## Physics (10th Edition)

The speed of the object is $0.5m/s$.
Initially, the spring is compressed by $x_0=-0.065m$, where $v_0=0$. Finally, the spring is stretched by $x_f=+0.048m$, where $v_f$ is the value we need to find. According to the conservation of energy, $$KE_0+PE_{elastic, 0}=KE_f+PE_{elastic, f}$$ $$0+\frac{1}{2}kx_0^2=\frac{1}{2}mv_f^2+\frac{1}{2}kx_f^2$$ $$k(x_0^2-x_f^2)=mv_f^2$$ $$v_f=\sqrt{\frac{k}{m}}\times\sqrt{x_0^2-x_f^2}=\omega\sqrt{x_0^2-x_f^2}$$ We are given $\omega=11.3rad/s$, so $$v_f=0.5m/s$$