TYPES OF ENERGY STORE  examples explained
(e) Kinetic energy stores and calculations
Doc Brown's GCSE 91 Physics Revision
Notes
Suitable for GCSE/IGCSE Physics/Science courses or
their equivalent
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How solve numerical problems involving ?
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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}

E_{ke} = ^{1}/_{2}
m v^{2}

kinetic energy, E_{ke}, 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. 2^{2} : 4^{2}
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}
E_{ke} = ^{1}/_{2}
m v^{2}
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 v^{2} = 0.5 x 70 x 1.4^{2} = 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 v^{2} , 10 g ≡
10/1000 = 0.01 kg
E_{ke} = 0.5 x 0.01 x 400^{2}
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.
(a) What is the kinetic energy
store of the car?
E_{ke} = ^{1}/_{2}
m v^{2} , 10 g ≡
10/1000 = 0.01 kg
E_{ke} = 0.5 x 1200 x 30^{2}
E_{ke} =
540,000 J of kinetic energy (5.40 x 10^{5}J,
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 v^{2} , rearranging: v^{2} = 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 v^{2} , rearranging: v^{2} = 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 v^{2} , rearranging: m = 2 x E_{ke} / v^{2}
m = 2 x E_{ke} / v^{2}
= 2 x 5000 / 500^{2} =
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 'takeoff' (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
takeoff by the pole vaulter? (take gravity strength as 9.8 N/kg).
KE = ^{1}/_{2}mv^{2},
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
TOP OF PAGE
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 transfersconversions, efficiency  calculations and
Sankey diagrams
Energy resources: uses, general survey & trends,
comparing renewables, nonrenewables, 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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