Question

It took 1700 J of work to stretch a spring from its natural length of 2m to a length of 5m. Find the spring's force constant.

Answer 1

The work to stretch a spring from its rest position is

               (1/2) (spring constant) (distance of the stretch)²

           E = 1/2 k x² .

You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write

                 1700 joules  =  1/2 k (3m)²

1 joule = 1 newton-meter

                 1700 N-m =  1/2 k  (3m)²

Multiply each side by 2:   3400 N-m  =  k · 9m²   

Divide each side by  9m²     k  =  3400 N-m / 9m²

                                                   =  (377 and 7/9)   newton per meter    

Answer 2

 W = (1/2)kx^2, where W=work required to stretch the spring from its equilibrium position
k = spring constant
x = displacement = 5 - 2 = 3m 
1800 J = (1/2)(k)(3)^2
by solving it we get
k = 400 answer.