4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
A spring of mass ms hangs from a ceiling, and a block of mass M hangs at rest from its lower end. The spring is not ideal, because its mass is not negligible. If g is the acceleration due to gravity, what is the magnitude of the force exerted by the ceiling on the spring?
Answer and reasoning
AMg A student who treats the spring as transmitting only the load, as an ideal spring would, picks this. The spring's own weight also hangs from the ceiling, so the force at the top exceeds the force at the bottom by ms g.
B(M + ms/2)g A student who thinks the spring's weight is shared between its two ends picks this. The lower end is attached to the block, which pulls down; only the ceiling holds the spring up, so it supports the whole of ms g.
Cms g A student who thinks the ceiling holds up only the spring it touches picks this. The block hangs from the spring, which hangs from the ceiling, so the ceiling supports both weights.
D(M + ms)gCorrect Take the spring and block together as the system. It is at rest, and the only forces on it are its total weight, (M + ms)g, and the ceiling's upward force, so the ceiling exerts (M + ms)g. The spring pulls on the block with only Mg, so the force differs from one end of the spring to the other.
Working System = spring + block, at rest: Fceiling − Mg − ms g = 0 → Fceiling = (M + ms)g. (Block alone: the spring's lower end pulls up with Mg; spring alone: Fceiling = Mg + ms g.)
An ideal spring has a relaxed length of 0.10 m. A force of 1.2 N holds it stretched to a length of 0.12 m. What force must be exerted on the spring to hold it stretched to a length of 0.16 m?
Answer and reasoning
A3.6 NCorrect Hooke's law uses the change in length from the relaxed length. k = 1.2 N/(0.12 m − 0.10 m) = 60 N/m. At 0.16 m the change in length is 0.06 m, so the force is (60 N/m)(0.06 m) = 3.6 N.
B1.6 N A student who takes the force to be proportional to the spring's total length picks this: (1.2 N)(0.16 m/0.12 m) = 1.6 N. The force is proportional to the change from the relaxed length, which triples from 0.02 m to 0.06 m.
C2.4 N A student who measures the further stretch from the 0.12 m length picks this: (60 N/m)(0.04 m) = 2.4 N. Δx is always measured from the relaxed length, 0.10 m, so it is 0.06 m.
D1.2 N A student who thinks a spring exerts a fixed force picks this. The spring's force grows in proportion to its change in length, so a larger stretch needs a larger force.
Working k = F₁/Δx₁ = 1.2 N/(0.12 m − 0.10 m) = 60 N/m. Δx₂ = 0.16 m − 0.10 m = 0.06 m. F₂ = kΔx₂ = (60 N/m)(0.06 m) = 3.6 N.
A block on a frictionless, horizontal surface is attached to an ideal spring whose other end is fixed to a wall. The diagram shows the block at x = 0, where the spring is relaxed, and three other positions of the block, P, Q and R; +x is to the right. FP, FQ and FR are the x-components of the spring force on the block at P, Q and R. Which ranking is correct?
Answer and reasoning
AFR > FQ > FP A student who takes the spring force to point the same way as the displacement, F = +kx, picks this. The spring force points back toward x = 0, opposite to the displacement.
BFP > FQ > FRCorrect Fx = −kx. At P (x = −0.10 m) the compressed spring pushes right: FP = +0.10k. At Q (+0.05 m) and R (+0.20 m) the stretched spring pulls left: FQ = −0.05k and FR = −0.20k. In signed values, FP > FQ > FR.
CFR > FP > FQ A student who ranks the sizes of the forces (0.20k, 0.10k, 0.05k) instead of their x-components picks this. At R the force points left, so its x-component, −0.20k, is the smallest of the three.
DFP = FQ = FR A student who thinks a spring exerts the same force wherever the block is picks this. The force is proportional to the change in the spring's length and points toward x = 0, so it differs at the three positions.
Working Fx = −kx: P (−0.10 m) → +0.10k; Q (+0.05 m) → −0.05k; R (+0.20 m) → −0.20k. Ranking FP > FQ > FR.
The diagram, not drawn to scale, shows three arrangements, X, Y and Z, of identical ideal springs supporting identical blocks at rest. ΔxX, ΔxY and ΔxZ are the distances each block is below the position it would have if its springs were relaxed. Which ranking is correct?
Answer and reasoning
AΔxZ > ΔxX > ΔxY A student who swaps the rules, adding constants in series and reciprocals in parallel, picks this: Y would be 2k and Z would be k/2. Springs side by side share the load, so Z is the stiffest arrangement.
BΔxX = ΔxY = ΔxZ A student who gives any combination of identical springs the average constant, k, picks this. Two springs in series make a softer combination (k/2) and two side by side a stiffer one (2k).
