3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
In a lab, a student models a spring as an ideal spring. Which statement describes how an ideal spring behaves?
Answer and reasoning
AIts force is proportional to its length, measured end to end. A student who thinks spring force depends on total length picks this. The force is proportional to the change in length from the relaxed length; at its relaxed length a spring exerts no force, however long it is.
BIts force doubles when its stretch doubles; its mass is negligible.Correct An ideal spring has negligible mass and exerts a force proportional to its change in length from its relaxed length, so doubling the stretch doubles the force.
CIts force is the same however far it is stretched or compressed. A student who treats a spring force as a fixed property of the spring picks this. The force of an ideal spring grows in proportion to its change in length and is zero at the relaxed length.
DIts force depends on how fast the attached object is moving. A student who links force with speed picks this. The force of an ideal spring depends only on its change in length at that instant, whatever the velocity of the object attached to it.
The graph shows the magnitude F of the force exerted by an ideal spring as a function of the spring's length L, from its relaxed length to a length of 20 cm. What is the spring constant of the spring?
Answer and reasoning
A15 N/mCorrect The spring constant is the slope of the line. The spring's relaxed length is 10 cm, so at 20 cm, Δx = 0.10 m and k = 1.5 N/0.10 m = 15 N/m.
B7.5 N/m A student who divides the force by the total length, 1.5 N/0.20 m, gets 7.5 N/m. The force is proportional to the change in length from 10 cm, not to the length, which is why the line does not pass through the origin.
C0.15 N/m A student who divides by the change in length in centimeters, 1.5 N/10 cm, gets 0.15. In SI units the change in length is 0.10 m, so k = 15 N/m.
D0.075 N/m A student who takes the area under the graph, ½(1.5 N)(0.10 m) = 0.075 N·m, as the spring constant picks this. The spring constant is the slope of the graph, 15 N/m.
Working The relaxed length is where F = 0: L = 10 cm. At L = 20 cm, Δx = 20 cm − 10 cm = 10 cm = 0.10 m and F = 1.5 N. k = F/Δx = 1.5 N/0.10 m = 15 N/m, which is the slope of the line.
A cart on a level track with negligible friction is attached to an ideal spring whose other end is fixed to a wall on the left. At one instant, the cart is to the right of its equilibrium position and is moving to the right, away from the wall. Which of the numbered free-body diagrams shown correctly shows the horizontal forces exerted on the cart at that instant?
Answer and reasoning
ADiagram 1 A student who thinks the force on a moving object points in its direction of motion picks this. The spring force points toward the equilibrium position, to the left, whichever way the cart is moving; the cart is slowing down.
BDiagram 4 A student who thinks a moving object carries a forward 'force of motion' picks this. No object exerts such a force: the cart keeps moving right because it already has a velocity, and the only horizontal force is the spring's pull to the left.
CDiagram 2 A student who thinks a spring being stretched by the cart exerts no force picks this. The stretched spring pulls on the cart with a force of magnitude k|Δx|, to the left, toward equilibrium.
DDiagram 3Correct The cart is to the right of equilibrium, so the spring is stretched and pulls the cart to the left, toward the equilibrium position, even though the cart is moving to the right. That leftward spring force is the only horizontal force; it is slowing the cart down.
Working Horizontal forces only (friction is negligible). The cart is to the right of its equilibrium position, so the spring is stretched and pulls the cart toward the wall: one force, Fs, to the left. Its direction is set by the displacement from equilibrium, not by the velocity, so it points left although the cart moves right (the cart is slowing down). No other object exerts a horizontal force on the cart, so there is no forward force. The diagram with a single leftward arrow is number 3.
In preparation: 0 of 3 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 model of a spring with negligible mass that exerts a force proportional to the change in its length from its relaxed length. On the AP exam, springs are ideal unless stated otherwise.
Relaxed length
The length of a spring when it is neither stretched nor compressed, so it exerts no force. Changes in length are measured from this length.
