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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    An object of mass m falls on to a spring of constant k from h. Then the spring undergoes compression by a length\[x\]. The maximum compression x is given by the equation:

    A)  \[mgh=\frac{1}{2}k{{x}^{2}}\]

    B)  \[mgh(h+x)=\frac{1}{2}k{{x}^{2}}\]

    C)  \[mg(h+x)=-kx\]

    D)  \[mgh=-kx\]

    E)  \[mgh=-\frac{1}{2}k{{x}^{2}}\]

    Correct Answer: B

    Solution :

    Conservation of energy yields \[mgh+mgx=\frac{1}{2}k{{x}^{2}}\] \[\Rightarrow \]               \[mg(h+x)=\frac{1}{2}k{{x}^{2}}\]


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