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JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Self Evaluation Test - Work, Energy and Power

  • question_answer
    A particle is placed at the origin and a force \[F=kx\] is acting on it (where k is positive constant). If\[U(0)=0\], the graph of \[U(x)\]versus x will be (where U is the potential energy function)

    A)

    B)

    C)

    D)

    Correct Answer: A

    Solution :

    [a] \[U=-\int_{0}^{x}{Fdx}=-\int_{0}^{x}{kxdx=-\frac{1}{2}k{{x}^{2}}}\] It is correctly drawn in [a]


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