Answer
An additional amount of work 3W.
Work Step by Step
Hooke’s Law tells us that the spring’s force is proportional to the distance stretched from equilibrium.
As seen in equation 6.9, the work done by the force is proportional to the square of the distance stretched, $W=\frac{1}{2}kx^2$.
If a final stretch of 2x is desired, then the total work done is $ \frac{1}{2}k(2x)^2$, or 4W.
An amount of W was already done in stretching the spring a distance x, so a work of 3W must be done to stretch the spring the additional distance of x to the final position.
