QUESTIONS on kinetic energy store problems to solve

Q1 A swimmer of mass 70 kg is moving at a speed of 1.4 m/s.

Calculate the kinetic energy of the swimmer in J (to 2 sf).

ANSWERS

 

Q2 A bullet with a mass of 10.0 g is travelling with an initial speed of 400 m/s.

(a) What is the bullet's Initial kinetic energy store?

(b) If the bullet embeds itself in a wooden plank, by how much will the planks thermal energy store be increased?

ANSWERS

 

Q3 A car of mass 1200 kg is travelling at a steady speed of 30.0 m/s (~67 mph, 108 km/hour).

(a) What is the kinetic energy store of the car?

(b) If on slowing down the kinetic energy of the car is halved, what speed is it then doing?

ANSWERS

 

Q4 The chemical potential energy store of a 100 g rocket, on firing, gives it an initial kinetic energy store of 200 J.

Calculate the initial vertical velocity of the rocket.

ANSWERS

 

Q5 A projectile is required to impart a kinetic energy blow of 5000 J at a speed of 500 m/s.

ANSWERS

 

Q6. A pole vaulter has a weight of 800 N and vaults to a height of 4.5 m.

(See gravitational potential energy calculation page)

(a) How much work does the pole vaulter do?

(c) At the maximum height reached, what gravitational potential energy does the pole vaulter possess?

(b) What kinetic energy did the pole vaulter impart to its body on 'take-off' (and what assumption are you making?)

(c) What is the kinetic energy of the pole vaulter immediately before impacting on the ground?

(d) What was the initial speed of take-off by the pole vaulter? (take gravity strength as 9.8 N/kg).

ANSWERS

 

Q7 A diver of mass 60 kg is perched on a 7.5 m high diving board. (First see GPE calculations page)

In your GPE calculations take gravitational field strength acceleration to be 9.8 m/s².

The 10 m diving board is the highest in Olympic sport, 7.5 m board is the next highest.

(a) Calculate the gravitational potential energy (GPE) of the diver with respect to the surface of the water in the pool.

(b) Just before the diver hits the water surface, what is the diver's maximum kinetic energy? Explain your answer.

(c) From your answers to (a) and (b) calculate the diver's speed on impacting on the surface of the water.

ANSWERS

 

Q8 A small bee of mass 0.06 g leaps up 10 cm above a flower. (g = 10 N/kg)

(a) Calculate the gain in gravitational potential energy.

(b) Assuming there is no further acceleration, and ignoring air resistance, what was the initial take off speed of the bee?

(c) If the same bee leaps from another flower and gains 2.0 x 10^-4 J, how high did the bee rise?

ANSWERS

 

Q9 In falling, an initially stationary 2.00 kg weight, loses 500 J of energy before hitting the floor.

(a) At what height was the weight dropped? (assume g = 10 N/kg)

(b) Neglecting air resistance, at what speed did the 2 kg weight hit the floor?

ANSWERS

 

Q10 An electric motor in a toy train does 0.2 J of work to accelerate it to a speed of 25 cm/s.

(a) If the toy train has a mass of 3500 g, what is the efficiency of the motor?

(b) Why is the efficiency much less than 100%?

ANSWERS

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Key points for Physics - kinetic energy store conversions - energy transfers involving kinetic energy

Information sources for Doc Brown's key points: IGCSE-GCSE physics are based on textbooks & syllabus-specifications for students taking the UK AQA, Edexcel, OCR 21st Century Science, OCR Gateway science suite, WJEC, CCEA & CIE GCSE physics 9-1 level science examinations.

A detailed, syllabus-aligned revision guide on Kinetic Energy Stores, tailored for GCSE/IGCSE Physics students across WJEC, CCEA, CIE, AQA, Edexcel, and OCR exam boards:

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Kinetic Energy Stores – GCSE Physics Revision Notes

What is Kinetic Energy?

• Kinetic energy is the energy an object possesses due to its motion.
• Any object that is moving has kinetic energy.
• The faster or more massive the object, the greater its kinetic energy.

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Kinetic Energy Formula

KE = ½mv²

• KE = kinetic energy (Joules)
• m = mass (kg)
• v = speed (m/s)

Speed is squared – doubling the speed quadruples the kinetic energy.

