Question

A spring of constant $5\times {10}^{3}N{m}^{-1}$ is stretched initially by $5cm$ from the unstretched position. Find the work required to stretch it further by another $5cm$.
A) $6.25Nm$
B) $12.5Nm$
C) $18.75Nm$
D) $25.00Nm$

Answer 1

Hint: Work done in stretching a spring is stored as the elastic potential energy given by the equation,
$U={\displaystyle \frac{1}{2}}k{x}^2$
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.
Therefore,
$W=U_f-U_i$
$U_i$ is the initial energy given as,
$U_i={\displaystyle \frac{1}{2}}k{x}_i^2$
Where $x_i$ is the initial displacement.
$U_f$ is the final energy given as,
$U_f={\displaystyle \frac{1}{2}}k{x}_f^2$
Where $x_f$ is the final displacement.

Complete step by step solution:
Work done in stretching a spring is stored as the elastic potential energy given by the equation,
$U={\displaystyle \frac{1}{2}}k{x}^2$
Where $k$ is the spring constant and $x$ is the displacement.
This potential energy can be said as the energy due to the deformation of the spring
Given,
Initial displacement,
$$\begin{gathered} x_i=5cm \\ \Rightarrow 5\times {10}^{-2}m \\ \Rightarrow 0.05m \end{gathered}$$
Final displacement,
$$\begin{gathered} \Rightarrow x_f=0.05+0.05 \\ \Rightarrow 0.1m \end{gathered}$$
Value of spring constant, $k=5\times {10}^{3}N{m}^{-1}$
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 further by another $5cm$ is the change in initial potential energy and final potential energy.
$W=U_f-U_i$ ………… (1)
Where $U_i$ is the initial energy given as,
$U_i={\displaystyle \frac{1}{2}}k{x}_i^2$
Substituting the given values, we get
$U_i={\displaystyle \frac{1}{2}}\times 5\times {10}^3\times {\left(0.05\right)}^2=6.25Nm$
$U_f$ is the final energy given as,
$U_f={\displaystyle \frac{1}{2}}k{x}_f^2$
Substituting the given values, we get
$U_f={\displaystyle \frac{1}{2}}\times 5\times {10}^3\times {\left(0.1\right)}^2=25Nm$
Now substitute the initial and final energies in equation (1).
$$\begin{gathered} W=U_f-U_i \\ \Rightarrow 25Nm-6.5Nm \\ \Rightarrow 18.5Nm \end{gathered}$$
Therefore, the work required to stretch it further by another $5cm$ is $18.5Nm$.

So, the answer is option C.

Note: Here, while substituting for the value of displacement in the final potential energy remember to substitute the value $0.1m$. The spring was already stretched by $0.05m$ and we need to find the work done when it is further stretched by $0.05m$.So the final displacement will be $0.05m+0.05m=0.1m$