A body of mass m dropped from a certain height strikes a light vertical fixed spring of stiffness k. The height of its fall before touching the spring if the maximum compression of the spring is equal to 3mgk is
A
3mg2k
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B
2mgk
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C
3mg4k
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D
mg4k
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Solution
The correct option is A3mg2k Change in height of the particle is h+x
where x is compression in the spring
Applying law of conservation of energy
work done by gravity = work stored in spring in the form of potential energy
mg(h+x)=12kx2 ( ∵x=3mgk maximum compression) (h+x)=kx22mg h=kx22mg−x=k(3mg/k)22mg−3mgkh=9mg2k−3mgk=3mg2k