Question

The work done by external agent in stretching a spring of force constant $k = 100\,N/cm$ from deformation ${x_1} = 10\,cm$ to deformation ${x_2} = 20\,cm$
A. $ - 150\,J$
B. $50\,J$
C. $150\,J$
D. None of these

Answer 1

Hint: Work done in stretching a spring is stored as the elastic potential energy given by the equation,
$\Delta PE = \dfrac{1}{2}k{x^2}$
Where, $k$ is the spring constant and $x$ is the displacement.
The change in potential energy of an object between two positions is equal to the work done in moving the object from first position to next. So, in order to calculate the work done to move spring from one position to another, it is enough to find the difference in potential energy between two positions. This work depends upon the spring constant and the distance stretched.

Complete step by step answer:
Work done in stretching a spring is stored as the elastic potential energy given by the equation,
$\Delta PE = \dfrac{1}{2}k{x^2}$
Where, $k$ is the spring constant and $x$ is the displacement.
This potential energy is the energy due to the deformation of the spring
Given,
Initial deformation,
${x_i} = 10\,cm$
$ \Rightarrow x = 10 \times {10^{ - 2}}\,m$
$\therefore x = 0.1\,m$
Final displacement,
${x_i} = 20\,cm$
$ \Rightarrow x = 20 \times {10^{ - 2}}\,m$
$\therefore x = 0.2\,m$
Value of spring constant,
$k = 100\,N/cm$
$ \Rightarrow k = 100\, \times {10^{2\,}}\,N/m$
The change in potential energy of an object between two positions is equal to the work done in moving the object from first position to next. So, in order to calculate the work done to move spring from one position to another, it is enough to find the difference in potential energy between two positions. This work depends upon the spring constant and the distance stretched.
Therefore, the work required to stretch it from deformation ${x_1} = 10\,cm$ to deformation ${x_2} = 20\,cm$ is the change in initial potential energy and final potential energy.
$W = {U_f} - {U_i}$ ………… (1)
Where ${U_i}$ is the initial potential energy given as,
${U_i} = \dfrac{1}{2}kx_i^2$
Substituting the given values, we get
${U_i} = \dfrac{1}{2} \times 10000\,N/m \times {\left( {0.1\,m} \right)^2}$
$\therefore {U_i} = 50\,Nm$
${U_f}$ is the final potential energy given as,
${U_f} = \dfrac{1}{2}kx_f^2$
Substituting the given values, we get
${U_f} = \dfrac{1}{2} \times 10000\,N/m \times {\left( {0.2\,m} \right)^2}$
$\therefore {U_f} = 200\,Nm$
Now, work done is a change in potential energy.
$W = {U_f} - {U_i}$
$ \Rightarrow W = 200\,Nm - 50\,Nm$
$\therefore W = 150\,Nm$
So, work done is $150\,J$. Correct answer is option C.

Note: Here spring constant is given in $N/cm$ remember to convert it into SI unit $N/m$. Also, while calculating work we need to subtract initial potential energy from final potential energy. If we do the reverse then we will get answer as option A. So, remember that work is
$W = {U_f} - {U_i}$
${U_f}$ is the final potential energy given as
${U_f} = \dfrac{1}{2}kx_f^2$
${U_i}$ is the initial potential energy given as
${U_i} = \dfrac{1}{2}kx_i^2$.