A 4.0\ kg object moves at 5.0\ m s^-1. Calculate its kinetic energy. [2 marks]

A 3.0\ kg box is lifted 2.0\ m (g = 10\ N kg^-1). Calculate its gain in gravitational potential energy. [2 marks]

SingaporePhysicsSyllabus dot point

How do we calculate the kinetic energy of a moving object and the potential energy of a raised one?

Apply the relationships for kinetic energy and gravitational potential energy and use energy conservation

A focused answer to the O-Level Physics outcome on kinetic and potential energy. The relationships for kinetic energy and gravitational potential energy, their units, and using conservation of energy to solve falling-object problems.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

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

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What this dot point is asking
The answer
Examples in context
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What this dot point is asking

SEAB wants you to calculate the kinetic energy of a moving object using $E_k = \tfrac{1}{2}mv^2$ and the gravitational potential energy of a raised object using $E_p = mgh$ , and to use conservation of energy to connect them, for example to find the speed of a falling object. The big idea is that energy of motion and energy of position can be swapped into each other.

The answer

Kinetic energy

Kinetic energy is the energy an object has because it is moving:

E_k = \tfrac{1}{2}mv^2

where $m$ is the mass in kilograms and $v$ the speed in $\text{m s}^{-1}$ , giving energy in joules. Because the speed is squared, doubling the speed makes the kinetic energy four times larger. This is why fast vehicles are so much harder to stop.

Gravitational potential energy

Gravitational potential energy is the energy an object has because of its height above the ground:

E_p = mgh

where $g$ is the gravitational field strength and $h$ the height in metres. Lifting an object higher increases its potential energy, because more work is done against gravity.

Swapping between the two

When an object falls, its gravitational potential energy is transferred to kinetic energy. Ignoring air resistance, the loss in potential energy equals the gain in kinetic energy:

mgh = \tfrac{1}{2}mv^2

This is the most useful equation in this dot point, because it lets you find the speed of a falling object without knowing the time.

Notice the mass cancels

In $mgh = \tfrac{1}{2}mv^2$ the mass appears on both sides and cancels, leaving $v = \sqrt{2gh}$ . So, ignoring air resistance, the final speed of a dropped object depends only on the height, not on its mass, matching the free-fall result.

Worked example

A $0.50\ \text{kg}$ ball is thrown straight up at $8.0\ \text{m s}^{-1}$ . Take $g = 10\ \text{N kg}^{-1}$ and ignore air resistance. Find the maximum height it reaches.

Step 1: Find the kinetic energy at launch

E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 0.50 \times 8.0^2 = \tfrac{1}{2} \times 0.50 \times 64 = 16\ \text{J}

Step 2: Apply conservation of energy

At the highest point the ball stops momentarily, so all the kinetic energy has become gravitational potential energy:

mgh = 16\ \text{J}

Step 3: Solve for the height

0.50 \times 10 \times h = 16 \quad\Rightarrow\quad 5.0\,h = 16 \quad\Rightarrow\quad h = 3.2\ \text{m}

The ball rises $3.2\ \text{m}$ . Energy conservation gives the height directly, with no need for the time of flight.

Examples in context

Example 1. Stopping distances. A car travelling twice as fast has four times the kinetic energy, and the brakes must do four times as much work to stop it. This is why braking distances rise sharply with speed and why speed limits matter so much for road safety.

Example 2. Hydroelectric power. Water stored high in a reservoir has large gravitational potential energy. As it falls through pipes to the turbines, this becomes kinetic energy of fast-moving water, which spins the turbines to generate electricity, a direct large-scale conversion of $mgh$ into useful energy.

Try this

Q1. A $4.0\ \text{kg}$ object moves at $5.0\ \text{m s}^{-1}$ . Calculate its kinetic energy. [2 marks]

Cue. $E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 4.0 \times 5.0^2 = \tfrac{1}{2} \times 4.0 \times 25 = 50\ \text{J}$ .

Q2. A $3.0\ \text{kg}$ box is lifted $2.0\ \text{m}$ ( $g = 10\ \text{N kg}^{-1}$ ). Calculate its gain in gravitational potential energy. [2 marks]

Cue. $E_p = mgh = 3.0 \times 10 \times 2.0 = 60\ \text{J}$ .

Q3. Explain, using energy, why a faster car needs a longer distance to stop. [2 marks]

Cue. Kinetic energy depends on speed squared, so a faster car has much more kinetic energy, and the brakes must do more work over a longer distance to remove it.

Exam-style practice questions

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

Original4 marksA car of mass

1000\ \text{kg}

travels at

20\ \text{m s}^{-1}

. (a) Calculate its kinetic energy. (b) The car speeds up to

40\ \text{m s}^{-1}

. State and explain how its kinetic energy changes.

Show worked answer →

(a) Kinetic energy: $E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 1000 \times 20^2 = \tfrac{1}{2} \times 1000 \times 400 = 200\,000\ \text{J}$ .

(b) Doubling the speed makes the kinetic energy four times as large, because $E_k$ depends on $v^2$ . At $40\ \text{m s}^{-1}$ : $\tfrac{1}{2} \times 1000 \times 40^2 = 800\,000\ \text{J}$ , four times the original.

Markers reward $E_k = \tfrac{1}{2}mv^2$ with $v$ squared, the correct value, and the key idea that doubling speed quadruples kinetic energy.

Original5 marksA

2.0\ \text{kg}

ball is dropped from a height of

5.0\ \text{m}

. Take

g = 10\ \text{N kg}^{-1}

and ignore air resistance. (a) Find its gravitational potential energy at the top. (b) Use energy conservation to find its speed just before it hits the ground.