IMAT Physics · Dynamics, Work and Energy

Work, Kinetic Energy, Potential Energy and Springs

A force does work when it moves its point of action along its own direction; that work becomes kinetic energy, gravitational potential energy or energy stored in a stretched spring.

All Dynamics, Work and Energy notes

Why this matters

The 2026 ministry paper asked for a kinetic energy from a mass and a speed in km/h. The older papers asked which quantity can be measured in joules per metre (2016) and for the mass of a block stopped by a resistive force over a given distance (2022).

Concept 1 of 3: Work done by a force

In physics, work is done only when a force moves something along the direction of the force. Holding a heavy bag still does no work on it, however tiring it feels. If the force is at an angle to the motion, only the part of the force along the motion counts.

Definition

Work $W = Fs\cos\theta$ , where $\theta$ is the angle between the force and the displacement. A scalar, in joules: $1\ \text{J} = 1\ \text{N m}$ .
$\theta = 0$ : $W = Fs$ . $\theta = 90^\circ$ : no work (for example the centripetal force, or the normal reaction on a body sliding on a level floor).
A force against the motion (friction, braking) does negative work: it takes energy away.
Rearranged, force = work ÷ distance, so a newton is a joule per metre.

Work

W = Fs\cos\theta

$F$ size of the force, in N
$s$ displacement, in m
$\theta$ angle between force and displacement

Worked example

A rope pulls a sledge 50 m along level snow with a force of 80 N at 60° above the horizontal. How much work does the rope do?

Practice this conceptself-check · 3 quick reps

No movement along the force means no work

A force that is perpendicular to the motion, or that moves nothing, does zero work, however large it is. Effort and tiredness are not work in the physics sense.

Concept 2 of 3: Kinetic and gravitational potential energy, and the work-energy theorem

Doing work on a body hands it energy. Speeding it up stores the energy as kinetic energy, which grows with the square of the speed. Lifting it stores the energy as gravitational potential energy, which grows with height. The work done by the resultant force equals exactly the change in kinetic energy.

Definition

Kinetic energy $E_k = \tfrac{1}{2}mv^2$ : double the speed, four times the energy.
Gravitational potential energy change $\Delta E_p = mg\Delta h$ near the Earth's surface (only changes in height matter).
Work-energy theorem: the total work done by the resultant force equals the change of kinetic energy, $W_{\text{net}} = \Delta E_k$ .
So a constant resistive force $F$ stopping a body over a distance $s$ satisfies $Fs = \tfrac{1}{2}mv^2$ .
Convert speeds in km/h to m/s first: divide by 3.6.

Kinetic and potential energy

E_k = \tfrac{1}{2}mv^2 \qquad \Delta E_p = mg\,\Delta h \qquad W_{\text{net}} = \Delta E_k

$m$ mass, in kg
$v$ speed, in m/s
$\Delta h$ change of height, in m

Worked example

A 0.50 kg ball is 12 m above the ground and moving at 8.0 m/s. Taking

g = 10\ \text{N/kg}

and the ground as zero height, find its kinetic and its gravitational potential energy.

Practice this conceptself-check · 3 quick reps

Kinetic energy goes with the square of the speed

Doubling the speed multiplies the kinetic energy by four, and a speed left in km/h makes the answer

3.6^2 \approx 13

times too large. Convert to m/s first, then square.

Concept 3 of 3: Hooke's law and elastic potential energy

Pull a spring twice as hard and it stretches twice as far, as long as you do not overstretch it. The work you do stretching it is stored, ready to be given back. Because the force grows from zero as you stretch, the stored energy is the average force times the extension, which is half the final force times the extension.

Definition

Hooke's law: the extension (or compression) $x$ of a spring is proportional to the force, $F = kx$ , up to the limit of proportionality.
$k$ is the spring constant in N/m: a stiffer spring has a larger $k$ .
Beyond the elastic limit the spring is permanently deformed and does not return to its original length.
Elastic potential energy stored: $E = \tfrac{1}{2}kx^2 = \tfrac{1}{2}Fx$ , the area under the force-extension graph.

Hooke's law and stored energy

F = kx \qquad E_{\text{el}} = \tfrac{1}{2}kx^2

$k$ spring constant, in N/m
$x$ extension or compression, in m

Worked example

A spring with

k = 200\ \text{N/m}

is stretched by 0.15 m. Find the force needed and the energy stored.

Practice this conceptself-check · 3 quick reps

Stored energy is half force times extension, not force times extension

The force grows from zero to

kx

during the stretch, so the work done is the area of a triangle:

\tfrac{1}{2}kx^2

. Using

Fx

with the final force doubles the answer.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Work done by a force
Work
$W = Fs\cos\theta$
Kinetic and gravitational potential energy, and the work-energy theorem
Kinetic and potential energy
$E_k = \tfrac{1}{2}mv^2 \qquad \Delta E_p = mg\,\Delta h \qquad W_{\text{net}} = \Delta E_k$
Hooke's law and elastic potential energy
Hooke's law and stored energy
$F = kx \qquad E_{\text{el}} = \tfrac{1}{2}kx^2$

Watch out for (3)

No movement along the force means no work
→ Work done by a force
Kinetic energy goes with the square of the speed
→ Kinetic and gravitational potential energy, and the work-energy theorem
Stored energy is half force times extension, not force times extension
→ Hooke's law and elastic potential energy

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