CΔxX = ΔxY > ΔxZ A student who thinks the load is shared between the springs in Y, as it is in Z, picks this: each spring of Y would stretch mg/(2k), a total of mg/k. Each spring in series carries all of mg, so Y stretches 2mg/k.
DΔxY > ΔxX > ΔxZCorrect With spring constant k for each spring, X has k, Y (series) has k/2 and Z (parallel) has 2k. For the same weight, Δx = mg/keq, so Y stretches 2mg/k, X mg/k and Z mg/(2k).
Working X: keq = k → mg/k. Y (series): 1/keq = 2/k → keq = k/2 → 2mg/k. Z (parallel): keq = 2k → mg/(2k). ΔxY > ΔxX > ΔxZ.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
2.8.A.1 Ideal spring Fix
Ideal spring
A spring that has negligible mass and exerts a force proportional to the change in its length measured from its relaxed length. Springs are assumed ideal unless stated otherwise.
Nonideal spring
A spring that has nonnegligible mass, or whose force is not proportional to the change in its length from its relaxed length (for example, a graph of force against stretch that curves or does not pass through the origin).
Relaxed length
The length of a spring when it exerts no force. Changes in length, Δx, are measured from it. Unit: meter (m).
Students often think A spring transmits only the load hung on it, so the force at the top equals the force at the bottom even when the spring has mass. In fact No. A spring with nonnegligible mass is nonideal: at rest and hanging, its upper end must also support the spring's own weight, so the force at the top exceeds the force at the bottom by the spring's weight.
Students often think The weight of a hanging spring is shared between its two ends, so the ceiling supports only half of it. In fact No. A hanging spring touches only the ceiling and the block below it, which pulls down on it. The ceiling therefore supports the spring's whole weight as well as the block's.
2.8.A.2 Hooke's law and the spring constant, k Fix
Hooke's law and the spring constant, k
For an ideal spring, F⃗s = −kΔx⃗, where Δx⃗ is the change in the spring's length from its relaxed length and k is the spring constant, the magnitude of the force per unit change in length. Unit of k: newton per meter (N/m).
Students often think The force a spring exerts is proportional to its length, so F = kL with L the spring's total length. In fact No. The force of an ideal spring is proportional to the change in its length from its relaxed length, Δx, not to its total length.
Students often think When a stretched spring is stretched further, Δx in Hooke's law is the extra stretch measured from its previous length. In fact From the spring's relaxed length, every time. A further stretch from an already-stretched length adds to Δx; it does not restart it.
2.8.A.3 Direction of the spring force Fix
Direction of the spring force
The force a spring exerts on an object points opposite to the displacement of the spring's end from its relaxed position, so it points back toward the equilibrium position of the object–spring system: a stretched spring pulls, a compressed spring pushes.
Students often think The spring force on an object points in the same direction as the object's displacement from the relaxed position, F = +kΔx. In fact No. The spring force points opposite to the displacement from the relaxed position: a block displaced to the right is pulled to the left, and a block displaced to the left is pushed to the right.
Students often think When forces are ranked by their x-components, only their sizes matter, so the largest force has the largest x-component. In fact No. An x-component carries a sign: a force of 5 N to the left has x-component −5 N, which is less than +1 N.
2.8.B.1 Equivalent spring constant, keqFix
Equivalent spring constant, keq
The spring constant of a single spring that, put in place of a collection of springs, would exert the same force on the object for the same displacement. Unit: N/m.
Springs in series
Springs joined end to end, so that each carries the same force and their changes in length add: 1/keq,series = Σi 1/ki.
Series combination is softer
Because every term 1/ki in the series sum is positive, 1/keq,series exceeds the largest 1/ki, so keq,series is less than the smallest ki.
Springs in parallel
Springs that act on an object side by side, so that each has the same change in length and their forces add: keq,parallel = Σi ki.
Students often think Spring constants in series add (keq = k₁ + k₂), and in parallel the reciprocals add. In fact No. For springs the rules are the reverse of the resistor rules: spring constants in parallel add, and in series the reciprocals add.
Students often think A combination of springs is only as stiff as its softest spring, so keq equals the smallest spring constant. In fact No. In series the equivalent constant is less than the softest spring's, and in parallel it is greater than the stiffest spring's.
8 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 8
The graph shows the magnitude F of the force exerted by each of three springs, X, Y and Z, as a function of the change Δx in its length from its relaxed length. Each spring has negligible mass. Which of the springs behave as ideal springs over the range shown?
Answer and reasoning
AX and Y, as both of their graphs are straight lines A student who takes 'linear' to mean 'proportional' picks this. Y's line does not pass through the origin: its force is 5 N at 0.10 m and 8 N at 0.20 m, so doubling the stretch does not double the force.
BX and Z, as both of their graphs pass through the origin A student who checks only that the force is zero at zero stretch picks this. Z's graph curves, so F/Δx is not constant: about 6.75 N at 0.10 m but only 9 N at 0.20 m.
