Multi-Spring Systems — Practice Problem Solutions

1. Both springs push the same direction, so friction must balance their combined force: $f=k_1\Delta x_1+k_2\Delta x_2=(170)(0.12)+(210)(0.12)=(380)(0.12)=45.6\text{ N}$.

2. Single spring: $\Delta x=\dfrac{mg}{k}=\dfrac{(5.0)(9.8)}{180}\approx0.272\text{ m}$. Two springs sharing $10.0\text{ kg}$ equally: each supports half the weight, $\dfrac{(10.0)(9.8)}{2}=49\text{ N}$, giving the same $\Delta x=\dfrac{49}{180}\approx0.272\text{ m}$ per spring. They're equal — each spring in the two-spring case ends up supporting exactly the same load (49 N) as the single spring did in the first case, so it stretches the identical amount.

3. a) $\Delta x_{air}=\dfrac{mg}{k}=\dfrac{(3.0)(9.8)}{120}=0.245\text{ m}$. b) Spring force while submerged: $F_s=k\Delta x=(120)(0.10)=12\text{ N}$. c) The block is still in equilibrium, so spring force plus buoyant force must equal the weight: $F_{buoyant}=mg-F_s=29.4-12=17.4\text{ N}$.

4. The acceleration points to the left (back toward equilibrium). A spring's restoring force always points back toward its natural (equilibrium) length — so even though the block is moving further right at that instant, the spring is pulling it back the other way, meaning its acceleration is directed opposite to its current motion.