(i) The increase in kinetic energy (KE) is given by:

\(KE = \frac{1}{2} \times 1100 \times v^2 = 550v^2\)

The increase in potential energy (PE) is given by:

\(PE = 1100 \times 9.8 \times \frac{160}{1760} \times x = 1000x\)

The work done by the driving force minus the work done against resistance is:

\(1800x - 700x = 1100x\)

Equating the work done to the increase in energy:

\(1100x = 550v^2 + 1000x\)

\(100x = 550v^2\)

\(x = 5.5v^2\)

Thus, \(k = 5.5\).

(ii) For \(x > 1760\), using the equation of motion:

At \(A\), \(5.5v^2 = 1760\) gives \(v^2 = 320\).

Using energy principles from \(A\) to \(B\):

\(550(v^2 - 320) = 1800(x - 1760) - 700(x - 1760)\)

\(550(v^2 - 320) = 1100(x - 1760)\)

\(v^2 - 320 = 2(x - 1760)\)

\(v^2 = 2x - 3200\)