A

Step 1: Identify the forces acting on the block.
The forces are:
- Weight of the block: $W = mg = 1 ext{ kg} \times 10 ext{ m/s}^2 = 10 ext{ N}$.
- The component of weight acting down the incline: $W_{x} = W \sin(37°) = 10 \sin(37°) = 10 \times 0.6 = 6 ext{ N}$.
- The normal force: $N = W \cos(37°) = 10 \cos(37°) = 10 \times 0.8 = 8 ext{ N}$.
- Spring force when the spring is stretched by 0.1 m (10 cm): $F_s = kx = 100 \times 0.1 = 10 ext{ N}$.
- Frictional force: $F_f = \mu N = \mu \times 8$.

Step 2: Apply the work-energy principle. The work done by the forces will equal the elastic potential energy stored in the spring when the block comes to rest:
$W_{x} + F_f + F_s = 0$.
Substitute the forces:
$6 + \mu \times 8 - 10 = 0$.
Which simplifies to:
$\mu \times 8 = 4$.
Therefore, $\mu = \frac{4}{8} = 0.5$.

Therefore, the coefficient of friction between the block and the incline is 0.5.