Change in length, Δx (m)
How far a spring is stretched or compressed from its relaxed length: Δx = (length) − (relaxed length). It is not the spring's total length. Measured in meters.
Students often think The force exerted by a spring is proportional to its total length (so doubling the length doubles the force, and a longer spring always exerts a larger force). In fact No. It is proportional to the change in length measured from the relaxed length. A spring at its relaxed length exerts no force, however long it is.
Students often think A spring exerts a force of fixed size, a property of the spring like its 'strength', whatever its change in length. In fact No. The force of an ideal spring is proportional to its change in length: doubling the stretch doubles the force, and at the relaxed length the force is zero.
2.8.A.2 Hooke's law Fix
Hooke's law
The force exerted by an ideal spring on an object is F⃗s = −kΔx⃗: its magnitude is k|Δx|, and the minus sign shows that it points opposite to the displacement of the spring's end from its relaxed position.
Spring constant, k (N/m)
The magnitude of the spring force per unit change in length, k = |Fs|/|Δx|, measured in N/m. A larger k means a stiffer spring: a larger force for the same stretch, or a smaller stretch for the same force. It is the slope of a graph of force magnitude against change in length (or against length).
Spring force, F⃗s (N)
The contact force exerted by a spring on an object attached to its end. A stretched spring pulls the object; a compressed spring pushes it. Measured in newtons.
Students often think A change in length read in centimeters can be divided into a force in newtons to give the spring constant in N/m. In fact No. To get k in N/m, the change in length must be in meters: 5.0 cm = 0.050 m.
Students often think The spring constant measures how easily a spring stretches, so a spring that stretches more under a given force has a larger k (k = Δx/F). In fact No. A larger k means a stiffer spring: it needs a larger force for each meter of stretch, so under the same force it stretches less. k = F/Δx, not Δx/F.
2.8.A.3 Equilibrium position (object–spring system) Fix
Equilibrium position (object–spring system)
The position at which the net force on the object is zero. For an object attached to a horizontal spring with no other horizontal forces exerted on it, it is the position at which the spring has its relaxed length.
Restoring force
A force that points toward the equilibrium position whichever way the object is displaced. For an object on a horizontal ideal spring with no other horizontal forces, the spring force is a restoring force: whichever way the object is displaced and whichever way it is moving, it points back toward equilibrium, where the spring has its relaxed length. For an object hanging from a vertical spring, it is the net force, spring force plus gravitational force, that points toward equilibrium.
Students often think The force on a moving object points in its direction of motion and is largest where the object moves fastest. In fact No. The spring force depends only on the spring's change in length, not on the object's velocity. For an object oscillating on a horizontal spring with no other horizontal forces, the spring force is zero where the object moves fastest (at equilibrium, where the spring has its relaxed length) and largest where the object is momentarily at rest (farthest from equilibrium).
Students often think A moving object carries a force in its direction of motion (a 'force of motion' or impetus) in addition to the forces exerted on it by other objects. In fact No. The only horizontal force on the cart is the spring force. The cart keeps moving forward because it already has a velocity, not because a forward force acts on it.
6 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 6
An ideal spring has a relaxed length of 10 cm. When it is stretched to a length of 14 cm, it exerts a force of magnitude 3.0 N on the object attached to it. What is the magnitude of the force it exerts when it is stretched to a length of 22 cm?
Answer and reasoning
A4.7 N A student who takes the force to be proportional to the total length scales by 22/14 and gets 4.7 N. The force is proportional to the change in length from 10 cm, which triples.
B6.0 N A student who measures the change in length from 14 cm uses Δx = 8 cm and gets 6.0 N. The change in length is measured from the relaxed length: 22 cm − 10 cm = 12 cm.
C3.0 N A student who thinks a spring exerts the same force however far it is stretched keeps 3.0 N. An ideal spring's force is proportional to its change in length, which has tripled.
