Virtual laboratory/Physics/Mechanics/Hooke's Law — elastic force and the formula F = kx

Hooke's Law — elastic force and the formula F = kx

What we will do: We will stretch a spring with different forces and watch how its extension changes, and what difference the stiffness of the spring makes.

What you see on screen: A spring with a weight, sliders for the force and the spring constant, and an F(x) graph rising along a straight line. A stiffer spring extends less; once the elastic limit is passed, the line bends. Readouts: force, the constant k, extension and elastic energy.

Hooke's Law — elastic force and the formula F = kx
Physics · Mechanics · Hooke's law (FIZ.DRGAN.PP.01)
Stretching force F5.0 N
Spring constant k50 N/m
You double the force stretching the spring (F: 5 N → 10 N, k = 50 N/m) — the extension:
force F
5.0 N
constant k
50 N/m
extension x
0.10 m
energy Ep
0.25 J
state
Elastic
🎯 Challenge: a force and an extension were measured — set the spring constant kk

What just happened?

You doubled the force and the spring extended exactly twice as much. The extension is directly proportional to the force.

Hooke's law:

F=k xF = k\,x

F — stretching force, x — extension, k — spring constant (the spring's stiffness). The larger k is, the stiffer the spring and the less it extends under the same force.

What does it mean?

Double the force → double the extension, triple → triple. On the F(x) graph this is a straight line, and its slope is exactly k: k=Fxk = \dfrac{F}{x}

Elastic energy:

a stretched spring stores energy Ep=12kx2E_p = \tfrac{1}{2}k x^2 — it is what launches the body once released.

Note — the elastic limit:

Hooke's law holds only up to a certain force. Beyond it the spring deforms permanently and no longer returns to its original length.

Why you need this for the exam:

F = kx and Ep = ½kx² are the basis of problems on springs, elastic force and oscillations. The trick: find k from the measured F and x, then compute the energy.

Remember: the extension is proportional to the force (F = kx), but only within the elastic range. A stiffer spring (larger k) extends less.

Illustrative model · NexTutor Virtual laboratory

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Theory & formulas

Hooke's law

Within the elastic range, a spring's extension is directly proportional to the applied force: F=k xF = k\,x The F(x)F(x) graph is a straight line whose slope equals the constant kk.

Spring constant

kk is the stiffness of the spring: the larger kk is, the harder it is to stretch (a smaller extension for the same force).

Elastic energy

A stretched spring stores elastic potential energy: E=12kx2E = \tfrac{1}{2}k x^2 Once the elastic limit is exceeded, the law no longer holds — the spring deforms permanently.

Notation: FF — force, kk — spring constant (in N/m\mathrm{N/m}), xx — extension, EE — elastic energy.

Frequently asked questions

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What does Hooke's law state?

A spring's extension is directly proportional to the applied force: F = kx (within the elastic range).

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What does the spring constant k mean?

The stiffness of the spring — the larger k, the harder it is to stretch; it is the slope of the F(x) graph.

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When does Hooke's law stop holding?

Once the elastic limit is exceeded — the spring then deforms permanently and its extension is no longer proportional to the force.

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