(i) Resolve the forces vertically. The vertical component of the pulling force is \(25 \sin \theta\). The normal force is given as 20 N. The weight of the block is \(2.7g\), where \(g\) is the acceleration due to gravity. At constant speed, the net vertical force is zero:

\(20 + 25 \sin \theta = 2.7g\)

Solving for \(\sin \theta\):

\(25 \sin \theta = 2.7g - 20\)

\(\sin \theta = \frac{2.7g - 20}{25}\)

Given \(g = 9.8\), substitute to find \(\sin \theta\):

\(\sin \theta = \frac{2.7 \times 9.8 - 20}{25} = 0.28\)

(ii) The work done by the pulling force is given by \(Fd \cos \theta\), where \(F = 25\) N, \(d = 5\) m, and \(\cos \theta = \sqrt{1 - \sin^2 \theta}\).

Calculate \(\cos \theta\):

\(\cos \theta = \sqrt{1 - 0.28^2} = \sqrt{1 - 0.0784} = \sqrt{0.9216}\)

\(\cos \theta \approx 0.96\)

Now calculate the work done:

\(\text{Work done} = 25 \times 5 \times 0.96 = 120 \text{ J}\)