Hooke’s law says an ideal spring exerts a restoring force proportional to its displacement from relaxed (equilibrium) length: F_s = −k Δx. Here k (N/m) is the spring constant (stiffness), Δx is the spring’s displacement vector (positive if stretched), and the minus sign means the force points toward equilibrium. Example: k = 200 N/m, stretch x = 0.05 m → magnitude F = 200·0.05 = 10 N directed back toward the unstretched position. Remember AP language: “ideal spring” means negligible mass and linear force (CED 2.8.A). Linearity holds only up to the spring’s elastic limit. For systems, springs in series combine by 1/k_eq = 1/k1 + 1/k2; in parallel k_eq = k1 + k2 (CED 2.8.B). You’ll see problems asking for net force, energy, or effective k on the exam—practice applying F = −kΔx, draw a free-body diagram, and use equivalent-k rules. For a focused review, check the Topic 2.8 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/8-spring-forces/study-guide/p65phze7OgnnyC9g) and more practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

The negative sign in Hooke’s law, F_s = −k Δx, just tells you direction: the spring’s force is always a restoring force, pointing opposite the displacement from equilibrium. kΔx (or |k||Δx|) gives the magnitude; the minus sign makes the force vector point back toward the relaxed length. So if you stretch the spring to +x, Δx is positive and F_s is negative (pulls left); if you compress it, Δx is negative and F_s is positive (pushes right). This is exactly what the CED says: an ideal spring exerts a force proportional to the change in length and directed toward the equilibrium position (2.8.A.1–2.8.A.3). For AP problems, treat Δx as a vector or choose a coordinate axis and apply the minus sign to get the correct force direction. For more practice and examples, check the Topic 2.8 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/8-spring-forces/study-guide/p65phze7OgnnyC9g) and lots of practice questions (https://library.fiveable.me/practice/ap-physics-1-revised).

Equilibrium position for a spring is the point where the net force on the object attached to the spring is zero—so the spring exerts no net push or pull there. For an ideal spring Hooke’s law is F_s = −k Δx, where Δx is the object’s displacement measured from that equilibrium position. The minus sign means the spring’s force is a restoring force: it always points toward equilibrium. Important distinctions: the relaxed length of a spring is its natural length with no load; for a horizontal spring with no other forces, relaxed length = equilibrium. But if you hang a mass, gravity shifts equilibrium downward (the new equilibrium is where the spring force balances mg). On the AP exam you’ll be asked to identify equilibrium when net force = 0 and to use Δx measured from that point (CED 2.8.A.2–3). For a quick refresher, see the Topic 2.8 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/8-spring-forces/study-guide/p65phze7OgnnyC9g) and more Unit 2 review (https://library.fiveable.me/ap-physics-1-revised/unit-2). For extra practice problems go here: (https://library.fiveable.me/practice/ap-physics-1-revised).

An ideal spring (what AP uses) is a simplified model: it has negligible mass and its restoring force is exactly proportional to its displacement from relaxed length, Fs = −kΔx (Hooke’s law). The force always points toward the equilibrium position (CED 2.8.A.1–3). A real spring can differ in several ways: it may have non-negligible mass, show a nonlinear force–displacement relationship (especially near or past its elastic limit), lose energy to internal friction (hysteresis) or air damping, and its k can change with wear or geometry. Real springs therefore can deviate from Fs ∝ Δx and won’t always behave perfectly elastically. For AP Physics 1 you should treat springs as ideal unless a problem explicitly gives nonideal behavior. Review the CED essentials and practice Hooke’s-law problems on the Topic 2.8 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/8-spring-forces/study-guide/p65phze7OgnnyC9g) and try more practice questions at (https://library.fiveable.me/practice/ap-physics-1-revised).

