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NEET Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति NEET PYQ-Work Energy Power and Collision

  • question_answer
    A force \[F=20+10y\]acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle from \[y=0\text{ }to\text{ }y=1\]m is:                [NEET 2019]

    A) 25 J     

    B) 20 J

    C) 30 J     

    D) 5 J

    Correct Answer: A

    Solution :

    [a] \[F=20+10y\]
    \[W=\int\limits_{0}^{1}{(20\text{ }10y)dy}\]
    \[=20[y]_{0}^{1}+\left[ \frac{10{{y}^{2}}}{2} \right]_{0}^{-1}\]
    \[=20+5\]
    \[=25\]


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