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BHU PMT BHU PMT (Mains) Solved Paper-2010

  • question_answer
    A block of mass 10 kg is moving in x-direction with a constant speed of\[10m{{s}^{-1}}\]. It is subjected    to    a    retarding    force\[F=-\text{ }0.1\text{ }x\,J\,{{m}^{-1}}\]during its travel from\[x=20\]m to\[x=30\text{ }m\]. Its final kinetic energy will be

    A)  475 J                                    

    B)  450 J

    C)  275 J                                    

    D)  250 J

    Correct Answer: A

    Solution :

                     The work, \[W=K{{E}_{F}}-{{K}_{{{E}_{I}}}}\] \[W=K{{E}_{F}}-\frac{1}{2}m{{v}^{2}}\] \[F.dx=K{{E}_{F}}-\frac{1}{2}\times 10\times {{10}^{2}}\]                 \[-0.1\,x\,dx\,=K{{E}_{F}}-500\] \[-0.1\int_{20}^{30}{x\,}dx=K{{E}_{F}}-500\] \[-0.1\left[ \frac{{{x}^{2}}}{2} \right]_{20}^{30}=K{{E}_{F}}-500\] \[-\frac{0.1}{2}[{{(30)}^{2}}-{{(20)}^{2}}]=K{{E}_{F}}-500\] \[-\frac{0.1}{2}[900-400]=K{{E}_{F}}-500\] \[=25=K{{E}_{F}}-500\] Final kinetic energy, \[K{{E}_{F}}=475J\]


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