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WB JEE Medical WB JEE Medical Solved Paper-2013

  • question_answer
    A uniform force of \[(3i+j)N\]acts on a particle of mass 2 kg. Hence the particle is displaced from position\[(2i+k)m\] to position \[(4i+3j-k)m.\] The work done by the force on the particle is

    A)  9J            

    B)  6J

    C)  13 J

    D)  15 J

    Correct Answer: A

    Solution :

     Given, force \[F=3i+j\] \[{{r}_{1}}=(2i+k)m\]and \[{{r}_{2}}=(4i+3j-k)m\] \[\therefore \]  \[s={{r}_{2}}-{{r}_{1}}=(4i+3j-k)-(2i+k)\] \[=(2i+3j-2k)m\] \[\therefore \] \[W=F.s=(3i+j).(2i+3j-2k)\] \[=3\times 2+3+0\] \[=6+3=9J\]


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