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EAMCET Medical EAMCET Medical Solved Paper-1997

  • question_answer
    A body of mass 10 kg lies on a rough horizontal surface. When a horizontal force of F newton acts on it. it gets an acceleration of \[5\text{ }m/{{s}^{2}}\]and when the horizontal force is doubled, it gets an acceleration of 18 m/s2. Then the coefficient of friction between the body and the horizontal surface is: (assume\[g=10\text{ }m{{s}^{-2}}\])

    A)  0.2                                        

    B)  0.8

    C)  0.4                                        

    D)  0.6

    Correct Answer: B

    Solution :

                     The resultant force on the body when force F is applied and the force of friction is\[f\]and its acceleration is \[5m/{{s}^{2}}\] is given by \[m{{a}_{1}}=F-f\] \[10\times 5=F-f\]                                           ?(i) Similarly, if the acceleration is \[18\,m/{{s}^{2}}\]and the applied force on the body is doubled IF and force of friction is\[f,\] then resultant force  on the body is given by \[m{{a}_{2}}=2F-f\] \[10\times 18=2F-f\] Solving Eqs. (i) and (ii) \[f=80\,N\] Also, the force of friction \[f=\mu R\](where \[\mu \]is the coefficient offriction) or                            \[\mu =\frac{f}{R}=\frac{f}{mg}\]                                 \[\mu =\frac{80}{10\times 10}=0.8\]


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