AP Physics 1 · Topic 2.8

Topic 2.8: Spring Forces

Unit 2: Force and Translational Dynamics18-23% of the multiple-choice section

Topic 2.8 is one equation and one idea. An ideal spring pulls or pushes with a force whose size is the spring constant times how far the spring has been stretched or compressed from its relaxed length, and that force always points back toward the equilibrium position of the object and spring.

AP Physics: Unit 2 (topics 2.8 Spring Forces). Topic 2.8 carries a single CED learning objective, 2.8.A, describe the force exerted on an object by an ideal spring, with three essential knowledge statements covering the ideal-spring model, Hooke's law, and the restoring direction of the force. The CED's suggested skills for this topic are 1.B, 2.A, 2.C, 3.A, and 3.B, two of which are graph and experimental-design skills. Unit 2 is weighted at 18 to 23 percent of the multiple-choice section, tied with Unit 3 for the largest share.

What Topic 2.8 requires

One learning objective, three essential knowledge statements, and one equation. LO 2.8.A asks you to describe the force exerted on an object by an ideal spring. That is the whole required content, which makes 2.8 the shortest topic in Unit 2.

Short does not mean lightweight. The three statements each carry a separate idea. EK 2.8.A.1 defines the ideal spring: negligible mass, and a force proportional to the change in its length as measured from its relaxed length. EK 2.8.A.2 gives Hooke's law for the magnitude of that force. EK 2.8.A.3 fixes the direction: the force exerted on an object by a spring is always directed toward the equilibrium position of the object and spring system.

The CED lists five suggested skills here, more than most topics in the unit: 1.B (create quantitative graphs with appropriate scales and units, including plotting data), 2.A (derive a symbolic expression), 2.C (compare physical quantities between scenarios), 3.A (create experimental procedures appropriate for a given scientific question), and 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim). Two of those, 1.B and 3.A, are graph and lab skills, which tells you where this topic tends to land on a test.

Hooke's law, and exactly what the minus sign does

The CED prints Hooke's law in vector form:

\vec{F}_s = -k \Delta \vec{x}

The same line appears on the AP Physics 1 equation sheet. Three separate claims are packed into those few symbols.

$\Delta \vec{x}$ is measured from the relaxed length of the spring. Not from the floor, not from the ceiling, not from wherever the object happened to start. EK 2.8.A.1 is explicit that the force is proportional to the change in length as measured from the relaxed length.

$k$ is the spring constant, in newtons per meter. It belongs to the spring, not to the object hanging on it, and it does not change as you stretch the spring further. Holding $k$ fixed over the whole range of the problem is a large part of what the word ideal is doing.

The minus sign is a statement about direction and nothing else. It says $\vec{F}_s$ points opposite $\Delta \vec{x}$ : stretch the spring right and it pulls left, compress it left and it pushes right. EK 2.8.A.3 says the same thing in words. What the minus sign does not do is make the force negative. Its magnitude is $k|\Delta x|$ , a positive number of newtons, and that positive number is what goes on a free-body diagram and into a net-force sum. Only when you pick a one-dimensional axis with the origin at the equilibrium position does the component form $F_s = -kx$ carry a usable sign, and there the sign is telling you which way the arrow points.

What the word ideal buys you

An ideal spring has two properties, and you get to assume both. The exam's own convention list, printed alongside the Table of Information, states that springs and strings are assumed to be ideal unless otherwise stated, so no problem has to tell you.

Negligible mass. The spring contributes nothing to the inertia of the system. That is why the forces at its two ends are equal in size and opposite in direction, exactly as for an ideal string, and why only the attached mass appears in the period expression you meet in Unit 7.

Force proportional to the change in length. Double the stretch and you double the force, over the whole range the problem uses. Plotted, that is a straight line through the origin, which is the entire reason the graph work in this topic is as clean as it is.

Neither property is free in a real spring, and the CED's model is deliberately the simple one. Nothing in Topic 2.8 asks you to reason about a spring that has stopped obeying the proportionality, so build your answers on the two assumptions above and say so when a derivation depends on them.

Measuring the spring constant from a graph

Skill 1.B and skill 3.A both point at the same experiment, and it is the standard way a lab or a data question reaches this topic.

