**
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 (~67 mph, 108 km/hour).

**(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. (**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}/_{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** 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/s^{2}.

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 = **441****0
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 mv^{2},
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}/_{2}mv^{2},
rearranging gives **v = √(2KE/m)**

v = √(2KE/m) = √(2 x 10^{-5}
/ 6 x 10^{-5}) = √0.3333 =
**0.5****8
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}/_{2}mv^{2},
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}/_{2}mv^{2
} = 0.5 x 3.5 x 0.25^{2} = 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 **
Reaction times, stopping distances, safety
aspects. calculations involving kinetic energy

TOP OF PAGE

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 *
Electrical
& electrostatic
energy stores

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

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

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

**See also **
More on methods of reducing heat transfer eg in a house, insulating properties of materials

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 - biomass - biofuels & alternative fuels,
hydrogen, biogas, biodiesel gcse chemistry notes

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

The absorption and emission of radiation by
materials - temperature & surface factors including global warming

The Usefulness of Electricity gcse
physics electricity revision notes

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