Spring Forces: Study Guide (Physics C: Mechanics 2.8)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 2.8 called Spring Forces. This guide is the calculus-based one. It covers the same ideal spring and Hooke’s law, then adds combinations of springs in series and in parallel, derived from first principles, and treats the spring constant as a slope that can change for a nonideal spring. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise. This topic follows Topic 2.7, Kinetic and Static Friction.

The ideal spring model

A spring has a relaxed length: its length when nothing stretches or squeezes it. An ideal spring has two properties:

1. Its mass is negligible.
2. The force it exerts is proportional to its change in length from the relaxed length.

A spring that breaks either rule is nonideal. A heavy spring hanging under its own weight breaks rule 1. A rubber cord, or a metal spring stretched too far, breaks rule 2.

The first property has a useful consequence. Newton’s second law for the spring itself is F_net = m_spring a. If m_spring ≈ 0, the net force on the spring must be zero, even when it accelerates. So an ideal spring pulls (or pushes) with the same size of force at both ends. That is why the same force passes through every spring in a series chain below.

Hooke’s law and its direction

For an ideal spring, the size of the force is

|F_s| = k|Δx|

• Δx is the change in length from the relaxed length: an extension or a compression.
• k is the spring constant, in N/m. A larger k means a stiffer spring.

Direction. A stretched spring pulls on the object; a compressed spring pushes on it. Either way, the force points back towards the position where the spring is relaxed. Put the origin at that position and take +x along the spring’s axis. Then

F_s,x = −kx

The minus sign carries the direction. If x > 0 (stretched to the right of relaxed), the force is towards −x. If x < 0, it is towards +x. For a block on a level, smooth surface attached to a horizontal spring, the relaxed position is also the block’s equilibrium position, so the spring force always points towards equilibrium.

A hanging block: two different “centres”

Hang a block of mass m from a vertical spring. Take +y upward. Let s be the stretch. The spring pulls up with ks and the weight pulls down with mg. Equilibrium is where they balance: s₀ = mg/k.

At any stretch s, the net force is F_net,y = ks − mg = k(s − s₀).

• The spring force always points towards the relaxed length. While the spring is stretched at all, it pulls up.
• The net force points towards the new equilibrium at s₀, with size k times the distance from it.

So between the relaxed length and the equilibrium position, the spring still pulls up, but the net force points down. Keep the two ideas separate. When a statement says a spring force points “towards equilibrium”, apply it to the spring’s own relaxed position, as for the horizontal spring above; with gravity also acting, it is the net force that points to the new equilibrium at s₀.

Graphs: k as a slope

Plot the size of the spring force F against the extension Δx. For an ideal spring the graph is a straight line through the origin, and

k = slope = dF/d(Δx)

For a nonideal spring the graph curves. Then there is no single k, but the local slope dF/d(Δx) still tells you how stiff the spring is near a given extension.

Testing whether a real spring is ideal

You can check the model with a simple experiment.

1. Hang the spring vertically next to a ruler and record the position of its lower end with nothing attached. This is the relaxed position.
2. Add known masses one at a time. For each, wait until the mass is at rest and record the new position. The extension is the difference.
3. At rest, the spring force equals the weight, so F = mg for each load.
4. Plot F on the vertical axis against extension in metres on the horizontal axis.
5. If the points lie on a straight line through the origin, the spring is ideal over that range and the slope is k. If the graph curves, or the spring does not return to its relaxed length when the masses are removed, it is not.

Two practical points help. Read the ruler at eye level to avoid parallax. Take readings while loading and again while unloading: a gap between the two sets is another sign of nonideal behaviour. Any real spring is only close to ideal over a limited range, so state the range when you quote k.

Representing spring forces

On a free-body diagram, draw the spring force from the point where the spring attaches, along the spring’s axis. Label it F_s (not “k” or “kx”), and decide its direction from whether the spring is stretched or compressed. A spring attached to a block and a wall pulls (or pushes) on both with equal-sized forces in opposite directions. A spring scale reads the size of that force. So a spring scale pulled by two people with 30 N each reads 30 N, not 60 N.

Combining springs

Several springs acting on one object can behave like a single spring with an equivalent spring constant k_eq. The course expects only two arrangements: all in series or all in parallel. You do not need mixed arrangements.

Springs in series (end to end)

The springs form one chain. Because each ideal spring is massless, the same force F acts through every spring. Each spring stretches by F/kᵢ, and the total stretch is the sum:

x = F/k₁ + F/k₂ + … = F(1/k₁ + 1/k₂ + …)

Write x = F/k_eq and compare:

1/k_eq = 1/k₁ + 1/k₂ + …

Each term 1/kᵢ is positive, so 1/k_eq is larger than any single 1/kᵢ. That means k_eq is smaller than the smallest kᵢ. A chain is softer than its softest link.

