2.8Force and Translational Dynamics

Spring Forces — AP PHYSICS 1 Revision Notes & Topical Questions

5 min read 8 parts Lesson 30 of 36 1 quiz

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Lesson summary

2.8 Spring Forces — AP PHYSICS 1 notes: Describe the force exerted by an ideal spring using Hooke's law, force–extension graphs and spring combinations.Ideal spring = a spring of negligible mass whose force is proportional to its change in length from the relaxed length.Relaxed (…

Relaxed (equilibrium) length
Extension x
Minus sign
Beyond the elastic limit
Area under graph
Many springs

Part 1 of 6Ideal spring

Ideal spring

Ideal spring a spring of negligible mass whose force is proportional to its change in length from the relaxed length.

Relaxed (equilibrium) length

Relaxed (equilibrium) length the length when the spring is neither stretched nor compressed, with zero net force.

Extension x

Extension x stretched length − relaxed length, measured from the relaxed length not from the total length.

Part 2 of 6Hooke's law

Fs = −kx

Fs = −kx the spring force is proportional to the displacement from the relaxed length.
F_s is the spring force in N.
k is the spring constant in N/m, the force per unit extension.
x is the change in length in m.

Minus sign

Minus sign the spring force is a restoring force, always directed toward the equilibrium position.

Magnitude

|F_s| = k|x|, valid for both extension and compression.

Stiffness

Stiffness a larger k means a stiffer spring needing more force per metre.

Part 3 of 6Force–extension graph

Ideal spring

Ideal spring a straight line through the origin.

Gradient

Gradient equals k when force is on the vertical axis.
- (1/k if the axes are swapped)

Beyond the elastic limit

Beyond the elastic limit the graph curves and Hooke's law fails.

Area under graph

Area under graph equals the elastic potential energy ½kx².

Part 4 of 6Springs in combination

Arrangement	k_eff	Result for identical springs
Single spring	k	Extension x
Series (end to end)	1/k_eff = 1/k₁ + 1/k₂, so k_eff = k₁k₂/(k₁ + k₂)	k/2, softer; extension doubles
Parallel (side by side)	k_eff = k₁ + k₂	2k, stiffer; extension halves
Spring cut in half	Each half is stiffer	2k

Series

Series the force in each spring is equal and the extensions add.

Parallel

Parallel the extension is the same and the forces add.

Many springs

Many springs the same rules extend to any number of springs.

Part 5 of 6Hanging mass at rest

Equilibrium

Equilibrium kx = mg, so x = mg/k.

Forces balance

Forces balance the upward spring force kx equals the downward weight mg, so the net force is zero.

Proportional

Proportional doubling the hanging mass doubles the extension, as long as the spring stays within its elastic limit.

Stiffness

Stiffness a stiffer spring (larger k) stretches less for the same mass.

Part 6 of 6Determining k experimentally

Measure the unstretched length of the spring.

Hang known masses and measure the new length for each.

Calculate extension = new length − unstretched length and force = mg.

Plot force (vertical) against extension (horizontal).

Draw a best-fit line; its slope is the spring constant k.

Key terms

Term	Meaning
Hooke's law	F_s = −kx; spring force proportional to extension.
Spring constant	Stiffness k in N/m.
Restoring force	Force directed toward equilibrium.
Elastic potential energy	Energy ½kx² stored in a stretched spring.
Ideal spring	Massless spring obeying Hooke's law.

Avoid these mistakes

Using the total length of the spring instead of the extension — measure from the relaxed length.

Reading k from a graph with swapped axes — extension against force has gradient 1/k.

Adding spring constants for springs in series — series springs combine through reciprocals.

Forgetting to convert cm to m.

Last updated: 14 October 2026