JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति JEE PYQ - Work Energy Power and Collision

  • question_answer
    A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is                   [AIEEE 2002]

    A) 16 J

    B) 8 J

    C) 32 J

    D) 24 J

    Correct Answer: B

    Solution :

    [b] The work is stored as the PE of the body and is given by \[\int_{{{x}_{1}}}^{{{x}_{2}}}{{{F}_{external}}}\,dx\]
                or \[U=\int_{{{x}_{1}}}^{{{x}_{2}}}{kx\,dx=\frac{1}{2}k({{x}_{2}}^{2}-{{x}_{1}}^{2})}\]       \[(\therefore \left| F \right|=kx)\]
                \[=\frac{800}{2}[{{(0.15)}^{2}}-{{(0.05)}^{2}}]\]       (\[\because k=800\], given)
                \[=400[0.2\times 0.1]=8\,J\]


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