P1.5 Kinetic Energy and Elastic Energy Stores

A free GCSE Combined Science: Trilogy lesson for AQA (8464) from OBA School: ek = ½ m v², Ee = ½ k e², spring constant and energy changes in falling objects. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

• State what the energy in a kinetic store depends on and calculate it using Ek = ½ m v²
• Rearrange the kinetic energy equation to find speed or mass
• Describe how to investigate kinetic energy using a ramp and light gates
• Calculate the energy in an elastic potential store using Ee = ½ k e², up to the limit of proportionality
• Use conservation of energy to find the speed of a falling or launched object

Key words

• Kinetic energy (Ek): Energy in the kinetic store of a moving object. It depends on mass and speed.
• Elastic potential energy (Ee): Energy stored in an object when work is done to stretch, squash or bend it.
• Extension (e): How much longer a spring is than its natural (unstretched) length, in metres.
• Spring constant (k): The force needed per metre of extension, in N/m. A stiffer spring has a bigger k.
• Limit of proportion- ality: The point beyond which force is no longer directly proportional to extension.
• Light gate: A sensor that times an object as it breaks a beam of light, connected to a timer or data logger.

Kinetic energy stores

Every moving object has energy in its kinetic store. The amount depends on its mass and its speed:

• Ek = ½ m v²; Symbols and units Ek = kinetic energy, in joules (J) m = mass, in kilograms (kg) v = speed, in metres per second (m/s); Rearranged v = √(2 Ek ÷ m) m = 2 Ek ÷ v²

Because the speed is squared, speed matters much more than mass. Doubling the mass doubles Ek; doubling the speed makes Ek four times bigger; tripling the speed makes it nine times bigger.

Worked example: a moving car

A car of mass 1400 kg travels at 12 m/s.
Ek = ½ × 1400 kg × (12 m/s)² = 0.5 × 1400 × 144

Answer: 100 800 J (about 101 kJ)

Worked example: a change in speed

A 1000 kg car slows down from 25 m/s to 15 m/s. Work out Ek at each speed, then subtract.
before: ½ × 1000 × 25² = 312 500 J; after: ½ × 1000 × 15² = 112 500 J
decrease = 312 500 − 112 500 = 200 000 J

Answer: the kinetic store decreases by 200 kJ (do not use the 10 m/s difference in speed)

Worked example: finding a speed

A 160 g hockey ball has 50 J in its kinetic store. How fast is it moving?
m = 160 g = 0.160 kg
v² = 2 Ek ÷ m = 2 × 50 ÷ 0.160 = 625, so v = √625

Answer: v = 25 m/s

Exam tip: Square the speed before multiplying by ½ m. When finding v, remember the last step: take the square root.

Investigating kinetic energy

• Release the ball from a measured height h. Its GPE store decreases by m g h. If friction is small, its kinetic store increases by the same amount.
• Light gates time the ball between X and Y. Speed = distance ÷ time.
• Repeat for different heights. Four times the height drop gives twice the speed, showing Ek ∝ v².
• Light gates are better than a stopwatch: they remove human reaction time and give times to 0.01 s or better.

Worked example: speed from a height drop

A 0.50 kg trolley rolls down a ramp from a height of 0.30 m. Ignore friction.
ΔEp = m g Δh = 0.50 × 9.8 × 0.30 = 1.47 J, so Ek = 1.47 J
v = √(2 × 1.47 ÷ 0.50) = √5.88 = 2.42 m/s

Answer: about 2.4 m/s

Elastic potential energy stores

When you stretch or squash a spring, the work you do is stored in its elastic potential store. The force needed obeys Hooke’s law, F = k e, up to the limit of proportionality. The spring constant k (N/m) tells you how stiff the spring is.

• Ee = ½ k e² (also F = k e); Symbols and units Ee = elastic potential energy, in joules (J) k = spring constant, in newtons per metre (N/m) e = extension, in metres (m)

Worked example: energy in a stretched spring

A spring with k = 60 N/m is stretched from 10 cm to 22 cm long.
e = 22 − 10 = 12 cm = 0.12 m
Ee = ½ × 60 × 0.12² = 0.5 × 60 × 0.0144

Answer: Ee = 0.432 J (0.43 J)

Exam tip: Use the extension, not the total length, and convert it to metres. The equation only works up to the limit of proportionality.

Energy changes when objects fall or are launched

If air resistance is small, energy lost from one store equals energy gained by another. For a falling object, the decrease in the GPE store equals the increase in the kinetic store: m g h = ½ m v². For a catapult, the elastic store of the band goes to the kinetic store of the stone, and then to its GPE store as it rises.

Worked example: speed of a dropped ball

A 0.20 kg ball is dropped from 4.0 m. Ignore air resistance.
GPE lost = 0.20 × 9.8 × 4.0 = 7.84 J = Ek gained
v = √(2 × 7.84 ÷ 0.20) = √78.4 = 8.9 m/s

Answer: 8.9 m/s just before it lands

Worked example: a spring launcher

A toy launcher spring (k = 600 N/m) is compressed by 0.040 m and fires a 0.015 kg ball straight up.
Ee = ½ × 600 × 0.040² = 0.48 J. At the top all 0.48 J is in the GPE store (ignoring air resistance).
h = 0.48 ÷ (0.015 × 9.8) = 3.3 m

Answer: the ball rises about 3.3 m

Common mistakes

• Forgetting to square the speed (or squaring ½ m v instead of just v).
• Forgetting the square root when finding v from Ek.
• Using the stretched length instead of the extension, or leaving it in cm.
• Using Ee = ½ k e² beyond the limit of proportionality.
• Saying “movement energy” instead of the kinetic energy store.

Revision questions

Try these, then check your answers below each one.

Question 1. Calculate the kinetic energy of a 3.0 kg ball moving at 4.0 m/s.

Answer: 24 J

Working: 0.5 × 3.0 × 4.0² = 0.5 × 3.0 × 16 = 24 J.

Question 2. A car of mass 1200 kg travels at 15 m/s. Calculate the energy in its kinetic store, in kJ.

Answer: 135 kJ

Working: 0.5 × 1200 × 15² = 135 000 J = 135 kJ.

Question 3. A cyclist doubles her speed. What happens to the energy in her kinetic store? (Choose from: It becomes 4 times bigger., It doubles., It halves., It stays the same..)

Answer: It becomes 4 times bigger.

Working: Eₖ ∝ v², so doubling v multiplies Eₖ by 2² = 4.

Related lessons

• Previous lesson: P1.4 Gravitational Potential Energy Stores
• Next lesson: P1.6–7 Energy Dissipation and Energy and Efficiency
• In this chapter: P1.1 Changes in Energy Stores
• In this chapter: P1.2 Conservation of Energy
• In this chapter: P1.3 Energy and Work

Continue in the full course

This lesson is section P1.5 (AGCSC3005) of our free GCSE Combined Science course for AQA. In the full course you get every video for this section, a self-marking practice quiz, a printable worksheet with a mark scheme, chapter tests, and your progress saved. Go to the full GCSE Combined Science course.