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JIPMER Jipmer Medical Solved Paper-2004

  • question_answer
    The work done in pulling up a block of wood weighing 2 kN for a length of 10 m on a smooth plane inclined at an angle of \[15{}^\circ \] with the horizontal is:

    A)  9.82 kJ                                 

    B)  89 Kj                    

    C)  4.35 kJ                 

    D)         5.17 kJ

    Correct Answer: D

    Solution :

    Here : Weight of block \[w=2kN,\] Distance d = 10 m Angle of inclination on the plane \[\alpha =15{}^\circ \] The block will be pulled up on a smooth plane Hence, force of resistance due to inclination \[F=\omega \,\,\sin \alpha =2\times {{10}^{3}}\sin 15{}^\circ \]     \[=2\times {{10}^{3}}\times 0.2588\]     \[=0.5176\,kN\] Now work done, \[W=Fd=0.5176\times {{10}^{3}}\times 10\]                      \[=5.17\,kN\]


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