# 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

Verified

126k+ views

**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: 29th Sep 2023

•

Total views: 126k

•

Views today: 3.26k

Recently Updated Pages

10 Examples of Evaporation in Daily Life with Explanations

10 Examples of Friction in Our Daily Life

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference between physical and chemical change class 11 chemistry CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

10 examples of law on inertia in our daily life

When was Bhagat Singh born A 25 September 1905 B 28 class 11 social science CBSE

How do I convert ms to kmh Give an example class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Explain zero factorial class 11 maths CBSE