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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

How solve numerical problems involving ?

How to explain examples of ?

How to use the formula and do calculations involving ? 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)

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

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

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Chemical  * Elastic potential energy  * Gravitational potential energy

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