The work done in stretching a linear spring by 4 cm from its...

The work done in stretching a spring by a distance $x$ from its natural length is given by $W = \dfrac{1}{2}kx^2$.

Given $W_1 = 10\text{ J}$ for $x_1 = 4\text{ cm}$.

$W_1 = \dfrac{1}{2}k(4)^2 = 10$

The final extension of the spring is $x_2 = 4\text{ cm} + 8\text{ cm} = 12\text{ cm}$.

The total work done to stretch the spring by $12\text{ cm}$ is $W_2 = \dfrac{1}{2}k(12)^2$.

Taking the ratio of $W_2$ to $W_1$:

$\dfrac{W_2}{W_1} = \left(\dfrac{12}{4}\right)^2 = 3^2 = 9$

$W_2 = 9 \times 10 = 90\text{ J}$

The additional work required is $\Delta W = W_2 - W_1 = 90\text{ J} - 10\text{ J} = 80\text{ J}$.

Answer: $80\text{ J}$