Coefficient of Friction Between Block and Moving Cube Surface

To find the coefficient of friction $\mu$, we analyze the forces acting on blocks $P$ and $Q$ in the non-inertial reference frame of the cube $(S)$, which is accelerating to the right with $a = \frac{g}{2}$.

In this non-inertial frame, a pseudo force $F_{\text{pseudo}} = m a$ acts on every mass towards the left.

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1. Analysis of Block $P$ ($m_P = 2\text{ kg}$):

Block $P$ lies on the frictionless top surface of the cube.

• Horizontal forces acting on $P$:
  • Tension force $T$ from the string towards the right (towards the pulley).
  • Pseudo force $F_{\text{pseudo}, P} = m_P a$ towards the left.

Since block $P$ is stationary relative to the cube: $T = m_P a = m_P \left(\frac{g}{2}\right)$ Substituting $m_P = 2\text{ kg}$: $T = 2 \cdot \left(\frac{g}{2}\right) = g$

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2. Analysis of Block $Q$ ($m_Q = 1.5\text{ kg}$):

Block $Q$ hangs vertically and touches the side surface of the cube.

• Horizontal direction:

  • The pseudo force $F_{\text{pseudo}, Q} = m_Q a$ pushes block $Q$ against the vertical side surface of the cube.
  • The normal force $N$ exerted by the vertical wall on block $Q$ is equal to this pseudo force: $N = m_Q a = m_Q \left(\frac{g}{2}\right) = 1.5 \left(\frac{g}{2}\right) = 0.75 g$

• Vertical direction:

  • Downward force due to gravity: $W_Q = m_Q g = 1.5 g$
  • Upward tension force from the string: $T = g$
  • Since $W_Q > T$, block $Q$ has a tendency to slide downwards. Therefore, the friction force $f$ acts vertically upwards.

For vertical equilibrium of block $Q$: $T + f = m_Q g$ $f = m_Q g - T = 1.5 g - 1.0 g = 0.5 g$

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3. Value of Coefficient of Friction $\mu$:

Using the relation for limiting friction force $f = \mu N$: $0.5 g = \mu (0.75 g)$

$\mu = \frac{0.5}{0.75} = \frac{2}{3} \approx 0.67$

Correct Answer: B ($0.67$)