Question

Two springs have force constant $K_{1}$ and $K_{2}\left(K_{1} > K_{2}\right)$. 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

Answer: (B)

$x=\frac{F}{K}$
$U=\frac{1}{2} K x^{2}=\frac{1}{2} K\left(\frac{F}{K}\right)^{2}$
$U=\frac{F^{2}}{2 K}$
$U \alpha \frac{1}{K}$
$\frac{U_{1}}{U_{2}}=\frac{K_{2}}{K_{1}}$