(a) The initial potential energy (PE) of the ball is given by:

\(\text{PE} = mgh = 1.6 \times 10 \times 5 = 80 \text{ J}\)

When the ball hits the ground, it loses 8 J of kinetic energy, so the kinetic energy (KE) just before hitting the ground is:

\(\text{KE} = 80 - 8 = 72 \text{ J}\)

Using the conservation of energy, the potential energy at the greatest height \(h\) after rebounding is equal to the kinetic energy just after the bounce:

\(mgh = 72\)

\(1.6 \times 10 \times h = 72\)

\(h = \frac{72}{16} = 4.5 \text{ m}\)

(b) For the downward motion, using the equation:

\(s = ut + \frac{1}{2}gt^2\)

\(5 = 0 + \frac{1}{2} \times 10 \times t^2\)

\(t^2 = 1\)

\(t = 1 \text{ s}\)

For the upward motion, using the equation:

\(s = vt - \frac{1}{2}gt^2\)

\(4.5 = 0 - \frac{1}{2} \times (-10) \times t^2\)

\(t^2 = 0.9\)

\(t = \sqrt{0.9} \approx 0.95 \text{ s}\)

Total time taken is:

\(1 + 0.95 = 1.95 \text{ s}\)

(a) The driving force \(DF\) is given by \(\frac{960,000}{30} = 32,000 \text{ N}\).

Resolve forces along the slope: \(DF - 75,000g \sin \alpha - R = 0\).

Substitute \(\sin \alpha = 0.01\) and \(g = 9.8 \text{ m s}^{-2}\):

\(32,000 - 75,000 \times 9.8 \times 0.01 - R = 0\)

\(R = 32,000 - 7,350 = 24,650 \text{ N}\)

Resistance force \(R = 24,500 \text{ N}\) (rounded to 3 significant figures).

(b) Work done by engine in 60 s is \(900,000 \times 60 = 54,000,000 \text{ J}\).

Initial kinetic energy \(KE_{\text{init}} = \frac{1}{2} \times 75,000 \times 30^2\).

Final kinetic energy \(KE_{\text{final}} = \frac{1}{2} \times 75,000 \times v^2\).

Using the work-energy principle:

\(54,000,000 + \frac{1}{2} \times 75,000 \times 30^2 = 46,500,000 + \frac{1}{2} \times 75,000 \times v^2\)

Solve for \(v\):

\(v = \sqrt{1100} = 33.2 \text{ m s}^{-1}\)

(a) Use conservation of energy between points A and B. The potential energy at A is converted into kinetic energy at B.

Potential energy at A: \(mgh = mg \times 1.8\)

Kinetic energy at B: \(\frac{1}{2} mv^2\)

Equating the energies: \(mg \times 1.8 = \frac{1}{2} mv^2\)

Cancel \(m\) and solve for \(v\):

\(g \times 1.8 = \frac{1}{2} v^2\)

\(9.8 \times 1.8 = \frac{1}{2} v^2\)

\(v^2 = 2 \times 9.8 \times 1.8\)

\(v^2 = 35.28\)

\(v = \sqrt{35.28} = 6 \text{ m s}^{-1}\)

(b) Use the work-energy principle from B to the point where the block comes to rest.

Initial kinetic energy at B: \(\frac{1}{2} m \times 6^2\)

Potential energy at the final point: \(mg \times 1.2\)

Work done against resistance: 4.5 J

Using the work-energy equation:

\(\frac{1}{2} m \times 6^2 = 4.5 + mg \times 1.2\)

\(18m = 4.5 + 1.2mg\)

\(18m = 4.5 + 1.2 \times 9.8m\)

\(18m = 4.5 + 11.76m\)

\(6.24m = 4.5\)

\(m = \frac{4.5}{6.24} = 0.75 \text{ kg}\)

(a) The change in potential energy (PE) is given by \(\Delta PE = mgh\), where \(h = s \sin \theta\). Here, \(s = 30 \times 20 = 600 \text{ m}\) and \(\sin \theta = 0.12\). Thus, \(h = 600 \times 0.12 = 72 \text{ m}\).

\(\Delta PE = 1600 \times 9.8 \times 72 = 1152000 \text{ J}\).

(b) The total work done \(W = 1960 \times 1000 = 1960000 \text{ J}\). The work done against resistance \(W_D = W - \Delta PE = 1960000 - 1152000 = 808000 \text{ J}\).

The force resisting motion \(R = \frac{W_D}{s} = \frac{808000}{600} = 1350 \text{ N}\).

(c) The power developed \(P = \frac{W}{t} = \frac{1960000}{30} = 65333.33 \text{ W} = 65.3 \text{ kW}\).

(d) The new power is \(0.85 \times 65333.33 = 55533.33 \text{ W}\). Using \(P = Fv\), the driving force \(DF = \frac{55533.33}{20} = 2776.67 \text{ N}\).

Using Newton's second law, \(DF - R - mg \sin \theta = ma\).

\(2776.67 - 1350 - 1600 \times 9.8 \times 0.12 = 1600a\).

\(a = \frac{-490}{1600} = -0.306 \text{ m s}^{-2}\).

(a) The total mass of the car and trailer is 1900 kg. The increase in kinetic energy (KE) is given by:

\(\frac{1}{2} \times 1900 \times 30^2 - \frac{1}{2} \times 1900 \times 20^2 = 475000 \text{ J}\)

The loss of potential energy (PE) is:

\(1900 \times g \times s \times \sin 5^{\circ}\)

Using the work-energy principle:

\(1900 \times g \times s \times \sin 5^{\circ} + 150000 = \frac{1}{2} \times 1900 \times 30^2 - \frac{1}{2} \times 1900 \times 20^2\)

Solving for \(s\):

\(s = 196 \text{ m}\)

(b) Using the equation \(v^2 = u^2 + 2as\):

\(30^2 = 20^2 + 2a \times 200\)

Solving for \(a\):

\(a = 1.25 \text{ m s}^{-2}\)

Applying Newton's second law to the trailer:

\(T - 100 + 500g \sin 5^{\circ} = 500a\)

Solving for \(T\):

\(T = 289 \text{ N}\)