Spring Forces — Practice Problem Solutions

1. $F_s=k\Delta x=(400)(0.05)=20\text{ N}$.

2. $k=\dfrac{mg}{\Delta x}=\dfrac{(3.0)(9.8)}{0.15}=196\text{ N/m}$.

3. $mg=k\Delta x\implies m=\dfrac{k\Delta x}{g}=\dfrac{(180)(0.10)}{9.8}\approx1.84\text{ kg}$.