The maximum potential energy of a block executing simple harmonic...

The maximum potential energy of a block executing simple harmonic motion is given by {U}_\max = \dfrac{1}{2}m{\omega}^{2}{A}^{2} = 25J

So, kinetic energy of block $KEat\dfrac{A}{2} = \dfrac{1}{2}m{v}_{1}^{2} = \dfrac{1}{2}m{\omega}^{2}\left({A}^{2} - \dfrac{{A}^{2}}{4}\right)$

$⇒ KE = \dfrac{1}{2}m{\omega}^{2}\dfrac{3{A}^{2}}{4} = \dfrac{3}{4}\left(\dfrac{1}{2}m{\omega}^{2}{A}^{2}\right)$

$⇒ KE = \dfrac{3}{4} \times 25 = 18 . 75J$