Question

An elastic spring of unstretched length L and force constant K is stretched by a small length x. It is further stretched by another small length y. Work done during the second stretching is
A. $\dfrac{{ky}}{2}\left( {x + 2y} \right)$
B. $\dfrac{k}{2}\left( {x + 2y} \right)$
C. $ky\left( {x + 2y} \right)$
D. $\dfrac{{ky}}{2}\left( {2x + y} \right)$

Answer 1

Hint: Conservative force is known to have a property that the total work done in moving a particle from one point to another is independent of the path taken. To solve the given question, we will find the work done at the time of stretching, then find the difference between work done by the spring in first and second stretching, which would result in our required answer.

Formula used:
$W = \dfrac{1}{2}k{x^2}$

Complete step-by-step answer:
Elastic potential energy is defined as the potential energy stored as a result of stretching of the spring. It is equal to the work done to stretch the spring, which depends upon the distance stretched and the spring constant.
Mathematically it can be written as, $W = \dfrac{1}{2}k{x^2}$
Where k is the spring constant and x denotes the distance stretched.
Work done in first stretching will be,
${W_1} = \dfrac{1}{2}k{x^2}$
Since, it's already stretched x times then in second stretching y is added.
So, work done in second stretching will be,
${W_2} = \dfrac{1}{2}k{\left( {x + y} \right)^2}$
Electric force in a string is conservative in nature. i.e. $W = \Delta U$
Where W is work done by elastic force of string and $\Delta U$ is the change in elastic potential energy
Thus, work done during the second stretching is,
$W = - \left( {{U_f} - {U_i}} \right) = {U_I} - {U_f}$
$ \Rightarrow W = \dfrac{1}{2}k{x^2} - \dfrac{1}{2}k{\left( {x + y} \right)^2}$
$ \Rightarrow W = \dfrac{1}{2}k{x^2} - \dfrac{1}{2}k\left( {{x^2} + {y^2} + 2xy} \right)$
$ \Rightarrow W = \dfrac{1}{2}k{x^2} - \dfrac{1}{2}k{x^2} - \dfrac{1}{2}k{y^2} - \dfrac{1}{2}k\left( {2xy} \right)$
$ \Rightarrow W = kxy - \dfrac{1}{2}k{y^2} = \dfrac{1}{2}ky\left( { - 2x - y} \right)$
Thus, the work done against elastic force is
${W_{external}} = - W = \dfrac{{ky}}{2}\left( {2x + y} \right)$

So, the correct answer is “Option D”.

Note: A compression spring has elasticity born within the spring wire used for storing mechanical energy. The force of a spring is calculated according to Hooke’s law which states that the force required to stretch the spring will be directly proportional to the amount stretched.