Cutting a spring. A spring of constant k is like n identical pieces in series. If each piece has constant k_p, then 1/k = n/k_p, so k_p = nk. Cut a spring in half and each half is twice as stiff.

Springs in parallel (side by side)

The springs share the load and all have the same extension x (for example, both attached to a bar that stays level). The forces add:

F = k₁x + k₂x + … = (k₁ + k₂ + …)x

k_eq = k₁ + k₂ + …

Parallel springs are stiffer than any one of them.

Worked example 1: the same springs, two arrangements

Question. Take +y upward. Springs with k₁ = 300 N/m and k₂ = 600 N/m are ideal. A 1.2 kg block hangs at rest from them, first (a) in series and then (b) in parallel, as in Figure 2. For each case find k_eq, the total extension and the force in each spring.

1. Weight: mg = 1.2 × 9.8 = 11.76 N. At rest, the springs together must pull up with 11.76 N.
2. (a) Series. 1/k_eq = 1/300 + 1/600 = 3/600, so k_eq = 200 N/m. Check: smaller than 300 N/m, as it must be.
3. Total extension: 11.76 ÷ 200 = 0.0588 m ≈ 5.9 cm.
4. Each spring carries the full 11.76 N. Extensions: 11.76 ÷ 300 = 0.0392 m and 11.76 ÷ 600 = 0.0196 m. Sum: 0.0588 m, which agrees with step 3.
5. (b) Parallel. k_eq = 300 + 600 = 900 N/m.
6. Common extension: 11.76 ÷ 900 = 0.01307 m ≈ 1.3 cm.
7. Forces: 300 × 0.01307 = 3.92 N and 600 × 0.01307 = 7.84 N. Sum: 11.76 N. The stiffer spring carries twice the load.

Factor of change. The extension in series is 0.0588 ÷ 0.01307 = 4.5 times the extension in parallel, because k_eq goes from 200 N/m to 900 N/m.

Worked example 2: a block between two springs

Question. Take +x to the right. A 0.50 kg block on a smooth, level surface is attached to a spring of k₁ = 80 N/m on its left and a spring of k₂ = 120 N/m on its right. The far ends are fixed to walls. Both springs are relaxed when the block is at x = 0. Find the net force and acceleration of the block at (a) x = +0.050 m and (b) x = −0.030 m. (c) Show that the pair acts like one spring and find k_eq.

1. (a) At x = +0.050 m the left spring is stretched by 0.050 m. It pulls the block left: F₁,x = −80 × 0.050 = −4.0 N.
2. The right spring is compressed by 0.050 m. It pushes the block left: F₂,x = −120 × 0.050 = −6.0 N.
3. Net force: −10 N. Acceleration: −10 ÷ 0.50 = −20 m/s² (towards the left, back towards x = 0).
4. (b) At x = −0.030 m the left spring is compressed and pushes right with 2.4 N; the right spring is stretched and pulls right with 3.6 N. Net force +6.0 N; acceleration +12 m/s².
5. (c) In general, F_net,x = −k₁x − k₂x = −(k₁ + k₂)x. That has the Hooke’s law form with k_eq = k₁ + k₂ = 200 N/m.

Check. The springs sit on opposite sides, yet they combine like parallel springs. What decides the rule is the physics, not the picture: both springs have the same change in length as the block moves, and their forces add. Check: −200 × 0.050 = −10 N, matching step 3.

Common misconceptions

• “Springs drawn side by side are parallel; springs on opposite sides are series.” Decide by asking: same force (series) or same extension (parallel)? Worked example 2 is parallel.
• “Series springs are stiffer because there are more of them.” k_eq in series is smaller than the smallest k.
• Using centimetres with k in N/m. Convert to metres first.
• Mixing up spring force and net force for a hanging block. The spring force points towards the relaxed length. Gravity moves the equilibrium down, and it is the net force that points towards that new equilibrium.
• “Δx is the spring’s length.” It is the change from the relaxed length.
• “Cutting a spring in half halves k.” Each half has 2k.
• “Any spring obeys F = kx.” Only an ideal one. Check that the force–extension graph is straight.

Where this leads

Next, Topic 2.9, Resistive Forces deals with forces that depend on velocity. In Unit 3 you will integrate F_s,x = −kx to find the work done by a spring and its potential energy, and in Unit 7 the restoring force −kx produces simple harmonic motion. Try the practice questions now, then use the revision notes and the checklist. You can also return to the course roadmap.