The work done by the tension can be calculated using the formula:

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

where:

• \(F = 25 \text{ N}\) (tension in the string)
• \(\alpha = 20^\circ\) (angle with the horizontal)
• \(d\) is the distance traveled by the block

The distance \(d\) can be calculated as:

\(d = \text{speed} \times \text{time} = 2 \times 8 = 16 \text{ m}\)

Substituting the values into the work done formula:

\(\text{WD} = 25 \times 16 \times \cos(20^\circ)\)

Calculating the above expression gives:

\(\text{WD} \approx 376 \text{ J}\)

The work done by the tension in the rope is given by the formula:

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

where \(F\) is the force (tension), \(d\) is the distance, and \(\alpha\) is the angle of inclination.

Substituting the given values:

\(8200 = 180 \times 50 \times \cos \alpha\)

Simplifying, we get:

\(8200 = 9000 \cos \alpha\)

\(\cos \alpha = \frac{8200}{9000}\)

\(\cos \alpha = 0.9111\)

Taking the inverse cosine, we find:

\(\alpha = \cos^{-1}(0.9111)\)

\(\alpha \approx 24.3^\circ\)

First, calculate the gravitational potential energy (PE) lost by the particle:

\(\text{Loss of PE} = mgh = 1.6 \times 9.8 \times 9 = 144 \text{ J}\)

Next, calculate the kinetic energy (KE) gained by the particle just before hitting the ground:

\(\text{KE} = \frac{1}{2}mv^2 = \frac{1}{2} \times 1.6 \times 12^2 = 115.2 \text{ J}\)

The work done against air resistance (WD) is the difference between the loss of potential energy and the gain in kinetic energy:

\(\text{WD} = \text{Loss of PE} - \text{KE} = 144 - 115.2 = 28.8 \text{ J}\)

(i) Use the equation of motion: \(v^2 = u^2 + 2as\).

Substitute \(u = 3 \text{ m/s}, a = 2.5 \text{ m/s}^2, s = 8 \text{ m}\).

\(v^2 = 3^2 + 2 \times 2.5 \times 8\)

\(v^2 = 9 + 40 = 49\)

\(v = 7 \text{ m/s}\)

(ii) Kinetic energy gain: \(\frac{1}{2} \times 0.8 \times (7^2 - 3^2) = 16 \text{ J}\)

Potential energy loss: \(16 + 7 = 23 \text{ J}\)

\(0.8 \times 10 \times 8 \sin \alpha = 23\)

\(\sin \alpha = \frac{23}{64}\)

\(\alpha = 21.1^\circ \text{ or } 0.368 \text{ radians}\)

(iii) Average speed \(= \frac{3 + 7}{2} = 5 \text{ m/s}\)

Use \(v^2 = u^2 + 2as\) for point X: \(5^2 = 3^2 + 2 \times 2.5 \times s\)

\(25 = 9 + 5s\)

\(s = 3.2 \text{ m}\)

Work done \(= \frac{7}{8} \times 3.2 = 2.8 \text{ J}\)

First, calculate the distance moved using the formula for average velocity:

\(s = \frac{(u + v)}{2} \times t\)

where \(u = 0\) m/s (initial speed), \(v = 0.5\) m/s (final speed), and \(t = 7\) s.

\(s = \frac{(0 + 0.5)}{2} \times 7 = 1.75\) m

Next, calculate the potential energy gain:

\(\text{PE gain} = mgh = 160 \times 9.8 \times 1.75 = 2800\) J

Calculate the kinetic energy gain:

\(\text{KE gain} = \frac{1}{2} mv^2 = \frac{1}{2} \times 160 \times (0.5)^2 = 20\) J

The total work done (WD) by the crane is the sum of the potential energy gain and the kinetic energy gain:

\(\text{WD} = 2800 + 20 = 2820\) J