Answer
The work must increase by a factor of 9.
Work Step by Step
As seen in equation 6.13, the work-energy theorem, the total work on the object is the change in its kinetic energy. We know that kinetic energy is $K=\frac{1}{2}mv^2$.
If the final speed $v_2$ is 3 times greater than $v_1$, then the final kinetic energy $K_2$ is 9 times greater than $K_1$. The work done on this object, starting from rest, gives the object its final kinetic energy.
The work $W_2$ is 9 times the work $W_1$.
