Analysis of Motion in the Frame of the Plank:

Let the mass of the bar be $M$ and the mass of the block be $m$. The plank is accelerating horizontally with acceleration $a$. It is convenient to analyze the motion of the block-bar system from the non-inertial reference frame of the plank.

In this frame, a pseudo-force equal to $(M+m)a$ acts on the system (bar + block) in the direction opposite to the plank’s acceleration. This force acts effectively like a constant horizontal gravitational field, shifting the equilibrium position of the spring.

Step 1: Dynamics of Oscillation

The system starts from rest (relative to the plank) with the spring at its natural length. Due to the pseudo-force $(M+m)a$, the system oscillates about a new mean position. The maximum extension of the spring $x_{max}$ occurs at the extreme position of this oscillation.

Using the Work-Energy Theorem in the plank frame (from initial position to max extension):

$$ \text{Work by Pseudo Force} = \Delta \text{Potential Energy in Spring} $$

$$ (M+m)a \cdot x_{max} = \frac{1}{2} k x_{max}^2 $$

$$ x_{max} = \frac{2(M+m)a}{k} $$

Step 2: Maximum Acceleration of the System

At the instant of maximum extension, the spring force is maximum and acts towards the nail (left in the diagram, assuming acceleration is to the right). The magnitude of the restoring force is:

$$ F_s = k x_{max} = k \left[ \frac{2(M+m)a}{k} \right] = 2(M+m)a $$

The acceleration of the bar-block system relative to the plank, $a_{rel}$, at this instant is:

$$ a_{rel} = \frac{F_s}{M+m} = \frac{2(M+m)a}{M+m} = 2a $$

Step 3: Condition for No Sliding

For the block not to slide on the bar, the static friction $f_s$ must provide the required relative acceleration $a_{rel}$ to the block. The maximum possible static friction is $f_{max} = \mu N = \mu mg$.

Applying Newton’s Second Law to the block in the plank frame:

$$ m a_{rel} \le f_{max} $$

$$ m (2a) \le \mu m g $$

$$ 2a \le \mu g $$

$$ a \le \frac{\mu g}{2} $$