Question

A person climbs up a conveyor belt with a constant acceleration. The speed of the belt is \( \sqrt{\frac{g h}{6}} \) and the coefficient of friction is \( \frac{5}{3\sqrt{3}} \). The time taken by the person to reach from A to B with maximum possible acceleration is: 

• A. \( \frac{\sqrt{h g}}{6} \)
• B. \( \sqrt{6gh} \)
• C. \( \frac{2h}{\sqrt{g}} \)
• D. \( \sqrt{\frac{6h}{g}}.\)

Hint:

When analyzing motion on a conveyor belt with friction, consider the forces acting due to friction and the maximum possible acceleration. Apply Newton's second law to solve for the time taken.

Answer 1

The Correct Option is D

We are given the following parameters:
- The velocity of the conveyor belt: \( v = \sqrt{\frac{gh}{6}} \),
- The coefficient of friction: \( \mu = \frac{5}{3\sqrt{3}} \),
- The person climbs with constant acceleration.
Step 1: Force analysis. The frictional force is given by: \[ f = \mu N = \mu mg. \] Step 2: Maximum possible acceleration. The maximum acceleration is found using the equation: \[ a_{\text{max}} = \frac{g}{6}. \] Step 3: Using the kinematic equation. Using the kinematic equation \( v^2 = u^2 + 2 a d \), we get: \[ \left(\sqrt{\frac{gh}{6}}\right)^2 = 0 + 2 \times \frac{g}{6} \times h. \] Simplifying, we find: \[ \frac{gh}{6} = \frac{gh}{3}. \] Step 4: Time taken. Using the equation \( v = u + at \), we solve for time \( t \): \[ \sqrt{\frac{gh}{6}} = \frac{g}{6} \times t, \] which simplifies to: \[ t = \sqrt{\frac{6h}{g}}. \] Final Answer: \( t = \sqrt{\frac{6h}{g}} \) .