Potential Energy of Spring: Overview, Questions, Preparation

Potential energy is a kind of energy stored in a system or object due to the virtue of position. For example, when you stand at some height (h), provided your mass is m, your potential energy will be mgh (that is equal to the work done to reach there).

As per the NCERT Textbooks," The term potential energy in simple terms means ‘stored’ energy. Physically, the notion of potential energy is applicable only to the class of forces where work done against the force gets ‘stored up’ as energy. When external constraints are removed,it manifests itself as kinetic energy.

Mathematically,

 

There are mainly three types of potential energy in the class 11 syllabus, namely;

• Gravitation Potential Energy
• Electrical Potential Energy
• Elastic Potential Energy

This article deals with elastic potential energy, especially the potential energy of a spring when stretched or compressed.

Whenever any elastic object is deformed, it stores energy to regain its original/equilibrium position. This stored energy is called Elastic potential energy. During the compressed or stretched condition, a spring also stores elastic potential energy and releases it as soon as the spring returns to its original shape.

This potential energy is always equivalent to the work done to deform its shape. Potential energy is denoted by U. The potential energy (U) stored in a spring, provided it is displaced by a distance x from the equilibrium position or original length, will be;

$U=\frac{1}{2}k{x}^2$

Where,

$k$ = spring constant (in N/m )

$x$ = change in length of spring (positive or negative)

The SI unit of potential energy is the joule (J), and the dimensional formula is ${\mathrm{M}}^1{\mathrm{L}}^2{\mathrm{T}}^{-2}$ .

The potential energy is always equal to the work done by the conservative force of the spring (equal and opposite to the applied external force).

Let's assume the length of the spring is displaced by $x\text{'}m$ (assuming in the x-axis) from initial position x. The conservative force exerted by the spring, also called the restoring force, is $\vec{F}$ .

According to Hooke's law:

$\vec{F}=k.\vec{x}=-kx\hat{i}$

We know, the work done is equal to the product of the force applied and the displacement due to the exerted force. The work done by the spring restoring force (using calculus method) is:

$W_{\text{spring }}={\int }_0^x \left(-k{x}^{\text{'}}\right)d{x}^{\text{'}}=-{\left[\frac{1}{2}k{x}^{\text{'}2}\right]}_0^x=-\frac{1}{2}k{x}^2$

Assuming initial potential energy: $U_i=0$ . The net change in potential energy (Negative of the work done by the conservative force):

$U_f-U_i=-W_{\text{spring }}=-(-\frac{1}{2}k{x}^2)\mathrm{}⟹U=\frac{1}{2}k{x}^2$

Here are a few important points related to the elastic potential energy of the system.

• The potential energy of a spring in its natural length is considered zero.
• The potential energy of a spring is always positive or zero. It can never be negative.
• The work done by the spring will only depend on the initial and final positions, not on the path/ process.
• The work done by the restoring force of the spring will always be negative or zero.
• The direction of the conservative force is always opposite to the displacement. That means it tends to restore the original shape and size.
• The potential energy of a spring (U) is directly proportional to the square of the displacement ${(\mathit{}\mathit{x}}^2)$ . That means the more the change, the larger the stored potential energy will be.

Work-energy theorem simply states that the net change in energy will be equal to the work done. So, for a system of springs, if there are no non-conservative forces involved, total mechanical energy will be conserved.

This concept states that the sum of kinetic and potential energy will remain the same for a spring system. The total mechanical energy at the initial and final positions will be equal:

$K_i+U_i=K_f+U_f⟹\frac{1}{2}m{v}_i^2+\frac{1}{2}k{x}_i^2=\frac{1}{2}m{v}_f^2+\frac{1}{2}k{x}_f^2$

This is a very useful concept for advanced-level problems asked in competitive exams like JEE and NEET.

The spring-mass system includes a spring attached to a block or several blocks & when stretched or compressed, the system starts oscillatory motion due to the continuous exchange of kinetic and potential energy.

Horizontal Spring-Mass Systems

The important fact to remember is that at equilibrium, potential energy is zero, and at maximum displacement, potential energy is maximum (kinetic energy is zero).

$U_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{1}{2}k{A}^2$

Vertical Spring-Mass Systems

The equilibrium position of the spring shifts lower due to gravitational pull in a vertical spring-mass system. The vertical energy systems are more complex since they include both elastic and gravitational forces. The formula for the shift of the equilibrium point in a vertical spring-mass system:

$x_0=\frac{mg}{k}$

Spring mass system with non-conservative forces

All non-conservative forces are dissipative. If any spring-mass system includes a non-conservative force, it dissipates the net energy. In most cases, friction is present as a non-conservative forces.  In this condition, the work done by friction must be included in the equation:

$W_f=-\mu mgs$

The work-energy theorem will include this work done by friction:

$W_{\text{spring }}+W_f=\mathrm{\Delta }K$

This friction reduces the kinetic energy, which affects the amplitude of the oscillations of the spring.

Question 1: A block of mass $2\mathrm{k}\mathrm{g}$ compresses a spring ( $k=100\mathrm{N}/\mathrm{m}$ ) by 0.3 m on a frictionless surface. If the spring is compressed by 0.1m, what will be the velocity once released?

Solution: Using conservation of mechanical energy:

$\frac{1}{2}k{x}_0^2=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$

$\frac{1}{2}\cdot 100\cdot {0.3}^2=\frac{1}{2}\cdot 100\cdot {0.1}^2+\frac{1}{2}\cdot 2\cdot v^2$

$4.5=0.5+v^2⟹v^2=4⟹v=2\mathrm{m}/\mathrm{s}$

Question 2:  A mass $m=1\mathrm{k}\mathrm{g}$ is attached to a spring $(k=50\mathrm{N}/\mathrm{m})$ and It stretches the spring by $x_0$ m. If pulled down an additional 0.1 m and released, find the maximum potential energy.

Solution:

The shift in the equilibrium of the spring: $x_0=\frac{mg}{k}=\frac{1\times 9.8}{50}=1.96m$

 

Net displacement: $x=0.196+0.1=0.296\mathrm{m}$

 

Hence, the elastic potential energy is:

$U=\frac{1}{2}k{x}^2=\frac{1}{2}\cdot 50\cdot {0.296}^2=2.19\mathrm{J}$