NEET Sample Paper NEET Sample Test Paper-14

  • question_answer 30) Two blocks of masses M = 5 kg and m = 3 kg are  placed on a horizontal surface as shown in the figure. The coefficient of friction between the blocks is 0.5 and that between the block M and the horizontal surface is 0.7. What is the maximum horizontal force F that can be applied to block M so that the two blocks move without slipping? Take \[g=10\,m{{s}^{-2}}.\]

    A)  4 N                            

    B) \[16\,N\]

    C)  24 N              

    D)  96 N

    Correct Answer: D

    Solution :

    The force of friction between the block m and the block\[M={{\mu }_{1}}\,mg.\]Force of friction between block M and the horizontal surface\[={{\mu }_{2}}(M+m)g.\]As there is no slipping between the blocks, so the entire system can be considered as a single system. Invoking Newton's second law on composite system. Thus,\[F=(M+m)a+{{\mu }_{2}}(M+m)g\]           (1) Now, since the force on block m is \[{{\mu }_{1}}mg,\] its acceleration is \[a=\frac{{{\mu }_{1}}mg}{m}={{\mu }_{1}}g\] Using Eqs. (1) and (2), \[F={{\mu }_{1}}(M+m)g+{{\mu }_{2}}(M+m)g\]\[=({{\mu }_{1}}+{{\mu }_{2}})(M+m)g\] \[=(0.5+0.7)\times (5+3)\times 10=96\,N.\] Hence, the correction option is [d].

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