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If the potential energy of a spring is U on stretching it by 2 cm, then its potential energy when it is stretched by 8 cm will be:

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$\frac{U}{4}$

$\frac{U}{16}$

16U

answer is D.

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Detailed Solution

$U = \frac{1}{2} (\frac{YA}{L}) l^{2} \Rightarrow U \propto l^{2}$

$\frac{U_{2}}{U_{1}} = {(\frac{l_{2}}{l_{1}})}^{2} = {(\frac{8}{2})}^{2} = 16 \Rightarrow U_{2} = 16 U_{1}$

i.e., potential energy of the spring will be 16U.

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