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. $\frac{E_A}{4}$
• C. $\frac{E_A}{2}$
• D. $4E_A$

Answer 1

The correct option is A $2E_A$

Let's consider the spring constant of two spring A and B are $K_A$ and $K_B$ respectively and are stretched by applying the force of the same magnitude.

Extension in spring is $x$, the spring force is

$f=-Kx$. . . . . .(1)

Energy stored in spring is,

$E=\frac{1}{2}K{x}^2$. . . . .(2)

From equation (1) and (2), we get

$E=\frac{f^2}{2K}$

from the above, $E\alpha \frac{1}{K}$. . . . . (3)

Given,

$K_A=2K_B$

$\frac{E_A}{E_B}=\frac{K_B}{K_A}=\frac{K_B}{2K_B}$

$K_B=2E_A$

The correct option is A.