(i) Apply Newton's second law to the lorry moving up the hill:

\(F - 24000g \sin 3^{\circ} - 3200 = 24000 \times 0.2\)

Solving for \(F\):

\(F = 24000 \times 9.8 \times \sin 3^{\circ} + 3200 + 24000 \times 0.2 = 20561 \text{ N}\)

The power developed by the engine is given by:

\(P = Fv = 20561 \times 25 = 514025 \text{ W} = 514 \text{ kW}\)

(ii) For steady speed, the net force is zero, so:

\(F = 3200 + 24000g \sin 3^{\circ} = 15761 \text{ N}\)

Using the power formula:

\(P = Fv \Rightarrow v = \frac{P}{F} = \frac{500000}{15761} = 31.7 \text{ m s}^{-1}\)

(i) The work done by the weightlifter is calculated using the formula for gravitational work done:

\(WD = mgh\)

where \(m = 200 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\), and \(h = 0.7 \text{ m}\).

Substituting the values, we get:

\(WD = 200 \times 9.8 \times 0.7 = 1400 \text{ J}\)

(ii) The average power developed is calculated using the formula:

\(\text{Power} = \frac{\text{Work Done}}{\text{Time}}\)

Substituting the values, we get:

\(\text{Power} = \frac{1400}{1.2} = 1166.67 \text{ W}\)

Rounding to the nearest whole number, the average power is \(1170 \text{ W}\).

Using the formula for power \(P = Fv\), where \(F\) is the force and \(v\) is the velocity, and applying Newton's second law \(F = ma\), we have:

At the first point:

\(\frac{P}{4.5} - R = 860 \times 4\)

At the second point:

\(\frac{P}{22.5} - R = 860 \times 0.3\)

Subtract the second equation from the first to eliminate \(R\):

\(\frac{P}{4.5} - \frac{P}{22.5} = 860(4 - 0.3)\)

Solving for \(P\):

\(P = 17900\) W

Substitute \(P\) back into one of the original equations to find \(R\):

\(\frac{17900}{4.5} - R = 860 \times 4\)

\(R = 537.5\) N

The work done by the tension is calculated using the formula:

\(ext{Work Done (WD)} = F imes s\)

where \(F = 500 \text{ N}\) is the tension and \(s\) is the distance traveled.

The distance \(s\) can be found using the speed \(v = 2.75 \text{ m/s}\) and time \(t = 40 \text{ s}\):

\(s = v imes t = 2.75 imes 40 = 110 \text{ m}\)

Thus, the work done is:

\(ext{WD} = 500 imes 110 = 55000 \text{ J}\)

The power applied by the tension is given by:

\(P = \frac{\text{WD}}{t} = \frac{55000}{40} = 1375 \text{ W}\)

Alternatively, power can also be calculated using:

\(P = F imes v = 500 imes 2.75 = 1375 \text{ W}\)

(a) The power of the winch is given as 4.5 kW, which is 4500 W. The work done by the winch is equal to the force times the distance, which is also equal to power times time. Therefore, we have:

\(4500 = \frac{600 \times 15}{t}\)

Solving for \(t\), we get:

\(t = \frac{600 \times 15}{4500} = 2 \text{ s}\)

(b) After the rope breaks, the only force acting on the crate is the resistance of 600 N. Using Newton's second law, \(F = ma\), we have:

\(600 = 200a\)

\(a = 3 \, \text{m/s}^2\)

The initial velocity \(u\) is the velocity at which the crate was moving before the rope broke. From part (a), the velocity \(v\) is:

\(600v = 4500 \Rightarrow v = 7.5 \, \text{m/s}\)

Using the equation \(v = u + at\) with \(v = 0\) (since the crate comes to rest), \(u = 7.5\), and \(a = -3\):

\(0 = 7.5 - 3t\)

Solving for \(t\), we get:

\(t = \frac{7.5}{3} = 2.5 \text{ s}\)