AP EAMCET · PHYSICS · Laws of Motion
A block of mass \(m\) is lying on a rough inclined plane having an inclination \(\alpha=\tan ^{-1}\left(\frac{1}{5}\right)\). The inclined plane is moving horizontally with a constant acceleration of \(a=2 \mathrm{~ms}^{-2}\) as shown in the figure. The minimum value of coefficient of friction, so that the block remains stationary with respect to the inclined plane is (Take, \(g=10 \mathrm{~ms}^{-2}\) )
- A \(\frac{2}{9}\)
- B \(\frac{5}{12}\)
- C \(\frac{1}{5}\)
- D \(\frac{2}{5}\)
Answer & Solution
Correct Answer
(B) \(\frac{5}{12}\)
Step-by-step Solution
Detailed explanation
The block-plane system is shown in the figure, So, from the above free body diagram (FBD), downward acceleration, Divided the Eq. (iii) by \(\sin \alpha\) and putting the values,…
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