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GCSE Physics notes: (e) Kinetic energy stores and calculations

TYPES OF ENERGY STORE - examples explained

(e) Kinetic energy stores and calculations

Doc Brown's GCSE 9-1 Physics Revision Notes

Suitable for GCSE/IGCSE Physics/Science courses or their equivalent

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Kinetic energy or movement energy (KE) and kinetic energy stores

  • Any moving object has kinetic energy and KE energy must be removed from the object to slow it down e.g. braking a moving car, fired bullet embedding in material on impact.

  • Kinetic energy is an example of a mechanical energy store.

  • Increasing or decreasing the speed of a moving object increases or decreases its kinetic energy store.

  • The greater the mass or the greater the speed of an object, the greater its kinetic energy store.

  • The kinetic energy of a moving object can be calculated using the equation:

    • kinetic energy (KE) = 0.5 mass (speed)2

    • Eke = 1/2 m v2

    • kinetic energy, Eke, in joules, J; mass, m, in kilograms, kg; speed/velocity, v, in metres per second, m/s

    • Note that by doubling the speed you quadruple the kinetic energy! (e.g. 22 : 42 is 4 : 16 or 1 : 4)

See also Reaction times, stopping distances, safety aspects. calculations involving kinetic energy


How to solve kinetic energy store problems.

kinetic energy (KE) = 0.5 mass (speed)2

Eke = 1/2 m v2

 

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).

Eke = 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?

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

Eke = 0.5 x 0.01 x 4002

Eke = 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.

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

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

Eke = 0.5 x 1200 x 302

Eke = 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?

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

v = √(2 x Eke / 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.

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

m = 100 g ≡ 100 / 1000 = 0.1 kg

v = √(2 x Eke / 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?

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

m = 2 x Eke / 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.

(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

 

See also Reaction times, stopping distances, safety aspects. calculations involving kinetic energy


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Energy resources, and transfers, work done and electrical power supply revision notes index

Types of energy & stores - examples compared/explained, calculations of mechanical work done and power

Chemical  * Elastic potential energy  * Electrical & electrostatic Gravitational potential energy

Kinetic energy store  *  Nuclear energy store  *  Thermal energy stores  * Light energy  * Sound energy

Conservation of energy, energy transfers-conversions, efficiency - calculations and Sankey diagrams

Energy resources: uses, general survey & trends, comparing renewables, non-renewables, generating electricity

Renewable energy (1) Wind power and solar power, advantages and disadvantages gcse physics revision notes

Renewable energy (2) Hydroelectric power and geothermal power, advantages and disadvantages gcse physics

Renewable energy (3) Wave power and tidal barrage power, advantages and disadvantages gcse physics

See also Renewable energy - biofuels & alternative fuels, hydrogen, biogas, biodiesel

Greenhouse effect, global warming, climate change, carbon footprint from fossil fuel burning gcse physics

The Usefulness of Electricity gcse physics electricity revision notes

and The 'National Grid' power supply, mention of small scale supplies, transformers gcse physics notes


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