Spring Forces: Study Guide (Physics 1 2.8)

This guide is for the algebra-based Physics 1 course. Everything here uses algebra, graphs and free-body diagrams. You need Newton’s second law (Topic 2.5) and weight, F_g = mg (Topic 2.6).

The ideal spring

Real springs have mass, can be overstretched, and wobble. In this course we use a simple model, the ideal spring:

• Its mass is negligible. That means it pulls (or pushes) with the same size of force on whatever is attached at each end.
• The force it exerts is proportional to how much its length has changed from its relaxed length (its natural length, when nothing stretches or squashes it).

Write Δx for the change in length. If a spring is 20.0 cm long when relaxed and 26.0 cm long when stretched, Δx = 6.0 cm = 0.060 m. If it is squashed to 17.0 cm, the size of Δx is 3.0 cm = 0.030 m, a compression.

Hooke’s law

The size of the spring force is

|F_s| = k|Δx|

With a sign for direction, along one axis:

F_s = −kΔx

• k is the spring constant, measured in newtons per metre (N/m). It tells you how many newtons it takes to change the length by one metre. A large k is a stiff spring; a small k is a soft one.
• Δx is measured from the relaxed length. Choose the axis so that Δx is positive when stretched in the +x direction.
• The minus sign means the force is opposite to the change in length. Stretch the spring towards +x and it pulls towards −x. Compress it towards −x and it pushes towards +x.

A spring force is a restoring force: it always acts to bring its own length back to the relaxed length. For a block attached to a horizontal spring on a smooth surface, the relaxed position is also the equilibrium position of the block–spring system, so the spring force always points back toward equilibrium. It does not matter which way the block is moving at that instant.

Representing a spring force

Graphs: finding k from data

Hooke’s law says force and change in length are directly proportional. So a graph of spring force (size) against change in length is a straight line through the origin, and its slope is k.

To find k in the lab, hang known masses from a spring and measure the stretch for each. While the mass hangs at rest, the spring force balances the weight, so |F_s| = mg.

Worked example 1: k from a graph of data

Question. A student hangs masses from a spring and records the extension Δx from the relaxed length. Take +y downward for the extension.

m (kg)	0.050	0.100	0.150	0.200	0.250
F = mg (N)	0.49	0.98	1.47	1.96	2.45
Δx (cm)	2.4	5.0	7.3	9.9	12.2

Plot the data and find k.

1. Convert extensions to metres: 0.024, 0.050, 0.073, 0.099 and 0.122 m.
2. Plot F (vertical axis, N) against Δx (horizontal axis, m). Choose scales that use most of the grid: 0 to 0.14 m and 0 to 3.0 N.
3. Draw one best-fit straight line. It should pass close to every point and through (or very near) the origin. Do not join the dots.
4. Take two points on the line, far apart: (0.020 m, 0.40 N) and (0.120 m, 2.40 N).
5. k = slope = (2.40 − 0.40) N ÷ (0.120 − 0.020) m = 2.00 N ÷ 0.100 m = 20 N/m.

Interpretation. The line passes through the origin and the points lie close to it, so force is proportional to extension: Hooke’s law holds over this range. Each F/Δx ratio is between 19.6 and 20.4 N/m, which agrees.

Check. Predict before you test: with 0.40 kg, the stretch should be mg/k = 3.92 N ÷ 20 N/m = 0.196 m. If a much larger stretch is measured, the spring may have gone past the range where it behaves ideally.

Designing the experiment well

• Measure the relaxed length first, and calculate every Δx from it. Plotting the total length instead of Δx gives a line that does not pass through the origin (it meets the length axis at the relaxed length).
• Read the ruler at eye level, at the same point on the spring each time.
• Use a range of masses large enough to give clear stretches, but stop before the spring stays permanently stretched.
• Let each mass come to rest before reading, so the spring force really equals mg.

Worked example 2: a spring on a slope

Question. A 0.80 kg block rests on a smooth (frictionless) slope at 30° to the horizontal. It is held by a spring with k = 40 N/m, fixed at the top of the slope and parallel to it. (a) Derive an expression for the stretch of the spring. (b) Calculate it. (c) Compare it with the stretch when the same block hangs vertically from the same spring. Take +x down the slope.

