Answer
The additional work required to stretch the spring another 4.0 cm is 48 J.
Work Step by Step
The work required to stretch a spring is equal to the potential energy stored in the spring.
Initially the work required to stretch the spring a distance of 2.0 cm is:
$Work_1 = 6~J$
$Work_1 = PE$
$Work_1 = \frac{1}{2}kx^2$
We increase the stretch distance x by a factor of 3.
$Work_2 = \frac{1}{2}k(3x)^2$
$Work_2 = 9\times\frac{1}{2}kx^2$
$Work_2 = 9\times Work_1$
$Work_2 = (9)(6 ~J)$
$Work_2 = 54 ~J$
$Work_2 = Work_1 + 48 ~J$
The additional work required to stretch the spring another 4.0 cm is 48 J.
