needed to change the shape of spring is proportional to the displacement of the spring. The displacement referred here is how far the spring is compressed or stretched from its normal shape. Mathematically, Hook’s Law can be summarised as F= – k x.

Here ‘k’ is the spring constant and ‘x’ is the displacement. This spring constant being unique for different springs depends on various factors like the material of the spring and the thickness of the coiled wire used in the spring. Hooke’s law is frequently represented in the negative form since the force is a restoring force, Â but the positive version of the law is also a valid representation.

**Browse more Topics under Work Energy And Power**

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- Work-Energy Theorem
- Various Forms of Energy: The Law of Conservation of Energy

### Spring potential energy

To find the Spring potential energy, we need to use the Hooke’s law. Since the potential energy is equal to the work done by a spring and work, in turn, is the product of forceÂ and distance, we get our force from Hooke’s law. Distance here is the displacement in the position of the spring.

In the figure, x is the displacement from the equilibrium position. When we pull the spring to a displacement of x as shown in the figure, the work done by the spring is :

WÂ =Â _{0}âˆ«^{xm}Â Fdx = -âˆ«kx dx = -k(x_{m})^{2}/2

Â The work done by pulling force F_{pÂ } is :

FpÂ = k (x_{m})^{2} /Â 2

The work done by the pulling force F_{pÂ }is in positive as it has overcome the force of spring. Therefore,

W=Â k *x*_{m2} /Â 2

When displacement is less than 0, the work done by the springs force is

W_{sÂ }= – kx_{cÂ }^{2}Â / 2

and the work done by the external force F is = + kx_{c}^{2}/2. In the process of displacement of the object from initial displacement x_{iÂ } to final displacement, x_{fÂ } the work done is,

W_{sÂ }= –Â _{x}_{f}âˆ«^{xiÂ }kx dx = Â k x_{i}^{2}/2 –Â k x_{f}Â ^{2}/2

From the equation, it is clear that the work done by the force of spring depends only on the endpoints of displacement. Also, we can see that in a cyclic process, the work done by the springs force is zero. Hence, we can say that the spring force is a conservative force because it depends on the initial and final positions only. Therefore, this work done is in the form of the Spring potential energy.

## Solved Examples For You

Q: If a spring extends by x on loading, then the energy stored by the spring is (Given T is the spring force and K is force constant):

A) 2x/T^{2}Â Â Â Â B) T^{2}/2kÂ Â Â Â C) 2k/T^{2}Â Â Â Â D) T^{2}/2x

Solution: B) Force or tension, T = kx [Hooke’s Law]. Hence, x = T/k

Energy stored in the spring = *k x ^{2}/2 = T^{2}/2k*

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