The potential energy of a long spring when stretched by 2 cm is U . If the spring is stretched by 8 cm , potential energy stored in it will be:

Question

Aicharya · Accepted Answer

Potential energy stored in spring is given by,

$U = \frac{1}{2} K x^{2} .$

First case:

$U = \frac{1}{2} K {(2 \times 10^{- 2})}^{2} . . . E q (1)$

Second case:

$U^{'} = \frac{1}{2} K {(8 \times 10^{- 2})}^{2} . . . E q (2)$

From both equations, we get

$\frac{U}{U^{'}} = \frac{1}{16} \Rightarrow U^{'} = 16 U$

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The potential energy of a long spring when stretched by 2 cm is U . If the spring is stretched by 8 cm ,…