Question

Two springs $A$ and $B$ are identical but $A$ is harder than $B\left(k_{A} >k_{B}\right)$. Let $W_{A}$ and $W_{B}$ represent the work done when the springs are stretched through the same distance and $W_{A}'$ and $W_{B}'$ are the work done when these are stretched by equal forces, then which of the following is true?

• A. $W_{A}>W_{B}$ and $W_{A}'=W_{B}'$
• B. $W_{A} >W_{B}$ and $W_{A}'< W_{B}'$
• C. $W_{A} > W_{B}$ and $W_{A}' > W_{B}'$
• D. $W_{A}< W_{B}$ and $W_{A}'< W_{B}'$

Answer 1

Answer: (B)

$W_{A} >W_{B}$ and $W_{A}'< W_{B}'$

Answer 2

Work done in stretching a spring $W=\frac{1}{2} k x^{2}$
Here, $W_{A}=\frac{1}{2} k_{A} x_{A}^{2}$
and $W_{B}=\frac{1}{2} k_{B} x_{B}^{2}$
As $x_{A}=x_{B}$
and $k_{A}>k_{B}$
$\therefore W_{A}>W_{B}$
Similarly, when forces are equal
$F_{A}=F_{B}$
$k_{A} x_{A}=k_{B} x_{B} ...$ (i)
As $k_{A}>k_{B}$
$\therefore x_{A}< x_{B} \ldots$ (ii)
Now, $W'_{A}=\frac{1}{2}\left(k_{A} x_{A}\right) x_{A}$
and $W'_{B}=\frac{1}{2}\left(k_{B} x_{B}\right) x_{B}$
From Eqs. (i) and (ii),
we conclude that $W'_{A}< W'_{B}$