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Spring energy at x = 10 cm,

$$\mathrm{U}_{\mathrm{i}} =\frac{1}{2} \mathrm{kx}_0^2 $$

Energy of the block at x = 10,

$$\mathrm{~K}_{\mathrm{i}} =0$$

JEE Main 2023 (Online) 1st February Evening Shift Physics - Work Power & Energy Question 70 English Explanation 2

Spring energy at x = 5 cm,

$$\mathrm{U}_{\mathrm{f}}=\frac{1}{2} \mathrm{k}\left(\frac{\mathrm{x}_0}{2}\right)^2 $$

Energy of the block at x = 5, (which is only kinetic energy, no potential energy of block presents as block is not moving in the vertical direction)

$$ \mathrm{~K}_{\mathrm{f}}=0.25 \mathrm{~J} $$

Applying energy conservation law,

Initial energy of Spring + Initial energy of Block = Final energy of Spring + Final energy of Block

$$ \frac{1}{2} \mathrm{kx}_0^2+0=\frac{1}{2} \mathrm{k} \frac{\mathrm{x}_0^2}{4}+0.25 $$

$$ \frac{1}{2} \mathrm{kx}_0^2 \frac{3}{4}=\frac{1}{4} $$

$$ \frac{1}{2} \mathrm{k} \frac{3}{100}=1 \Rightarrow \mathrm{k}=\frac{200}{3} \mathrm{~N} / \mathrm{m} $$

$$ =67 \mathrm{~N} / \mathrm{m} $$

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