(i) The kinetic energy (KE) at A is equal to the loss of potential energy (PE) from O to A. The vertical height from O to A is given by:

\(h = 2.4 \sin 50^\circ\)

The loss of PE is:

\(\text{PE loss} = mgh = 0.8 \times 9.8 \times 2.4 \sin 50^\circ\)

\(\text{KE at A} = \text{PE loss} = 14.7 \text{ J}\)

(ii) Using the kinetic energy at C, which is equal to the kinetic energy at A:

\(\frac{1}{2} mv^2 = 14.7\)

\(v^2 = \frac{2 \times 14.7}{0.8}\)

\(v = \sqrt{36.75} \approx 6.06 \text{ m s}^{-1}\)

(iii) Using the principle of conservation of energy, the greatest speed is 8 m s^-1:

\(\frac{1}{2} m (8^2) = mgh + \frac{1}{2} m (6.06^2)\)

\(32 = gh + 18.36\)

\(gh = 32 - 18.36 = 13.64\)

\(h = \frac{13.64}{9.8} \approx 1.36 \text{ m}\)

(i) Using the principle of conservation of energy, the kinetic energy (KE) gained by the particle is equal to the potential energy (PE) lost. The equation is:

\(\frac{1}{2} m v^2 - \frac{1}{2} m (7^2) = m g \times 5\)

Substituting the given values, \(m = 0.35\) kg, \(g = 10\) m/s²:

\(\frac{1}{2} \times 0.35 \times v^2 - \frac{1}{2} \times 0.35 \times 49 = 0.35 \times 10 \times 5\)

Solving for \(v\):

\(0.175 v^2 - 8.575 = 17.5\)

\(0.175 v^2 = 26.075\)

\(v^2 = 149\)

\(v = 12.2\) m/s

(ii) The work done against resistance is the difference between the initial mechanical energy and the final mechanical energy. The work done (WD) is given by:

\(\text{WD} = \text{PE loss} - \text{KE gain}\)

\(\text{WD} = m g h - \frac{1}{2} m (v_B^2 - v_A^2)\)

Substituting the values, \(h = 5\) m, \(v_A = 7\) m/s, \(v_B = 11\) m/s:

\(\text{WD} = 0.35 \times 10 \times 5 - \frac{1}{2} \times 0.35 \times (11^2 - 7^2)\)

\(\text{WD} = 17.5 - \frac{1}{2} \times 0.35 \times (121 - 49)\)

\(\text{WD} = 17.5 - 12.6\)

\(\text{WD} = 4.9\) J

(i) To find the work done by the driving force from A to B, we first calculate the increase in kinetic energy (KE) of the lorry:

Increase in KE = \(\frac{1}{2} \times 12500 \times (25^2 - 17^2)\)

\(= \frac{1}{2} \times 12500 \times (625 - 289)\)

\(= \frac{1}{2} \times 12500 \times 336\)

\(= 2100 \text{ kJ}\)

The total work done by the driving force is the sum of the work done against resistance and the increase in KE:

\(Work done by driving force = 2100 kJ + 5000 kJ = 7100 kJ\)

(ii) For the section from B to C, we use the work-energy principle. The work done by the driving force is 3300 kJ, and the work done against resistance is \(4800 \times 500 \div 1000 = 2400 \text{ kJ}\).

The potential energy (PE) gain is given by:

\(PE gain = (3300 + 2100) - 2400\)

\(= 3000 \text{ kJ}\)

Using \(mgh = \text{PE gain}\), we find the height \(h\):

\(12500 \times 10 \times h = 3000 \times 1000\)

\(125000h = 3000000\)

\(h = \frac{3000000}{125000} = 24 \text{ m}\)

(a) Using Newton's Second Law, the net force is given by:

\(F - 30 = 70 \times 0.3\)

\(F = 51 \text{ N}\)

The power developed by the cyclist is:

\(P = Fv = 51 \times 4 = 204 \text{ W}\)

(b) The change in kinetic energy is calculated as:

\(\text{Change in KE} = \frac{1}{2} \times 70 \times 12^2 - \frac{1}{2} \times 70 \times 6^2\)

\(= 3780 \text{ J}\)

(c) Using the work-energy principle:

\(70g \times d \sin 5 - 30d = 3780\)

Solving for \(d\):

\(d = 122 \text{ m}\)

(i) The greatest speed of the particle occurs at the lowest point M. Using conservation of energy, the loss of potential energy (PE) is converted into kinetic energy (KE):

\(Loss of PE = KE gain\)

\(mgh = \frac{1}{2} mv^2\)

\(0.5 \times 9.8 \times 2.45 = \frac{1}{2} \times 0.5 \times v^2\)

\(v^2 = 24.5\)

\(v = 7 \text{ m/s}\)

(ii) The kinetic energy of the particle at N is calculated by considering the height difference from L to N:

\(\text{KE at } N = 0.5 \times g \times (2.45 - 1.2)\)

\(= 0.5 \times 9.8 \times 1.25\)

\(= 6.25 \text{ J}\)

(iii) To find the least value of v for which the particle will reach L, use the kinetic energy at N:

\(\frac{1}{2} mv^2 = 6.25\)

\(\frac{1}{2} \times 0.5 \times v^2 = 6.25\)

\(v^2 = 25\)

\(v = 5 \text{ m/s}\)