Two blocks (\(P\) and \(Q\)) with respectively masses \(2\) kg and \(1.5\) kg are joined by a massless thread. These blocks are mounted on a frictionless pulley which is fixed on the edge of a cube (\(S\)), as shown in the figure below. Block \(P\) is positioned on the top surface which has no friction and block \(Q\) is in contact with side-surface, having coefficient friction \(\mu\). The cube (\(S\)) moves towards the right with acceleration of \(\dfrac{g}{2}\), where \(g\) is gravitational acceleration. During this movement the block \(P\) and \(Q\) remain stationary. The value of \(\mu\) is _______. (take \(g = 10\) m/s\(^2\))

Let us analyze the forces acting on blocks \(P\) and \(Q\) in the non-inertial frame of reference of the cube \(S\), which is moving to the right with an acceleration \(a = \dfrac{g}{2} = 5 \text{ m/s}^2\).

In this frame, a pseudo force acts on both blocks towards the left.

For block \(P\) (mass \(m_P = 2 \text{ kg}\)):
The forces in the horizontal direction are the tension \(T\) in the string (acting towards the right) and the pseudo force \(F_{pP}\) (acting towards the left). Since block \(P\) is stationary relative to the cube and there is no friction on the top surface:
\(T = m_P a = 2 \times 5 = 10 \text{ N}\)

For block \(Q\) (mass \(m_Q = 1.5 \text{ kg}\)):
The horizontal forces are the normal reaction \(N\) from the side surface of the cube (acting towards the right) and the pseudo force \(F_{pQ}\) (acting towards the left). Since block \(Q\) is stationary horizontally relative to the cube:
\(N = m_Q a = 1.5 \times 5 = 7.5 \text{ N}\)

The vertical forces acting on block \(Q\) are the downward gravitational force \(m_Q g\), the upward tension \(T\), and the frictional force \(f\).
The downward force is \(m_Q g = 1.5 \times 10 = 15 \text{ N}\).
Since \(m_Q g > T\) (\(15 \text{ N} > 10 \text{ N}\)), the block has a tendency to slide downwards, so the static friction \(f\) must act upwards.
For vertical equilibrium of block \(Q\):
\(T + f = m_Q g\)
\(10 + f = 15 \Rightarrow f = 5 \text{ N}\)

To keep the block stationary, the required friction must be less than or equal to the maximum static friction (\(f \le \mu N\)). Assuming the limiting case to find the value of \(\mu\):
\(\mu N = f\)
\(\mu (7.5) = 5\)
\(\mu = \dfrac{5}{7.5} = \dfrac{2}{3} \approx 0.67\)

Answer: \(0.67\)