Calculate the work done when a force of 40 N moves a box 3.0 m. [2 marks]

A 0.20 kg ball moves at 6.0 m s^-1. Calculate its kinetic energy. [2 marks]

ScotlandPhysicsSyllabus dot point

How do we account for energy when forces do work, and how is energy conserved as it changes form?

Energy: work done by a force, gravitational potential energy and kinetic energy, the conservation of energy, and using energy changes to solve motion problems such as a falling or braking object.

An SQA National 5 Physics answer on energy in dynamics, covering work done by a force, gravitational potential energy and kinetic energy, the conservation of energy, and how to combine these to solve problems such as a falling object or a braking car.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this key area is asking
Work done by a force
Gravitational potential energy
Kinetic energy
Conservation of energy
Energy and braking
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What this key area is asking

The SQA wants you to calculate the work done by a force, the gravitational potential energy of a raised object and the kinetic energy of a moving object, and to use the conservation of energy to link these together in problems such as a falling object or a braking car.

Work done by a force

If you lift a bag, the work you do against gravity equals the gravitational potential energy the bag gains. If you push a box against friction, the work done is transferred to heat in the surfaces.

Gravitational potential energy

This is the energy stored when an object is lifted. Only the vertical height change matters, not the path taken, so a ball carried up a ramp gains the same potential energy as one lifted straight up to the same height.

Kinetic energy

Because the speed is squared, doubling the speed gives four times the kinetic energy. This is why stopping distances grow so quickly with speed and why fast collisions are so dangerous.

Conservation of energy

Speed of a falling object using energy

A stone of mass $2.0 \text{ kg}$ falls from a cliff of height $5.0 \text{ m}$ . Assuming no energy is lost to air resistance, find the speed of the stone just before it lands. Take $g = 9.8 \text{ N kg}^{-1}$ .

Step 1: find the potential energy lost

$E_p = mgh = 2.0 \times 9.8 \times 5.0 = 98 \text{ J}$ .

Step 2: equate it to the kinetic energy gained

By conservation of energy, $E_k = E_p = 98 \text{ J}$ , so $\tfrac{1}{2}mv^2 = 98$ .

Step 3: solve for the speed

$\tfrac{1}{2}(2.0)v^2 = 98$ , so $v^2 = 98$ , giving $v = \sqrt{98} = 9.9 \text{ m s}^{-1}$ .

Energy and braking

A car braking is a common SQA context. The kinetic energy of the car is transferred to heat in the brakes by the work done by the braking force: $E_k = E_w$ , so $\frac{1}{2}mv^2 = Fd$ . This lets you find the braking distance $d$ for a given braking force, and it shows why doubling the speed quadruples the braking distance.

Try this

Q1. Calculate the work done when a force of $40 \text{ N}$ moves a box $3.0 \text{ m}$ . [2 marks]

Cue. $E_w = Fd = 40 \times 3.0 = 120 \text{ J}$ .

Q2. A $0.20 \text{ kg}$ ball moves at $6.0 \text{ m s}^{-1}$ . Calculate its kinetic energy. [2 marks]

Cue. $E_k = \frac{1}{2}(0.20)(6.0)^2 = 3.6 \text{ J}$ .

Q3. State the principle of conservation of energy. [1 mark]

Cue. Energy cannot be created or destroyed, only changed from one form to another.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA N5 style3 marksA crane lifts a 250 kg load to a height of 12 m. Calculate the gravitational potential energy gained by the load (g = 9.8 N per kg).

Show worked answer →

Use the relationship for gravitational potential energy.

Relationship: $E_p = mgh$ .

Substitution: $E_p = 250 \times 9.8 \times 12 = 29\,400 \text{ J}$ .

Markers reward selecting $E_p = mgh$ , correct substitution of mass, gravitational field strength and height, and a final answer in joules ( $\text{J}$ ).

SQA N5 style4 marksA ball of mass 0.50 kg is dropped from rest from a height of 1.8 m. Assuming no energy is lost, calculate the speed of the ball just before it hits the ground (g = 9.8 N per kg).

Show worked answer →

All the gravitational potential energy is converted to kinetic energy, so $E_k = E_p$ .

Potential energy lost: $E_p = mgh = 0.50 \times 9.8 \times 1.8 = 8.82 \text{ J}$ .

Set this equal to kinetic energy: $\frac{1}{2}mv^2 = 8.82$ , so $\frac{1}{2}(0.50)v^2 = 8.82$ , giving $v^2 = \dfrac{8.82}{0.25} = 35.28$ .

So $v = \sqrt{35.28} = 5.9 \text{ m s}^{-1}$ .

Markers reward equating $E_p$ and $E_k$ , correct substitution, and a final speed with the unit $\text{m s}^{-1}$ . (Note the mass cancels, so all objects reach the same speed.)

Related dot points

Sources & how we know this

SQA National 5 Physics Course Specification — SQA (2019)