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Real-World Applications of Kinetic Energy

Application	Description
Vehicles	Cars, bikes, and trains store kinetic energy while moving
Wind turbines	Wind’s kinetic energy turns blades to generate electricity
Sports	Balls, athletes, and equipment in motion store kinetic energy
Roller coasters	Gain kinetic energy as they descend from a height
Bullets/projectiles	Store large amounts of kinetic energy due to high speed
Hydroelectric power	Water falling from height gains speed → kinetic energy turns turbines

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Energy Transfers Involving Kinetic Energy

Scenario	Energy Transfer Pathway
Object falling	Gravitational Potential → Kinetic
Car braking	Kinetic → Thermal (brake pads)
Throwing a ball	Chemical (muscles) → Kinetic
Wind turbine spinning	Kinetic (wind) → Mechanical → Electrical
Pendulum swinging	Kinetic ↔ Gravitational Potential

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Exam Tips for All Boards

Use correct terminology

• Say “energy is transferred to the kinetic store” not “energy is made.”

Understand energy conservation

• In a closed system, total energy is conserved: energy is transferred, not lost.

Practice rearranging the formula

• You may be asked to find mass or speed, so be confident with algebra.

Draw energy transfer diagrams

• Use bar charts or Sankey diagrams to show energy flow between stores.

Know the 8 energy stores

• Kinetic, Gravitational Potential, Elastic Potential, Thermal, Chemical, Magnetic, Electrostatic, Nuclear

Be aware of units

• Mass in kg, speed in m/s, energy in J. Watch out for conversions!

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Worked Example

Q: A 1.5 kg ball is moving at 4 m/s. Calculate its kinetic energy.

A: E = ½mv² = ½ x 1.5 x 4² = 12 J

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Keywords, phrases and learning objectives on kinetic energy stores

Be able to explain with examples what a kinetic energy store is.

Be able to explain how a kinetic energy store is converted into another energy store e.g. thermal energy from friction, electrical energy from a generator.

Know how to do calculations and solve problems using the formula for kinetic energy.

ANSWERS to he kinetic energy store problems

Q1 A swimmer of mass 70 kg is moving at a speed of 1.4 m/s.

Calculate the kinetic energy of the swimmer in J (to 2 sf).

E_ke = ¹/₂ m v² = 0.5 x 70 x 1.4² = 68.6 = 69 J (2 sf)

 

Q2 A bullet with a mass of 10.0 g is travelling with an initial speed of 400 m/s.

(a) What is the bullet's Initial kinetic energy store?

E_ke = ¹/₂ m v² , 10 g ≡ 10/1000 = 0.01 kg

E_ke = 0.5 x 0.01 x 400²

E_ke = 800 J of kinetic energy

(b) If the bullet embeds itself in a wooden plank, by how much will the planks thermal energy store be increased?

All of the 800 J of kinetic energy will end up as heat. There will be some loss due to friction (air resistance) and sound energy, but most of the 800 J will end up as heat energy in the plank.

 

Q3 A car of mass 1200 kg is travelling at a steady speed of 30.0 m/s (~67 mph, 108 km/hour).

(a) What is the kinetic energy store of the car?

E_ke = ¹/₂ m v² , 10 g ≡ 10/1000 = 0.01 kg

E_ke = 0.5 x 1200 x 30²

E_ke = 540,000 J of kinetic energy (5.40 x 10⁵J, 0.540 MJ, 3 s.f.)

(b) If on slowing down the kinetic energy of the car is halved, what speed is it then doing?

E_ke = ¹/₂ m v² , rearranging:   v² = 2E_ke / m,  so v = √(2E_ke / m),  KE = 540,000/2 = 270,000 J

v = √(2 x E_ke / m) = √(2 x 270,000 / 1200) = 21.2 m/s (3 s.f.)

 

Q4 The chemical potential energy store of a 100 g rocket, on firing, gives it an initial kinetic energy store of 200 J.

Calculate the initial vertical velocity of the rocket.

E_ke = ¹/₂ m v² , rearranging:   v² = 2E_ke / m,  so v = √(2E_ke / m)

m = 100 g ≡ 100 / 1000 = 0.1 kg

v = √(2 x E_ke / m) = √(2 x 200 / 0.10) = 63.2 m/s (3 s.f.)

 

Q5 A projectile is required to impart a kinetic energy blow of 5000 J at a speed of 500 m/s.

What mass of projectile is required?

E_ke = ¹/₂ m v² , rearranging:   m = 2 x E_ke / v²

m = 2 x E_ke / v² = 2 x 5000 / 500² = 0.040 kg (40.0 g, 3 s.f.)

 

Q6. A pole vaulter has a weight of 800 N and vaults to a height of 4.5 m. (See gravitational potential energy calculation page)

(a) How much work does the pole vaulter do?

work (J) = force (N) x distance of action (m)

work done = 800 x 4.5 = 3600 J

(c) At the maximum height reached, what gravitational potential energy does the pole vaulter possess?

