Question

Two springs $A$ and $B$ having spring constant $K_{A}$ and $K_{B}\left(K_{A}=2 K_{B}\right)$ are stretched by applying force of equal magnitude. If energy stored in spring $A$ is $E_{A}$ then energy stored in $B$ will be

• A. $2E_A$
• B. $E_A/4$
• C. $E_A/2$
• D. $4E_A$

Answer 1

Answer: (A)

Energy $=\frac{1}{2} K x^{2}=\frac{1}{2} \frac{F^{2}}{K} .$
$\frac{K_{A}}{K_{B}}=2$
$\therefore \frac{E_{A}}{E_{B}}=\frac{1}{2} .$
or $E_{B}=2 E_{A} .$