If springs act like one spring, use an equivalent spring constant keq so Hooke’s law still applies: F = −keq Δx. - Parallel (springs attached side-by-side to the same object, share the same Δx): forces add, so k_eq,parallel = k1 + k2 + … . The system is stiffer (keq larger than any individual k). - Series (springs end-to-end, share the same force F but Δx splits): inverse add, so 1/k_eq,series = 1/k1 + 1/k2 + … , or for two springs k_eq = (k1 k2)/(k1 + k2). keq,series is smaller than the smallest k. Why: derive from Hooke’s law for each spring and add displacements (series) or forces (parallel). This is exactly what the AP CED expects for Topic 2.8 (equivalent spring constant; 2.8.B.1.i & ii). Want more examples and practice? See the Topic 2.8 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/8-spring-forces/study-guide/p65phze7OgnnyC9g), the Unit 2 overview (https://library.fiveable.me/ap-physics-1-revised/unit-2), and tons of practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Think about what a spring constant k measures: how much force you need to get a given change in length (Hooke’s law, Fs = −kΔx). For two springs in series the same force F passes through both, and each spring stretches: Δx1 = F/k1 and Δx2 = F/k2. The total stretch is Δx = Δx1 + Δx2 = F(1/k1 + 1/k2). Define an equivalent spring k_eq so F = k_eq Δx, then 1/k_eq = 1/k1 + 1/k2. That algebra shows k_eq is smaller than either k1 or k2—series springs are “softer” because the total system stretches more for the same force (each spring adds extra displacement). AP CED calls this out (2.8.B.1.i and 2.8.B.1.ii), so on the exam you can justify it with Hooke’s law and the sum of displacements. For a quick review, see the Topic 2.8 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/8-spring-forces/study-guide/p65phze7OgnnyC9g) and practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Saying the spring force is "proportional to displacement" means the force the spring exerts changes in direct, linear proportion to how far the spring is stretched or compressed from its relaxed (equilibrium) length. Mathematically: Fs = −k Δx. So if Δx doubles, the force doubles; if Δx = 0, the spring force is zero. The minus sign just tells you the force is a restoring force (it points toward the equilibrium position). Example: a spring with k = 200 N/m stretched 0.10 m gives |Fs| = 200·0.10 = 20 N back toward equilibrium. This proportional (linear) behavior is true for an ideal spring within its elastic limit and is what you’ll use on AP problems (Hooke’s law, finding keq for series/parallel springs). For a quick topic review, see the Fiveable Topic 2.8 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/8-spring-forces/study-guide/p65phze7OgnnyC9g) and practice questions at (https://library.fiveable.me/practice/ap-physics-1-revised).

Think of k as how much force a spring gives per unit stretch: F = kΔx. Which combination shares the same Δx or the same F? - Parallel springs: they’re clipped side-by-side so when you pull the block they both stretch the same amount Δx. Each spring provides its own force: Ftotal = F1 + F2 = k1Δx + k2Δx = (k1 + k2)Δx. So the effective k adds—more springs in parallel → more force for the same stretch. - Series springs: they’re end-to-end so the same force F runs through each spring, but the total stretch is the sum: Δxtot = Δx1 + Δx2 = F/k1 + F/k2. Rearranging F = keq Δxtot gives 1/keq = 1/k1 + 1/k2. So the effective k is smaller because one spring’s stretch adds to the other’s. This is all just Hooke’s law and equilibrium of forces—exactly what the CED calls for (2.8.A and 2.8.B). For a quick refresher and examples, check the Topic 2.8 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/8-spring-forces/study-guide/p65phze7OgnnyC9g). For more practice, try problems at (https://library.fiveable.me/practice/ap-physics-1-revised).

Look at how the springs connect to the object and to each other. - Series: springs are end-to-end (like one hanging from the other) so the same force runs through each spring and their extensions add. Use 1/k_eq = 1/k1 + 1/k2 + … (k_eq is smaller than the smallest k). - Parallel: springs attach to the same two points (they share the same displacement of those points), so each spring feels the same Δx and their forces add. Use k_eq = k1 + k2 + … (k_eq is larger). Quick checks: if the mass’s displacement stretches each spring by the same amount → parallel. If the same spring force flows through each spring and total stretch is sum of individual stretches → series. Draw a simple schematic or free-body diagram to see shared nodes (same endpoints = parallel; one-after-another = series). AP only expects you to handle pure series or pure parallel cases (CED 2.8.B). For more review and examples see the Topic 2.8 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/8-spring-forces/study-guide/p65phze7OgnnyC9g) and extra practice (https://library.fiveable.me/practice/ap-physics-1-revised).

If you compress or stretch an ideal spring, the spring force always tries to pull the object back toward the spring’s relaxed (equilibrium) length. Hooke’s law describes both cases: F_s = −k Δx (vector form). The magnitude |F_s| = k|Δx| grows linearly with how far you compress or stretch it. The sign (direction) changes: when you stretch the spring (Δx positive, away from equilibrium) the spring force points back toward the center (a restoring pull). When you compress it (Δx negative, shorter than relaxed) the spring force points outward toward equilibrium (a restoring push). Remember AP expects you to state magnitude ∝ displacement and that the force is directed toward the equilibrium position (CED 2.8.A). For more review and practice on these ideas, see the Topic 2.8 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/8-spring-forces/study-guide/p65phze7OgnnyC9g) and plenty of practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Because an ideal spring stores elastic potential energy whenever it’s stretched or compressed from its relaxed (equilibrium) length, it exerts a restoring force that tries to bring the system back to that length. Mathematically (CED 2.8.A.2–3), Hooke’s law gives Fs = −k Δx: the magnitude of the force is k|Δx| and the minus sign means the force points opposite the displacement. If you pull the mass right (positive Δx), the spring force points left; if you compress it (negative Δx), the force points right. That direction is what “toward the equilibrium position” means in the CED. Physically, springs resist change in length because the internal material wants to return to its relaxed configuration; the energy you put in shows up as elastic potential energy and the spring converts that back to kinetic energy as it returns. For AP review and practice on this Topic 2.8 idea, see Fiveable’s Topic 2.8 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/8-spring-forces/study-guide/p65phze7OgnnyC9g) and try related practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

