AP EAMCET · PHYSICS · Laws of Motion
A box of mass \(2 \mathrm{~kg}\) is placed on a inclined plane that makes \(30^{\circ}\) with the horizontal. The coefficient of friction between the box and inclined plane is 0.2 . A force \(F\) is applied on the box perpendicular to incline to prevent the box from sliding down. The minimum value of \(F\) is
(acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A \(28.6 \mathrm{~N}\)
- B \(22.8 \mathrm{~N}\)
- C \(32.7 \mathrm{~N}\)
- D \(44.6 \mathrm{~N}\)
Answer & Solution
Correct Answer
(A) \(28.6 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
As, block has a tendency to slide downwards, friction acts in upward direction. Box will not slide if friction balances component of weight acting in downward direction. \(\Rightarrow \quad f \geq m g \sin \theta\) Here, \(f=\) friction \(=\mu N\) or…
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