First, calculate the work done by the 120 N force:

\(\text{Work done} = 120 \times 5 \times \cos 20^{\circ} = 563.81557 \ldots\)

Next, calculate the change in potential energy:

\(\text{PE change} = 10g \times 5 \times \sin 30^{\circ} = 250\)

Apply the work-energy principle:

\(120 \times 5 \cos 20^{\circ} - 10g \times 5 \sin 30^{\circ} - 200 = \frac{1}{2} \times 10 \times v^2\)

\(563.815 \ldots - 250 - 200 = 5v^2\)

Solve for \(v\):

\(v^2 = \frac{113.815 \ldots}{5}\)

\(v = \sqrt{22.763 \ldots} \approx 4.77 \text{ ms}^{-1}\)

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

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

\(\text{KE gain} = \frac{1}{2} \times 80 \times (5.5^2 - 4^2) = 570 \text{ J}\)

The work done against resistance is:

\(\text{WD against Res} = 60P\)

According to the work-energy equation:

\(\frac{1}{2} \times 80 \times (5.5^2 - 4^2) + 60P = 1200\)

Substituting the KE gain:

\(570 + 60P = 1200\)

Solving for \(P\):

\(60P = 1200 - 570\)

\(60P = 630\)

\(P = \frac{630}{60} = 10.5\)

(i) The work done by the man is calculated using the formula for work done by a force at an angle:

\(WD = F \cdot d \cdot \cos \theta\)

where \(F = 35 \text{ N}\), \(d = 12 \text{ m}\), and \(\theta = 20^\circ\).

Thus,

\(WD = 35 \cdot 12 \cdot \cos 20^\circ = 395 \text{ J}\)

(ii) To find the speed attained by the wheelbarrow, we can use the work-energy principle or Newton's second law.

Using the work-energy principle:

The work done against resistance is \(15 \times 12 = 180 \text{ J}\).

The net work done is used to increase the kinetic energy:

\(35 \cdot 12 \cdot \cos 20^\circ - 180 = \frac{1}{2} \cdot 25 \cdot v^2\)

Solving for \(v\):

\(395 - 180 = \frac{1}{2} \cdot 25 \cdot v^2\)

\(215 = 12.5 \cdot v^2\)

\(v^2 = \frac{215}{12.5}\)

\(v = \sqrt{17.2} \approx 4.14 \text{ m/s}\)

Alternatively, using Newton's second law:

The net force is \(35 \cos 20^\circ - 15 = 25a\).

Solving for \(a\):

\(a = \frac{35 \cos 20^\circ - 15}{25} \approx 0.716 \text{ m/s}^2\)

Using the equation \(v^2 = u^2 + 2as\) with \(u = 0\):

\(v^2 = 2 \cdot 0.716 \cdot 12\)

\(v = \sqrt{17.2} \approx 4.14 \text{ m/s}\)

To find \(\theta\), we use the formula for work done by a force at an angle:

\(\text{Work Done (WD)} = Fd \cos \theta\)

Given: \(F = 20 \text{ N}\), \(d = 1.5 \times 12 \text{ m} = 18 \text{ m}\), and \(\text{WD} = 50 \text{ J}\).

Substitute these values into the formula:

\(50 = 20 \times 18 \times \cos \theta\)

Simplify to find \(\cos \theta\):

\(\cos \theta = \frac{50}{360}\)

\(\cos \theta = 0.1388\)

Find \(\theta\) using the inverse cosine function:

\(\theta = \cos^{-1}(0.1388)\)

\(\theta \approx 82^\circ\)

The potential energy (PE) lost by the particle is given by:

\(\text{PE loss} = mgh = 0.6 \times 10 \times 8 = 48 \text{ J}\)

The kinetic energy (KE) gained by the particle is given by:

\(\text{KE gain} = \frac{1}{2} mv^2 = \frac{1}{2} \times 0.6 \times 10^2 = 30 \text{ J}\)

The work done against air resistance is the difference between the potential energy lost and the kinetic energy gained:

\(\text{Work done against resistance} = 48 - 30 = 18 \text{ J}\)