(i) The work done by the pulling force is calculated using the formula:

\(\text{Work Done} = Fd \cos \beta\)

where \(F = 30 \text{ N}\), \(d = 20 \text{ m}\), and \(\beta = 10^{\circ}\).

\(\text{Work Done} = 30 \times 20 \times \cos 10^{\circ} = 591 \text{ J}\)

(ii) The gain in potential energy is given by:

\(\text{PE gain} = mgh\)

where \(m = 8 \text{ kg}\), \(g = 10 \text{ m/s}^2\), and \(h = d \sin \alpha\) with \(\alpha = 15^{\circ}\).

\(h = 20 \times \sin 15^{\circ}\)

\(\text{PE gain} = 8 \times 10 \times 20 \times \sin 15^{\circ} = 414 \text{ J}\)

(iii) The work done against the resistance to motion of the box is the difference between the work done by the pulling force and the gain in potential energy:

\(\text{Work done against resistance} = 591 \text{ J} - 414 \text{ J} = 177 \text{ J}\)

(i) The loss in kinetic energy (KE) is calculated using the formula:

\(\text{Loss in KE} = \frac{1}{2} m (v^2 - u^2)\)

where \(m = 50 \text{ kg}\), \(u = 7 \text{ m/s}\), and \(v = 3 \text{ m/s}\).

\(\text{Loss in KE} = \frac{1}{2} \times 50 \times (3^2 - 7^2) = 1000 \text{ J}\)

(ii) The gain in potential energy (PE) is given by:

\(\text{Gain in PE} = mgh\)

where \(h = 15 \text{ m}\) and \(g = 10 \text{ m/s}^2\).

\(\text{Gain in PE} = 50 \times 10 \times 15 = 7500 \text{ J}\)

(iii) The work done (WD) by the pulling force is calculated by considering the work done against resistance and the change in energy:

\(\text{WD by resistance} = 7.5 \times 200 = 1500 \text{ J}\)

\(\text{WD by pulling force} = \text{Gain in PE} + \text{Loss in KE} + \text{WD by resistance}\)

\(\text{WD by pulling force} = 7500 + 1500 - 1000 = 8000 \text{ J}\)

(iv) To find \(\alpha\), use the formula for work done by the force:

\(\text{WD} = F \cdot d \cdot \cos \alpha\)

where \(F = 45 \text{ N}\) and \(d = 200 \text{ m}\).

\(8000 = 45 \times 200 \times \cos \alpha\)

\(\cos \alpha = \frac{8000}{45 \times 200}\)

\(\alpha = \cos^{-1} \left( \frac{8000}{9000} \right) \approx 27.3 \degree\)

(i) Apply Newton's second law to both particles. For particle A:

\(0.3g - T = 0.3a\)

For particle B:

\(T - 0.2g = 0.2a\)

Adding these equations gives:

\(0.3g - 0.2g = 0.3a + 0.2a\)

Solving for a gives:

\(a = 2\)

(ii) Using the equation of motion:

\(v = u + at\)

\(where initial velocity u = 0, a = 2 m s^-2, and t = 1.2 s, we find:\)

\(v = 0 + 2 imes 1.2 = 2.4 ext{ m s}^{-1}\)

(iii) The gain in potential energy while the string is slack is given by:

\(PE = \frac{1}{2} imes 0.2 imes (2.4)^2 = 0.576 ext{ J}\)

The total distance moved by B is:

\(s_1 + s_2 = 1.728 ext{ m}\)

Total potential energy gain is:

\(PE = 0.2g imes 1.728 = 3.456 ext{ J}\)

(i) The increase in kinetic energy \(\Delta KE\) is given by:

\(\Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2\)

Since the crate starts from rest, \(u = 0\), and \(v = 2\) m/s:

\(\Delta KE = \frac{1}{2} \times 50 \times 2^2 = 100 \text{ J}\)

(ii) The work done against the resistance \(W_R\) is:

\(W_R = \text{Resistance} \times \text{Distance} = 250 \times 20 = 5000 \text{ J}\)

(iii) The work done by the force \(W_F\) is:

\(W_F = F \cdot d \cdot \cos \alpha = 400 \times 20 \times \cos \alpha\)

Using the work-energy principle, the net work done is equal to the change in kinetic energy:

\(W_F - W_R = \Delta KE\)

\(400 \times 20 \times \cos \alpha - 5000 = 100\)

\(8000 \cos \alpha = 5100\)

\(\cos \alpha = \frac{5100}{8000} = 0.6375\)

\(\alpha = \cos^{-1}(0.6375) \approx 50.4^\circ\)

(i) To find the acceleration at A, use Newton's second law. The driving force provided by the engine is given by:

\(\text{Driving force} = \frac{20,000}{10} = 2000 \text{ N}\)

Using the equation \(DF - R = ma\), where \(R = 500 \text{ N}\), we have:

\(2000 - 500 = 1200a\)

\(1500 = 1200a\)

\(a = \frac{1500}{1200} = 1.25 \text{ m/s}^2\)

(ii) To find the distance AB, consider the work-energy principle. The change in kinetic energy (KE) is:

\(\Delta KE = \frac{1}{2} \times 1200 \times (25^2 - 10^2) = 315,000 \text{ J}\)

The work done by the car's engine is:

\(\text{Work done} = \frac{20,000 \times 30.5}{1} = 610,000 \text{ J}\)

The work done by the engine is equal to the increase in KE plus the work done against resistance:

\(610,000 = 315,000 + \text{Work done against resistance}\)

\(\text{Work done against resistance} = 610,000 - 315,000 = 295,000 \text{ J}\)

Since \(\text{Work done against resistance} = R \times AB\), we have:

\(500 \times AB = 295,000\)

\(AB = \frac{295,000}{500} = 590 \text{ m}\)