JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

  • question_answer
    A block rests on a rough inclined plane making an angle of\[30{}^\circ \]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\[g=10\text{ }m/{{s}^{2}}\])                      [AIEEE 2004]

    A) 2.0

    B) 4.0

    C) 1.6

    D) 2.5

    Correct Answer: A

    Solution :

    [a] For a body or system to be in equilibrium net force acting on the body or system should be, zero. i.e., sum of component of all the forces acting in a particular direction is zero.
    Let the mass of block be m.
    Frictional force in rest position          
    \[F=mg\text{ }\sin \text{ }30{}^\circ \]
    (this is static frictional force and may be less than the limiting frictional force)
    At the condition of equilibrium block on incline.
    \[10=m\times 10\times \frac{1}{2}\]
    \[\therefore \]      \[m=\frac{2\times 10}{10}=2\,kg\]

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