D9.0 NCorrect The change in length triples, from 4 cm to 12 cm (both measured from the 10 cm relaxed length), so the force triples: 3 × 3.0 N = 9.0 N.
Working Change in length from the relaxed length: Δx₁ = 14 cm − 10 cm = 4 cm and Δx₂ = 22 cm − 10 cm = 12 cm. The force is proportional to the change in length, so F₂ = (12 cm/4 cm)(3.0 N) = 3 × 3.0 N = 9.0 N.
An object of mass 0.20 kg hangs at rest from a vertical ideal spring, which is stretched 5.0 cm from its relaxed length. Use g = 10 m/s². What is the spring constant of the spring?
Answer and reasoning
A4.0 N/m A student who uses the mass, 0.20 kg, in place of the weight gets 0.20/0.050 = 4.0. The spring force balances the weight, mg = 2.0 N, so k = 40 N/m.
B0.40 N/m A student who divides the weight by the stretch in centimeters, 2.0 N/5.0 cm, gets 0.40. The stretch must be in meters, 0.050 m, giving 40 N/m.
C40 N/mCorrect The object is at rest, so the spring force balances the gravitational force: kΔx = mg = 2.0 N. With Δx = 0.050 m, k = 2.0 N/0.050 m = 40 N/m.
D0.025 N/m A student who thinks k measures how easily a spring stretches divides the stretch by the force: 0.050 m/2.0 N = 0.025. The spring constant is force per unit stretch, k = F/Δx = 40 N/m.
Working At rest, the spring force balances the gravitational force: kΔx = mg = (0.20 kg)(10 m/s²) = 2.0 N. Δx = 5.0 cm = 0.050 m, so k = 2.0 N/0.050 m = 40 N/m.
A block on a level surface with negligible friction is attached to an ideal spring whose other end is fixed to a wall. The spring has its relaxed length when the block is at position Q. The block is pushed toward the wall to position P, a distance d from Q, and released from rest. As it moves, it passes position R, a distance d/2 beyond Q, as shown in the diagram. Which correctly ranks the magnitudes FP, FQ and FR of the force exerted on the block by the spring at P, Q and R?
Answer and reasoning
AFP > FR > FQCorrect The magnitude of the spring force is k|Δx|, with Δx measured from Q: kd at P, kd/2 at R and zero at Q. Compression and stretch count alike, so FP > FR > FQ.
BFR > FQ > FP A student who takes the force to be proportional to the spring's total length ranks the longest spring (at R) first and the shortest (at P) last. The force depends on the change in length from Q: it is largest at P, compressed by d.
CFQ > FR > FP A student who links the spring force with the block's speed ranks Q, where the block is fastest, first, and P, where it is at rest, last. The spring force at Q is zero because the spring is at its relaxed length.
DFP = FQ = FR A student who thinks a spring exerts the same force at any length picks this. An ideal spring's force is proportional to its change in length: kd at P, zero at Q and kd/2 at R.
Working Hooke's law: |Fs| = k|Δx|, with Δx measured from the relaxed length at Q. At P the spring is compressed by d: FP = kd. At Q, Δx = 0: FQ = 0. At R the spring is stretched by d/2: FR = kd/2. So FP > FR > FQ. (The block's speed, zero at P and greatest at Q, does not enter.)
A block of mass m on a level surface with negligible friction is attached to a horizontal ideal spring with spring constant k and relaxed length L. The block is pulled until the length of the spring is L + d and is then released from rest. What is the magnitude of the block's acceleration immediately after it is released?
Answer and reasoning
Aa=k(L+d)/m A student who uses the spring's total length instead of its change in length gets k(L + d)/m. The spring's force depends only on how far it is stretched from its relaxed length, d.
Ba=k·d/(m·g) A student who divides the force by the block's weight instead of its mass gets kd/(mg), which is not even an acceleration (it has no units). Newton's second law divides the net force by the mass: a = kd/m.