Think of a spring force like any other contact force: draw every force acting on the object, then add the spring force directed toward the spring’s equilibrium position. Quick steps: - Identify the object and draw it as a dot or box. - Add all external forces: gravity (mg down), normal (N), any applied push/pull, tension if there’s a string. - Add the spring force Fs starting at the object, pointing toward the spring’s relaxed (equilibrium) position. Label Fs = −kΔx (magnitude k|Δx|), where Δx is extension (+) or compression (−). The minus sign shows it’s a restoring force. - If there are two springs on one object, draw both spring forces with their directions and either combine them (vector sum) or use keq (parallel: keq = k1 + k2; series: 1/keq = 1/k1 + 1/k2) before applying Newton’s 2nd law. - Use your FBD to write ΣF = ma in the chosen axis; for equilibrium set ΣF = 0. On AP free-response you’ll earn points for clear, labeled FBDs and correct use of Hooke’s law/Newton’s 2nd law. For extra practice, check the Topic 2.8 study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/8-spring-forces/study-guide/p65phze7OgnnyC9g) and lots of practice problems (https://library.fiveable.me/practice/ap-physics-1-revised).

The spring constant k is a measure of how stiff an ideal spring is. In Hooke’s law F_s = −k Δx, k (units N/m) tells you how much restoring force the spring produces per meter of displacement: a larger k → more force for the same Δx → a stiffer spring; a smaller k → less force → a more compliant spring. The force is always directed toward the equilibrium position (restoring force). For combinations: springs in parallel add (k_eq = k1 + k2) so the system is stiffer; springs in series add inverses (1/k_eq = 1/k1 + 1/k2) so the system is less stiff than the softest spring. This idea (Hooke’s law, restoring force, k units, series/parallel combos) is exactly what AP Topic 2.8 tests—review the Topic 2.8 study guide for examples (https://library.fiveable.me/ap-physics-1-revised/unit-2/8-spring-forces/study-guide/p65phze7OgnnyC9g) and practice problems at (https://library.fiveable.me/practice/ap-physics-1-revised).

You can find the spring constant k with a simple Hooke’s-law lab. Procedure (two common ways): 1) Hanging-masses method - Measure the spring’s relaxed length x0. - Hang a known mass m, wait until it stops, measure new length x. Repeat for several masses (0.05–0.5 kg steps). - For each mass, calculate force F = mg and extension Δx = x − x0. - Plot F (vertical axis, N) vs Δx (m). The slope = k (N/m). Do multiple trials, use small masses so the spring stays in the linear (elastic) region. 2) Force-pull method (if you have a force sensor) - Attach spring to a hook; pull to different known displacements Δx and record force from the sensor. Plot F vs Δx; slope = k. Tips to reduce error: zero length carefully, take multiple measurements and average, stay within elastic limit, use a ruler with mm resolution or a force sensor, include units. This matches CED Topic 2.8 (Hooke’s law F = −kΔx) and the AP lab/experimental-design skills; see the Topic 2.8 study guide for review (https://library.fiveable.me/ap-physics-1-revised/unit-2/8-spring-forces/study-guide/p65phze7OgnnyC9g). For extra practice problems, check Fiveable’s practice page (https://library.fiveable.me/practice/ap-physics-1-revised).

Short answer: k is always positive for an ideal (linear, stable) spring. In Hooke’s law F_s = −k Δx (CED 2.8.A.2) the minus sign gives the restoring direction (force toward equilibrium); k is the positive proportionality constant (spring stiffness). Why not negative? A negative k would make the spring force point the same direction as the displacement (force pushes away from equilibrium), which is unstable and not an ideal Hookean spring. In real systems you might get effective “negative stiffness” only in engineered, non-linear or active systems (outside AP Physics 1 scope). Equivalent spring constants computed for series/parallel combinations (CED 2.8.B.1) are positive if the individual k’s are positive. For AP prep, remember: use the minus sign to show direction (restoring), and treat k as a positive scalar. For more review and practice on Topic 2.8, see the Fiveable study guide (https://library.fiveable.me/ap-physics-1-revised/unit-2/8-spring-forces/study-guide/p65phze7OgnnyC9g) and Unit 2 resources (https://library.fiveable.me/ap-physics-1-revised/unit-2).