1. Forces along the slope: the component of gravity, mg sin θ, down the slope (+x), and the spring force, k Δx, up the slope (−x). The normal force is perpendicular to the slope and has no component along it.
2. (a) The block is at rest, so the forces along the slope balance: kΔx = mg sin θ, giving Δx = mg sin θ ÷ k.
3. (b) Δx = (0.80 × 9.8 × sin 30°) ÷ 40 = 3.92 N ÷ 40 N/m = 0.098 m (9.8 cm).
4. (c) Hanging vertically, kΔx = mg, so Δx = mg ÷ k = 7.84 N ÷ 40 N/m = 0.196 m. The stretch on the slope is exactly half, because sin 30° = 0.5.

Check with limiting cases. If θ = 0 (a level surface), Δx = 0: nothing pulls the block along, so the spring stays relaxed. If θ = 90°, sin θ = 1 and the expression becomes the vertical result, mg/k. Both make sense.

Worked example 3: force and acceleration at different positions

Question. The block in Figure 1 has mass 0.60 kg and slides on a smooth horizontal surface. The spring has k = 120 N/m. Take +x to the right, with x = 0 where the spring is relaxed. Find the spring force and the block’s acceleration when the block is at (a) x = +5.0 cm, (b) x = −3.0 cm and (c) x = 0.

1. (a) Δx = +0.050 m. F_s = −kΔx = −120 × 0.050 = −6.0 N (to the left). a_x = F_s ÷ m = −6.0 ÷ 0.60 = −10 m/s².
2. (b) Δx = −0.030 m. F_s = −120 × (−0.030) = +3.6 N (to the right). a_x = 3.6 ÷ 0.60 = +6.0 m/s².
3. (c) Δx = 0, so F_s = 0 and a_x = 0.

Interpretation. The spring force and acceleration always point toward x = 0 and are largest furthest from it. At x = 0 the acceleration is zero, but the block may still be moving fast: zero force does not mean zero velocity. This back-and-forth pattern is the basis of oscillations in Unit 7.

Comparing springs and situations

Many questions ask you to compare two cases rather than calculate one. Start from kΔx = F and ask what stays fixed.

• Same force, different springs. Δx = F ÷ k, so the stiffer spring stretches less. A spring with twice the k stretches half as far under the same load.
• Same spring, different forces. Δx is proportional to F. Three times the hanging mass gives three times the stretch, as long as the spring stays in its Hooke’s-law range.
• Both ends of one spring. Because an ideal spring has negligible mass, the net force on it must be zero even when things accelerate. So it pulls with the same size of force on the objects at both ends. If a 4.0 N pull stretches it, it pulls back with 4.0 N on your hand and 4.0 N on the wall.
• Stretched or compressed by the same amount. The size of the force is the same, k|Δx|; only the direction changes.

Common misconceptions

• “Δx is the length of the spring.” It is the change in length from the relaxed length.
• “The spring force points in the direction the block moves.” It points toward the relaxed position, whichever way the block is moving (Worked example 3).
• “The minus sign means the force is negative in size.” The minus sign only gives the direction. The size is k|Δx|.
• “A bigger k is a softer spring.” A bigger k is a stiffer spring: more force per metre.
• “k changes with the mass hung on it.” k is a property of the spring. The stretch changes; k does not (Worked example 1).
• “Use centimetres with k in N/m.” Convert to metres first, or your force will be 100 times too large.
• “A compressed spring pulls.” A stretched spring pulls on what is attached to it; a compressed spring pushes.

Where this leads

Unlike kinetic friction in Topic 2.7, which stays the same as an object slides, a spring force changes as the object moves. The next topic, Circular Motion, returns to forces that change direction. Springs return in Unit 3 (energy stored in a spring) and Unit 7 (oscillations). Try the practice questions now, then use the revision notes and the checklist to consolidate. You can also return to the course roadmap.