How Does a Spring's Force Depend on How Far You Stretch It?

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Spring Forces: The Homesick Spring and Hooke's Law

Spring Forces: The Homesick Spring and Hooke's Law

|Fs| = k|Δx| = 100 × 0.000 = 0.00 N
Δx = 0 — spring relaxed, Fs = 0
k = 100 N/m
PHYSICS INSIGHTS

Δx is measured from the relaxed length. Relaxed, this spring is L0 = 20.0 cm long. Stretched to 24.0 cm, Δx = +4.0 cm, and it is 4.0 cm, not 24.0 cm, that goes into |Fs| = k|Δx|.

Equal stretch or squeeze, equal size of force. Both give the same |Fs|, pointing back toward the relaxed length. The minus sign in Fs = −kΔx only sets that direction, so +3 cm and −3 cm land on the same point of the graph.

k is the slope of the line. A stiffer spring needs more force for the same Δx, so its line is steeper. Read k as rise ÷ run, with the run in meters: 4.00 N ÷ 0.040 m = 100 N/m.

Simulation by The Science Cube — https://www.thesciencecube.com/
Spring Forces: Measuring k with Hanging Masses

Spring Forces: Measuring k with Hanging Masses

k — spring constant200 N/m
Fₛ − mg = 0⇒ kΔx = mg:200 × 0.000 = 0.0 N
Simulation by The Science Cube — https://www.thesciencecube.com/

How does a spring’s force depend on how far you stretch or squeeze it?

In short: The spring force is proportional to the change in length: Fs = −kΔx, so its size is |Fs| = k|Δx|. Here Δx is measured from the relaxed length, not from zero. The minus sign only says the force points back toward the relaxed length. The spring constant k, in N/m, is the slope of the force–stretch graph, so a stiffer spring has a steeper line.

Common doubts

  • “Is Δx the length of the spring?” No. Δx is the change from the relaxed length. A spring with a relaxed length of 20 cm that is stretched to 24 cm has Δx = 4 cm, and only that 4 cm goes into k|Δx|.
  • “Does squeezing give a different force from stretching by the same amount?” No. The size of the force is the same, and in both cases it points back toward the relaxed length.
  • “How do you measure k with hanging masses?” At rest the spring force balances the weight, so kΔx = mg. Add masses, record Δx each time, and plot mg against Δx. The best-fit slope is k.

Use the first simulation to see Fs = −kΔx, then add masses in the second one and record points to measure k.

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