# Expression of Potential Energy stored in a Spring

Statement: Derive the expression of Potential Energy stored in a spring, U = \frac{1}{2} k x^2

Solution:

Elastic potential energy (U) is potential energy stored as a result of deformation of an elastic object, such as the stretching of a spring. It is equal to the work done (W) to stretch the spring, which depends upon the spring constant k as well as the distance stretched i.e. U = W

Let the spring be stretched trough a small distance dx.
Then work done in stretching the spring through a distance dx is,

dW = F dx ...................................... (1)

Where, F is the force applied to the stretch the spring.

Total work done in stretching the spring from the interval x = 0 to x = x is obtained by integrating the expression:

\int dW=\int_{0}^{x}Fdx..............................(2)

Hooke’s Law states that, the elongation produced in an ideal spring is directly proportional to the spring force. That is,

F = - kx ......................................(3)

Here, k is called the spring constant.

Substituting equation (iii) in (ii) we get,

Work done by spring force,

W=\int_{0}^{x} -kxdx=-k\int_{0}^{x}xdx=-k [\frac{x^2}{2}]_0^x=-\frac{1}{2}kx^2 ............................ (4)

And the work done by external force = \frac{1}{2} kx^2

This work done to deform the spring is nothing but the elastic potential energy of the spring.

Hence, U = \frac{1}{2} k x^2

[Derived/Proved]