JEE Main & Advanced Physics NLM, Friction, Circular Motion Sample Paper Topic Test - Friction

  • question_answer
    A block of mass 15 kg is resting on a rough inclined plane as shown in figure. The block is tied up by a horizontal string which has a tension of 50 N. The coefficient of friction between the surfaces of contact is \[\left( g=10\text{ }m/{{s}^{2}} \right)\]

    A) 1/2                   

    B) 2/3

    C) 3/4                               

    D) ¼

    Correct Answer: A

    Solution :

    The free body diagram of the block is
    N is the normal reaction exerted by inclined plane on the block.
    Applying Newton's second law to the block along and normal to the incline.    
    \[mg\,\sin \,\,45{}^\circ =T\,\,\cos \,45{}^\circ +\mu \,N\]           ...…(1)
    \[N=mg\,\cos \,\,45{}^\circ +T\,\,sin\,45{}^\circ \]          ...…(2)
    Solving we get
    \[\mu ={1}/{2}\;\]

You need to login to perform this action.
You will be redirected in 3 sec spinner