JCECE Engineering JCECE Engineering Solved Paper-2004

  • question_answer
    A particle is displaced from the point\[A(5,\,\,-5,\,\,-7)\]to the point\[B(6,\,\,2,\,\,-2)\]under the influence of the forces\[{{P}_{1}}=10\widehat{\mathbf{i}}-\widehat{\mathbf{j}}+11\widehat{\mathbf{k}}\],\[{{P}_{2}}=4\widehat{\mathbf{i}}+5\widehat{\mathbf{j}}+6\widehat{\mathbf{k}}\]\[{{P}_{3}}=-2\widehat{\mathbf{i}}+\widehat{\mathbf{j}}-9\widehat{\mathbf{k}}\], the work done is:

    A) \[87\,\,unit\]                    

    B) \[85\,\,unit\]

    C) \[81\,\,unit\]                    

    D)  none of these

    Correct Answer: A

    Solution :

    \[\therefore \]Displacement,                 \[\overset{\to }{\mathop{\mathbf{d}}}\,=(6-5)\widehat{\mathbf{i}}+(2+5)\widehat{\mathbf{j}}+(-2+7)\widehat{\mathbf{k}}\]                 \[=\widehat{\mathbf{i}}+7\widehat{\mathbf{j}}+5\widehat{\mathbf{k}}\] And total force\[\overset{\to }{\mathop{\mathbf{P}}}\,=\overset{\to }{\mathop{{{\mathbf{P}}_{\mathbf{1}}}}}\,+\overset{\to }{\mathop{{{\mathbf{P}}_{\mathbf{2}}}}}\,+\overset{\to }{\mathop{{{\mathbf{P}}_{\mathbf{3}}}}}\,\]                 \[=(10\widehat{\mathbf{i}}-\widehat{\mathbf{j}}+11\widehat{\mathbf{k}})+(4\widehat{\mathbf{i}}+5\widehat{\mathbf{j}}+6\mathbf{\hat{k}})\]                 \[+(-2\widehat{\mathbf{i}}+\widehat{\mathbf{j}}-9\widehat{\mathbf{k}})\]                 \[=12\widehat{\mathbf{i}}+5\widehat{\mathbf{j}}+8\widehat{\mathbf{k}}\] \[\therefore \]Work done\[W=\overset{\to }{\mathop{\mathbf{P}}}\,\cdot \overset{\to }{\mathop{\mathbf{d}}}\,\]                 \[=(12\widehat{\mathbf{i}}+5\widehat{\mathbf{j}}+8\widehat{\mathbf{k}})\cdot (\widehat{\mathbf{i}}+7\widehat{\mathbf{j}}+5\widehat{\mathbf{k}})\]                 \[=12+35+40=87\,\,unit\]


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