Hang a known mass on the spring and let it come to rest. Two forces act on the mass: gravity pulling down with $mg$ , and the spring pulling up with $k\Delta x$ . It is not accelerating, so those balance:

k \Delta x = mg

Measure $\Delta x$ from the relaxed length, repeat with several masses, then plot the applied force on the vertical axis against the stretch on the horizontal. The slope of the best fit line is $k$ , in newtons per meter. Put force on the vertical axis rather than the other way round, or your slope comes out as $1/k$ .

Two diagnostics come free with the graph. A line that curves away from straight at the far end means the spring has left the proportional range the model assumes. A line that is straight but crosses the vertical axis well above zero usually means the relaxed length was recorded with the hanger already attached, so every stretch reading is short by the same amount.

The area under that same graph has a meaning too. EK 3.2.A.5, over in Unit 3, states that work is equal to the area under the curve of a graph of force as a function of displacement. Under a straight line of slope $k$ the area is a triangle, $\tfrac{1}{2}k(\Delta x)^2$ , which is exactly the elastic potential energy the equation sheet prints as $U_s = \tfrac{1}{2}k(\Delta x)^2$ .

A hanging spring moves the equilibrium, not the stiffness

Hang a mass on a vertical spring and let it settle. The spring is now permanently stretched by $\Delta x = mg/k$ , and that stretched position, not the relaxed length, is the equilibrium position of the object and spring system that EK 2.8.A.3 refers to.

Pull the mass down a further distance $A$ and release it. The extra spring force is $kA$ upward, gravity is unchanged, and the two combine into a restoring force of size $kA$ directed back at the new equilibrium. Gravity has shifted where the center of the motion sits. It has not changed $k$ , and it does not appear in the restoring force at all once you measure displacement from the new equilibrium.

That is worth holding onto because problems switch between the two reference points without warning. Stretch usually means the total extension from the relaxed length, which for a hanging system is $mg/k$ plus whatever extra you added. Displacement from equilibrium means the extra alone. Reading the wrong one into $F_s = -k\Delta x$ is the quiet way to get a spring problem wrong while every step looks correct.

A horizontal spring on a frictionless surface avoids the whole issue: there the relaxed position and the equilibrium position are the same point.

Spring force is not constant, so kinematics is out

The three constant-acceleration equations from Topic 1.3 require the acceleration along that axis to be constant. A spring force changes with position, so the acceleration changes with position too: halve the displacement and you halve the force, and therefore halve the acceleration. That rules out $v_x^2 = v_{x0}^2 + 2a_x(x - x_0)$ and its two companions for anything attached to a spring.

Two legitimate routes remain, and a problem's wording usually tells you which one it wants.

Energy, when the question asks for a speed at a position and never mentions time. Set $\tfrac{1}{2}k(\Delta x)^2$ against $\tfrac{1}{2}mv^2$ and read the answer off. Conservation of energy walks through the bookkeeping, and the work-energy theorem is the same idea written as work.
Simple harmonic motion, when the question asks for a time, a period, or a graph. If the spring force is the only force along the direction of motion, the motion is simple harmonic, and Unit 7 supplies the period and the sinusoidal position functions. See simple harmonic motion.

Newton's second law still works at any single instant, which is what makes the second worked example below possible. What you cannot do is treat the acceleration it gives you as valid over an interval.

How Topic 2.8 is tested

In the CED's sample free-response set, Question 2, the Translation Between Representations question, aligns to LO 2.8.A among several other objectives, with skills 1.A, 1.C, 2.A, 2.D, 3.B, and 3.C. The whole question runs on a block attached to an ideal spring, and it never asks for a single plug-in answer.

The parts, in order, ask a student to complete energy bar charts for spring potential energy and kinetic energy at three positions of the oscillation, read a spring constant out of a pair of graphs showing position against time and spring force against time, derive an expression for the block's mass starting from the period equation, sketch velocity against time on given axes, and finally judge whether a supplied free-body diagram is consistent with that velocity sketch at a stated instant.

