Elastic force. Hooke's law

Deformation (elastic and plastic), the elastic force, spring stiffness, Hooke's law $F = k\Delta l$.

Terms

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Formulas

Hooke's law

F = k\,\Delta l

Quantities
Symbol	Name	Unit
$F$	magnitude of the elastic force	N
$k$	stiffness of the spring (body)	N/m
$\Delta l$	extension (change in length)	m

Solved for another quantity:

k = \dfrac{F}{\Delta l}

\Delta l = \dfrac{F}{k}

Note. Here $F$ is the elastic force ( $F_{\text{el}}$ ). The formula gives the magnitude; the force points against the deformation. It holds for an elastic deformation. Substitute the extension in metres: $2\ \text{cm} = 0.02\ \text{m}$ . For example, with $k = 100\ \text{N/m}$ and $\Delta l = 0.05\ \text{m}$ we get $F_{\text{el}} = 5\ \text{N}$ . A weight of mass $m$ hanging at rest: $k\Delta l = mg$ .

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

A weight on a spring

The weight is at rest: the elastic force (up) balances the force of gravity mg (down), so the arrows are equal in length. The dashed line is the length l₀ of the unstretched spring. The deformation is elastic and the mass of the spring is neglected.

Graphs

The red line is the chosen spring, the others are for comparison. Click the graph to set the mass of the weight.

F_{\text{el}}(\Delta l)

k = 20 N/mk = 50 N/mk = 100 N/mk = 200 N/m

\Delta l

5 cm = 0.05 m

F

5 N

k

100 N/m

Formulas with your numbers

F = mg = 0.5\cdot 10 = 5\ \text{N}

\Delta l = \dfrac{F}{k} = \dfrac{5}{100} = 0.05\ \text{m} = 5\ \text{cm}

F = k\Delta l = 100\cdot 0.05 = 5\ \text{N}

How to read the graphs

$F(\Delta l)$ is a straight line through the origin: the elastic force is directly proportional to the extension, and it is zero at $\Delta l = 0$ .
The slope of the line is the stiffness: $k = \dfrac{F}{\Delta l}$ . The steeper the line, the stiffer the spring.
At the same extension (a vertical line) the stiffer spring gives the greater force.
Under the same force (a horizontal line) the stiffer spring stretches less.
On the graph the extension is in centimetres, but in $F = k\Delta l$ it must be in metres: $1\ \text{cm} = 0.01\ \text{m}$ .
The point on the graph is the state of the weight in the lab: its height equals the force of gravity $mg$ , because the weight is at rest.

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Terms

Studied 0 of 7

Formulas

Hooke's law

F = k\,\Delta l

Quantities
Symbol	Name	Unit
$F$	magnitude of the elastic force	N
$k$	stiffness of the spring (body)	N/m
$\Delta l$	extension (change in length)	m

Solved for another quantity:

k = \dfrac{F}{\Delta l}

\Delta l = \dfrac{F}{k}

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

A weight on a spring

Graphs

The red line is the chosen spring, the others are for comparison. Click the graph to set the mass of the weight.

F_{\text{el}}(\Delta l)

k = 20 N/mk = 50 N/mk = 100 N/mk = 200 N/m

\Delta l

5 cm = 0.05 m

F

5 N

k

100 N/m

Formulas with your numbers

F = mg = 0.5\cdot 10 = 5\ \text{N}

\Delta l = \dfrac{F}{k} = \dfrac{5}{100} = 0.05\ \text{m} = 5\ \text{cm}

F = k\Delta l = 100\cdot 0.05 = 5\ \text{N}

How to read the graphs

F(\Delta l)

is a straight line through the origin: the elastic force is directly proportional to the extension, and it is zero at

\Delta l = 0

The slope of the line is the stiffness:

k = \dfrac{F}{\Delta l}

. The steeper the line, the stiffer the spring.

At the same extension (a vertical line) the stiffer spring gives the greater force.

Under the same force (a horizontal line) the stiffer spring stretches less.

On the graph the extension is in centimetres, but in

F = k\Delta l

it must be in metres:

1\ \text{cm} = 0.01\ \text{m}

The point on the graph is the state of the weight in the lab: its height equals the force of gravity

mg

, because the weight is at rest.