VerifiedHint: Total work done by a body is equal to change in the kinetic energy of the body equals change in potential energy. In the case of spring, the work done in stretching it from equilibrium position is directly proportional to the square of the elongation in the spring from its point of equilibrium.
Formula used: The formula that we will be using to solve the given question is that of the work done in stretching the spring, i.e.,
$W=\dfrac{1}{2}k{{x}^{2}}$
Complete step by step answer:
For solving the given question, let us assume the work done by the spring be equal to “W” when spring is stretched by 1 cm
Now by using the formula that is given for work done by spring, i.e.,
$W=\dfrac{1}{2}k{{x}^{2}}$
Where, k is the value of the spring constant and x is the stretch in the spring from the equilibrium point
So, according to the question, when $x=5cm=5\times {{10}^{-2}}m$
$\Rightarrow W=\dfrac{1}{2}\times5\times {{10}^{3}}{{(5\times {{10}^{-2}})}^{2}}$
$\Rightarrow W=6.25Nm$
Now, when the spring is further stretched by 5 cm
\[x=10cm=10\times {{10}^{-2}}m\]
Let the work done now be equal to W’
Now, again applying the above given formula, i.e.,
$W=\dfrac{1}{2}k{{x}^{2}}$
$\Rightarrow {W}'=\dfrac{1}{2}\times 5\times {{10}^{3}}{{(10\times {{10}^{-2}})}^{2}}$
$\Rightarrow {W}'=\dfrac{1}{2}\times5\times {{10}^{3}}{{({{10}^{-1}})}^{2}}$
$\Rightarrow {W}'=25$
So, work done in stretching the spring further by 1 cm will be
\[W'-W=25-6.25\]
$W'-W=18.75Nm$
So, the correct answer is “Option C”.
Note: You can simply solve such question by applying the formula \[Work=\dfrac{1}{2}k(x_{2}^{2}-x_{1}^{2})\], where ${{x}_{2}}$and ${{x}_{1}}$ are the stretched lengths. Also, remember the origin of this formula, i.e., Hooke’s Law which states that the force needed to extend or compress a spring by some distance (x) scales linearly with respect to that distance.
