Spring Force and Hooke’s Law

A spring is a useful model because its force depends on the spring’s deformation rather than simply on where an object happens to be in space. Within the linear elastic range, the spring force is proportional to the displacement from the spring’s undeformed configuration and points in the direction that tends to restore the spring toward that configuration.

This article develops the one-dimensional spring-force law, its vector interpretation, the meaning and units of the spring constant, static equilibrium with gravity, Friction thresholds, and equivalent stiffness for simple combinations of ideal springs. The central modeling rule is

(1)

The minus sign is not an extra force. It records direction: the spring force opposes the signed deformation x.

1 Deformation is measured from a reference length

Let an ideal spring have natural length L₀. If its instantaneous length is L, define the signed deformation

(2)

Then

(3)

means the spring is stretched, while

(4)

means it is compressed.

The coordinate x here is a deformation coordinate. It need not be the same as the object’s coordinate measured from an arbitrary laboratory origin.

2 Hooke’s law and the restoring direction

For an ideal linear spring,

(5)

where k is the spring constant. Since force has SI units of newtons and deformation has units of metres,

(6)

The spring constant measures stiffness. A larger k means a larger force is required to produce the same deformation.

If x > 0, the spring is stretched and

(7)

so the force points toward decreasing x. If x < 0, the spring is compressed and

(8)

so the force points toward increasing x.

In both cases, the force points toward x = 0.

3 Force-displacement graph

Hooke’s law is linear. A graph of spring force versus deformation is therefore a straight line through the origin:

(9)

Its slope is

(10)

Thus the magnitude of the slope gives the spring constant.

The model is usually an approximation. Real springs are approximately linear only over a finite range of deformation. If the spring is stretched or compressed far enough, the force-displacement curve can become nonlinear, and sufficiently large deformation can produce permanent changes.

4 Newton’s second law with a spring force

For a block of mass m moving along the spring axis, suppose x = 0 is the natural-length configuration and the spring is the only horizontal force. Newton’s second law gives

(11)

Therefore

(12)

The acceleration points toward the undeformed configuration. At larger deformation, its magnitude is larger.

This equation is the dynamical seed of simple harmonic motion, but the full time-dependent solution is deferred to the oscillations sequence.

5 A useful vector form

For a spring constrained to act along a known axis with unit vector e, the force can be written

(13)

For a spring joining two particles, a more geometric statement is useful. Let

(14)

If the natural length is L₀, then the force on particle 2 due to the spring is

(15)

The force on particle 1 is equal and opposite:

(16)

This form makes clear that the spring force acts along the spring and satisfies Newton’s third law.

6 Worked example 1: horizontal spring acceleration

A 2.00 kg block is attached to a horizontal spring with spring constant k = 120 N∕m. The spring is stretched 0.150 m from its natural length. Neglect friction. Find the spring force and the block’s instantaneous acceleration.

Using Hooke’s law,

(17)

Thus the force has magnitude 18.0 N and points toward the natural-length position.

Newton’s second law gives

(18)

Therefore

(19)

7 Vertical springs and equilibrium

Consider a mass hanging from a vertical spring. Let x be the downward extension from the spring’s natural length. The forces on the mass are its Weight downward and the spring force upward.

At static equilibrium,

(20)

Hence

(21)

The spring is still stretched at equilibrium. Equilibrium does not mean the spring force is zero; it means the net force is zero.

8 Worked example 2: hanging mass equilibrium

A 0.750 kg mass hangs from a vertical spring with k = 95.0 N∕m. Find the equilibrium extension.

From

(22)

we obtain

(23)

Thus

(24)

Notice that the mass changes the equilibrium position but does not change the spring constant.

9 Displacement measured from equilibrium

For a vertical spring, it is often convenient to define a new coordinate y measured from the static equilibrium position rather than from natural length.

If the downward extension from natural length is

(25)

then Newton’s second law gives

(26)

Because

(27)

the constant terms cancel:

(28)

Gravity shifts the equilibrium position. Once displacement is measured from that equilibrium, the net restoring force has the same linear form as for a horizontal spring.

10 Worked example 3: instantaneous acceleration below equilibrium

A 1.50 kg mass hangs from a vertical spring with k = 60.0 N∕m. It is momentarily located 0.0800 m below its static equilibrium position. Find its instantaneous acceleration.

Using displacement y from equilibrium,

(29)

Therefore

(30)

Thus

(31)

The weight need not be inserted again after the coordinate has been shifted to equilibrium; its effect has already been absorbed into the equilibrium extension.