CX only, as its force alone is proportional to its stretchCorrect An ideal spring's force is proportional to Δx: doubling Δx doubles F. Only X does this (3 N at 0.10 m, 6 N at 0.20 m). Y is straight but its force is 2 N at zero stretch, so doubling Δx does not double F; Z's ratio F/Δx falls as it stretches.
DX, Y and Z, as each force grows as the stretch grows A student who takes 'pulls harder when stretched more' as the definition of an ideal spring picks this. All three forces increase, but only X's increases in proportion to Δx.
A 4.0 kg block hangs from an ideal spring of spring constant 100 N/m attached to the ceiling of an elevator. The block is at rest relative to the elevator, which has an upward acceleration of 3.0 m/s². Use g = 10 m/s². By how much is the spring stretched from its relaxed length?
Answer and reasoning
A0.40 m A student who takes the spring force to equal the weight picks this: (40 N)/(100 N/m). That holds only when the block does not accelerate; here the spring must also supply the net upward force ma.
B0.28 m A student who thinks an upward acceleration lightens the block picks this: (4.0 kg)(10 − 3.0) m/s²/(100 N/m). An upward acceleration needs a net upward force, so the spring pulls harder than the weight.
C0.52 mCorrect The block accelerates upward with the elevator, so the net force on it is up: kΔx − mg = ma. Δx = m(g + a)/k = (4.0 kg)(13 m/s²)/(100 N/m) = 0.52 m.
D0.12 m A student who sets the spring force equal to ma picks this: (4.0 kg)(3.0 m/s²)/(100 N/m). ma is the net force; the spring must also balance the weight, mg.
Working Up positive: kΔx − mg = ma → Δx = m(g + a)/k = (4.0 kg)(10 m/s² + 3.0 m/s²)/(100 N/m) = 0.52 m.
Two ideal springs, of spring constants k and 3k, are joined end to end. The upper spring is attached to a ceiling, and a block of mass m hangs at rest from the lower spring. If g is the acceleration due to gravity, how far below the position it would have if both springs were relaxed is the block?
Answer and reasoning
A(1/4)mg/k A student who adds spring constants in series, as resistances add, picks this: keq = 4k. Joining springs end to end adds their stretches, so the combination is softer than either spring, not stiffer.
B(4/3)mg/kCorrect Each spring in series carries the full weight, mg. The springs stretch by mg/k and mg/(3k), and the block's displacement is the sum: mg/k + mg/(3k) = (4/3)mg/k. Equivalently, 1/keq = 1/k + 1/(3k), so keq = (3/4)k.
C(1/2)mg/k A student who uses the average spring constant, 2k, picks this. In series the reciprocals add: 1/keq = 1/k + 1/(3k), so keq = (3/4)k.
D(2/3)mg/k A student who thinks each spring carries half the weight picks this: (mg/2)/k + (mg/2)/(3k) = (2/3)mg/k. The junction between the springs is massless, so the two spring forces on it are equal: each spring carries all of mg.
Working Junction (massless): both springs carry F = mg. Δx₁ = mg/k, Δx₂ = mg/(3k). Total = mg(1/k + 1/(3k)) = (4/3)mg/k; keq = (3/4)k.
Two ideal springs, of spring constants k and 2k, are joined end to end, and a load hung from the pair stretches it by a total of Δx₁. A third spring, of spring constant 2k, is added to the end of the chain, and the same load stretches the three springs by a total of Δx₂. What is Δx₂/Δx₁?
Answer and reasoning
A0.60 A student who adds spring constants in series picks this: keq goes from 3k to 5k, so the stretch falls to 3/5. Adding a spring in series adds another positive term to 1/keq, which makes the chain softer.
B1.00 A student who sets keq equal to the softest spring's constant, k, in both chains picks this. The added spring stretches too, by F/(2k), so the total stretch grows.
C0.89 A student who shares the load equally among the springs picks this: (F/3)(2/k) compared with (F/2)(1.5/k) gives 0.89. Each spring in series carries the whole load, so the added spring adds F/(2k) to the stretch.
D1.33Correct In series every spring carries the full load F, and the stretches add. Δx₁ = F/k + F/(2k) = 1.5F/k; Δx₂ = F/k + F/(2k) + F/(2k) = 2F/k. So Δx₂/Δx₁ = 2/1.5 = 1.33: adding a spring in series always increases the total stretch.
A student claims that when two or more ideal springs are joined end to end (in series), the equivalent spring constant is less than the smallest of their spring constants. Which reasoning correctly supports the claim?
Answer and reasoning
AEach spring carries the full load, so the chain stretches more than its softest spring would alone.Correct In series the force is the same in every spring, so each stretches by F/ki and the stretches add. The total, F Σ 1/ki, is greater than F/kmin, the stretch of the softest spring alone, so keq = F/(total stretch) < kmin.
