Question

Two springs have force constant $K_1$ and $K_2(K_1>K_2)$. Each spring is extended by same force $F$. It their elastic potential energy are $E_1$ and $E_2$ then $\frac{E_1}{E_2}$ is

• A. $\frac{K_1}{K_2}$
• B. $\frac{K_2}{K_1}$
• C. $\sqrt{\frac{K_1}{K_2}}$
• D. $\sqrt{\frac{K_2}{K_1}}$

Answer 1

The correct option is B $\frac{K_2}{K_1}$

Potential energy $E=\frac{1}{2}K{x}^2$

Force due to sprinf $F=Kx$

$x=\frac{F}{K}$..(1)

Using (1), $E=\frac{1}{2}\frac{F^2}{K}$

$\frac{E_1}{E_2}=\frac{\frac{1}{2}\frac{F^2}{K_1}}{\frac{1}{2}\frac{F^2}{K_2}}=\frac{K_2}{K_1}$