A long spring is stretched by 1 cm. If the work done in this process is W, then find out the work done in further stretching it by 1 cm.

Hint: Total work done by a body is equals to change in the kinetic energy of the body equals change in potential energy. In the case of spring, the work done is directly proportional to the stretch 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 by 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=1 cm
$W=\dfrac{1}{2}k{{(1)}^{2}}$
\[W=\dfrac{1}{2}k\]

Now, when the spring is further stretched by 1 cm
\[x=2cm\]
Let us assume the work done now be equal to W'
Now, again applying the above given formula, i.e.,
$\begin{align}
  & W=\dfrac{1}{2}k{{x}^{2}} \\
 & {W}'=\dfrac{1}{2}k{{(2)}^{2}} \\
 & {W}'=\dfrac{4}{2}k \\
 & {W}'=2k \\
\end{align}$
 So, work done in stretching the spring further by 1 cm will be
\[W'-W=2k-\dfrac{k}{2}\]
$W'-W=\dfrac{3k}{2}$
 So, work done in stretching the spring further by 1 cm will be \[\dfrac{3k}{2}\]

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, make sure to remember the origin of this formula, i.e. Hooke's Law which states “Force needed to extend or compress a spring by some distance (x) scales linearly with respect to that distance”.