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AMU Medical AMU Solved Paper-2003

  • question_answer
    A force \[\overset{\to }{\mathop{F}}\,~=\text{ 6}\overset{\hat{\ }}{\mathop{\text{i}}}\,\text{ }+\text{ 5}\overset{\hat{\ }}{\mathop{\text{J}}}\,\]newton acts and displaces  the body through \[\overset{\to }{\mathop{s}}\,~=\text{ 3}\overset{\hat{\ }}{\mathop{\text{i}}}\,\text{ }+\text{ 2}\overset{\hat{\ }}{\mathop{k}}\,\]metre. Then the work done is given by

    A)  8J                                          

    B)  28 J

    C)  18 J                                       

    D)  10 J

    Correct Answer: C

    Solution :

                     Work done (IV) = Force (F) ? displacement Given,   \[\vec{F}=6\,\hat{i}+5\hat{j}\], \[d=3\,\hat{i}-2\hat{k}\] Taking scalar product, we get                 \[\vec{W}=\vec{F}.\,\,\vec{d}\] \[\therefore \]  \[W=(6\,\hat{i}+5\hat{j}).\,\,(3\hat{i}-2\hat{k})\] Using,   \[\hat{i}\,.\,\,\hat{i}=1\],             \[\hat{i}\,.\,\,\hat{k}=0\]                             \[\hat{j}\,.\,\,\hat{i}=0\],             \[\hat{j}\,.\,\,\hat{j}=1\], \[\hat{k}\,.\,\,\hat{k}=1\]


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