(a) Use the conservation of energy principle. The initial kinetic energy (KE) at A is \(\frac{1}{2} \times 0.2 \times 5^2\). The final KE at B is \(\frac{1}{2} \times 0.2 \times 3^2\). The change in potential energy (PE) is \(0.2gh\).

Equating initial and final energies:

\(\frac{1}{2} \times 0.2 \times 5^2 = 0.2gh + \frac{1}{2} \times 0.2 \times 3^2\)

Solving for \(h\):

\(h = 0.8\) m

(b) Apply the work-energy principle from A to C. The initial KE is \(\frac{1}{2} \times 0.2 \times 5^2\). The work done against resistance is 3.1 J. The change in PE is \(0.2g \times 0.5\). The final KE at C is \(\frac{1}{2} \times 0.2 \times v^2\).

Using the work-energy equation:

\(\frac{1}{2} \times 0.2 \times 5^2 - 3.1 + 0.2g \times 0.5 = \frac{1}{2} \times 0.2 \times v^2\)

Solving for \(v\):

\(Speed = 2 m/s\)

The work-energy principle states that the work done on the train is equal to the change in kinetic energy plus the change in potential energy.

The change in kinetic energy (KE) is given by:

\(\Delta KE = \frac{1}{2} \times 150,000 \times 42^2 - \frac{1}{2} \times 150,000 \times 45^2 = -19,575,000 \text{ J}\)

The potential energy (PE) gain is:

\(\Delta PE = 150,000g \times 500 \sin \alpha\)

The total work done by the driving force and against resistance is:

\(16,000 \times 500 - 4 \times 10^6\)

Using the work-energy equation:

\(150,000g \times 500 \sin \alpha = 19,575,000 + 16,000 \times 500 - 4 \times 10^6\)

Solving for \(\sin \alpha\):

\(150,000g \times 500 \sin \alpha = 19,575,000 + 8,000,000 - 4,000,000\)

\(150,000g \times 500 \sin \alpha = 23,575,000\)

\(\sin \alpha = \frac{23,575,000}{150,000 \times 500 \times 9.8}\)

\(\alpha = \arcsin(0.0314) \approx 1.8^\circ\)

(a) The potential energy lost by the child from X to Y is given by:

\(PE_{\text{lost}} = mgh = 25 \times 9.8 \times 1.8 = 450 \text{ J}\)

The work-energy principle states that the change in kinetic energy is equal to the work done against resistance plus the potential energy lost:

\(\frac{1}{2} mv^2 = PE_{\text{lost}} - \text{Work done}\)

\(\frac{1}{2} \times 25 \times v^2 = 450 - 50\)

\(\frac{1}{2} \times 25 \times v^2 = 400\)

\(v^2 = \frac{400 \times 2}{25} = 32\)

\(v = \sqrt{32} = 4 \text{ m/s}\)

(b) Using energy methods, the potential energy gained from Y to Z is:

\(PE_{\text{gained}} = 25 \times g \times 2 \times 0.28 = 140 \text{ J}\)

The kinetic energy at Y is:

\(KE = \frac{1}{2} \times 25 \times (4)^2 = 200 \text{ J}\)

The work-energy principle for Y to Z gives:

\(F \times 2 = 25 \times g \times 2 \times 0.28 + 200\)

\(F \times 2 = 270\)

The normal reaction \(R\) is:

\(R = 25 \times g \times \cos \alpha = 25 \times 9.8 \times 0.96 = 240 \text{ N}\)

The frictional force \(F = \mu R\):

\(\mu = \frac{F}{R} = \frac{270}{240} = \frac{9}{8}\)

(i) The power required to maintain a constant speed is given by the formula:

\(P = Fv\)

where \(F = 3000 \text{ N}\) and \(v = 30 \text{ m s}^{-1}\).

Thus, \(P = 3000 \times 30 = 90,000 \text{ W} = 90 \text{ kW}\).

(ii) Using the energy method, the potential energy gained is:

\(PE = 25000gh\)

The initial kinetic energy at A is:

\(KE_{\text{initial}} = \frac{1}{2} \times 25000 \times 30^2 = 11,250,000 \text{ J}\)

The final kinetic energy at B is:

\(KE_{\text{final}} = \frac{1}{2} \times 25000 \times 25^2 = 7,812,500 \text{ J}\)

The loss in kinetic energy is:

\(KE_{\text{loss}} = 11,250,000 - 7,812,500 = 3,437,500 \text{ J}\)

Using the work-energy principle:

\(KE_{\text{initial}} = KE_{\text{final}} + 25000gh + \frac{3000h}{\sin 2^\circ}\)

Solving for \(h\):

\(3,437,500 = 25000gh + \frac{3000h}{\sin 2^\circ}\)

\(h = 10.2 \text{ m}\)

First, calculate the initial kinetic energy (KE) at the bottom of the hill:

\(\text{Initial KE} = \frac{1}{2} \times 75 \times 10^2 = 3750 \text{ J}\)

Next, calculate the final kinetic energy (KE) at the top of the hill:

\(\text{Final KE} = \frac{1}{2} \times 75 \times 5^2 = 937.5 \text{ J}\)

Calculate the potential energy (PE) gained as the cyclist ascends the hill:

\(\text{PE gained} = 75g \times 700 \times \sin 1.5^{\circ} \approx 13743 \text{ J}\)

Calculate the work done by the force \(F\):

\(\text{WD by } F = F \times 700\)

Using the work-energy principle:

\(\text{WD by } F + \text{Initial KE} = \text{Final KE} + \text{PE gain} + 2000\)

Substitute the known values:

\(F \times 700 + 3750 = 937.5 + 13743 + 2000\)

\(F \times 700 = 16480.5 - 3750\)

\(F \times 700 = 12730.5\)

\(F = \frac{12730.5}{700} \approx 18.5 \text{ N}\)