# A spring of spring constant \[5{\rm{ }} \times {\rm{ }}{10^3}\;N{m^{ - 1}}\;\]is stretched initially by \[5cm\] from the unstretched position. Then the work required to stretch it further by another \[5cm\]is:

1. \[12.50{\rm{N - m}}\]

2. \[{\rm{18}}{\rm{.75N - m}}\]

3. \[{\rm{25N - m}}\]

4. \[{\rm{6}}{\rm{.25N - m}}\]

Answer

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**Hint:**Here, the concept that we are going to use is of oscillation in the springs, but here it is not mentioned that the spring is oscillating. It's only one time from an unstretched position, and we have to calculate the other stretch of \[5cm\].

**Formula used:**

\[{W_i} = \dfrac{1}{2}k{x_i}^2\], \[{W_f} = \dfrac{1}{2}k{x_f}^2\], \[{W_{net}} = {W_f} - {W_i}\]

**Complete answer:**

Let us begin by sorting out the given data in the question. Here the initial position is taken as \[5cm\] away from unstretched position, therefore, by considering \[{x_i}\] as initial position we have

\[{x_i} = 5cm = 0.05m\]

Similarly, final position is at another \[5cm\] away from the initial position, therefore, by considering \[{x_f}\]as final position we have

\[{x_f} = 5cm + 5cm = 10cm = 0.1m\]

Also, spring constant \[5{\rm{ }} \times {\rm{ }}{10^3}\;N{m^{ - 1}}\;\]

Work done at initial position is calculated by

\[{W_i} = \dfrac{1}{2}k{x_i}^2\]

Let us put all the given values in the above formula for work done

\[{W_i} = \dfrac{1}{2} \times 5{\rm{ }} \times {\rm{ }}{10^3}\;N{m^{ - 1}}\; \times {\left( {0.05} \right)^2}{m^2}\]

\[\therefore {W_i} = 6.25N - m\]

Similarly, for final position work done is given by

\[{W_f} = \dfrac{1}{2}k{x_f}^2\]

Now, let us put all the values given in the above formula, we get

\[{W_f} = \dfrac{1}{2} \times 5{\rm{ }} \times {\rm{ }}{10^3}\;N{m^{ - 1}} \times {\left( {0.1} \right)^2}{m^2}\]

\[\therefore {W_f} = 25N - m\]

Now, the work done required for another \[5cm\] stretch is

Net work done, \[{W_{net}} = {W_f} - {W_i}\]

\[{W_{net}} = 25N - 6.25N\] … (from above calculated terms)

\[\therefore {W_{net}} = 18.75~N - m\]

**Thus, the total work done is given by \[18.75N - m\], i.e., option 2.**

**Note:**When we talk about net work done we have to calculate it with respect to initial and final work done then only we can be able to solve this type of questions. In this question you must have observed that the initial position does not start from zero because the question itself says consider the first stretch as the initial position.

Last updated date: 02nd Jun 2023

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