(a) Use conservation of energy between points A and B. The potential energy at A is converted into kinetic energy at B.

Potential energy at A: \(mgh = mg \times 1.8\)

Kinetic energy at B: \(\frac{1}{2} mv^2\)

Equating the energies: \(mg \times 1.8 = \frac{1}{2} mv^2\)

Cancel \(m\) and solve for \(v\):

\(g \times 1.8 = \frac{1}{2} v^2\)

\(9.8 \times 1.8 = \frac{1}{2} v^2\)

\(v^2 = 2 \times 9.8 \times 1.8\)

\(v^2 = 35.28\)

\(v = \sqrt{35.28} = 6 \text{ m s}^{-1}\)

(b) Use the work-energy principle from B to the point where the block comes to rest.

Initial kinetic energy at B: \(\frac{1}{2} m \times 6^2\)

Potential energy at the final point: \(mg \times 1.2\)

Work done against resistance: 4.5 J

Using the work-energy equation:

\(\frac{1}{2} m \times 6^2 = 4.5 + mg \times 1.2\)

\(18m = 4.5 + 1.2mg\)

\(18m = 4.5 + 1.2 \times 9.8m\)

\(18m = 4.5 + 11.76m\)

\(6.24m = 4.5\)

\(m = \frac{4.5}{6.24} = 0.75 \text{ kg}\)