#### Answer

a) The spring stretches by $9cm$
b) The speed of the object is $2.1m/s$

#### Work Step by Step

a) The amount by which the spring is stretched can be found by balancing the weight of the object and the restoring force from the spring: $$mg=ky$$ $$y=\frac{mg}{k}=\frac{1.1\times9.8}{120}=0.09m=9cm$$
b) The object goes up to its original position, which is $0.2m$ above the point it is released, so $h_f-h_0=0.2m$. The spring's displacement before it is released $y_0=0.09+0.2=0.29m$ and original position $y_f=0.09m$. We also know $v_0=0$
$$\frac{1}{2}ky_0^2+mgh_0=\frac{1}{2}ky_f^2+mgh_f+\frac{1}{2}mv_f^2$$ $$\frac{1}{2}mv_f^2=\frac{1}{2}k(y_0^2-y_f^2)+mg(h_0-h_f)$$ $$v_f=\sqrt{\frac{k(y_0^2-y_f^2)+2mg(h_0-h_f)}{m}}$$
We have $k=120N/m, y_0^2-y_f^2=0.076, m=1.1kg$ and $h_0-h_f=-0.2m$ $$v_f=2.1m/s$$