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:
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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.
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:
\[
a_{{max}} = \frac{g}{6}.
\]
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: \( t = \sqrt{\frac{6h}{g}} \) .
