Step 1: Analyzing Relative Motion & Slipping

The bar accelerates with magnitude $A$ at angle $\theta$. Its components are:

• $A_x = A \cos\theta$
• $A_y = A \sin\theta$

Forces on Block P:

• Y-axis: The bar pushes block P in the $y$-direction. Since they must stay in contact, P acquires the bar’s y-acceleration. $$ N = m A_y = m A \sin\theta $$
• X-axis: Friction $f$ acts to pull P along with the bar in the $x$-direction. $$ f_{req} = m A_x = m A \cos\theta $$

Checking for Slip:

Max static friction $f_{max} = \mu N = \mu (m A \sin\theta)$.

We compare required friction vs. max friction:

$$ \frac{f_{req}}{f_{max}} = \frac{m A \cos\theta}{\mu m A \sin\theta} = \frac{\cot\theta}{\mu} $$

Using values $\cos\theta = 0.8$, $\sin\theta = 0.6 \implies \cot\theta = 1.33$. Given $\mu = 0.75$.

Since $\cot\theta > \mu$ (i.e., $1.33 > 0.75$), friction is insufficient. The block slips relative to the bar in the x-direction.

Step 2: Kinematics of Block P

Since slipping occurs, the friction is Kinetic: $f_k = \mu N$.

Acceleration of P ($a_P$):

• $y$-component: Driven by Normal force. $$ a_{Py} = A_y = A \sin\theta $$
• $x$-component: Driven by Kinetic Friction. $$ a_{Px} = \frac{f_k}{m} = \frac{\mu (m A \sin\theta)}{m} = \mu A \sin\theta $$

Total Acceleration Magnitude: $$ a_P = \sqrt{a_{Px}^2 + a_{Py}^2} $$ $$ a_P = \sqrt{(\mu A \sin\theta)^2 + (A \sin\theta)^2} $$ $$ a_P = A \sin\theta \sqrt{\mu^2 + 1} $$

Step 3: Calculating Distance

Both the Bar and Block P start from rest. The distance covered is proportional to acceleration ($s = \frac{1}{2}at^2$).

Given the Bar moves distance $l = 40$ cm, the distance $s$ moved by P is:

$$ s = l \times \frac{a_P}{A_{bar}} $$ $$ s = l \times \frac{A \sin\theta \sqrt{\mu^2 + 1}}{A} $$ $$ s = l \sin\theta \sqrt{1 + \mu^2} $$

Step 4: Final Calculation

Substitute values: $l = 40$ cm, $\sin\theta = 0.6$, $\mu = 0.75 = 3/4$.

$$ \sqrt{1 + \mu^2} = \sqrt{1 + \left(\frac{3}{4}\right)^2} = \sqrt{\frac{16+9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} = 1.25 $$ $$ s = 40 \times 0.6 \times 1.25 $$ $$ s = 24 \times 1.25 $$ $$ s = 30 \text{ cm} $$