First, calculate the frictional force \(F\) using the formula:

\(F = 0.25 \left( 6.1 \times \frac{60}{61} \right) = 1.5 \text{ N}\)

Using Newton's second law, the equation of motion along the slope is:

\(6.1 \times \frac{11}{61} - 0.25 \times 6.1 \times \frac{60}{61} = 0.61a\)

Solving for \(a\):

\(a = \frac{-40}{61} = -0.656 \text{ m/s}^2\)

Using the equation \(0 = v_A^2 + 2as\) to find the distance \(s\):

\(0 = 2^2 + 2 \times \left(-0.656\right) \times s\)

\(s = \frac{4}{2 \times 0.656} = 3.05 \text{ m}\)

Thus, the block moves 3.05 m from A before it comes to rest.