Q.Two springs of force constants $K_1$ and $K_2$ ($K_1 > K_2$) are stretched by the same force. If $W_1$ and $W_2$ be the work done stretching the springs then ______. (A) $W_1 = W_2$
(B) $W_1 < W_2$
(C) $W_1 > W_2$
(D) $W_1 = W_2 = 0$

The elastic potential energy stored in a spring is PE = (1/2) k x^2, so quadrupling the extension multiplies the stored energy by 16.

PE = (1/2) k x^2.

At x1 = 2 cm, PE1 = (1/2) k (2)^2 = u (given).

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Work done in stretching a spring by force $F$ is $W=\dfrac{F^2}{2K}$, which is inversely proportional to the spring constant.

When a spring of force constant $K$ is stretched by an external force $F$, the extension produced is $x=F/K$ (Hooke's law), and the work done in stretching it (equal to the elastic PE stored) is

$W=\frac12 Kx^2=\frac12 K\left(\frac{F}{K}\right)^2=\frac{F^2}{2K}$

For the same applied force $F$ on both springs, $W\propto \dfrac1K$.

Given $K_1>K_2$: …