That last part is where the sign convention earns its keep. A diagram showing the spring force pointing one way is consistent with a velocity graph only if the block is on the matching side of equilibrium at that instant, which is EK 2.8.A.3 used as a test rather than as a fact to recite.

A four-function, scientific, or graphing calculator is allowed on both sections of the exam, so arithmetic is never what makes a spring question hard. Reading a graph correctly and keeping the reference point straight is.

Errors that cost points on spring questions

Five failures show up repeatedly in work on this topic, and four of them are bookkeeping rather than physics.

Treating the minus sign as part of the magnitude. The size of the force is $k|\Delta x|$ , always positive. The sign lives in the direction of the arrow.
Measuring the displacement from the wrong place. It comes from the relaxed length in Hooke's law, and from the loaded equilibrium in a hanging oscillator. Decide which one the problem means before you substitute.
Confusing the spring constant with a force. Written $k$ , it is in newtons per meter and describes the spring. It is a slope, not a reading.
Reaching for the kinematic equations. The acceleration is not constant, so use energy or the Unit 7 machinery instead.
Squaring the wrong thing in the energy expression. $U_s = \tfrac{1}{2}k(\Delta x)^2$ squares the displacement, so doubling the stretch quadruples the stored energy while only doubling the force.

Topic 2.8 also feeds forward more than its length suggests. The spring force is the restoring force behind every oscillation in Unit 7, and elastic potential energy is one of the terms in the Unit 3 energy accounting. Getting the sign and the reference point right here saves the same work twice later.

Spring constant from a hanging-mass data set

A student hangs masses from a vertical spring and measures the stretch from the spring's relaxed length each time: 0.100 kg gives 3.3 cm, 0.200 kg gives 6.4 cm, 0.300 kg gives 9.9 cm, and 0.400 kg gives 13.0 cm. Find the spring constant, and then find how far the spring stretches under a 0.250 kg mass.

Set the convention. Take down as the positive direction for the hanging mass. At rest the spring pulls up with $k\Delta x$ and gravity pulls down with $mg$ , and since the mass is not accelerating those two balance: $k\Delta x = mg$ .
Convert each mass to the force it applies, using $g = 9.8 \, \mathrm{m/s^2}$ : $0.100 \, \mathrm{kg}$ gives $0.980 \, \mathrm{N}$ , $0.200 \, \mathrm{kg}$ gives $1.960 \, \mathrm{N}$ , $0.300 \, \mathrm{kg}$ gives $2.940 \, \mathrm{N}$ , and $0.400 \, \mathrm{kg}$ gives $3.920 \, \mathrm{N}$ .
Plot force on the vertical axis against stretch on the horizontal, then take the slope between two well-separated points on the best fit line. Using the first and last: $k = \frac{3.920 \, \mathrm{N} - 0.980 \, \mathrm{N}}{0.130 \, \mathrm{m} - 0.033 \, \mathrm{m}} = \frac{2.940 \, \mathrm{N}}{0.097 \, \mathrm{m}} = 30.3 \, \mathrm{N/m}$ .
Check that the data really are proportional before trusting the slope. Point by point, $F/\Delta x$ gives 29.7, 30.6, 29.7, and 30.2 N/m. That is a spread of about 1 N/m, roughly 3 percent, with no drift in one direction, which is what a straight line through the origin looks like with millimeter readings.
The stretch measurements are read to the nearest millimeter, so the run of the slope, 9.7 cm, carries two significant figures and the spring constant does too: $k = 30 \, \mathrm{N/m}$ .
Now use it, keeping the unrounded slope. For a 0.250 kg mass, $\Delta x = mg/k = (0.250 \, \mathrm{kg})(9.8 \, \mathrm{m/s^2}) / (30.3 \, \mathrm{N/m}) = 2.45 \, \mathrm{N} / 30.3 \, \mathrm{N/m} = 0.0808 \, \mathrm{m}$ .

The spring constant is 30 N/m, and a 0.250 kg mass stretches the spring 8.1 cm. The prediction sits between the 6.4 cm and 9.9 cm readings and close to their midpoint, which is what a proportional spring should give for a mass halfway between 0.200 kg and 0.300 kg. That bracket is the fastest check available on an answer like this.

