A block is fastened to a horizontal spring. The block is pulled to...

At first the block is pulled to a distance ${x}_{0} = 10cm$(extreme position).

Potential energy of block is ${U}_{i} = \dfrac{1}{2}k{x}_{0}^{2}$ and kinetic energy os block, ${K}_{i} = 0$.

Now, when the block is at distance $5cm$.

Potential energy is ${U}_{f} = \dfrac{1}{2}k{\left(\dfrac{{x}_{0}}{2}\right)}$^{2}$and kinetic energy is${K}_{f} = 0 . 25J$</p><p>Total energy of block will be conserved, so </p><p>$\dfrac{1}{2}k{x}_{0}^{2} + 0 = \dfrac{1}{2}k{x}_{0}^{2}4 + 0 . 25$</p><p>$⇒ \dfrac{1}{2}k{x}_{0}^{2}\dfrac{3}{4} = \dfrac{1}{4}$</p><p>$⇒ \dfrac{1}{2}k\dfrac{3}{100} = 1 ⇒ k = \dfrac{200}{3}N{m}^{- 1}$</p><p>$= 67N{m}^{- 1}