BEach spring stretches by the same amount, so the chain stretches several times as far as one spring. A student who thinks springs in series stretch equally picks this. The springs carry equal forces, so a spring stretches by F/ki: the softer spring stretches more, the stiffer less.
CThe stiffer spring pulls back on the softer one, so their forces partly cancel and the chain is softer. A student who pictures the springs pulling against each other picks this. At each junction the two spring forces are equal and opposite because the junction is massless; nothing cancels in the force the chain exerts on the load.
DThe softest spring sets the stretch of the chain, and the other springs barely stretch at all. A student who pictures the softest spring as the weak link picks this. Every spring stretches by F/ki, and if the others did not stretch, the chain would have keq equal to, not less than, the smallest constant.
Two ideal springs, of spring constants 300 N/m and 600 N/m, are joined end to end. One free end is held fixed, and a 12 N force is exerted on the other free end along the springs. By how much does the pair stretch in total?
Answer and reasoning
A0.013 m A student who adds the spring constants, 900 N/m, picks this. Springs in series combine by their reciprocals, so keq is 200 N/m and the stretch is larger, not smaller, than either spring's alone.
B0.060 mCorrect Each spring carries the full 12 N: the 300 N/m spring stretches 0.040 m and the 600 N/m spring 0.020 m, a total of 0.060 m. Equivalently, keq = (300)(600)/(900) N/m = 200 N/m, less than either spring's constant.
C0.040 m A student who takes the pair to behave like its softer spring, 300 N/m, picks this. The 600 N/m spring also carries 12 N and stretches another 0.020 m.
D0.027 m A student who uses the average spring constant, 450 N/m, picks this. In series 1/keq = 1/300 + 1/600, giving 200 N/m.
Working 1/keq = 1/(300 N/m) + 1/(600 N/m) → keq = 200 N/m. Δx = F/keq = 12 N/(200 N/m) = 0.060 m (= 0.040 m + 0.020 m).
The diagram shows a cart on a frictionless, level floor between two ideal springs, each attached to a wall, with the spring constants labeled. Both springs are relaxed when the cart is at x = 0. The cart is moved a small distance Δx to the right, as shown, where +x is to the right. What is the x-component of the net force exerted on the cart by the springs?
Answer and reasoning
A−1.0kΔx A student who thinks springs on opposite sides always oppose each other picks this: 2kΔx − kΔx. The compressed right spring pushes the cart to the left, the same way as the stretched left spring pulls it.
B−2.0kΔx A student who thinks a spring can only pull counts only the stretched left spring. The compressed right spring also exerts a force, kΔx, pushing the cart back toward x = 0.
C−3.0kΔxCorrect The left spring (2k) is stretched by Δx and pulls the cart left with 2kΔx; the right spring (k) is compressed by Δx and pushes the cart left with kΔx. The forces add: Fx = −(2k + k)Δx = −3.0kΔx. The springs act like springs in parallel.
D−1.5kΔx A student who gives the pair the average spring constant, (2k + k)/2, picks this. Both springs exert forces on the cart in the same direction, so their constants add: keq = 3k.
Working Left spring (2k), stretched by Δx: F₁ = −2kΔx. Right spring (k), compressed by Δx: F₂ = −kΔx (pushes left). Fnet,x = −3kΔx = −3.0kΔx (coefficient form).
A block hangs at rest from an ideal spring of spring constant k, which is stretched by Δx₁. The spring is then cut into two equal halves, and the two halves are hung side by side from the ceiling, both attached to the same block. At rest, each half is stretched by Δx₂. What is Δx₂/Δx₁?
Answer and reasoning
A0.25Correct The whole spring is its two halves in series: 1/k = 2/khalf, so each half has khalf = 2k. Side by side, the halves act in parallel: keq = 2k + 2k = 4k. The same weight gives Δx₂ = mg/(4k), a quarter of Δx₁ = mg/k.
B4.00 A student who swaps the series and parallel rules picks this: the halves would have k/2 each, and side by side they would combine to k/4. Halves of a spring are stiffer than the whole, and springs side by side add their constants.
C0.50 A student who takes each half to keep the spring constant k picks this: side by side they give 2k. Each half stretches half as much under the same force, so its constant is 2k.
D1.00 A student who thinks half a spring is half as stiff picks this: two halves of k/2 side by side give k again. Under the same force half the coils stretch half as far, so each half has constant 2k.
Working Whole spring = two halves in series: 1/k = 1/kh + 1/kh → kh = 2k. Halves in parallel: keq = 4k. Δx₂/Δx₁ = (mg/4k)/(mg/k) = 0.25.
Compiled from the AP Physics C: Mechanics Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
