Question
A long spring , when stretched by `x` cm has a potential energy `u` . On increasing the length of spring by stretching to `nx` cm , the potential energy stored in the spring will be
Answer: Option C
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Initially potential energy = `1/2 kx^2`
`rArr u = 1/2 kx^2`
or ` 2u = kx^2`
`rArr k = (2u)/(x^2)`
when it is stretched to `nx` cm
`:.` `P.E. = 1/2 kx_1^2 = 1/2 xx (2u)/(x^2) xx n^2 x^2 = n^2 u`
`:. ` potential energy stored in the spring = `n^2 u`
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