First, calculate the change in potential energy (PE) as the block moves from A to B:

\(\text{PE change} = 60g \times 17.5 = 10500 \text{ J}\)

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

\(\text{KE change} = \frac{1}{2} \times 60 \times (8.5^2 - 3.5^2) = 1800 \text{ J}\)

The work done (WD) against resistance is:

\(\text{WD against resistance} = 6 \times 250 = 1500 \text{ J}\)

The total work done by the pulling force is the sum of the changes in PE, KE, and the work done against resistance:

\(\text{WD by pulling force} = 10500 + 1800 + 1500 = 13800 \text{ J}\)

Using the formula for work done by the pulling force:

\(\text{WD by pulling force} = 50 \cos \alpha \times 250\)

Equating the two expressions for work done:

\(50 \cos \alpha \times 250 = 13800\)

Solve for \(\cos \alpha\):

\(\cos \alpha = \frac{13800}{12500} = \frac{102}{125}\)

Finally, calculate \(\alpha\):

\(\alpha = \cos^{-1} \left( \frac{102}{125} \right) \approx 35.3^\circ\)

(i) Using Newton's second law and the equation for friction \(F = \mu R\), we have:

\(-\left(\frac{1}{3}\right)(W \cos \alpha) - W \sin \alpha = \left(\frac{W}{g}\right)a\)

Substituting \(\sin \alpha = 0.28\) and \(\cos \alpha = \sqrt{1 - \sin^2 \alpha} \approx 0.96\), we get:

\(-(0.32 - 0.28)g = a\)

\(a = -6\).

(ii) Using the equation \(0 = u^2 + 2as\) where \(u = 5.4 \text{ m s}^{-1}\) and \(a = -6 \text{ m s}^{-2}\):

\(0 = (5.4)^2 + 2(-6)s\)

Solving for \(s\), we find:

\(s = 2.43 \text{ m}\)

(i) Apply Newton's second law to the cyclist and bicycle system. The net force equation along the incline is:

\(F - 780 \times \frac{36}{325} - 32 = 78 \times (-0.2)\)

Solving for \(F\):

\(F = 78 \times (-0.2) + 780 \times \frac{36}{325} + 32\)

\(F = -15.6 + 86.4 + 32\)

\(F = 102.8\)

Rounding to the nearest whole number, \(F = 103\) N.

(ii) Use the equation of motion:

\(0 = 7^2 + 2(-0.2)s\)

\(0 = 49 - 0.4s\)

\(0.4s = 49\)

\(s = \frac{49}{0.4}\)

\(s = 122.5 \text{ m}\)

(i) Resolve forces parallel to the plane: The force equation is \(F = 5.9 - 6.1 \sin \alpha\). The normal reaction is \(R = 6.1 \cos \alpha\). For equilibrium, \(5.9 - 6.1 \sin \alpha \leq \mu (6.1 \cos \alpha)\). Given \(\tan \alpha = \frac{11}{60}\), \(\sin \alpha = \frac{11}{61}\) and \(\cos \alpha = \frac{60}{61}\). Substituting, \(5.9 - 6.1 \times \frac{11}{61} \leq \mu \times 6.1 \times \frac{60}{61}\). Simplifying gives \(\mu \geq \frac{4}{5}\).

(ii) When the force acts down the plane, the net force is \(6.1 \times \frac{11}{61} + 5.9 - \mu \times 6.1 \times \frac{60}{61} > 0\). Simplifying gives \(\mu < \frac{7}{6}\).

(iii) Using Newton's second law, \(6.1 \times \frac{11}{61} + 5.9 - \mu \times 6.1 \times \frac{60}{61} = 6.1 \times 1.7\). Solving gives \(\mu = 0.994\).

(i) Resolve the forces normal to the plane: \(R = mg \cos \alpha\).

Given \(R = 1.2\) N and \(\cos \alpha = 0.96\), we have:

\(1.2 = mg \times 0.96\)

\(mg = \frac{1.2}{0.96} = 1.25\)

\(m = \frac{1.25}{10} = 0.125\) kg

(ii) Use Newton's second law along the plane:

\(-mg \sin \alpha - F = ma\)

\(-0.125 \times 10 \times 0.28 - 0.4 = 0.125a\)

\(-0.35 - 0.4 = 0.125a\)

\(a = -6\)

Deceleration is 6 m/s²

(iii) Compare magnitudes of \(\mu R\) and \(mg \sin \alpha\):

\(\mu R = 0.4\) and \(mg \sin \alpha = 0.35\)

Since \(\mu R > mg \sin \alpha\), the particle remains at rest.