11 Spring force and static friction

A spring can gradually increase the tangential force on a body resting on a rough surface. Static friction adjusts as needed until its maximum magnitude is reached.

For a horizontal surface,

(32)

so the limiting static friction is

(33)

If a spring pulls horizontally, impending slip occurs when

(34)

Therefore the threshold deformation is

(35)

12 Worked example 4: spring extension required to start sliding

A 4.00 kg block rests on a horizontal surface with coefficient of static friction μ_s = 0.350. It is attached to a horizontal spring with k = 180 N∕m. How far must the spring be stretched before the block is on the verge of sliding?

At impending motion,

(36)

Hence

(37)

Therefore

(38)

For smaller extensions, the static friction force is simply equal in magnitude to kx; it is not yet at its maximum value.

13 Springs in parallel

Suppose two ideal springs with constants k₁ and k₂ are attached in parallel so that each experiences the same deformation x.

Their restoring forces add:

(39)

Thus

(40)

The equivalent spring constant is

(41)

Parallel springs are stiffer than either spring alone.

14 Springs in series

For two ideal massless springs in series under a static load, the force magnitude is the same through both springs. Let that magnitude be F. Their extensions are

(42)

The total extension is

(43)

Defining F = k_eqx gives

(44)

For two springs,

(45)

Series springs are less stiff than either spring alone.

15 Limits of the ideal spring model

Hooke’s law is a constitutive model, not a universal law valid for every deformation. Important limitations include the following:

• The relation F_s = −kx is usually accurate only over a finite linear elastic range.
• The spring constant can depend on geometry and material.
• Real springs have mass; when their inertia matters, the force need not be identical at every point along the spring.
• Real systems can dissipate mechanical energy through internal damping and friction.
• A coil spring may have geometric limits that prevent unlimited compression or extension.

The ideal model is powerful because it isolates the leading linear behavior near an equilibrium configuration.

16 Common mistakes

• Using the object’s coordinate from an arbitrary origin as though it were automatically the spring deformation.
• Dropping the minus sign without separately tracking the force direction.
• Assuming a spring always pulls. An ideal compression spring can push when compressed.
• Assuming static equilibrium means the spring force is zero.
• Setting static friction equal to μ_sN before checking whether the limiting value is actually required.
• Adding spring constants for springs in series. Direct addition applies to the parallel case.

17 Practice problems

Use g = 9.81 m∕s² unless otherwise stated.

1. A spring with k = 250 N∕m is stretched 0.0400 m. Find the magnitude and direction of the spring force.
2. A 3.00 kg block is attached to a horizontal spring with k = 150 N∕m. At an instant when the spring is compressed 0.120 m, find the block’s acceleration if friction is negligible.
3. A spring requires a 24.0 N force to hold it stretched 0.0800 m from natural length. Find k.
4. A graph of F_s versus x is a straight line through the origin with slope −420 N∕m. What is the spring constant?
5. A 1.20 kg mass hangs at rest from a vertical spring with k = 75.0 N∕m. Find the equilibrium extension.
6. The mass in Problem 5 is displaced 0.0500 m downward from its static equilibrium position. Find its instantaneous acceleration.
7. A 5.00 kg block rests on a horizontal surface with μ_s = 0.300 and is attached to a horizontal spring with k = 200 N∕m. Find the largest extension that can be maintained without slipping.
8. Two springs with constants 120 N∕m and 180 N∕m are connected in parallel. Find the equivalent spring constant.
9. The same two springs are connected in series. Find the equivalent spring constant.
10. Two springs in series have k₁ = 100 N∕m and k₂ = 300 N∕m. A static force of 12.0 N is applied. Find the extension of each spring and the total extension.

18 Answers

1. 10.0 N, toward the natural-length position.
2. 6.00 m∕s² toward the natural-length position.
3. 300 N∕m.
4. 420 N∕m.
5. 0.157 m.
6. 3.13 m∕s² upward.
7. 0.0736 m.
8. 300 N∕m.
9. 72.0 N∕m.
10. x₁ = 0.120 m, x₂ = 0.0400 m, total 0.160 m.

19 Summary

For an ideal linear spring,

(46)

The force is proportional to deformation and points toward the undeformed configuration. The spring constant is the magnitude of the slope of the force-displacement graph. For a vertical hanging mass,

(47)

When displacement is measured from that equilibrium position, gravity cancels from the net restoring-force equation.

For simple ideal spring combinations,

(48)

and

(49)

These results provide the force-law foundation for later work on elastic potential energy and oscillations.

References