(a) For the motion from A to B, the potential energy (PE) loss is converted into kinetic energy (KE):

\(0.2 \times 10 \times 0.5 = \frac{1}{2} \times 0.2 \times v_B^2\)

Solving gives \(v_B^2 = 10\).

For the motion from B to C, the frictional force \(F\) is given by:

\(R = 0.2 \times 10 \times \cos 30 = \sqrt{3}\)

\(F = \mu R = \frac{\sqrt{3}}{2} \times \sqrt{3} = 1.5\)

The work done against friction is \(1.5 \times 1\).

Using the work-energy principle:

\(\frac{1}{2} \times 0.2 \times 10 + 0.2 \times 10 \times 0.5 = 1.5 \times 1 + \frac{1}{2} \times 0.2 \times v_c^2\)

Solving gives \(v_c = \sqrt{5} \text{ m/s}\).

(b) If the particle comes to rest at C, then:

\(0 = 10 + 2a\)

\(a = -5\)

Using Newton's second law:

\(0.2 \times 10 \times \sin 30 - F = 0.2 \times -5\)

\(2 = \mu \sqrt{3}\)

Solving gives \(\mu = \frac{2}{\sqrt{3}}\).