A 0.5kg block moving at a speed of 12ms1 compresses a spring through a distance 30cm when its speed is halved. The spring constant of the spring will be…………Nm1.

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