Spring Question

[discowisco]

Suppose the spring has a force constant k = 500 N/m and it is compressed 2.5 m in the pre- launch poisition. When the cart is launched about how much work does the spring do on the cart?

TBR said basically W = change in PE

so basically 1/2 kx^2 = W and the answer is 1600 J

However as an alternative method, why doesnt this work:

W = f x d
Fspring = kx
W = (kx) x d
W= (500 x 2.5 ) x 2.5

This doesnt give me the right answer

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[Sterling Archer]

The second method doesn't work because the force of the spring changes with its displacement.