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NEET Sample Paper NEET Sample Test Paper-40

  • question_answer
    A body of mass 1 kg is thrown upwards with a velocity \[20\,m{{s}^{-1}}.\]It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? \[(g=10\,m{{s}^{-2}})\]

    A)  20 J

    B)  30 J

    C)  40 J

    D)  10 J

    Correct Answer: A

    Solution :

    The energy lost due to air friction is equal to difference of initial kinetic energy and final potential energy. Initially body possesses only kinetic energy and after attaining a height the kinetic energy is zero Therefore, loss of energy = KE - PE \[=\frac{1}{2}m{{v}^{2}}-mgh\] \[=\frac{1}{2}\times 1\times 400-1\times 18\times 10\] \[=200-180=20\,J\]


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