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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Motion on Inclined Surface

  • question_answer
    A block rests on a rough inclined plane making an angle of \[{{30}^{o}}\] with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (take \[g=10\,\,m/{{s}^{2}})\]                                  [AIEEE 2004]

    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 \[=mg\sin \theta =10\ N\] \[\therefore \] \[m=\frac{10}{9\times \sin 30{}^\circ }=2\ kg\]  


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