To solve this problem, we apply Newton's second law to both masses.

For block B on the table, the frictional force is given by:

\(F = 0.5 imes 0.6g\)

For particle A hanging vertically, the forces are:

\(0.4g - T = 0.4a\)

For block B on the table, the forces are:

\(T - F = 0.6a\)

Substituting the expression for frictional force:

\(T - 0.5 imes 0.6g = 0.6a\)

We solve these two equations simultaneously:

1. \(0.4g - T = 0.4a\)

2. \(T - 0.3g = 0.6a\)

Adding the two equations to eliminate T:

\(0.4g - 0.3g = 0.4a + 0.6a\)

\(0.1g = a\)

Thus, the acceleration \(a = 1 \text{ m/s}^2\).

Substitute \(a = 1\) into equation 1:

\(0.4g - T = 0.4 \times 1\)

\(T = 0.4g - 0.4\)

\(T = 3.6 \text{ N}\)

For part (ii), using the equation of motion:

\(s = ut + \frac{1}{2}at^2\)

Where \(s = 3 \text{ m}\), \(u = 0\), \(a = 1 \text{ m/s}^2\):

\(3 = 0 + \frac{1}{2} \times 1 \times t^2\)

\(t^2 = 6\)

\(t = \sqrt{6} \approx 2.45 \text{ s}\)

(i) For particle A in limiting equilibrium, the tension \(T\) in the string is equal to the weight of A, so \(T = 0.2g\).

For particle B, the tension \(T\) is equal to the frictional force \(F\), so \(T = F\).

The normal reaction \(R\) on B is \(R = 0.3g\).

Using the friction formula \(F = \mu R\), we have \(0.2g = \mu \times 0.3g\).

Solving for \(\mu\), we get \(\mu = \frac{2}{3}\).

(ii) With the additional force on B, the vertical component is 1.8 N upwards, so the new normal reaction is \(R = 0.3g - 1.8\).

The frictional force is \(F = \frac{2}{3}(0.3g - 1.8)\).

In limiting equilibrium, the horizontal force \(X\) is equal to the sum of the tension and the frictional force, so \(X = T + F\).

Substituting \(T = 0.2g\) and \(F = \frac{2}{3}(0.3g - 1.8)\), we find \(X = 2.8\).

Let the acceleration of the system be a and the tension in the string be T.

For particle P on the table, applying Newton's second law:

\(T = 1.7a\)

For particle Q hanging vertically, applying Newton's second law:

\(0.3g - T = 0.3a\)

Substitute the expression for T from the first equation into the second equation:

\(0.3g - 1.7a = 0.3a\)

Combine and solve for a:

\(0.3g = 2a\)

\(a = \frac{0.3g}{2}\)

Using \(g = 9.8 \text{ m/s}^2\):

\(a = \frac{0.3 \times 9.8}{2} = 1.47 \text{ m/s}^2\)

Rounding to one decimal place, \(a = 1.5 \text{ m/s}^2\).

Substitute \(a = 1.5 \text{ m/s}^2\) back into the equation for T:

\(T = 1.7 \times 1.5 = 2.55 \text{ N}\)

To solve this problem, we apply Newton's second law to both particles.

For particle A on the table:

\(T - 0.6 = 0.4a\)

For particle B hanging vertically:

\(0.1g - T = 0.1a\)

Adding these two equations to eliminate \(T\):

\(0.1g - 0.6 = 0.5a\)

Solving for \(a\):

\(a = \frac{0.1 \times 9.8 - 0.6}{0.5} = 0.8 \text{ m/s}^2\)

Substitute \(a = 0.8\) back into the first equation to find \(T\):

\(T - 0.6 = 0.4 \times 0.8\)

\(T = 0.92 \text{ N}\)

For part (ii), using \(v = at\) to find the speed of B after 1.5 s:

\(v = 0.8 \times 1.5 = 1.2 \text{ m/s}\)

(i) For the smooth table, apply Newton's second law to the system:

\(2.4g - T = 2.4a\)

\(T = 1.6a\)

Combine to solve for \(a\):

\(2.4g = (1.6 + 2.4)a\)

\(a = 6 \text{ m/s}^2\)

Using \(s = \frac{1}{2} a t^2\):

\(0.5 = \frac{1}{2} \times 6 \times t^2\)

\(t = 0.408 \text{ s}\)

(ii) For the rough table, the frictional force \(F = \frac{3}{8} \times 1.6g = 6 \text{ N}\).

Apply Newton's second law:

\(2.4g - 6 = (1.6 + 2.4)a\)

\(a = 4.5 \text{ m/s}^2\)

Find velocity \(v\) as B reaches the ground:

\(v^2 = 2 \times 4.5 \times 0.5\)

\(v = 2.12 \text{ m/s}\)

Deceleration of A:

\(-6 = 1.6a - a\)

\(a = -3.75 \text{ m/s}^2\)

Using \(v^2 = u^2 + 2as\) to find \(s\):

\(0 = 4.5 + 2 \times (-3.75) \times (s - 0.5)\)

\(s = 1.1 \text{ m}\)