JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Sample Paper Topic Test - Work Energy Power

  • question_answer
    Two springs of force constant 100 N/m and 150 N/m are in series as shown. The block is pulled by a distance of 2.5 cm to the right from equilibrium position. What is the ratio of work done by the spring at left to the work done by the spring at right.

    A) \[\frac{3}{2}\]

    B) \[\frac{2}{3}\]

    C) 0.2

    D) None of these

    Correct Answer: A

    Solution :

    [a] \[{{k}_{1}}{{x}_{1}}={{k}_{2}}{{x}_{2}}\] \[100{{x}_{1}}=150{{x}_{2}}\]
    and\[{{x}_{1}}+{{x}_{2}}=2.5\Rightarrow {{x}_{2}}=1cm{{x}_{1}}=1.5cm\]
    \[W.D.=-\frac{1}{2}\left( x_{f}^{2}-x_{1}^{2} \right)\];Ratio\[\frac{-\frac{1}{2}100\times {{(1.5)}^{2}}}{-\frac{1}{2}150\times {{(1)}^{2}}}=\frac{3}{2}\]

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