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

  • question_answer
    The coefficient of friction between a body and the surface of an inclined plane at 45° is 0.5. If \[g=9.8\,m/{{s}^{2}}\], the acceleration of the body downwards in \[m/{{s}^{2}}\] is                                                             [EAMCET 1994]            

    A)                         \[\frac{4.9}{\sqrt{2}}\]                      

    B)                         \[4.9\sqrt{2}\]            

    C)                         \[19.6\sqrt{2}\]            

    D)                         4.9

    Correct Answer: A

    Solution :

                                \[a=g(\sin \theta -\mu \cos \theta )=9.8(\sin 4{{5}^{o}}-0.5\cos {{45}^{o}})\]                \[=\frac{4.9}{\sqrt{2}}\ m/{{\sec }^{2}}\]            


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