AP EAMCET · PHYSICS · Laws of Motion
A person climbs up a conveyor belt with a constant acceleration. The speed of the belt is \(\sqrt{\frac{g h}{6}}\) and 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 \(\sqrt{\frac{h g}{6}}\)
- B \(\sqrt{6 g h}\)
- C \(\sqrt{\frac{2 h}{g}}\)
- D \(\sqrt{\frac{6 h}{g}}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{\frac{6 h}{g}}\)
Step-by-step Solution
Detailed explanation
Maximum possible acceleration of the person on the conveyor belt is \(a_{\max }=\frac{\mu m g \cos \theta-m g \sin \theta}{m}=g(\mu \cos \theta-\sin \theta)\)…
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