(i) First, calculate the normal reaction force, \(R\), using:

\(R = 0.8g \cos 10^{\circ} \approx 7.88 \text{ N}\)

The frictional force, \(F\), is given by:

\(F = \mu R = 0.4 \times 0.8g \cos 10^{\circ} \approx 3.15 \text{ N}\)

Using Newton's second law along the plane:

\(-0.8g \sin 10^{\circ} - F = 0.8a\)

\(-0.8g \sin 10^{\circ} - 3.15 = 0.8a\)

Solving for \(a\):

\(a = \frac{-0.8g \sin 10^{\circ} - 3.15}{0.8} \approx -5.68 \text{ m s}^{-2}\)

(ii) Using the equation of motion \(v^2 = u^2 + 2as\) with final velocity \(v = 0\), initial velocity \(u = 12 \text{ m s}^{-1}\), and acceleration \(a = -5.68 \text{ m s}^{-2}\):

\(0 = 12^2 + 2(-5.68)s\)

\(s = \frac{144}{2 \times 5.68} \approx 12.7 \text{ m}\)

(i) Using Newton's second law for the system:

\(2500 - 2000g \times 0.1 - 250 = 2000a\)

\(a = \frac{1}{8} = 0.125 \text{ m/s}^2\)

For tension \(T\):

\(2500 - T - 100 - 1200g \times 0.1 = 1200 \times 0.125\)

or

\(T - 150 - 800g \times 0.1 = 800 \times 0.125\)

Solving gives \(T = 1050 \text{ N}\).

(ii) With no driving force, using Newton's second law:

\(-2000g \times 0.1 - 250 = 2000a\)

\(a = -1.125 \text{ m/s}^2\)

Using \(v = u + at\):

\(0 = 30 - 1.125t\)

\(t = 26.7 \text{ s}\)

(a) The diagram should show three forces: the gravitational force acting downwards, the normal force perpendicular to the plane, and the frictional force opposing the motion.

(b) Resolve forces perpendicular to the plane: \(R = 8g \cos 30^{\circ} = 40\sqrt{3} \approx 69.282\). Resolve forces parallel to the plane and apply Newton's second law: \(8g \sin 30^{\circ} - F = 8 \times 2.4\). This gives \(F = 20.8\). Use \(F = \mu R\) to find \(\mu\):

\(8g \sin 30^{\circ} - 8g \mu \cos 30^{\circ} = 8 \times 2.4\)

\(40 - 40\sqrt{3} \mu = 19.2\)

\(\mu = \frac{20.8}{40\sqrt{3}} \approx 0.3\)

(c) Use the equation \(v^2 = u^2 + 2as\) with \(u = 0\), \(a = 2.4\), and \(s = 3\):

\(v^2 = 2 \times 2.4 \times 3\)

\(v = \sqrt{14.4} \approx 3.79 \text{ m/s}\)

(i) The acceleration \(a\) down the plane is given by \(a = g \sin \alpha = g \times \frac{16}{65}\).

Using the equation \(v^2 = u^2 + 2as\), where \(u = 0\), \(v = 8\), and \(s = S\), we have:

\(8^2 = 2 \times \frac{16}{65}g \times S\)

\(64 = \frac{32}{65}gS\)

\(S = \frac{64 \times 65}{32g} = 13\) m

To find the speed when \(s = \frac{1}{2}S = 6.5\) m, use \(v^2 = 2as\):

\(v^2 = 2 \times \frac{16}{65}g \times 6.5\)

\(v = \sqrt{\frac{32}{65}g \times 6.5} = 5.66\) m s^-1

(ii) The time \(T\) to travel distance \(S\) is given by \(v = u + at\), so \(8 = 0 + \frac{16}{65}gT\).

\(T = \frac{8 \times 65}{16g}\)

For \(\frac{1}{2}T\), the distance \(s\) is given by \(s = \frac{1}{2}a(\frac{T}{2})^2\):

\(s = \frac{1}{2} \times \frac{16}{65}g \times (\frac{8 \times 65}{32g})^2\)

\(s = 3.25\) m

(i) The acceleration for \(t < 0.8\) is calculated using the formula \(a = \frac{v}{t} = \frac{4}{0.8} = 5 \text{ m s}^{-2}\). Since the plane is smooth between A and B, the acceleration is due to gravity: \(a = g \sin \theta\). Therefore, \(5 = 10 \sin \theta\), giving \(\sin \theta = 0.5\). Thus, \(\theta = 30^\circ\).

(ii) For \(0.8 < t < 4.8\), the acceleration is \(\frac{-4}{4.8 - 0.8} = -1 \text{ m s}^{-2}\). Using Newton's second law, \(mg \sin 30^\circ - F = m(-1)\), where \(F\) is the frictional force. The frictional force \(F = \mu R\), where \(R = mg \cos 30^\circ\). Therefore, \(\mu = \frac{mg \sin 30^\circ + m}{mg \cos 30^\circ}\). Simplifying gives \(\mu = 0.693\).