(i) To find the car's gain in kinetic energy, we use the formula for kinetic energy gain:

\(\text{KE gain} = \frac{1}{2} m (v_{\text{top}}^2 - v_{\text{bottom}}^2)\)

First, find the velocities at the bottom and top using the power formula \(P = Fv\):

\(30000/v - 1000 - 1250g \times 30/500 = 1250a\)

At the bottom, \(a = 4 \text{ m/s}^2\):

\(v_{\text{bottom}} = \frac{30000}{1250 \times 4 + 1000 + 750}\)

At the top, \(a = 0.2 \text{ m/s}^2\):

\(v_{\text{top}} = \frac{30000}{1250 \times 0.2 + 1000 + 750}\)

Substitute these velocities into the kinetic energy formula to find the increase in KE:

\(\text{Increase in KE} = 128000 \text{ J}\)

(ii) To find the work done by the car's engine, consider the total work done:

\(\text{WD}_{\text{car}} = \text{KE gain} + \text{PE gain} + \text{WD against resistance}\)

Potential energy gain:

\(\text{PE gain} = 1250g \times 30\)

Work done against resistance:

\(\text{WD against resistance} = 1000 \times 500\)

Substitute these values:

\(\text{WD}_{\text{car}} = 128000 + 375000 + 500000 = 1000000 \text{ J}\)

(i) Using the principle of conservation of energy from A to B and A to C:

\(\frac{1}{2} m v_B^2 = \frac{1}{2} m v_A^2 - mg \times 2.7\)

\(\frac{1}{2} m v_C^2 = \frac{1}{2} m v_A^2 - mg \times 3\)

Substituting for \(v_A = 8\) m s^-1:

\(v_B^2 = 8^2 - 20 \times 2.7\)

\(v_C^2 = 8^2 - 20 \times 3\)

Loss of speed = \(\sqrt{v_B^2} - \sqrt{v_C^2} = 10^{1/2} - 2 = 1.16\) m s^-1

(ii) Work done against friction from C to D:

\(\text{WD} = \frac{1}{2} \times 0.2 \times 2^2 + 0.2 \times g \times 3 = 6.4\)

For the mid-point M of CD:

\(\frac{1}{2} (0.4 + 6) = \frac{1}{2} \times 0.2 \times v_M^2 + 0.2g \times 1.5\)

Solving gives speed at midpoint \(v_M = 1.41\) m s^-1

(i) The work done against resistance is \(800 \times 500 = 400,000 \text{ J}\).

The total work done by the driving force is \(2,800,000 \text{ J}\).

Using the equation: \(\text{Work done by driving force} = \text{Potential energy gain} + \text{Work done against resistance}\)

\(2,800,000 = \text{PE gain} + 400,000\)

\(\text{PE gain} = 2,400,000 \text{ J}\)

\(\text{PE gain} = mgh = 16,000 \times 9.8 \times 500 \sin \alpha\)

\(2,400,000 = 16,000 \times 9.8 \times 500 \sin \alpha\)

Solving for \(\alpha\), \(\alpha = 1.7^\circ\).

(ii) The kinetic energy gain is given by:

\(\text{KE gain} = \text{Work done by driving force} + \text{PE loss} - \text{Work done against resistance}\)

\(\text{KE gain} = 2,400,000 + 2,400,000 - 800,000 = 4,000,000 \text{ J}\)

Using the kinetic energy formula: \(\frac{1}{2} m (v^2 - 20^2) = 4,000,000\)

\(\frac{1}{2} \times 16,000 \times (v^2 - 400) = 4,000,000\)

\(8,000 \times (v^2 - 400) = 4,000,000\)

\(v^2 - 400 = 500\)

\(v^2 = 900\)

\(v = 30 \text{ m s}^{-1}\)

(i) Using Newton's second law, the acceleration is given by:

\(-0.12 = 0.15a\)

\(a = -0.8 \text{ m/s}^2\)

Using \(v = u + at\) to find the speed of approach:

\(v = 3 - 0.8 \times 2 = 1.4 \text{ m/s}\)

When the block hits the wall, it loses 0.072 J of kinetic energy:

\(\frac{1}{2} \times 0.15 \times (1.4^2 - v_r^2) = 0.072\)

Solving for \(v_r\), the velocity when leaving Y is:

\(v_r = -1 \text{ m/s}\)

For the block to come to rest at Z:

\(0 = v_r + a(t - 2)\)

\(t = 3.25 \text{ s}\)

(ii) Using the area property to find distances XY and YZ:

\(XY = \frac{1}{2} (3 + 1.4) \times 2 = 4.4 \text{ m}\)

\(YZ = \frac{1}{2} \times 1.25 \times 1 = 3.775 \text{ m}\)

The displacement at Y is 4.4 m, and at Z is 3.775 m.

(i) The potential energy (PE) gain is calculated as:

\(\text{PE gain} = 1250 \times 10 \times 400 \times 0.125 = 625000 \text{ J}\)

The work done against resistance is:

\(\text{WD against resistance} = 800 \times 400 = 320000 \text{ J}\)

The total work done by the car's engine is the sum of the PE gain and the work done against resistance:

\(\text{WD by car's engine} = 625000 + 320000 = 945000 \text{ J} = 945 \text{ kJ}\)

(ii) Using the power and force relationship \(P = Fv\), we have:

\(\frac{v_2}{v_1} = \frac{P_2}{P_1} \times \frac{F_1}{F_2}\)

Given \(v_1 = 6 \text{ m/s}\), \(\frac{P_2}{P_1} = 5\), \(\frac{F_1}{F_2} = \frac{1}{3}\), we find \(v_2 = 10 \text{ m/s}\).

The kinetic energy (KE) gain is:

\(\text{KE gain} = \frac{1}{2} \times 1250 \times (10^2 - 6^2) = 40000 \text{ J}\)

The total work done by the car's engine is:

\(\text{WD by car's engine} = 945000 + 40000 = 985000 \text{ J} = 985 \text{ kJ}\)