Question

Two springs A and B having spring constant K$_A$ and K$_B$ (K$_A$ = 2K$_B$) 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. 4$E_A$

Answer 1

The Correct Option is A

Energy =$\frac{1}{2}Kx^2 =\frac{1}{2}\frac{F^2}{K}.$
$\frac{K_A}{K_B}=2$
$\therefore \, \, \, \, \frac{E_A}{E_B}=\frac{1}{2}$. or $\, \, \, E_B=2E_A$.