c
The free body diagram of the block of mass \(2 kg\) is shown below.

From the work energy theorem,

\(W=K E_{2}-K E_{1}\)

\(m g x=\frac{1}{2} k x^{2}-0\)

\(kx=2 m g\)

\(=2 \times 0.25 \times 10\)

\(=5 N\)

Taking the equilibrium of forces in vertical direction,

\(N=k x+m g\)

\(=5+2 \times 10\)

\(=25 N\)