(i) Using the work-energy principle, the work done by the pulling force is equal to the increase in kinetic energy minus the decrease in potential energy plus the work done against resistance.

Given: Work done by pulling force = 1150 J, mass \(m = 16\) kg, distance \(s = 50\) m, \(\sin \theta = 0.05\), final speed \(v = 10\) m/s.

Potential energy change \(= mgh = 16 \times g \times 50 \times 0.05\).

Kinetic energy change \(= \frac{1}{2} m v^2 = \frac{1}{2} \times 16 \times 10^2\).

1150 = \(\frac{1}{2} \times 16 \times 10^2 - 16 \times g \times 50 \times 0.05 + \text{WD by resistance}\).

Solving for work done by resistance, \(\text{WD by resistance} = 750\) J.

(ii) When the block is pulled up from B to A, the work done by the pulling force is equal to the gain in kinetic energy plus the gain in potential energy plus the work done against resistance.

1150 = increase in kinetic energy + \(16 \times g \times 50 \times 0.05 + 750\).

Since the work done by the pulling force and the work done against resistance are the same as in the downward motion, the gain in kinetic energy is zero.

Thus, the speed at A is the same as the speed at B.

First, calculate the increase in potential energy (PE):

\(\text{Increase in PE} = 1250 \times 10 \times 600 \times \sin 2.5^\circ\)

Next, calculate the decrease in kinetic energy (KE):

\(\text{Decrease in KE} = \frac{1}{2} \times 1250 \times (30^2 - v_{\text{top}}^2)\)

Calculate the work done against resistance:

\(\text{WD against resistance} = 400 \times 600\)

Using the work-energy principle:

\(450000 = 327145 + 240000 - 562500 + 625v_{\text{top}}^2\)

Solve for \(v_{\text{top}}\):

\(v_{\text{top}} = 26.7 \text{ m s}^{-1}\)

The work done by a force is given by the formula:

\(WD = Fd \cos \alpha\)

where \(F\) is the force, \(d\) is the distance, and \(\alpha\) is the angle between the force and the direction of motion.

Here, \(F = 45 \text{ N}\), \(d = 25 \text{ m}\), and \(\alpha = 14^{\circ}\).

Substituting these values into the formula gives:

\(WD = 45 \times 25 \times \cos 14^{\circ}\)

Calculating this gives:

\(WD = 1090 \text{ J}\)

Therefore, the work done is 1090 J (1.09 kJ).

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

\(WD = Fd \cos \alpha\)

where \(F = 6 \text{ N}\) is the tension, \(d\) is the distance traveled, and \(\alpha = 24^\circ\) is the angle.

The distance \(d\) is given by the speed multiplied by time:

\(d = 0.5 \times 8 = 4 \text{ m}\)

Substituting the values into the work done formula:

\(WD = 6 \times 4 \times \cos 24^\circ\)

\(WD = 6 \times 4 \times 0.9135 \approx 21.9 \text{ J}\)

The work done by a force is given by the formula:

\(W = Fd \cos \alpha\)

where \(W\) is the work done, \(F\) is the magnitude of the force, \(d\) is the distance moved, and \(\alpha\) is the angle between the force and the direction of movement.

Given:

• \(W = 75 \text{ J}\)
• \(d = 5 \text{ m}\)
• \(\alpha = 60^{\circ}\)

Substitute these values into the formula:

\(75 = F \times 5 \times \cos 60^{\circ}\)

\(\cos 60^{\circ} = 0.5\)

\(75 = F \times 5 \times 0.5\)

\(75 = 2.5F\)

\(F = \frac{75}{2.5}\)

\(F = 30 \text{ N}\)