The GPE equals the work done in raising the pole vaulter to a height of 4.5 m, that is 3600 J

(b) What kinetic energy did the pole vaulter impart to its body on 'take-off' (and what assumption are you making?)

If you ignore air resistance i.e. no energy lost - wasted, the initial KE = maximum GPE = 3600 J

(c) What is the kinetic energy of the pole vaulter immediately before impacting on the ground?

maximum KE = maximum GPE gained = 3600 J (neglecting air resistance energy losses)

(d) What was the initial speed of take-off by the pole vaulter? (take gravity strength as 9.8 N/kg).

KE = ¹/₂mv², rearranging gives v = √(2KE/m)

initial KE = 3600 J, mass = 800 / 9.8 = 81.63 kg

v = √(2KE/m) = √(2 x 3600/81.63) = 9.4 m/s (2sf)

 

Q7 A diver of mass 60 kg is perched on a 7.5 m high diving board. (First see GPE calculations page)

In your GPE calculations take gravitational field strength acceleration to be 9.8 m/s².

The 10 m diving board is the highest in Olympic sport, 7.5 m board is the next highest.

(a) Calculate the gravitational potential energy (GPE) of the diver with respect to the surface of the water in the pool.

GPE = m g h = 60 x 9.8 x 7.5 = 4410 J

(b) Just before the diver hits the water surface, what is the diver's maximum kinetic energy? Explain your answer.

From the law of conservation of energy, the GPE is converted into kinetic energy (neglecting air resistance).

Therefore the diver's maximum kinetic energy is also 4410 J

(c) From your answers to (a) and (b) calculate the diver's speed on impacting on the surface of the water.

KE = ½ mv², rearranging gives: v = √(2KE / m)

v = √(2 x 4410 / 60) = √147 = 12 m/s (2 s.f.)

 

Q8 A small bee of mass 0.06 g leaps up 10 cm above a flower. (g = 10 N/kg)

(a) Calculate the gain in gravitational potential energy.

∆E = mgh = (0.0.06/1000) x 10 x (10/100)

= 6 x 10^-5 x 10 x 10^-1 =  6.0 x 10^-5 J  (2 sf)

(b) Assuming there is no further acceleration, and ignoring air resistance, what was the initial take off speed of the bee?

As the bee ascends, the kinetic energy store of the bee is converted into the GPE store of the bee,

Therefore final maximum ∆GPE = initial maximum ∆KE

So, ∆GPE = ∆KE = 6.0 x 10^-5 J

KE = ¹/₂mv², rearranging gives v = √(2KE/m)

v = √(2KE/m) = √(2 x 10^-5 / 6 x 10^-5) = √0.3333 = 0.58 m/s  (2 sf)

(c) If the same bee leaps from another flower and gains 2.0 x 10^-4 J, how high did the bee rise?

∆GPE = mgh, rearranging gives h = ∆GPE / mg

h = 2.0 x 10^-4 / (6.0 x 10^-5 x 10) = 0.33 m  (33 cm, 2 sf)

 

Q9 In falling, an initially stationary 2.00 kg weight, loses 500 J of energy before hitting the floor.

(a) At what height was the weight dropped? (assume g = 10 N/kg)

The loss of energy is equal to the loss of GPE as it is converted to KE.

∆GPE = mgh, rearranging gives:

h = ∆GPE / (g x m) = 500 / (10 x 2) = 500 / 20 = 25 m

(b) Neglecting air resistance, at what speed did the 2 kg weight hit the floor?

Assuming maximum ∆GPE = ∆KE on impact

KE = ¹/₂mv², rearranging gives v = √(2KE/m)

v = √(2KE/m) = √(2 x 500 /2) = √500 = 22.4 m/s  (3 sf)

Q10 An electric motor in a toy train does 0.2 J of work to accelerate it to a speed of 25 cm/s.

(a) If the toy train has a mass of 3500 g, what is the efficiency of the motor?

First calculate the kinetic energy store of the train.

mass = 3500/1000 = 3.50 kg, speed = (25/100) = 0.25 m/s

KE = ¹/₂mv²  = 0.5 x 3.5 x 0.25² = 0.1094 J

Efficiency = useful energy out / total energy input = 0.1094 / 0.20 = 0.55 (2 s.f.)

Efficiency = 0.55 x 100 = 55% (2 s.f.)

(b) Why is the efficiency much less than 100%?

Apart from a little sound, most energy is lost (dissipated) from the friction of the moving parts of the electric motor and friction between the wheels and the rails - all energy lost ends up increasing the thermal energy store of the surroundings.

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