QJEEM23Ph2997: JEE Main 2023 Physics - Class 11 JEEM23Ph2997 Solution: A block is fastened to a horizontal spring. The block is pulled …

Spring energy at x = 10 cm,

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

Energy of the block at x = 10,

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


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

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

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

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

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