Dimensional Formula of Spring Constant: [M¹ L⁰ T⁻²]

The dimensional formula of spring constant comes to [M¹ L⁰ T⁻²]; hide the zero power of length and you have [MT⁻²]. It is the same bracket for force constant and spring stiffness, which are other names for k in Hooke's law.

Hooke's law says the restoring force of a spring is proportional to its extension, F = kx, provided the elastic limit is not crossed. So k is force per unit length. Force has one power of length and the extension removes it, leaving mass over time squared. A stiff car suspension spring and a soft pen spring differ hugely in value, yet both carry this bracket. See the spring force formula for problems on F = kx.

Quick facts

Dimensional formula
[M¹ L⁰ T⁻²], or [MT⁻²]
Defining formula
$k = \frac{F}{x}$
SI unit
N/m, equal to kg s⁻²
Other names
Force constant, spring stiffness
Type
Scalar
Same bracket
Surface tension, surface energy

Deriving the dimensional formula of spring constant

Hooke's law: $F = kx$ , taking the size of the force and dropping the minus sign that shows direction.
Rearrange: $k = \frac{F}{x}$ .
Force: $[MLT^{-2}]$ . Extension: [L].
Divide: $[k] = [ML^{1-1}T^{-2}] = [MT^{-2}]$ , or [M¹ L⁰ T⁻²] in full form.

Energy offers a second path. Elastic potential energy is $U = \frac{1}{2}kx^{2}$ , so $k = \frac{2U}{x^{2}}$ . That gives $\frac{[ML^{2}T^{-2}]}{[L^{2}]} = [MT^{-2}]$ again, which is a reassuring check. See spring potential energy for that formula in use.

N/m and kg/s² are the same unit

From k = F/x, the natural unit is newton per metre. Replace the newton with kg m s⁻² and the metre cancels, leaving kg s⁻². The two look unrelated, but the bracket [MT⁻²] proves they are equal.

Example: a 2.0 N pull stretches a spring by 4.0 cm. Find k in SI units.

Convert: 4.0 cm = 0.040 m.

$k = \frac{2.0\ \text{N}}{0.040\ \text{m}} = 50\ \text{N/m}$ , the same as 50 kg s⁻².

Do not leave the extension in centimetres. 2.0/4.0 = 0.50 N/cm is correct, but it is not the SI value.

Testing the spring oscillation formula

Example: check $T = 2\pi\sqrt{\frac{m}{k}}$ , then use it for a 0.50 kg block on the spring above.

Dimensions: $\frac{[M]}{[MT^{-2}]} = [T^{2}]$ , and the square root is [T], a time. The formula is consistent.

Numbers: $T = 2\pi\sqrt{\frac{0.50}{50}} = 2\pi\sqrt{0.010} = 2\pi(0.10) = 0.628\ \text{s}$ .

A common wrong version is $T = 2\pi\sqrt{k/m}$ . Its bracket is [T⁻¹], a frequency, so dimensions reject it at once. The angular frequency is $\omega = \sqrt{k/m}$ , and that one does carry [T⁻¹].

Quantities with the bracket [MT⁻²]

Force per length and energy per area
Quantity	Relation	Usual unit
Spring constant	$F/x$	N/m
Surface tension	Force per length of edge	N/m
Surface energy	Energy per area	J/m²
Force gradient	$dF/dx$	N/m

Spring constant and surface tension share both a bracket and a unit, even though one describes a coil of metal and the other a liquid surface. Dimensions cannot tell them apart; the context of the question must. Do not confuse k with tension in a string, which is a force, [MLT⁻²].

Spring constant bracket slips

Writing [MLT⁻²]. That is force. Dividing by extension removes the L.
Writing [ML⁻¹T⁻²]. That is stress, force per area, not force per length.
Keeping the minus sign of Hooke's law in the bracket. Signs show direction and never enter dimensions.
Confusing k for springs with k for Boltzmann or Coulomb. The same letter names three different constants.

A wide range of mechanics brackets, from force to stress, is set out on the dimensional formulas hub.

Frequently asked questions

What is the dimensional formula of spring constant?

The bracket is [M¹ L⁰ T⁻²], shortened to [MT⁻²]. It comes from k = F/x, force [MLT⁻²] divided by extension [L].

What is the dimension formula of spring constant, also called force constant?

Force constant is another name for spring constant, so the answer is the same, [MT⁻²], with SI unit N/m.

Which SI unit does spring constant use?

Newton per metre, N/m. In base units this becomes kg s⁻² because the metre in the newton cancels.

Do spring constant and surface tension have the same dimensions?

Correct, they match: each is a force divided by a length, giving [MT⁻²] and the unit N/m.

Does cutting a spring in half change its dimensions?

No. Each half becomes twice as stiff, so the value of k doubles, but the bracket stays [MT⁻²].