(i) The work done by the pulling force is calculated using the formula for work: \(W = F \cdot d \cdot \cos(\theta)\), where \(F = 25 \text{ N}\), \(d = 2 \text{ m}\), and \(\theta = 15^\circ\). Thus, \(W = 25 \times 2 \times \cos(15^\circ) = 48.3 \text{ J}\).

(ii) To find the normal component of the contact force, resolve the forces vertically. The equation is \(N + 25 \sin(15^\circ) = 3 \times 10\). Solving for \(N\), we get \(N = 30 - 25 \sin(15^\circ) = 23.5 \text{ N}\).