(i) The increase in kinetic energy is given by the change in kinetic energy from A to B:

\(\Delta KE = \frac{1}{2} m (v_B^2 - v_A^2)\)

Substituting the given values:

\(\Delta KE = \frac{1}{2} \times 12 \times (7^2 - 3^2) = 240 \text{ J}\)

(ii) Using the principle of conservation of energy, the gain in kinetic energy is equal to the loss in potential energy:

\(mgh = \Delta KE\)

\(12g \times AB \times \sin 10^{\circ} = 240\)

Solving for \(AB\):

\(AB = \frac{240}{12g \sin 10^{\circ}} \approx 11.5 \text{ m}\)

(iii) To find the magnitude of the horizontal force \(F\), use the equation:

\(F \times AB \times \cos 10^{\circ} = \Delta KE\)

\(F \times 11.5 \times \cos 10^{\circ} = 240\)

Solving for \(F\):

\(F = \frac{240}{11.5 \times \cos 10^{\circ}} \approx 21.2 \text{ N}\)