(a)(i) The power developed by the engine is given by \(P = Fv\), where \(F\) is the total resistance force and \(v\) is the velocity. The total resistance force is \(440 + 280 = 720 \text{ N}\). Therefore, \(P = 720 \times 30 = 21600 \text{ W} = 21.6 \text{ kW}\).

(a)(ii) The power is decreased by 8 kW, so the new power is \(21600 - 8000 = 13600 \text{ W}\). The new driving force (DF) is \(\frac{13600}{30} = 453.33 \text{ N}\). Using Newton's second law for the system: \(DF - (440 + 280) = 2050a\). Solving for \(a\), we get \(a = \frac{453.33 - 720}{2050} = -0.13 \text{ m s}^{-2}\). For the tension \(T\) in the tow-bar, use \(T - 280 = 800a\), giving \(T = 176 \text{ N}\).

(b)(i) The driving force up the incline is \(DF = 720 + 2050 \times 0.06 = 1950 \text{ N}\). Using \(P = DF \times v\), we have \(28000 = 1950 \times v\), so \(v = \frac{28000}{1950} = 14.4 \text{ m s}^{-1}\).

(b)(ii) The increase in potential energy (PE) is \(PE = mgh\), where \(h\) is the height gained in 60 s. Using \(v = 14.4 \text{ m s}^{-1}\), the distance traveled is \(14.4 \times 60\). The height \(h = 14.4 \times 60 \times 0.06\). Therefore, \(PE = 800 \times 9.8 \times 14.4 \times 60 \times 0.06 = 414000 \text{ J} = 414 \text{ kJ}\).

Given:

• Mass \(m = 1.6 \text{ kg}\)
• Initial speed \(u = 20 \text{ m/s}\)
• \(\tan \alpha = \frac{3}{4} \Rightarrow \sin \alpha = \frac{3}{5}\)

Using the conservation of energy, the initial kinetic energy is converted into gravitational potential energy:

\(\frac{1}{2} m u^2 = m g x \sin \alpha\)

Substitute the known values:

\(\frac{1}{2} \times 1.6 \times 20^2 = 1.6 \times 9.8 \times x \times \frac{3}{5}\)

\(320 = 9.6x\)

Solve for \(x\):

\(x = \frac{320}{9.6} = 33.3 \text{ m}\)

Initial kinetic energy (KE) is given by:

\(\frac{1}{2} \times 0.6 \times 4^2 = 4.8 \text{ J}\)

Potential energy (PE) loss is given by:

\(0.6 \times g \times 15 \times \sin 10^{\circ} = 15.628 \text{ J}\)

Final kinetic energy (KE) is given by:

\(\frac{1}{2} \times 0.6 \times v^2\)

Using the energy conservation equation:

\(0.6 \times g \times 15 \times \sin 10^{\circ} + \frac{1}{2} \times 0.6 \times 4^2 = \frac{1}{2} \times 0.6 \times v^2\)

Solving for \(v\):

\(v = 8.25 \text{ m s}^{-1}\)

(a)(i) To find the speed at the bottom of the slope, use conservation of energy. The potential energy (PE) at the top is converted to kinetic energy (KE) at the bottom:

\(\text{PE} = 35g \times 2.5 \sin 30^{\circ}\)

\(\frac{1}{2} \times 35 \times v^2 = 35g \times 2.5 \sin 30^{\circ}\)

Solving for \(v\), we get \(v = 5 \text{ m/s}\).

(a)(ii) Use work-energy principle for the horizontal section:

\(\frac{1}{2} \times 35 \times 5^2 = 250d\)

Solving for \(d\), we get \(d = 1.75 \text{ m}\).

(b) For the rough slope, use work-energy principle and Newton's second law:

\(\frac{1}{2} \times 35 \times v^2 = 250 \times 1.05\)

\(v^2 = 15\)

\(R = 35g \cos 30^{\circ}\)

Using \(v^2 = u^2 + 2as\) to find \(a\), we have \(a = 3\).

Using Newton's second law down the slope:

\(35g \sin 30^{\circ} - F = 35a\)

\(F = \mu R\)

Solving for \(\mu\), we get \(\mu = 0.231\).

(a) Apply Newton's second law for particle P:

\(0.8 + 0.5g \sin 30^{\circ} - T = 0.5a\)

For particle Q:

\(T - 0.3g \sin 45^{\circ} = 0.3a\)

For the system:

\(0.8 + 0.5g \sin 30^{\circ} - 0.3g \sin 45^{\circ} = 0.8a\)

Solving for T using the equations:

\(T = 2.56 \text{ N (3sf)}\)

(b) Using energy method:

Kinetic energy (KE) and potential energy (PE) for m kg particle:

\(\frac{1}{2} m \times 0.6^2 = 0.18m\)

\(mg \sin 45^{\circ} = 5\sqrt{2}m\)

KE and PE for 0.5 kg particle:

\(\frac{1}{2} \times 0.5 \times 0.6^2 = 0.09\)

\(0.5g \sin 30^{\circ} = 2.5\)

Apply the work-energy equation:

\(0.5g \times 1 \times \sin 30^{\circ} - mg \times 1 \times \sin 45^{\circ} + 0.8 \times 1 = \frac{1}{2} \times (0.5 + m) \times 0.6^2 + 0.5\)

Solving gives:

\(m = 0.374 \text{ kg}\)