The potential energy of 'a long spring when it is stretched by 3 cm is ' U '. If the spring is stretched by 9 cm , potential energy stored in it will be

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The potential energy of a stretched spring is given by $U = \frac{1}{2} k x^{2}$ .

For an extension $x_{1} = 3$ cm, the potential energy is:

$U_{1} = \frac{1}{2} k (3)^{2} = \frac{9}{2} k = U$

For an extension $x_{2} = 9$ cm, the potential energy is:

$U_{2} = \frac{1}{2} k (9)^{2} = \frac{81}{2} k$

Taking the ratio of $U_{2}$ to $U_{1}$ :

$\frac{U _{2}}{U _{1}} = \frac{81}{9} = 9$

$U_{2} = 9 U$

Answer: $9 U$

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