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: \( t = \sqrt{\frac{6h}{g}} \) .

Answer 2

Given:
- Speed of the belt, \( v = \sqrt{\frac{gh}{6}} \)
- Coefficient of friction, \( \mu = \frac{5}{3 \sqrt{3}} \)
- Inclination angle, \( \theta = 30^\circ \)
- Height of conveyor, \( h \)

Step 1: Find the length \( L \) of the conveyor belt:
\[ L = \frac{h}{\sin 30^\circ} = \frac{h}{\frac{1}{2}} = 2h \]

Step 2: Let the person accelerate with maximum acceleration \( a \) up the belt. The frictional force provides the maximum possible acceleration.

Step 3: Using friction:
The maximum acceleration due to friction:
\[ a_{\max} = \mu g \cos \theta - g \sin \theta \] where the component \( g \sin \theta \) acts down the incline, and friction acts up the incline opposing slipping.

Step 4: Substitute values:
\[ a_{\max} = \mu g \cos 30^\circ - g \sin 30^\circ = \frac{5}{3 \sqrt{3}} \times g \times \frac{\sqrt{3}}{2} - g \times \frac{1}{2} = \frac{5g}{6} - \frac{g}{2} = \frac{5g - 3g}{6} = \frac{2g}{6} = \frac{g}{3} \]

Step 5: Using kinematic equation:
\[ L = v t + \frac{1}{2} a t^2 \] where initial speed \( u = v = \sqrt{\frac{gh}{6}} \), acceleration \( a = \frac{g}{3} \), and distance \( L = 2h \).

Step 6: Substitute values:
\[ 2h = \sqrt{\frac{gh}{6}} \cdot t + \frac{1}{2} \times \frac{g}{3} \times t^2 = \sqrt{\frac{gh}{6}} \cdot t + \frac{g t^2}{6} \]

Step 7: Multiply both sides by 6 to clear denominator:
\[ 12h = 6 \sqrt{\frac{gh}{6}} t + g t^2 \] Rewrite \( 6 \sqrt{\frac{gh}{6}} \):
\[ 6 \sqrt{\frac{gh}{6}} = 6 \times \sqrt{\frac{g h}{6}} = 6 \times \frac{\sqrt{g h}}{\sqrt{6}} = \sqrt{6} \sqrt{g h} \]

Step 8: So:
\[ 12h = \sqrt{6 g h} \, t + g t^2 \] Rewrite:
\[ g t^2 + \sqrt{6 g h} \, t - 12 h = 0 \]

Step 9: Solve quadratic for \( t \):
\[ t = \frac{-\sqrt{6 g h} \pm \sqrt{6 g h + 48 g h}}{2g} = \frac{-\sqrt{6 g h} \pm \sqrt{54 g h}}{2g} \] \[ t = \frac{-\sqrt{6 g h} \pm 3 \sqrt{6 g h}}{2g} \] Take positive root:
\[ t = \frac{-\sqrt{6 g h} + 3 \sqrt{6 g h}}{2g} = \frac{2 \sqrt{6 g h}}{2g} = \frac{\sqrt{6 g h}}{g} = \sqrt{\frac{6 h}{g}} \]

Therefore, the time taken by the person to reach from A to B is:
\[ \boxed{ \sqrt{\frac{6 h}{g}} } \]