**Hooke's law** is written as `F = kx`

where:

`F` = force

`k ` = proportionality constant or spring constant

`x` = length displacement from its natural length

Apply Hooke's Law to the integral application for work: `W = int_a^b F dx` , we get:

`W = int_a^b kx dx`

`W = k * int_a^b x dx`

Apply Power rule for integration: `int x^n(dx) = x^(n+1)/(n+1).`

`W = k * x^(1+1)/(1+1)|_a^b`

`W = k * x^2/2|_a^b`

From the required work 18 ft-lbs, note that the units has "ft" instead of inches. To be consistent, apply the **conversion factor: 12 inches = 1 foot** then:

`4` inches = `1/3` ft

`7` inches = `7/12` ft

To solve for k, we consider the initial condition: W =18 ft-lbs to stretch a spring `4` inches or `1/3` ft from its natural length. Stretching `1/3` ft of it natural length implies the boundary values: `a=0` to `b=1/3` ft.

Applying `W = k * x^2/2|_a^b` , we get:

`18= k * x^2/2|_0^(1/3)`

Apply definite integral formula: `F(x)|_a^b = F(b)-F(a)` .

`18 =k [(1/3)^2/2-(0)^2/2]`

`18 = k * [(1/9)/2 -0]`

`18 = k *[1/18]`

`18 = k/18`

`k =18*18`

`k= 324`

To solve for the work need to stretch the spring with **additional 3 inches**, we plug-in: `k =324` , `a=1/3,` and `b = 7/12` on `W = k * x^2/2|_a^b` .

Note that stretching "additional 3 inches" from its initial stretch of 4 inches is the same as stretching 7 inches from its natural length.

`W= 324 * x^2/2|_((1/3))^((7/12))`

`W =324 [ (7/12)^2/2 -(1/3)^2/2]`

`W = 324 [ 49/288 -1/18]`

`W = 324[11/96]`

`W=297/8` or `37.125` ft-lbs