The spring force at two positions, and why kinematics fails

A 0.60 kg block sits on a frictionless horizontal surface against an ideal spring of spring constant $k = 45$ N/m. The block is pushed 0.12 m to the left of the spring's relaxed position and released. Find the spring force and the block's acceleration at the release point, and again when the block has returned to 0.060 m left of the relaxed position.

Set the convention. Let positive $x$ point right, with the origin at the relaxed position of the spring, which on a frictionless horizontal surface is also the equilibrium position of the block and spring system.
At release the block is at $x = -0.12 \, \mathrm{m}$ , so $\Delta x = -0.12 \, \mathrm{m}$ and the component form of Hooke's law gives $F_s = -k\Delta x = -(45 \, \mathrm{N/m})(-0.12 \, \mathrm{m}) = +5.4 \, \mathrm{N}$ . The plus sign says the force points right, back toward equilibrium, which matches EK 2.8.A.3.
Newton's second law at that instant: $a_x = F_s/m = 5.4 \, \mathrm{N} / 0.60 \, \mathrm{kg} = +9.0 \, \mathrm{m/s^2}$ , directed right.
Repeat at $x = -0.060 \, \mathrm{m}$ : $F_s = -(45 \, \mathrm{N/m})(-0.060 \, \mathrm{m}) = +2.7 \, \mathrm{N}$ , and $a_x = 2.7 \, \mathrm{N} / 0.60 \, \mathrm{kg} = +4.5 \, \mathrm{m/s^2}$ .
Compare the two. Halving the displacement halved the force and halved the acceleration. Both values are correct at their own instant and neither describes the interval between them, so $v_x^2 = v_{x0}^2 + 2a_x(x-x_0)$ has no constant $a_x$ to take.
To get the speed at the halfway point you need energy instead: $\tfrac{1}{2}k(0.12)^2 = \tfrac{1}{2}k(0.060)^2 + \tfrac{1}{2}mv^2$ , which is a Unit 3 calculation rather than a Unit 2 one.

At 0.12 m from equilibrium the spring pushes with 5.4 N and the block accelerates at $9.0 \, \mathrm{m/s^2}$ ; at 0.060 m the spring pushes with 2.7 N and the acceleration is $4.5 \, \mathrm{m/s^2}$ . Both point back toward equilibrium. Because the acceleration tracks the displacement rather than staying fixed, the constant-acceleration equations do not apply anywhere on this motion.

Frequently asked questions

What does AP Physics 1 Topic 2.8 cover?

One learning objective, 2.8.A: describe the force exerted on an object by an ideal spring. Three essential knowledge statements sit under it. An ideal spring has negligible mass and exerts a force proportional to the change in its length from its relaxed length. Hooke's law gives the magnitude of that force. And the force is always directed toward the equilibrium position of the object and spring system.

What does the negative sign in Hooke's law mean?

It records direction, not size. The spring force points opposite the displacement of the spring from its relaxed length, so a stretch to the right produces a pull to the left. The magnitude is the spring constant times the size of the displacement, and that is always a positive number of newtons. Put the positive magnitude on a free-body diagram and let the arrow carry the direction.

How do you find a spring constant from experimental data?

Hang known masses, measure the stretch from the relaxed length for each, and plot the applied force on the vertical axis against the stretch on the horizontal. The slope of the best fit line is the spring constant in newtons per meter. A nonzero vertical intercept usually means the relaxed length was measured with the hanger already attached, and curvature at the top end means the spring has left the proportional range the model assumes.

Where do you measure the spring's displacement from?

For Hooke's law itself, from the relaxed length of the spring. For a mass hanging on a vertical spring and oscillating, the useful reference is the loaded equilibrium position, which already sits mg over k below the relaxed length. Both are legitimate; the failure is switching between them mid-problem. On a horizontal frictionless surface they coincide, which is why that setup is used so often.

Can you use the kinematic equations on a mass attached to a spring?

No. Those three equations need a constant acceleration on the axis, and the spring force changes with position, so the acceleration does too. Use conservation of energy when the question asks for a speed at a position, or the simple harmonic motion results from Unit 7 when it asks for a time or a period. Newton's second law still gives you the acceleration at any single instant.