Springs: force, extension, and stored energy

Key ideas

A spring stretched (or compressed) by a distance $x$ pulls (or pushes) back with a restoring force proportional to the extension — Hooke's law:

$F = -kx$

$k$ is the spring constant (N m⁻¹) — how stiff the spring is.
The minus sign says the spring's force is opposite to the extension: stretch it right, it pulls left. (When you just need sizes, use $F = kx$ .)

The energy stored in a stretched spring is the elastic potential energy:

$E_p = \tfrac{1}{2}kx^2$

Force–extension graph for a spring: a straight line through the origin whose gradient is the spring constant k, with the triangular area under the line shaded and labelled as the stored elastic potential energy.

On a force–extension graph, the gradient is $k$ and the area under the line is the stored energy — the same graph-reading rules as motion graphs, wearing different units.

Worked ExampleSpring force and stored energy

A spring with $k = 200$ N m⁻¹ is stretched $0.15$ m. Find the force needed and the energy stored.

Step 1 — Force, from Hooke's law

$F = kx = (200)(0.15) = 30 \text{ N}$

Step 2 — Stored elastic potential energy

$E_p = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)(0.15)^2 = 2.25 \approx 2.3 \text{ J}$

Practice question

A spring stretches $0.040$ m when a $10$ N weight hangs from it. Find the spring constant.

Worked solution: $k = \dfrac{F}{x} = \dfrac{10}{0.040} = 250$ N m⁻¹.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A spring has a spring constant of $200$ N m⁻¹.

Calculate the force needed to stretch it by $0.25$ m.

Merit

A spring stretches $0.08$ m when a $12$ N weight hangs from it.

Calculate the spring constant, then the elastic potential energy stored in the spring at that stretch.

Excellence

A spring of spring constant $300$ N m⁻¹ is compressed by $0.20$ m and used to launch a $0.40$ kg trolley along a frictionless track.

Calculate the launch speed of the trolley, and describe the energy transfer involved, using physics principles.

Springs: force, extension, and stored energy

Covers AS91171

Key ideas

A spring stretched (or compressed) by a distance $x$ pulls (or pushes) back with a restoring force proportional to the extension — Hooke's law:

$F = -kx$

$k$ is the spring constant (N m⁻¹) — how stiff the spring is.
The minus sign says the spring's force is opposite to the extension: stretch it right, it pulls left. (When you just need sizes, use $F = kx$ .)

The energy stored in a stretched spring is the elastic potential energy:

$E_p = \tfrac{1}{2}kx^2$

On a force–extension graph, the gradient is $k$ and the area under the line is the stored energy — the same graph-reading rules as motion graphs, wearing different units.

Worked ExampleSpring force and stored energy

A spring with $k = 200$ N m⁻¹ is stretched $0.15$ m. Find the force needed and the energy stored.

Step 1 — Force, from Hooke's law

$F = kx = (200)(0.15) = 30 \text{ N}$

Step 2 — Stored elastic potential energy

$E_p = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)(0.15)^2 = 2.25 \approx 2.3 \text{ J}$

Practice question

A spring stretches $0.040$ m when a $10$ N weight hangs from it. Find the spring constant.

Worked solution: $k = \dfrac{F}{x} = \dfrac{10}{0.040} = 250$ N m⁻¹.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A spring has a spring constant of $200$ N m⁻¹.

Calculate the force needed to stretch it by $0.25$ m.

Merit

A spring stretches $0.08$ m when a $12$ N weight hangs from it.

Calculate the spring constant, then the elastic potential energy stored in the spring at that stretch.

Excellence

A spring of spring constant $300$ N m⁻¹ is compressed by $0.20$ m and used to launch a $0.40$ kg trolley along a frictionless track.

Calculate the launch speed of the trolley, and describe the energy transfer involved, using physics principles.