Answer:
0.375 feet-lb
Explanation:
We have been given that the work required to stretch a spring 2 ft beyond its natural length is 6 ft-lb. We are asked to find the work needed to stretch the spring 6 in. beyond its natural length.
We can represent our given information as:

We will use Hooke's Law to solve our given problem.

Substituting this value in our integral, we will get:

Using power rule, we will get:
![6=\left[ (kx^2)/(2) \right ]^2_0](https://img.qammunity.org/2021/formulas/mathematics/college/ddptfmk0xusmpf1so0fmu3vs5s5myouuas.png)


We know that 6 inches is equal to 0.5 feet.
Work needed to stretch it beyond 6 inches beyond its natural length would be

Using power rule, we will get:
![\int\limits^(0.5)_0 {3x} \, dx = \left [(3x^2)/(2)\right]^(0.5)_0](https://img.qammunity.org/2021/formulas/mathematics/college/hh4yq6j1budaxolsy9mhr8rhgya2yl12h3.png)

Therefore, 0.375 feet-lb work is needed to stretch it 6 in. beyond its natural length.