NEET 2023 — Physics solution

NEET2023Physics

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:

Solution

Elastic PE $U = \frac{1}{2} k x^{2} \propto x^{2}$ .

Stretching from $2$ cm to $8$ cm increases $x$ by a factor of $4$ , so $U$ increases by $4^{2} = 16$ .

New PE $= 16 U$ .

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