Kinetic Energy and Potential Energy

by conservation of mechanical energy

              K.E_i + P.E_i = K.E_f + P.E_f

              $\frac{1}{2}*0.5v^2+0=\frac{1}{2}*0.5{\left(\frac{v}{2}\right)}^2+\frac{1}{2}k{x}^2$  

              $\Rightarrow \frac{1}{2}*0.5\left[v^2-\frac{v^2}{4}\right]=\frac{1}{2}k{x}^2$  

              $\Rightarrow \frac{1.5}{4}12*12=\frac{k*9}{100}$  

              $\frac{1.5*36}{9}*100=k$  

              k = 600 N/m¹

K? = ½ mv² + ½ Iω²
Since I =? mr² and v=rω,
K? = ½ mv² + ½ (? mr²) (v/r)² = ½ mv² +? mv² = (7/10)mv² = 140 J

K_f = 0.05 K?
(7/10)mv_f² = 0.05 * (140)
(7/10)mv_f² = 7
v_f² = 20
v_f = √20 ≈ 4.47 m/s