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Question

A ball of mass m is dropped from a height h on a platform fixed at the top of a vertical spring. The platform is displaced by a distance x. The spring constant is (Take acceleration due to gravity =g):

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A
2mgx
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B
2mghx2
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C
2mg(h+x)x2
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D
2mg(h+x)h2
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Solution

The correct option is C 2mg(h+x)x2
The particle is dropped from height h and the spring be compressed by x.
Loss in PE of the particle = gain in elastic PE of the spring
mg(h+x)=12kx2k=2mg(h+x)x2

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