JCECE Engineering JCECE Engineering Solved Paper-2004

  • question_answer
    A force of \[7\widehat{\mathbf{i}}+6\widehat{\mathbf{k}}\] makes a body to move on a plane with the velocity \[3\widehat{\mathbf{j}}+4\widehat{\mathbf{k}}\]. The power developed is:

    A) \[\sqrt{45}\]                                      

    B) \[45\]

    C) \[24\]                                   

    D) \[\sqrt{24}\]

    Correct Answer: C

    Solution :

    Key Idea: Power developed is equal to scalar product of force and velocity vector. Power is given by                 \[P=\overset{\to }{\mathop{\mathbf{F}}}\,\cdot \overset{\to }{\mathop{\mathbf{v}}}\,\] Given,   \[\overset{\to }{\mathop{\mathbf{F}}}\,=7\widehat{\mathbf{i}}+6\widehat{\mathbf{k}},\,\,\overset{\to }{\mathop{\mathbf{v}}}\,=3\widehat{\mathbf{j}}+4\widehat{\mathbf{k}}\] \[\therefore \]  \[P=\overset{\to }{\mathop{\mathbf{F}}}\,\cdot \overset{\to }{\mathop{\mathbf{v}}}\,\]                 \[P=(7\widehat{\mathbf{i}}+6\widehat{\mathbf{k}})\cdot (3\widehat{\mathbf{i}}+4\widehat{\mathbf{k}})\]                 \[P=6\times 4\]                 \[P=24\,\,unit\] Note: Power is a scalar quantity.


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