Ca=k·d/mCorrect The spring is stretched by d from its relaxed length, so it exerts a force kd, the only horizontal force on the block. By Newton's second law, a = kd/m.
Da=d/(k·m) A student who thinks k measures how easily a spring stretches writes the spring force as d/k instead of kd, and gets d/(km). The force of a spring stretched by d is kd: a stiffer spring (larger k) pulls harder, so a = kd/m.
Working Immediately after release the only horizontal force on the block is the spring force, of magnitude k|Δx| with Δx = (L + d) − L = d, so Fnet = kd. Newton's second law: a = Fnet/m = kd/m.
A block on a level surface with negligible friction is attached to two identical ideal springs; the other end of each spring is fixed to a wall, one on each side of the block. When the block is at its equilibrium position, both springs have their relaxed lengths. The block is pulled a distance d to the right and released from rest. Which statement describes the forces exerted on the block by the two springs just after it is released?
Answer and reasoning
AThey point in opposite directions and add up to zero. A student who thinks a spring always pulls toward its fixed end has the left spring pulling left and the right spring pulling right. The right spring is compressed, so it pushes the block away from the right wall, to the left.
BBoth point to the left, toward the block's equilibrium position.Correct The left spring is stretched by d, so it pulls the block to the left; the right spring is compressed by d, so it pushes the block to the left. Each force has magnitude kd, and both point toward the equilibrium position, giving a net spring force of 2kd to the left.
CBoth point to the right, the direction of the displacement. A student who drops the minus sign in F⃗s = −kΔx⃗ picks this. Each spring force points opposite to the block's displacement from equilibrium, so both point to the left.
DBoth are zero, as the block is at rest at that instant. A student who thinks an object at rest has no force exerted on it picks this. At release the block is at rest, but the springs are stretched and compressed by d, so each exerts a force of magnitude kd.
A block of mass m is at rest on a ramp inclined at an angle θ above the horizontal. Friction between the block and the ramp is negligible. The block is attached to an ideal spring of spring constant k and relaxed length L. The spring is parallel to the ramp, and its upper end is fixed to a post at the top of the ramp. Which expression gives the length of the spring?
Answer and reasoning
AL + mg cosθ/k A student who takes the component of the gravitational force along the ramp to be mg cos θ picks this. The component along the ramp is mg sin θ (it is zero on a level surface, θ = 0, as it must be); mg cos θ is the component perpendicular to the ramp, which the normal force balances.
BL − mg sinθ/k A student who thinks a spring's force points the same way as the displacement of its end picks this: the spring must pull the block up the ramp, so this student moves the block's end up the ramp, toward the post, and shortens the spring. In F⃗s = −kΔx⃗ the force is opposite to the displacement: a spring pulling toward the post has been stretched down the ramp, so its length is L + mg sin θ/k.
CL + mg sinθ/kCorrect The block is at rest, so the spring's pull up the ramp balances the component of the gravitational force down the ramp: kΔx = mg sin θ. The spring pulls toward its fixed end, so it is stretched by mg sin θ/k beyond its relaxed length L.
DL + k·mg sinθ A student who thinks k measures how easily a spring stretches (k = Δx/F) multiplies the force by k to find the stretch. Hooke's law gives the stretch as the force divided by k, Δx = mg sin θ/k, so a stiffer spring stretches less; k·mg sin θ does not even have units of length.
Working Along the ramp two forces act on the block: the component of the gravitational force down the ramp, mg sin θ, and the spring force. (The normal force is perpendicular to the ramp.) The block is at rest, so the spring force must point up the ramp, toward the post, with magnitude kΔx = mg sin θ; the stretch is Δx = mg sin θ/k. A spring that pulls the block toward its fixed end is stretched, not compressed (F⃗s = −kΔx⃗: the block's end is displaced down the ramp). Length = relaxed length + stretch = L + mg sin θ/k.
Compiled from the AP Physics 1 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
