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MGIMS WARDHA MGIMS WARDHA Solved Paper-2011

  • question_answer
     A block of mass 2 kg is kept on the floor. The coefficient of static friction is 0.4. If a force F of 2.5 N is applied on the block as shown in the figure, the frictional force between the block and the floor will be

    A)  2.5 N                                    

    B) b) 5N

    C)   7.84 N                                

    D)  10N

    Correct Answer: A

    Solution :

                     Limiting frictional force \[{{F}_{s}}={{\mu }_{s}}R\] \[={{\mu }_{s}}mg=0.4\times 2\times 9.8=7.84N\] Applied force \[2.5\text{ }N<7.84\text{ }N\] So, block cannot move and force of friction will be equal to applied force. Force of friction\[=2.5\text{ }N\].


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