Question

Two similar spring \[P\] and $Q$ have spring constants ${{K}_{P}}$ and ${{K}_{Q}}$ such that ${{K}_{P}} > {{K}_{Q}}$. They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the spring ${{W}_{P}}$ and ${{W}_{Q}}$ are related as, in case (a) and case(b), respectively:
A. ${{W}_{P}} > {{W}_{Q}};{{W}_{Q}} > {{W}_{P}}$
B. ${{W}_{P}} < {{W}_{Q}};{{W}_{Q}} < {{W}_{P}}$
C. ${{W}_{P}} = {{W}_{Q}};{{W}_{P}} > {{W}_{Q}}$
D. ${{W}_{P}} = {{W}_{Q}};{{W}_{P}} = {{W}_{Q}}$

Answer 1

Hint: The work done in stretching a spring is equal to the change in its potential energy. The force applied on the spring is proportional to the amount of stretch produced in the spring. Determine the relationship between the displacement for two springs when the force applied is the same in case (b).
Formula Used:
$W=\dfrac{1}{2}K{{x}^{2}}$
$F=Kx$

Complete answer:
When a force is applied on a spring with spring constant $K$, it changes the potential energy of the spring and the work done is determined as the change in potential energy of the spring. The work done can be mathematically written as,
$W=\dfrac{1}{2}K{{x}^{2}}$
Here, $W$ is the work done, $K$ is the spring constant and \[x\] is the amount of stretch.
Now, for case(a), it is mentioned that both the springs are stretched by the same amount. Thus, the work done for both the springs can be written as,
${{W}_{P}}=\dfrac{1}{2}{{K}_{P}}{{x}^{2}};{{W}_{Q}}=\dfrac{1}{2}{{K}_{Q}}{{x}^{2}}$
Now, the question mentions that ${{K}_{P}} > {{K}_{Q}}$. Thus, from the above equation, ${{W}_{P}} > {{W}_{Q}}$.
Now, let us consider case (b). In this case, both the springs are stretched by the same force. Now, the force applied on a spring can be written as,
$F=Kx$
Given,
$\begin{align}
  & {{F}_{P}}={{F}_{Q}} \\
 & {{K}_{P}}{{x}_{P}}={{K}_{Q}}{{x}_{Q}}
\end{align}$
From the above equation, ${{x}_{P}} < {{x}_{Q}}$ since ${{K}_{P}} > {{K}_{Q}}$.
The work done on both springs can be written as,
${{W}_{P}}=\dfrac{1}{2}\left( {{K}_{P}}{{x}_{P}} \right){{x}_{P}}$ and ${{W}_{Q}}=\dfrac{1}{2}\left( {{K}_{Q}}{{x}_{Q}} \right){{x}_{Q}}$.
Now,
$\begin{align}
  & \dfrac{{{W}_{P}}}{{{W}_{Q}}}=\dfrac{\dfrac{1}{2}\left( {{K}_{P}}{{x}_{P}} \right){{x}_{P}}}{\dfrac{1}{2}\left( {{K}_{Q}}{{x}_{Q}} \right){{x}_{Q}}} \\
 & =\dfrac{{{x}_{P}}}{{{x}_{Q}}}
\end{align}$
Since we calculated ${{x}_{P}} < {{x}_{Q}}$, ${{W}_{P}} < {{W}_{Q}}$.

Thus, the correct option is (A).

Note:
The work done on both springs is determined as change in potential energy. Take care of the relationship between the spring constant of two springs before deducing further relationships. Determine work done in terms of force applied so as to reach to a conclusion in part (b).