Given:

The angle of inclination, $ \theta = 30^\circ $.

The speed of the conveyor belt, $ v_{\text{belt}} = \sqrt{\frac{g h}{6}} $.

The coefficient of friction, $ \mu = \frac{5}{3\sqrt{3}} $.

AP EAPCET 2024 - 21th May Evening Shift Physics - Laws of Motion Question 19 English Explanation

Calculation of Maximum Possible Acceleration:

The maximum force of friction that can act on the person while climbing is given by:

$ F_{\max} = \mu m g \cos \theta $

The component of gravitational force acting parallel to the incline is:

$ F_{\text{parallel}} = m g \sin \theta $

Using Newton’s second law, the maximum acceleration $ a $ is:

$ a = \frac{F_{\max} - F_{\text{parallel}}}{m} $

Substitute the given values:

$ \begin{aligned} a & = \frac{\mu m g \cos \theta - m g \sin \theta}{m} \\ & = \frac{\frac{5}{3\sqrt{3}} m g \times \frac{\sqrt{3}}{2} - m g \times \frac{1}{2}}{m} \\ & = \frac{\frac{5}{6} m g - \frac{1}{2} m g}{m} \\ & = \frac{g}{3} \end{aligned} $

Solving for Time $ t $:

The displacement from point $ A $ to point $ B $ is $ AB = \frac{OB}{\sin \theta} = \frac{h}{1/2} = 2h $. The equation of motion is:

$ AB = v_{\text{belt}} \cdot t + \frac{1}{2} a t^2 $

Substituting the known values, we have:

$ \begin{aligned} 2h & = \sqrt{\frac{g h}{6}} \cdot t + \frac{1}{2} \cdot \frac{g}{3} \cdot t^2 \\ & = \sqrt{\frac{g h}{6}} \cdot t + \frac{g}{6} \cdot t^2 \end{aligned} $

Reorganizing terms gives:

$ 2h = \sqrt{\frac{g h}{6}} \cdot t + \frac{g}{6} \cdot t^2 $

Solving this quadratic equation in terms of $ t $, we find:

$ t = \sqrt{\frac{6h}{g}} $

Thus, the time taken for the person to travel from $ A $ to $ B $ with maximum possible acceleration is:

$ t = \sqrt{\frac{6h}{g}} $

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