For block A, the force triangle involves the weight of A, the normal reaction, and the 18 N force acting parallel to the slope. The angle between the weight and the 18 N force is 30°.

Using the sine rule for the triangle of forces, we have:

\(W_A \sin 30^{\circ} = 18\)

\(W_A = \frac{18}{\sin 30^{\circ}} = 36 \text{ N}\)

For block B, the force triangle involves the weight of B, the normal reaction, and the 18 N force acting horizontally. The angle between the weight and the 18 N force is 30°.

Using the sine rule for the triangle of forces, we have:

\(W_B \sin 30^{\circ} = 18 \cos 30^{\circ}\)

\(W_B = \frac{18 \cos 30^{\circ}}{\sin 30^{\circ}} = 31.2 \text{ N}\)

To find the tension in the string, we resolve the forces parallel to the line of greatest slope. The weight component along the slope is given by \(W \sin \alpha\), where \(W = 5.1\) N and \(\sin \alpha = \frac{8}{17}\).

The tension \(T\) in the string has a component \(T \cos \beta\) along the slope, where \(\cos \beta = \sqrt{1 - \sin^2 \beta} = \sqrt{1 - \left(\frac{7}{25}\right)^2} = \frac{24}{25}\).

Equating the components along the slope, we have:

\(T \cos \beta = W \sin \alpha\)

\(T \left(\frac{24}{25}\right) = 5.1 \left(\frac{8}{17}\right)\)

\(T = \frac{5.1 \times 8 \times 25}{17 \times 24}\)

\(T = 2.5 \text{ N}\)

(i) When the force is parallel to the plane, resolve the forces parallel to the plane. The component of the weight acting down the plane is given by:

\(W \sin 40^{\circ} = 12 \sin 40^{\circ}\)

Equating this to the force P, we have:

\(P = 12 \sin 40^{\circ}\)

Calculating gives:

\(P = 7.71 \text{ N}\)

(ii) When the force is horizontal, resolve the forces horizontally and vertically. The horizontal component of P is \(P \cos 40^{\circ}\) and the vertical component of the weight is \(W \sin 40^{\circ}\). Equating the horizontal component of P to the vertical component of the weight gives:

\(P \cos 40^{\circ} = 12 \sin 40^{\circ}\)

Solving for P gives:

\(P = \frac{12 \sin 40^{\circ}}{\cos 40^{\circ}}\)

Calculating gives:

\(P = 10.1 \text{ N}\)

(i) When the force is parallel to the plane, resolve the forces parallel to the plane. The component of the weight down the plane is \(18 \sin 30^{\circ}\). For equilibrium, this must equal the force \(P\), so:

\(P = 18 \cos 60^{\circ}\)

\(P = 9\)

(ii) When the force is horizontal, resolve the forces perpendicular to the plane. The horizontal force \(P\) has a component \(P \cos 30^{\circ}\) parallel to the plane. For equilibrium, this must balance the component of the weight down the plane:

\(P \cos 30^{\circ} = 18 \cos 60^{\circ}\)

\(P = \frac{18 \cos 60^{\circ}}{\cos 30^{\circ}}\)

\(P = 10.4\)

(a) Apply Newton's second law to block B:

\(T - 40g \sin 20^{\circ} - 50 = 40a\)

Solving gives:

\(T = 40a + 136.8\)

Apply Newton's second law to block A:

\(500 \cos 15^{\circ} - 80g \sin 20^{\circ} - T = 80a\)

Substitute for \(T\):

\(500 \cos 15^{\circ} - 80g \sin 20^{\circ} - (40a + 136.8) = 80a\)

Simplify and solve for \(a\):

\(482.96 - 273.61 - 136.8 = 120a\)

\(a = 0.188 \text{ m/s}^2\)

Substitute \(a\) back to find \(T\):

\(T = 40(0.188) + 136.8 = 194 \text{ N}\)

(b) Use the equation \(v = u + at\) with \(u = 0\), \(v = 1.2 \text{ m/s}\), and \(a = 0.188 \text{ m/s}^2\):

\(1.2 = 0 + 0.188t\)

Solve for \(t\):

\(t = \frac{1.2}{0.188} = 6.39 \text{ s}\)