A) 2.0
B) 4.0
C) 1.6
D) 2.5
Correct Answer: A
Solution :
Angle of repose \[\alpha ={{\tan }^{-1}}(\mu )={{\tan }^{-1}}(0.8)=38.6{}^\circ \] Angle of inclined plane is given \[\theta =30{}^\circ \]. It means block is at rest therefore, Static friction = component of weight in downward direction \[=mg\sin \theta =10\ N\] \[\therefore \] \[m=\frac{10}{9\times \sin 30{}^\circ }=2\ kg\]You need to login to perform this action.
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