Spring constant, \(k=100 N m ^{-1}\)

Displacement in the block, \(x=10 cm =0.1 m\)

The given situation can be shown as in the following figure.

At equilibrium:

Normal reaction, \(R=m g \cos 37^{\circ}\)

Frictional force, \(f{=} \mu_{R}=m g \sin 37^{\circ}\)

Where, \(\mu\) is the coefficient of friction

Net force acting on the block \(=m g \sin 37^{\circ}-f\)

\(=m g \sin 37^{\circ}-\mu m g \cos 37^{\circ}\)

\(=m g\left(\sin 37^{\circ}-\mu \cos 37^{\circ}\right)\)

At equilibrium, the work done by the block is equal to the potential energy of the spring, i.e.,

\(m g\left(\sin 37^{\circ}-\mu \cos 37^{\circ}\right) x=\frac{1}{2} k x^{2}\)

\(1 \times 9.8\left(\sin 37^{\circ}-\mu \cos 37^{\circ}\right)=\frac{1}{2} \times 100 \times 0.1\)

\(0.602-\mu \times 0.799=0.510\)

\(\therefore \mu=\frac{0.092}{0.799}=0.115\)