School Physics revision notes: (e) Kinetic energy stores and calculations

TYPES OF ENERGY STORE - examples explained

(e) Kinetic energy stores and calculations

Doc Brown's school physics revision notes: GCSE physics, IGCSE physics, O level physics,  ~US grades 8, 9 and 10 school science courses or equivalent for ~14-16 year old students of physics

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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 or a 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)

• Rearrangements

• v = √(2Eke / m)  to calculate the velocity for a known mass and kinetic energy

• m = 2 x Eke / v2  to calculate mass for a known kinetic energy and velocity

See also

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 (~67 mph, 108 km/hour).

(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. (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.

For more on efficiency see Energy transfers-conversions, efficiency - calculations

Q11

See also

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Energy resources, energy 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 energy stores  * Elastic potential energy stores, calculations  *

Gravitational potential energy and calculations  *  Kinetic energy stores and calculations  *  Nuclear energy store

Thermal energy stores  *  Light energy  * Sound energy  * Magnetic energy stores

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